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Solutions Manual A Glencoe Program Student Edition Teacher Wraparound Edition Teacher Chapter Resources Mini Lab Worksheets Physics Lab Worksheets Study Guide Section Quizzes Reinforcement Enrichment Transparency Masters Transparency Worksheets Chapter Assessment Teacher Classroom Resources Teaching Transparencies Laboratory Manual, Student Edition Laboratory Manual, Teacher Edition Probeware Laboratory Manual, Student Edition Probeware Laboratory Manual, Teacher Edition Forensics Laboratory Manual, Student Edition Forensics Laboratory Manual, Teacher Edition Supplemental Problems Additional Challenge Problems Pre-AP/Critical Thinking Problems Physics Test Prep: Studying for the End-of-Course Exam, Student Edition Physics Test Prep: Studying for the End-of-Course Exam, Teacher Edition Connecting Math to Physics Solutions Manual Technology Answer Key Maker ExamView® Pro Interactive Chalkboard McGraw-Hill Learning Network StudentWorks™ CD-ROM TeacherWorks™ CD-ROM physicspp.com Web site Copyright © by The McGraw-Hill Companies, Inc. All rights reserved. Permission is granted to reproduce the material contained herein on the condition that such material be reproduced only for classroom use; be provided to students, teachers, and families without charge; and be used solely in conjunction with the Physics: Principles and Problems program. Any other reproduction, for use or sale, is prohibited without prior written permission of the publisher. Send all inquiries to: Glencoe/McGraw-Hill 8787 Orion Place Columbus, Ohio 43240 ISBN 0-07-865893-4 Printed in the United States of America 1 2 3 4 5 6 7 8 9 045 09 08 07 06 05 04 Contents Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. To the Teacher . . . . . . . . . . . . . . . . . . . . . . iv Chapter 1 A Physics Toolkit . . . . . . . . . . . . . . . . . . . . 1 Chapter 17 Reflection and Mirrors . . . . . . . . . . . . . . 357 Chapter 2 Representing Motion . . . . . . . . . . . . . . . . 15 Chapter 18 Refraction and Lenses . . . . . . . . . . . . . . 377 Chapter 3 Accelerated Motion . . . . . . . . . . . . . . . . . 29 Chapter 19 Interference and Diffraction . . . . . . . . . 399 Chapter 4 Forces in One Dimension . . . . . . . . . . . . 61 Chapter 20 Static Electricity . . . . . . . . . . . . . . . . . . . 413 Chapter 5 Forces in Two Dimensions . . . . . . . . . . . 87 Chapter 21 Electric Fields . . . . . . . . . . . . . . . . . . . . . 427 Chapter 6 Motion in Two Dimensions . . . . . . . . . 115 Chapter 22 Current Electricity . . . . . . . . . . . . . . . . . 445 Chapter 7 Gravitation . . . . . . . . . . . . . . . . . . . . . . . 141 Chapter 23 Series and Parallel Circuits . . . . . . . . . . 463 Chapter 8 Rotational Motion . . . . . . . . . . . . . . . . . 169 Chapter 24 Magnetic Fields . . . . . . . . . . . . . . . . . . . 485 Chapter 9 Momentum and Its Conservation . . . . 193 Chapter 25 Electromagnetic Induction . . . . . . . . . . 501 Chapter 10 Energy, Work, and Simple Machines . . 225 Chapter 26 Electromagnetism . . . . . . . . . . . . . . . . . . 517 Chapter 11 Energy and Its Conservation . . . . . . . . . 247 Chapter 27 Quantum Theory . . . . . . . . . . . . . . . . . . 531 Chapter 12 Thermal Energy . . . . . . . . . . . . . . . . . . . 271 Chapter 28 The Atom . . . . . . . . . . . . . . . . . . . . . . . . 545 Chapter 13 States of Matter . . . . . . . . . . . . . . . . . . . 287 Chapter 29 Solid-State Electronics . . . . . . . . . . . . . . 559 Chapter 14 Vibrations and Waves . . . . . . . . . . . . . . 311 Chapter 30 Nuclear Physics . . . . . . . . . . . . . . . . . . . 573 Chapter 15 Sound . . . . . . . . . . . . . . . . . . . . . . . . . . . 329 Appendix B Additional Problems . . . . . . . . . . . . . . . 591 Chapter 16 Fundamentals of Light . . . . . . . . . . . . . 345 Physics: Principles and Problems Contents iii To the Teacher The Solutions Manual is a comprehensive guide to the questions and problems in the Student Edition of Physics: Principles and Problems. This includes the Practice Problems, Section Reviews, Chapter Assessments, and Challenge Problems for each chapter, as well as the Additional Problems that appear in Appendix B of the Student Edition. The Solutions Manual restates every question and problem so that you do not have to look back at the text when reviewing problems with students. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. iv To the Teacher Physics: Principles and Problems CHAPTER 1 A Physics Toolkit Practice Problems 1.1 Mathematics and Physics pages 3–10 page 5 For each problem, give the rewritten equation you would use and the answer. 1. A lightbulb with a resistance of 50.0 ohms is used in a circuit with a 9.0-volt battery. What is the current through the bulb? V 9.0 volt I 0.18 ampere R 50.0 ohms 2. An object with uniform acceleration a, starting from rest, will reach a speed of v in time t according to the formula v at. What is the acceleration of a bicyclist who accelerates from rest to 7 m/s in 4 s? v t 7 m/s 4s Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. a 1.75 m/s2 3. How long will it take a scooter accelerating at 0.400 m/s2 to go from rest to a speed of 4.00 m/s? v a 4.00 m/s 0.400 m/s t 2 10.0 s 4. The pressure on a surface is equal to the force divided by the area: P F/A. A 53-kg woman exerts a force (weight) of 520 Newtons. If the pressure exerted on the floor is 32,500 N/m2, what is the area of the soles of her shoes? F P 520 N 32,500 N/m 2 A 2 0.016 m page 7 Use dimensional analysis to check your equation before multiplying. 6. Convert 5021 centimeters to kilometers. 5021 cm 1m 1 km 100 cm 1000 m 5.021102 km 7. How many seconds are in a leap year? 366 days 24 h 60 min 60 s 1h 1 day 1 min 31,622,400 s 8. Convert the speed 5.30 m/s to km/h. 5.30 m 60 s 60 min 1 km 1 s 1 min 1 h 1000 m 19.08 km/h page 8 Solve the following problems. 9. a. 6.201 cm 7.4 cm 0.68 cm 12.0 cm 6.201 7.4 0.68 12.0 cm cm cm cm 26.281 cm 26.3 cm after rounding b. 1.6 km 1.62 m 1200 cm 1.6 km 1600 m 1.62 m 1.62 m 1200 cm 12 m 1613.62 m 1600 m or 1.6 km after rounding 10. a. 10.8 g 8.264 g 10.8 g 8.264 g 2.536 g 2.5 g after rounding 5. How many megahertz is 750 kilohertz? 750 kHz 1 MHz 1000 Hz 1,000,000 Hz 1 kHz 0.75 MHz Physics: Principles and Problems Solutions Manual 1 Chapter 1 continued b. 4.75 m 0.4168 m 4.75 m 0.4168 m a. Substitute the values into the equation you will use. Are the units correct? F Bqv 4.3332 m 4.33 m after rounding 11. a. 139 cm 2.3 cm 320 cm2 or 3.2102 cm2 b. 3.2145 km 4.23 km 13.6 km2 12. a. 13.78 g 11.3 mL 1.22 g/mL b. 18.21 g 4.4 cm3 4.1 g/cm3 (4.5 kg/As2)(1.601019 As) (2.4105 m/s) Force will be measured in kgm/s2, which is correct. b. The values are written in scientific notation, m10n. Calculate the 10n part of the equation to estimate the size of the answer. 1019105 1014; the answer will be about 201014, or 21013. c. Calculate your answer. Check it against your estimate from part b. 1.71013 kgm/s2 Section Review 1.1 Mathematics and Physics pages 3–10 page 10 13. Math Why are concepts in physics described with formulas? The formulas are concise and can be used to predict new data. F F Bqv, so B qv kgm/s2 (As)(m/s) kg As T 2 1 T 1 kg/As2 15. Magnetism A proton with charge 1.601019 As is moving at 2.4105 m/s through a magnetic field of 4.5 T. You want to find the force on the proton. 2 Solutions Manual The least-precise value is 4.5 T, with 2 significant digits, so the answer is rounded to 2 significant digits. 16. Magnetism Rewrite F Bqv to find v in terms of F, q, and B. F Bq v 17. Critical Thinking An accepted value for the acceleration due to gravity is 9.801 m/s2. In an experiment with pendulums, you calculate that the value is 9.4 m/s2. Should the accepted value be tossed out to accommodate your new finding? Explain. No. The value 9.801 m/s2 has been established by many other experiments, and to discard the finding you would have to explain why they were wrong. There are probably some factors affecting your calculation, such as friction and how precisely you can measure the different variables. Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 14. Magnetism The force of a magnetic field on a charged, moving particle is given by F Bqv, where F is the force in kgm/s2, q is the charge in As, and v is the speed in m/s. B is the strength of the magnetic field, measured in teslas, T. What is 1 tesla described in base units? d. Justify the number of significant digits in your answer. Chapter 1 continued d. How precise is the measure of the height of one box? Of 12 boxes? Section Review 1.2 Measurement pages 11–14 page 14 18. Accuracy Some wooden rulers do not start with 0 at the edge, but have it set in a few millimeters. How could this improve the accuracy of the ruler? As the edge of the ruler gets worn away over time, the first millimeter or two of the scale would also be worn away if the scale started at the edge. 19. Tools You find a micrometer (a tool used to measure objects to the nearest 0.01 mm) that has been badly bent. How would it compare to a new, high-quality meterstick in terms of its precision? Its accuracy? It would be more precise but less accurate. 20. Parallax Does parallax affect the precision of a measurement that you make? Explain. 21. Error Your friend tells you that his height is 182 cm. In your own words, explain the range of heights implied by this statement. His height would be between 181.5 and 182.5 cm. Precision of a measurement is one-half the smallest division on the instrument. The height 182 cm would range 0.5 cm. 22. Precision A box has a length of 18.1 cm and a width of 19.2 cm, and it is 20.3 cm tall. a. What is its volume? 23. Critical Thinking Your friend states in a report that the average time required to circle a 1.5-mi track was 65.414 s. This was measured by timing 7 laps using a clock with a precision of 0.1 s. How much confidence do you have in the results of the report? Explain. A result can never be more precise than the least precise measurement. The calculated average lap time exceeds the precision possible with the clock. Practice Problems 1.3 Graphing Data pages 15–19 page 18 24. The mass values of specified volumes of pure gold nuggets are given in Table 1-4. Table 1-4 Mass of Pure Gold Nuggets Volume (cm3) Mass (g) 1.0 19.4 2.0 38.6 3.0 58.1 4.0 77.4 5.0 96.5 a. Plot mass versus volume from the values given in the table and draw the curve that best fits all points. 7.05103 cm3 b. How precise is the measure of length? Of volume? nearest tenth of a cm; nearest 10 cm3 c. How tall is a stack of 12 of these boxes? 243.6 cm Physics: Principles and Problems 100 Mass (g) Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. No, it doesn’t change the fineness of the divisions on its scale. nearest tenth of a cm; nearest tenth of a cm 80 60 40 20 0 1 2 3 4 5 Volume (cm3) Solutions Manual 3 Chapter 1 continued b. Describe the resulting curve. a straight line c. According to the graph, what type of relationship exists between the mass of the pure gold nuggets and their volume? The relationship is linear. d. What is the value of the slope of this graph? Include the proper units. 19.3 g/cm3 e. Write the equation showing mass as a function of volume for gold. m (19.3 g/cm3)V Write a word interpretation for the slope of the line. The mass for each cubic centimeter of pure gold is 19.3 g. 28. Predict Use the relation in Figure 1-18 to predict the current when the resistance is 16 ohms. 29. Critical Thinking In your own words, explain the meaning of a shallower line, or a smaller slope than the one in Figure 1-16, in the graph of stretch versus total mass for a different spring. The spring whose line has a smaller slope is stiffer, and therefore requires more mass to stretch it one centimeter. Chapter Assessment Concept Mapping Section Review 1.3 16 g 7.5 A 96.5 g 19.4 g y slope x 5.0 cm3 1.0 cm3 f. 27. Predict Use the relation illustrated in Figure 1-16 to determine the mass required to stretch the spring 15 cm. Graphing Data pages 15–19 hypothesis 0 Time (s) 5 Speed (m/s) 12 10 10 15 20 25 30 35 8 6 4 2 2 2 experiment Speed (m/s) 12 8 measurement 4 0 10 20 30 40 dependent variable independent variable Time (s) 26. Interpret a Graph What would be the meaning of a nonzero y-intercept to a graph of total mass versus volume? There is a nonzero total mass when the volume of the material is zero. This could happen if the mass value includes the material’s container. 4 Solutions Manual graph mathematical model Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. page 19 25. Make a Graph Graph the following data. Time is the independent variable. page 24 30. Complete the following concept map using the following terms: hypothesis, graph, mathematical model, dependent variable, measurement. Chapter 1 continued Mastering Concepts page 24 31. Describe a scientific method. (1.1) Identify a problem; gather information about it by observing and experimenting; analyze the information to arrive at an answer. 32. Why is mathematics important to science? (1.1) Mathematics allows you to be quantitative, to say “how fast,” not just “fast.” 33. What is the SI system? (1.1) The International System of Units, or SI, is a base 10 system of measurement that is the standard in science. The base units are the meter, kilogram, second, kelvin, mole, ampere, and candela. 34. How are base units and derived units related? (1.1) The derived units are combinations of the base units. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 35. Suppose your lab partner recorded a measurement as 100 g. (1.1) a. Why is it difficult to tell the number of significant digits in this measurement? Zeros are necessary to indicate the magnitude of the value, but there is no way of knowing whether or not the instrument used to measure the values actually measured the zeros. The zeros may serve only to locate the 1. b. How can the number of significant digits in such a number be made clear? Write the number in scientific notation, including only the significant digits. 36. Give the name for each of the following multiples of the meter. (1.1) 1 a. m 100 centimeter Physics: Principles and Problems 1 b. m 1000 millimeter c. 1000 m kilometer 37. To convert 1.8 h to minutes, by what conversion factor should you multiply? (1.1) 60 min , because the units will cancel 1h correctly. 38. Solve each problem. Give the correct number of significant digits in the answers. (1.1) a. 4.667104 g 3.02105 g 3.49105 g b. (1.70102 J) (5.922104 cm3) 2.87105 J/cm3 39. What determines the precision of a measurement? (1.2) the precision of a measuring device, which is limited by the finest division on its scale 40. How does the last digit differ from the other digits in a measurement? (1.2) The final digit is estimated. 41. A car’s odometer measures the distance from home to school as 3.9 km. Using string on a map, you find the distance to be 4.2 km. Which answer do you think is more accurate? What does accurate mean? (1.2) The most accurate measure is the measure closest to the actual distance. The odometer is probably more accurate as it actually covered the distance. The map is a model made from measurements, so your measurements from the map are more removed from the real distance. 42. How do you find the slope of a linear graph? (1.3) The slope of a linear graph is the ratio of the vertical change to the horizontal change, or rise over run. Solutions Manual 5 Chapter 1 continued 43. For a driver, the time between seeing a stoplight and stepping on the brakes is called reaction time. The distance traveled during this time is the reaction distance. Reaction distance for a given driver and vehicle depends linearly on speed. (1.3) 46. Given the equation F mv2/R, what relationship exists between each of the following? (1.3) a. F and R inverse relationship b. F and m a. Would the graph of reaction distance versus speed have a positive or a negative slope? linear relationship c. F and v Positive. As speed increases, reaction distance increases. Larger. The driver who was distracted would have a longer reaction time and thus a greater reaction distance at a given speed. Temperature is the independent variable; volume is the dependent variable. 45. What type of relationship is shown in Figure 1-20? Give the general equation for this type of relation. (1.3) y pages 25–26 47. Figure 1-21 gives the height above the ground of a ball that is thrown upward from the roof of a building, for the first 1.5 s of its trajectory. What is the ball’s height at t 0? Predict the ball’s height at t 2 s and at t 5 s. Height of Ball v. Time 25 20 15 10 5 0 1 2 3 4 Time (s) ■ Figure 1-21 When t 0 and t 2, the ball’s height will be about 20 m. When t 5, the ball will have landed on the ground so the height will be 0 m. 48. Is a scientific method one set of clearly defined steps? Support your answer. x ■ Figure 1-20 quadratic; y ax2 bx c 6 Solutions Manual There is no definite order of specific steps. However, whatever approach is used, it always includes close observation, controlled experimentation, summarizing, checking, and rechecking. Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 44. During a laboratory experiment, the temperature of the gas in a balloon is varied and the volume of the balloon is measured. Which quantity is the independent variable? Which quantity is the dependent variable? (1.3) Applying Concepts Height (m) b. A driver who is distracted has a longer reaction time than a driver who is not. Would the graph of reaction distance versus speed for a distracted driver have a larger or smaller slope than for a normal driver? Explain. quadratic relationship Chapter 1 continued 49. Explain the difference between a scientific theory and a scientific law. A scientific law is a rule of nature, where a scientific theory is an explanation of the scientific law based on observation. A theory explains why something happens; a law describes what happens. 50. Density The density of a substance is its mass per unit volume. a. Give a possible metric unit for density. possible answers include g/cm3 or kg/m3 b. Is the unit for density a base unit or a derived unit? derived unit 51. What metric unit would you use to measure each of the following? a. the width of your hand cm b. the thickness of a book cover mm c. the height of your classroom Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. m d. the distance from your home to your classroom km 52. Size Make a chart of sizes of objects. Lengths should range from less than 1 mm to several kilometers. Samples might include the size of a cell, the distance light travels in 1 s, and the height of a room. sample answer: radius of the atom, 51011 m; virus, 107 m; thickness of paper, 0.1 mm; width of paperback book, 10.7 cm; height of a door, 1.8 m; width of town, 7.8 km; radius of Earth, 6106 m; distance to the Moon, 4108 m 53. Time Make a chart of time intervals. Sample intervals might include the time between heartbeats, the time between presidential elections, the average lifetime of a human, and the age of the United Physics: Principles and Problems States. Find as many very short and very long examples as you can. sample answer: half-life of polonium-194, 0.7 s; time between heartbeats, 0.8 s; time to walk between physics class and math class, 2.4 min; length of school year, 180 days; time between elections for the U.S. House of Representatives, 2 years; time between U.S. presidential elections, 4 years; age of the United States, (about) 230 years 54. Speed of Light Two students measure the speed of light. One obtains (3.001 0.001)108 m/s; the other obtains (2.999 0.006)108 m/s. a. Which is more precise? (3.001 0.001)108 m/s b. Which is more accurate? (2.999 0.006)108 m/s 55. You measure the dimensions of a desk as 132 cm, 83 cm, and 76 cm. The sum of these measures is 291 cm, while the product is 8.3105 cm3. Explain how the significant digits were determined in each case. In addition and subtraction, you ask what place the least precise measure is known to: in this case, to the nearest cm. So the answer is rounded to the nearest cm. In multiplication and division, you look at the number of significant digits in the least precise answer: in this case, 2. So the answer is rounded to 2 significant digits. 56. Money Suppose you receive $5.00 at the beginning of a week and spend $1.00 each day for lunch. You prepare a graph of the amount you have left at the end of each day for one week. Would the slope of this graph be positive, zero, or negative? Why? negative, because the change in vertical distance is negative for a positive change in horizontal distance Solutions Manual 7 Chapter 1 continued 57. Data are plotted on a graph, and the value on the y-axis is the same for each value of the independent variable. What is the slope? Why? How does y depend on x? Zero. The change in vertical distance is zero. y does not depend on x. 58. Driving The graph of braking distance versus car speed is part of a parabola. Thus, the equation is written d av2 bv c. The distance, d, has units in meters, and velocity, v, has units in meters/second. How could you find the units of a, b, and c? What would they be? The units in each term of the equation must be in meters because distance, d, is measured in meters. av 2 a(m/s)2, so a is in s2/m; bv b(m/s), so b is in s1. 59. How long is the leaf in Figure 1-22? Include the uncertainty in your measurement. c. What is the number of significant digits for the total mass? four d. Why is the number of significant digits different for the total mass and the individual masses? When adding measurements, the precision matters: both masses are known to the nearest hundredth of a gram, so the total should be given to the nearest hundredth of a gram. Significant digits sometimes are gained when adding. 61. History Aristotle said that the speed of a falling object varies inversely with the density of the medium through which it falls. a. According to Aristotle, would a rock fall faster in water (density 1000 kg/m3), or in air (density 1 kg/m3)? Lower density means faster speed, so the rock falls faster in air. b. How fast would a rock fall in a vacuum? Based on this, why would Aristotle say that there could be no such thing as a vacuum? ■ Figure 1-22 8.3 cm 0.05 cm, or 83 mm 0.5 mm 60. The masses of two metal blocks are measured. Block A has a mass of 8.45 g and block B has a mass of 45.87 g. a. How many significant digits are expressed in these measurements? A: three; B: four b. What is the total mass of block A plus block B? 62. Explain the difference between a hypothesis and a scientific theory. A scientific theory has been tested and supported many times before it becomes accepted. A hypothesis is an idea about how things might work—it has much less support. 63. Give an example of a scientific law. Newton’s laws of motion, law of conservation of energy, law of conservation of charge, law of reflection 54.32 g 8 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Because a vacuum would have a zero density, the rock should fall infinitely fast. Nothing can fall that fast. Chapter 1 continued 64. What reason might the ancient Greeks have had not to question the hypothesis that heavier objects fall faster than lighter objects? Hint: Did you ever question which falls faster? Air resistance affects many light objects. Without controlled experiments, their everyday observations told them that heavier objects did fall faster. 65. Mars Explain what observations led to changes in scientists’ ideas about the surface of Mars. As telescopes improved and later probes were sent into space, scientists gained more information about the surface. When the information did not support old hypotheses, the hypotheses changed. 66. A graduated cylinder is marked every mL. How precise a measurement can you make with this instrument? 0.5 mL Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Mastering Problems pages 26–28 1.1 Mathematics and Physics 67. Convert each of the following measurements to meters. a. 42.3 cm 0.423 m b. 6.2 pm 6.21012 m 68. Add or subtract as indicated. a. 5.80109 s 3.20108 s 6.12109 s b. 4.87106 m 1.93106 m 2.94106 m c. 3.14105 kg 9.36105 kg 1.250104 kg d. 8.12107 g 6.20106 g 7.50107 g 69. Rank the following mass measurements from least to greatest: 11.6 mg, 1021 g, 0.000006 kg, 0.31 mg. 0.31 mg, 1021 µg, 0.000006 kg, 11.6 mg 70. State the number of significant digits in each of the following measurements. a. 0.00003 m 1 b. 64.01 fm 4 c. 80.001 m 5 d. 0.720 g 3 e. 2.40106 kg 3 f. 6108 kg 1 g. 4.071016 m 3 c. 21 km 2.1104 m d. 0.023 mm 2.3105 m e. 214 m 2.14104 m f. 57 nm 5.7108 m 71. Add or subtract as indicated. a. 16.2 m 5.008 m 13.48 m 34.7 m b. 5.006 m 12.0077 m 8.0084 m 25.022 m c. 78.05 cm2 32.046 cm2 46.00 cm2 d. 15.07 kg 12.0 kg 3.1 kg Physics: Principles and Problems Solutions Manual 9 Chapter 1 continued 72. Multiply or divide as indicated. a. (6.21018 2.9109 m)(4.71010 m) 78. How precise a measurement could you make with the scale shown in Figure 1-23? m2 b. (5.6107 m)/(2.81012 s) 2.0105 m/s c. (8.1104 km)(1.6103 km) 1.3106 km2 d. (6.5105 kg)/(3.4103 m3) 1.9102 kg/m3 73. Gravity The force due to gravity is F mg where g 9.80 m/s2. a. Find the force due to gravity on a 41.63-kg object. 408 kgm/s2 b. The force due to gravity on an object is 632 kgm/s2. What is its mass? ■ Figure 1-23 0.5 g 79. Give the measure shown on the meter in Figure 1-24 as precisely as you can. Include the uncertainty in your answer. 64.5 kg 3 2 4 0 5A 74. Dimensional Analysis Pressure is measured in pascals, where 1 Pa 1 kg/ms2. Will the following expression give a pressure in the correct units? 1 A CLASS A (0.55 kg)(2.1 m/s) 9.8 m/s2 1.2 Measurement 75. A water tank has a mass of 3.64 kg when it is empty and a mass of 51.8 kg when it is filled to a certain level. What is the mass of the water in the tank? 48.2 kg 76. The length of a room is 16.40 m, its width is 4.5 m, and its height is 3.26 m. What volume does the room enclose? 2.4102 m3 77. The sides of a quadrangular plot of land are 132.68 m, 48.3 m, 132.736 m, and 48.37 m. What is the perimeter of the plot? ■ Figure 1-24 3.6 0.1 A 80. Estimate the height of the nearest door frame in centimeters. Then measure it. How accurate was your estimate? How precise was your estimate? How precise was your measurement? Why are the two precisions different? A standard residential doorframe height is 80 inches, which is about 200 cm. The precision depends on the measurement instrument used. 362.1 m 10 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. No; it is in kg/s3 Chapter 1 continued 81. Base Units Give six examples of quantities you might measure in a physics lab. Include the units you would use. Mass of Three Substances 800 Sample: distance, cm; volume, mL; mass, g; current, A; time, s; temperature, °C Mass (g) 700 82. Temperature The temperature drops from 24°C to 10°C in 12 hours. 600 C 500 400 B 300 200 a. Find the average temperature change per hour. 100 0 1.2°C/h A 10 20 30 40 50 Volume (cm3) b. Predict the temperature in 2 more hours if the trend continues. ■ Figure 1-25 8°C c. Could you accurately predict the temperature in 24 hours? No. Temperature is unlikely to continue falling sharply and steadily that long. Table 1-5 Distance Traveled with Different Forces a. What is the mass of 30 cm3 of each substance? Force (N) 24 10.0 49 15.0 75 20.0 99 25.0 120 30.0 145 b. If you had 100 g of each substance, what would be their volumes? (a) 36 cm3, (b) 11 cm3, (c) 7 cm3 c. In one or two sentences, describe the meaning of the slopes of the lines in this graph. The slope represents the increased mass of each additional cubic centimeter of the substance. d. What is the y-intercept of each line? What does it mean? The y-intercept is (0,0). It means that when V 0 cm3, there is none of the substance present (m 0 g). Distance (cm) 5.0 (a) 80 g, (b) 260 g, (c) 400 g a. Plot the values given in the table and draw the curve that best fits all points. 160 Distance (cm) Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1.3 Graphing Data 83. Figure 1-25 shows the masses of three substances for volumes between 0 and 60 cm3. 84. During a class demonstration, a physics instructor placed a mass on a horizontal table that was nearly frictionless. The instructor then applied various horizontal forces to the mass and measured the distance it traveled in 5 seconds for each force applied. The results of the experiment are shown in Table 1-5. 120 80 40 0.0 10.0 20.0 30.0 Force (N) Physics: Principles and Problems Solutions Manual 11 Chapter 1 continued b. Describe the resulting curve. b. Describe the resulting curve. a hyperbola a straight line c. Use the graph to write an equation relating the distance to the force. d 4.9F d. What is the constant in the equation? Find its units. The constant is 4.9 and has units cm/N. e. Predict the distance traveled when a 22.0-N force is exerted on the object for 5 s. 108 cm or 110 cm using 2 significant digits 85. The physics instructor from the previous problem changed the procedure. The mass was varied while the force was kept constant. Time and distance were measured, and the acceleration of each mass was calculated. The results of the experiment are shown in Table 1-6. Table 1-6 c. According to the graph, what is the relationship between mass and the acceleration produced by a constant force? Acceleration varies inversely with mass. d. Write the equation relating acceleration to mass given by the data in the graph. 12 m a e. Find the units of the constant in the equation. kgm/s2 f. Predict the acceleration of an 8.0-kg mass. 1.5 m/s2 86. During an experiment, a student measured the mass of 10.0 cm3 of alcohol. The student then measured the mass of 20.0 cm3 of alcohol. In this way, the data in Table 1-7 were collected. Acceleration of Different Masses 1.0 12.0 2.0 5.9 Volume (cm3) Mass (g) 3.0 4.1 10.0 7.9 4.0 3.0 20.0 15.8 5.0 2.5 30.0 23.7 6.0 2.0 40.0 31.6 50.0 39.6 12.0 The Mass Values of Specific Volumes of Alcohol a. Plot the values given in the table and draw the curve that best fits all the points. 40.0 Mass (g) 9.0 6.0 3.0 0.0 30.0 20.0 10.0 2.0 4.0 Mass (kg) 6.0 0.0 10.0 20.0 30.0 40.0 50.0 60.0 Volume (cm3) 12 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Acceleration (m/s2) a. Plot the values given in the table and draw the curve that best fits all points. Acceleration (m/s2) Table 1-7 Mass (kg) Chapter 1 continued b. Describe the resulting curve. a straight line c. Use the graph to write an equation relating the volume to the mass of the alcohol. m 0.79V d. Find the units of the slope of the graph. What is the name given to this quantity? g/cm3; density e. What is the mass of 32.5 cm3 of alcohol? 25.7 g Mixed Review page 28 87. Arrange the following numbers from most precise to least precise 0.0034 m 45.6 m 1234 m 0.0034 m, 45.6 m, 1234 m Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 88. Figure 1-26 shows the toroidal (doughnutshaped) interior of the now-dismantled Tokamak Fusion Test Reactor. Explain why a width of 80 m would be an unreasonable value for the width of the toroid. What would be a reasonable value? 90. You are given the following measurements of a rectangular bar: length 2.347 m, thickness 3.452 cm, height 2.31 mm, mass 1659 g. Determine the volume, in cubic meters, and density, in g/cm3, of the beam. Express your results in proper form. volume 1.87104 m3, or 187 cm3; density 8.87 g/cm3 91. A drop of water contains 1.71021 molecules. If the water evaporated at the rate of one million molecules per second, how many years would it take for the drop to completely evaporate? 1.71021 molecules 1.71015 s 1,000,000 molecules 1s 1 day 1y 1h (1.71015 s) 3600 s 24 h 365 days 5.4107 y 92. A 17.6-gram sample of metal is placed in a graduated cylinder containing 10.0 cm3 of water. If the water level rises to 12.20 cm3, what is the density of the metal? m V density 17.6 g 12.20 cm 10.0 cm 3 3 8.00 g/cm3 Thinking Critically ■ Figure 1-26 80 meters is equivalent to about 260 feet, which would be very large. 10 meters would be a more reasonable value. 89. You are cracking a code and have discovered the following conversion factors: 1.23 longs 23.0 mediums, and 74.5 mediums 645 shorts. How many shorts are equal to one long? 1 long 162 shorts 23.0 med 645 short 1.23 long 74.5 med Physics: Principles and Problems page 28 93. Apply Concepts It has been said that fools can ask more questions than the wise can answer. In science, it is frequently the case that one wise person is needed to ask the right question rather than to answer it. Explain. The “right” question is one that points to fruitful research and to other questions that can be answered. 94. Apply Concepts Find the approximate mass of water in kilograms needed to fill a container that is 1.40 m long and 0.600 m wide to a depth of 34.0 cm. Report your result to one significant digit. (Use a reference source to find the density of water.) Solutions Manual 13 Chapter 1 continued Vw (140 cm)(60.0 cm)(34.0 cm) 285,600 cm3. Because the density of water is 1.00 g/cm3, the mass of water in kilograms is 286 kg. 95. Analyze and Conclude A container of gas with a pressure of 101 kPa has a volume of 324 cm3 and a mass of 4.00 g. If the pressure is increased to 404 kPa, what is the density of the gas? Pressure and volume are inversely proportional. Pressure and volume are inversely proportional. Since the pressure is 1 4 times greater, the volume will be 4 of the original volume. 324 cm3 81.0 cm3 4 4.00 g 0.0494 g/cm3 81.0 cm3 99. Explain how improved precision in measuring time would have led to more accurate predictions about how an object falls. Answers will vary. For example, students might suggest that improved precision can lead to better observations. Challenge Problem page 17 An object is suspended from spring 1, and the spring’s elongation (the distance it stretches) is X1. Then the same object is removed from the first spring and suspended from a second spring. The elongation of spring 2 is X2. X2 is greater than X1. 1. On the same axes, sketch the graphs of the mass versus elongation for both springs. 96. Design an Experiment How high can you throw a ball? What variables might affect the answer to this question? mass of ball, footing, practice, and conditioning d v 1.4961011 m 3.0010 m /s E tE 8 499 s 8.31 min dJ 7.781011 m tJ v 3.00108 m/s 2593 s 43.2 min Writing in Physics page 28 98. Research and describe a topic in the history of physics. Explain how ideas about the topic changed over time. Be sure to include the contributions of scientists and to evaluate the impact of their contributions on scientific thought and the world outside the laboratory. X X1 F 2. Is the origin included in the graph? Why or why not? Yes; the origin corresponds to 0 elongation when the force is 0. 3. Which slope is steeper? The slope for X2 is steeper. 4. At a given mass, X2 1.6 X1. If X2 5.3 cm, what is X1? X2 1.6X1 5.3 cm 1.6X1 3.3 cm X1 Answers will vary. 14 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 97. Calculate If the Sun suddenly ceased to shine, how long would it take Earth to become dark? (You will have to look up the speed of light in a vacuum and the distance from the Sun to Earth.) How long would it take the surface of Jupiter to become dark? X2 CHAPTER 2 Representing Motion Section Review 2.1 Picturing Motion pages 31–33 page 33 1. Motion Diagram of a Runner Use the particle model to draw a motion diagram for a bike rider riding at a constant pace. 4. Critical Thinking Use the particle model to draw motion diagrams for two runners in a race, when the first runner crosses the finish line as the other runner is threefourths of the way to the finish line. Runner 2 t0 Runner 1 t0 t1 t2 t1 t3 t2 t4 t3 t4 Start 2. Motion Diagram of a Bird Use the particle model to draw a simplified motion diagram corresponding to the motion diagram in Figure 2-4 for a flying bird. What point on the bird did you choose to represent it? Finish Section Review 2.2 Where and When? pages 34–37 page 37 5. Displacement The particle model for a car traveling on an interstate highway is shown below. The starting point is shown. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Here ■ Figure 2-4 3. Motion Diagram of a Car Use the particle model to draw a simplified motion diagram corresponding to the motion diagram in Figure 2-5 for a car coming to a stop at a stop sign. What point on the car did you use to represent it? There Make a copy of the particle model, and draw a vector to represent the displacement of the car from the starting time to the end of the third time interval. Here There 6. Displacement The particle model for a boy walking to school is shown below. Home School Make a copy of the particle model, and draw vectors to represent the displacement between each pair of dots. Home ■ School Figure 2-5 Physics: Principles and Problems Solutions Manual 15 Chapter 2 continued A position vector goes from the origin to the object. When the origins are different, the position vectors are different. On the other hand, a displacement vector has nothing to do with the origin. 8. Critical Thinking A car travels straight along the street from the grocery store to the post office. To represent its motion you use a coordinate system with its origin at the grocery store and the direction the car is moving in as the positive direction. Your friend uses a coordinate system with its origin at the post office and the opposite direction as the positive direction. Would the two of you agree on the car’s position? Displacement? Distance? The time interval the trip took? Explain. 2.3 Position-Time Graphs pages 38–42 page 39 For problems 9–11, refer to Figure 2-13. 150.0 100.0 50.0 0.0 1.0 3.0 50.0 5.0 7.0 Time (s) ■ Figure 2-13 9. Describe the motion of the car shown by the graph. The car begins at a position of 125.0 m and moves toward the origin, arriving at the origin 5.0 s after it begins moving. The car continues beyond the origin. 10. Draw a motion diagram that corresponds to the graph. t0 0.0 s t5 5.0 s 125.0 m 0.0 m d 11. Answer the following questions about the car’s motion. Assume that the positive d-direction is east and the negative d-direction is west. a. When was the car 25.0 m east of the origin? at 4.0 s b. Where was the car at 1.0 s? 100.0 m 12. Describe, in words, the motion of the two pedestrians shown by the lines in Figure 2-14. Assume that the positive direction is east on Broad Street and the origin is the intersection of Broad and High Streets. 16 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. The two students should agree on the displacement, distance, and time interval for the trip, because these three quantities are independent of where the origin of the coordinate system is placed. The two students would not agree on the car’s position, because the position is measured from the origin of the coordinate system to the location of the car. Practice Problems Position (m) 7. Position Two students compared the position vectors they each had drawn on a motion diagram to show the position of a moving object at the same time. They found that their vectors did not point in the same direction. Explain. Chapter 2 continued page 41 For problems 14–17, refer to the figure in Example Problem 2. Broad St. Position (m) East 200.0 Position (m) A High St. B ■ 50.0 0.0 2.0 4.0 6.0 8.0 10.0 12.0 Time (s) Figure 2-14 ■ Pedestrian A starts west of High Street and walks east (the positive direction). Pedestrian B begins east of High Street and walks west (the negative direction). Sometime after B crosses High Street, A and B pass each other. Sometime after they pass, Pedestrian A crosses High Street. 13. Odina walked down the hall at school from the cafeteria to the band room, a distance of 100.0 m. A class of physics students recorded and graphed her position every 2.0 s, noting that she moved 2.6 m every 2.0 s. When was Odina in the following positions? a. 25.0 m from the cafeteria 19 s Example Problem 2 Figure 14. What event occurred at t 0.0 s? Runner A passed the origin. 15. Which runner was ahead at t 48.0 s? runner B 16. When runner A was at 0.0 m, where was runner B? at 50.0 m 17. How far apart were runners A and B at t 20.0 s? approximately 30 m 18. Juanita goes for a walk. Sometime later, her friend Heather starts to walk after her. Their motions are represented by the positiontime graphs in Figure 2-16. b. 25.0 m from the band room 6.0 58 s 100.0 80.0 Ju an ita Position (km) 5.0 c. Create a graph showing Odina’s motion. Distance from cafeteria (m) 100.0 Time (s) West Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 150.0 4.0 3.0 a He 2.0 er 1.0 60.0 0.0 40.0 0.5 1.5 1.0 20.0 0.00 th 2.0 Time (h) 10.0 30.0 50.0 Time (s) 70.0 ■ Figure 2-16 a. How long had Juanita been walking when Heather started her walk? 6.0 min Physics: Principles and Problems Solutions Manual 17 Chapter 2 continued b. Will Heather catch up to Juanita? How can you tell? t0 0.0 s t7 7.0 s 0m 140 m No. The lines representing Juanita’s and Heather’s motions get farther apart as time increases. The lines will not intersect. For problems 21–23, refer to Figure 2-18. 21. Time Use the position-time graph of the hockey puck to determine when it was 10.0 m beyond the origin. Section Review Position-Time Graphs pages 38–42 2.3 0.5 s page 42 19. Position-Time Graph From the particle model in Figure 2-17 of a baby crawling across a kitchen floor, plot a position-time graph to represent his motion. The time interval between successive dots is 1 s. 0 20 40 60 80 100 120 140 160 Position (cm) Position (m) ■ 100 m 23. Time Interval Use the position-time graph for the hockey puck to determine how much time it took for the puck to go from 40 m beyond the origin to 80 m beyond the origin. 2.0 s Figure 2-17 24. Critical Thinking Look at the particle model and position-time graph shown in Figure 2-19. Do they describe the same motion? How do you know? Do not confuse the position coordinate system in the particle model with the horizontal axis in the position-time graph. The time intervals in the particle model are 2 s. 12 345 6 78 Time (s) 20. Motion Diagram Create a particle model from the position-time graph of a hockey puck gliding across a frozen pond in Figure 2-18. 0 10 Position (m) 12 140 Position (m) Position (m) 120 100 80 60 40 8 4 20 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 0 1 2 18 Solutions Manual Figure 2-18 4 5 Time (s) Time (s) ■ 3 ■ Figure 2-19 Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 160 140 120 100 80 60 40 20 0 22. Distance Use the position-time graph of the hockey puck to determine how far it moved between 0.0 s and 5.0 s. Chapter 2 continued Practice Problems 2.4 How Fast? pages 43–47 page 45 25. The graph in Figure 2-22 describes the motion of a cruise ship during its voyage through calm waters. The positive d-direction is defined to be south. 26. Describe, in words, the motion of the cruise ship in the previous problem. The ship is moving to the north at a speed of 0.33 m/s. 27. The graph in Figure 2-23 represents the motion of a bicycle. Determine the bicycle’s average speed and average velocity, and describe its motion in words. 20 Position (km) No, they don’t describe the same motion. Although both objects are traveling in the positive direction, one is moving more quickly than the other. Students can cite a number of different specific examples from the graph and particle model to back this up. 15 10 5 0 5 Time (min) ■ Time (s) Position (m) 1 2 3 4 0 1 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. ■ Figure 2-22 a. What is the ship’s average speed? Using the points (0.0 s, 0.0 m) and (3.0 s, 1.0 m) d v t d d t t 1.0 m 0.0 m 3.0 s 0.0 s 2 Figure 2-23 Because the bicycle is moving in the positive direction, the average speed and average velocity are the same. Using the points (0.0 min, 0.0 km) and (15.0 min, 10.0 km), d v t 2 2 10 15 20 25 30 1 1 d d t t 10.0 km 0.0 km 15.0 min 0.0 min 2 2 1 1 0.67 km/min v 0.67 km/min in the positive direction The bicycle is moving in the positive direction at a speed of 0.67 km/min. 0.33 m/s 0.33 m/s b. What is its average velocity? The average velocity is the slope of the line, including the sign, so it is 0.33 m/s or 0.33 m/s north. Physics: Principles and Problems Solutions Manual 19 Chapter 2 continued 28. When Marilyn takes her pet dog for a walk, the dog walks at a very consistent pace of 0.55 m/s. Draw a motion diagram and position-time graph to represent Marilyn’s dog walking the 19.8-m distance from in front of her house to the nearest fire hydrant. Motion Diagram t0 0 s t6 36 s 0.0 m House 19.8 m Hydrant Position from house (m) Position-Time Graph 3 2 SlopeB SlopeC 1 SlopeD 1 30. Average Velocity Rank the graphs according to average velocity, from greatest average velocity to least average velocity. Specifically indicate any ties. B, D, C, A SlopeA 2 3 2 SlopeB 20.0 SlopeC 1 15.0 SlopeD 1 10.0 5.0 0 6 12 18 24 30 36 Time (s) Section Review 2.4 SlopeA 2 How Fast? pages 43–47 29. Average Speed Rank the position-time graphs according to the average speed of the object, from greatest average speed to least average speed. Specifically indicate any ties. Position (m) B A, C, B, D. Yes, the ranking from greatest to least distance would be A, C, D, B. 32. Average Speed and Average Velocity Explain how average speed and average velocity are related to each other. Average speed is the absolute value of the average velocity. Speed is only a magnitude, while velocity is a magnitude and a direction. D A C Time (s) ■ Figure 2-25 For speed use the absolute value, therefore A, B, C D 20 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. page 47 For problems 29–31, refer to Figure 2-25. 31. Initial Position Rank the graphs according to the object’s initial position, from most positive position to most negative position. Specifically indicate any ties. Would your ranking be different if you had been asked to do the ranking according to initial distance from the origin? Chapter 2 continued 33. Critical Thinking In solving a physics problem, why is it important to create pictorial and physical models before trying to solve an equation? Answers will vary, but here are some of the important points. Drawing the models before writing down the equation helps you to get the problem situation organized in your head. It’s difficult to write down the proper equation if you don’t have a clear picture of how things are situated and/or moving. Also, you choose the coordinate system in this step, and this is essential in making sure you use the proper signs on the quantities you will substitute into the equation later. Chapter Assessment Concept Mapping Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. page 52 34. Complete the concept map below using the following terms: words, equivalent representations, position-time graph. words equivalent representations data table position-time graph motion diagram 37. The following quantities describe location or its change: position, distance, and displacement. Briefly describe the differences among them. (2.2) Position and displacement are different from distance because position and displacement both contain information about the direction in which an object has moved, while distance does not. Distance and displacement are different from position because they describe how an object’s location has changed during a time interval, where position tells exactly where an object is located at a precise time. 38. How can you use a clock to find a time interval? (2.2) Read the clock at the beginning and end of the interval and subtract the beginning time from the ending time. 39. In-line Skating How can you use the position-time graphs for two in-line skaters to determine if and when one in-line skater will pass the other one? (2.3) Draw the two graphs on the same set of axes. One inline skater will pass the other if the lines representing each of their motions intersect. The position coordinate of the point where the lines intersect is the position where the passing occurs. page 52 35. What is the purpose of drawing a motion diagram? (2.1) 40. Walking Versus Running A walker and a runner leave your front door at the same time. They move in the same direction at different constant velocities. Describe the position-time graphs of each. (2.4) A motion diagram gives you a picture of motion that helps you visualize displacement and velocity. Both are straight lines that start at the same position, but the slope of the runner’s line is steeper. Mastering Concepts 36. Under what circumstances is it legitimate to treat an object as a point particle? (2.1) An object can be treated as a point particle if internal motions are not important and if the object is small in comparison to the distance it moves. Physics: Principles and Problems 41. What does the slope of a position-time graph measure? (2.4) velocity Solutions Manual 21 Chapter 2 continued It is possible to calculate the average velocity from the information given, but it is not possible to find the instantaneous velocity. 46. Figure 2-26 is a graph of two people running. Position (m) 42. If you know the positions of a moving object at two points along its path, and you also know the time it took for the object to get from one point to the other, can you determine the particle’s instantaneous velocity? Its average velocity? Explain. (2.4) R 45. When can a football player be treated as a point particle? A football player can be treated as a point particle if his or her internal motions are not important and if he or she is small in comparison to the distance he or she moves — for distances of several yards or more. 22 Solutions Manual Figure 2-26 Runner B is faster, as shown by the steeper slope. c. What occurs at point P and beyond? Runner B passes runner A at point P. Be ss ie 47. The position-time graph in Figure 2-27 shows the motion of four cows walking from the pasture back to the barn. Rank the cows according to their average velocity, from slowest to fastest. da olin Mo lly Do Time (s) ■ Figure 2-27 Moolinda, Dolly, Bessie, Elsie Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. A football can be treated as a point particle if its rotations are not important and if it is small in comparison to the distance it moves — for distances of 1 yard or more. A b. Which runner is faster? ie 44. Football When can a football be considered a point particle? ner Run Runner A has a head start by four units. E ls d t increases when either term increases. The sign of d t depends upon the relative sizes of d and t. d t increases when either increases. t /d decreases with increasing displacement and increases with increasing time interval, which is backwards from velocity. B a. Describe the position of runner A relative to runner B at the y-intercept. Position (m) page 52 43. Test the following combinations and explain why each does not have the properties needed to describe the concept of velocity: d t, d t, d t, t/d. er Time (s) ■ Applying Concepts n un Chapter 2 continued 48. Figure 2-28 is a position-time graph for a rabbit running away from a dog. Position (m) 3 2 1 0 1 2 3 Time (s) ■ Figure 2-28 a. Describe how this graph would be different if the rabbit ran twice as fast. The only difference is that the slope of the graph would be twice as steep. b. Describe how this graph would be different if the rabbit ran in the opposite direction. The magnitude of the slope would be the same, but it would be negative. Mastering Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 2.4 How Fast? page 53 Level 1 49. A bike travels at a constant speed of 4.0 m/s for 5.0 s. How far does it go? d vt (4.0 m/s)(5 s) 20 m 50. Astronomy Light from the Sun reaches Earth in 8.3 min. The speed of light is 3.00108 m/s. How far is Earth from the Sun? d vt (3.00108 m/s)(8.3 min) 1.51011 m Physics: Principles and Problems 60 s 1 min Solutions Manual 23 Chapter 2 continued Level 2 51. A car is moving down a street at 55 km/h. A child suddenly runs into the street. If it takes the driver 0.75 s to react and apply the brakes, how many meters will the car have moved before it begins to slow down? d vt (55 km/h)(0.75 s) 1h 1000 m 3600 s 1 km 11 m d vt d v a. Plot a position-time graph of the cyclist’s location from point A at 10.0-s intervals for 60.0 s. 550 500 Position (m) 52. Nora jogs several times a week and always keeps track of how much time she runs each time she goes out. One day she forgets to take her stopwatch with her and wonders if there’s a way she can still have some idea of her time. As she passes a particular bank, she remembers that it is 4.3 km from her house. She knows from her previous training that she has a consistent pace of 4.0 m/s. How long has Nora been jogging when she reaches the bank? t pages 53–54 Level 1 54. Cycling A cyclist maintains a constant velocity of 5.0 m/s. At time t 0.0 s, the cyclist is 250 m from point A. 450 400 350 300 250 200 0.0 10.0 20.0 30.0 40.0 50.0 60.0 Time (s) b. What is the cyclist’s position from point A at 60.0 s? 550 m (4.3 km) 1000m 1 km 4.0 m/s c. What is the displacement from the starting position at 60.0 s? 1075 s 550 m 250 m 3.0102 m 1 min 60 s 18 min Level 3 53. Driving You and a friend each drive 50.0 km. You travel at 90.0 km/h; your friend travels at 95.0 km/h. How long will your friend have to wait for you at the end of the trip? 55. Figure 2-29 is a particle model for a chicken casually walking across the road. Time intervals are every 0.1 s. Draw the corresponding position-time graph and equation to describe the chicken’s motion. This side The other side Time intervals are 0.1 s. ■ d vt d v d 50.0 km 90.0 km/h t1 The other side 0.556 h d v Figure 2-29 50.0 km 95.0 km/h t2 0.526 h t1 t2 (0.556 h 0.526 h) 60 min 1h 1.8 min Mixed Review 24 Solutions Manual This side 0 t 1.9 s 56. Figure 2-30 shows position-time graphs for Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. (1075 s) Chapter 2 continued Joszi and Heike paddling canoes in a local river. Car B 16 Joszi 14 Heike 12 10 Position (km) 200 18 Position (km) 250 170 150 Car A 120 100 50 8 6 0 4 1.0 1.6 1.4 2.0 3.0 2 0 0.5 1.0 1.5 2.0 Time (h) ■ Time (h) 2.5 Figure 2-30 a. At what time(s) are Joszi and Heike in the same place? 1.0 h b. How much time does Joszi spend on the river before he passes Heike? b. Both cars passed a gas station 120 km from the school. When did each car pass the gas station? Calculate the times and show them on your graph. Level 2 57. Driving Both car A and car B leave school when a stopwatch reads zero. Car A travels at a constant 75 km/h, and car B travels at a constant 85 km/h. a. Draw a position-time graph showing the motion of both cars. How far are the two cars from school when the stopwatch reads 2.0 h? Calculate the distances and show them on your graph. dA vAt (75 km/h)(2.0 h) 150 km d B 120 km 85 km/h 58. Draw a position-time graph for two cars traveling to the beach, which is 50 km from school. At noon, Car A leaves a store that is 10 km closer to the beach than the school is and moves at 40 km/h. Car B starts from school at 12:30 P.M. and moves at 100 km/h. When does each car get to the beach? 50 40 Position (m) Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. from 6.0 to 9.0 km from the origin 120 km 75 km/h tB 1.4 h 45 min c. Where on the river does it appear that there might be a swift current? d A tA 1.6 h 30 Car A Car B 20 10 0 Noon 1210 1220 1230 1240 1250 100PM Time Both cars arrive at the beach at 1:00 P.M. dB vBt (85 km/h)(2.0 h) 170 km Physics: Principles and Problems Level 3 59. Two cars travel along a straight road. When a stopwatch reads t 0.00 h, car A is at dA 48.0 km moving at a constant 36.0 km/h. Later, when the watch reads t 0.50 h, car B is at dB 0.00 km moving Solutions Manual 25 Chapter 2 continued at 48.0 km/h. Answer the following questions, first, graphically by creating a position-time graph, and second, algebraically by writing equations for the positions dA and dB as a function of the stopwatch time, t. 60. Figure 2-31 shows the position-time graph depicting Jim’s movement up and down the aisle at a store. The origin is at one end of the aisle. 14.0 12.0 a. What will the watch read when car B passes car A? Position (m) 300.0 200.0 Position (km) 10.0 150.0 Car B 100.0 6.0 4.0 2.0 Car A 150.0 8.0 0.00 10.0 20.0 30.0 40.0 50.0 60.0 Time (s) 50.0 ■ 1.00 0.00 1.00 2.00 3.00 4.00 5.00 6.00 7.00 Time (h) Cars pass when the distances are equal, dA dB dA 48.0 km (36.0 km/h)t and dB 0 (48.0 km/h)(t 0.50 h) (48.0 km) (36.0 km/h)t (48.0 km/h)t 24 km 72 km (12.0 km/h)t t 6.0 h b. At what position will car B pass car A? dA 48.0 km (36.0 km/h)(6.0 h) 2.6102 km c. When the cars pass, how long will it have been since car A was at the reference point? a. Write a story describing Jim’s movements at the store that would correspond to the motion represented by the graph. Answers will vary. b. When does Jim have a position of 6.0 m? from 8.0 to 18.0 s, 53.0 to 56.0 s, and at 43.0 s c. How much time passes between when Jim enters the aisle and when he gets to a position of 12.0 m? What is Jim’s average velocity between 37.0 s and 46.0 s? t 33.0 s Using the points (37.0 s, 12.0 m) and (46.0 s, 3.00 m) df di 3.00 m 12.0 m v 46.0 s 37.0 s tf ti 1.00 m/s d vt d v 48.0 km 36.0 km/h so t 1.33 h Car A has started 1.33 h before the clock started. t 6.0 h 1.33 h 7.3 h 26 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. so 48.0 km (36.0 km/h)t (48.0 km/h)(t 0.50 h) Figure 2-31 Chapter 2 continued Thinking Critically page 54 61. Apply Calculators Members of a physics class stood 25 m apart and used stopwatches to measure the time which a car traveling on the highway passed each person. Their data are shown in Table 2-3. Table 2-3 Position v. Time Position (m) 0.0 0.0 1.3 25.0 2.7 50.0 3.6 75.0 5.1 100.0 5.9 125.0 7.0 150.0 8.6 175.0 10.3 200.0 Use a graphing calculator to fit a line to a position-time graph of the data and to plot this line. Be sure to set the display range of the graph so that all the data fit on it. Find the slope of the line. What was the speed of the car? 200.0 Position (m) Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Time (s) 150.0 100.0 50.0 0.0 2.0 4.0 6.0 8.0 10.0 12.0 Time (s) The slope of the line and the speed of the car are 19.7 m/s. 62. Apply Concepts You plan a car trip for which you want to average 90 km/h. You cover the first half of the distance at an average speed of only 48 km/h. What must your average speed be in the second half of the trip to meet your goal? Is this reasonable? Note that the velocities are based on half the distance, not half the time. 720 km/h; No Physics: Principles and Problems Explanation: Assume you want to travel 90 km in 1 h. If you cover the first half of the distance at 48 km/h, then you’ve gone 45 km in d v 0.9375 h (because t ). This means you have used 93.75% of your time for the first half of the distance leaving 6.25% of the time to go the remaining 45 km. 45 km 0.0625 h v 720 km/h 63. Design an Experiment Every time a particular red motorcycle is driven past your friend’s home, his father becomes angry because he thinks the motorcycle is going too fast for the posted 25 mph (40 km/h) speed limit. Describe a simple experiment you could do to determine whether or not the motorcycle is speeding the next time it is driven past your friend’s house. There are actually several good possibilities for answers on this one. Two that should be among the most popular are briefly described here. 1) Get several people together and give everyone a watch. Synchronize the watches and stand along the street separated by a consistent distance, maybe 10 m or so. When the motorcycle passes, have each person record the time (at least to an accuracy of seconds) that the motorcycle crossed in front of them. Plot a position time graph, and compute the slope of the best-fit line. If the slope is greater than 25 mph, the motorcycle is speeding. 2) Get someone with a driver’s license to drive a car along the street at 25 mph in the same direction as you expect the motorcycle to go. If the motorcycle gets closer to the car (if the distance between them decreases), the motorcycle is speeding. If the distance between them stays the same, the motorcycle is driving at the speed limit. If the distance increases, the motorcycle is driving less than the speed limit. 64. Interpret Graphs Is it possible for an Solutions Manual 27 Chapter 2 continued object’s position-time graph to be a horizontal line? A vertical line? If you answer yes to either situation, describe the associated motion in words. It is possible to have a horizontal line as a position-time graph; this would indicate that the object’s position is not changing, or in other words, that it is not moving. It is not possible to have a position-time graph that is a vertical line, because this would mean the object is moving at an infinite speed. Writing in Physics page 54 65. Physicists have determined that the speed of light is 3.00108 m/s. How did they arrive at this number? Read about some of the series of experiments that were done to determine light’s speed. Describe how the experimental techniques improved to make the results of the experiments more accurate. 66. Some species of animals have good endurance, while others have the ability to move very quickly, but for only a short amount of time. Use reference sources to find two examples of each quality and describe how it is helpful to that animal. page 54 67. Convert each of the following time measurements to its equivalent in seconds. (Chapter 1) a. 58 ns 5.8108 s b. 0.046 Gs 4.6107 s c. 9270 ms 9.27 s d. 12.3 ks 1.23104 s 68. State the number of significant digits in the following measurements. (Chapter 1) a. 3218 kg 4 b. 60.080 kg 5 c. 801 kg 3 d. 0.000534 kg 3 69. Using a calculator, Chris obtained the following results. Rewrite the answer to each operation using the correct number of significant digits. (Chapter 1) a. 5.32 mm 2.1 mm 7.4200000 mm 7.4 mm b. 13.597 m 3.65 m 49.62905 m2 49.6 m2 c. 83.2 kg 12.804 kg 70.3960000 kg 70.4 kg Answers will vary. Examples of animals with high endurance to outlast predators or prey include mules, bears, and coyotes. Animals with the speed to quickly escape predators or capture prey include cheetahs, antelopes and deer. 28 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Answers will vary. Galileo attempted to determine the speed of light but was unsuccessful. Danish astronomer Olaus Roemer successfully measured the speed of light in 1676 by observing the eclipses of the moons of Jupiter. His estimate was 140,000 miles/s (225,308 km/s). Many others since have tried to measure it more accurately using rotating toothed wheels, rotating mirrors and the Kerr cell shutter. Cumulative Review CHAPTER 3 Accelerated Motion 3. Refer to the v-t graph of the toy train in Figure 3-6 to answer the following questions. Practice Problems Acceleration pages 57–64 12.0 page 61 1. A dog runs into a room and sees a cat at the other end of the room. The dog instantly stops running but slides along the wood floor until he stops, by slowing down with a constant acceleration. Sketch a motion diagram for this situation, and use the velocity vectors to find the acceleration vector. Time interval Velocity 1 2 3 v1 v2 v3 10.0 8.0 6.0 4.0 2.0 0.0 10.0 20.0 30.0 40.0 Time (s) ■ Figure 3-6 a. When is the train’s speed constant? Position Start Stop v2 v1 a 5.0 to 15.0 s b. During which time interval is the train’s acceleration positive? 0.0 to 5.0 s c. When is the train’s acceleration most negative? 2. Figure 3-5 is a v-t graph for Steven as he walks along the midway at the state fair. Sketch the corresponding motion diagram, complete with velocity vectors. 15.0 to 20.0 s 4. Refer to Figure 3-6 to find the average acceleration of the train during the following time intervals. a. 0.0 s to 5.0 s v v Velocity (m/s) Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Velocity (m/s) 3.1 2 1 a t t 2 1 10.0 m/s 0.0 m/s 5.0 s 0.0 s 2.0 m/s2 0 1 2 3 4 5 6 7 8 9 10 b. 15.0 s to 20.0 s Time (s) ■ v v 2 1 a t t Figure 3-5 2 Time (s) 0 1 2 3 4 56 8 7 9 10 1 4.0 m/s 10.0 m/s 20.0 s 15.0 s 1.2 m/s2 c. 0.0 s to 40.0 s v v 2 1 a t t 2 Physics: Principles and Problems 1 Solutions Manual 29 Chapter 3 continued 0.0 m/s 0.0 m/s a. What is the average acceleration of the bus while braking? 40.0 s 0.0 s 0.0 m/s2 v a t 5. Plot a v-t graph representing the following motion. An elevator starts at rest from the ground floor of a three-story shopping mall. It accelerates upward for 2.0 s at a rate of 0.5 m/s2, continues up at a constant velocity of 1.0 m/s for 12.0 s, and then experiences a constant downward acceleration of 0.25 m/s2 for 4.0 s as it reaches the third floor. Velocity (m/s) 1.0 0.0 m/s 25 m/s 8.3 m/s2 3.0 s b. If the bus took twice as long to stop, how would the acceleration compare with what you found in part a? half as great (4.2 m/s2) 10. Rohith has been jogging to the bus stop for 2.0 min at 3.5 m/s when he looks at his watch and sees that he has plenty of time before the bus arrives. Over the next 10.0 s, he slows his pace to a leisurely 0.75 m/s. What was his average acceleration during this 10.0 s? v a t 0.0 5.0 10.0 15.0 20.0 0.75 m/s 3.5 m/s 10.0 s Time (s) 0.28 m/s2 11. If the rate of continental drift were to abruptly slow from 1.0 cm/yr to 0.5 cm/yr over the time interval of a year, what would be the average acceleration? v 36 m/s 4.0 m/s a 8.0 m/s2 t 4.0 s 7. The race car in the previous problem slows from 36 m/s to 15 m/s over 3.0 s. What is its average acceleration? v 15 m/s 36 m/s a 7.0 m/s2 t 3.0 s 8. A car is coasting backwards downhill at a speed of 3.0 m/s when the driver gets the engine started. After 2.5 s, the car is moving uphill at 4.5 m/s. If uphill is chosen as the positive direction, what is the car’s average acceleration? 4.5 m/s (3.0 m/s) v a 3.0 m/s2 t 2.5 s 9. A bus is moving at 25 m/s when the driver steps on the brakes and brings the bus to a stop in 3.0 s. 30 Solutions Manual 0.5 cm/yr 1.0 cm/yr v a t 1.0 yr 0.5 cm/yr2 Section Review 3.1 Acceleration pages 57–64 page 64 12. Velocity-Time Graph What information can you obtain from a velocity-time graph? The velocity at any time, the time at which the object had a particular velocity, the sign of the velocity, and the displacement. 13. Position-Time and Velocity-Time Graphs Two joggers run at a constant velocity of 7.5 m/s toward the east. At time t 0, one Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. page 64 6. A race car’s velocity increases from 4.0 m/s to 36 m/s over a 4.0-s time interval. What is its average acceleration? Chapter 3 continued is 15 m east of the origin and the other is 15 m west. a. What would be the difference(s) in the position-time graphs of their motion? Both lines would have the same slope, but they would rise from the d-axis at different points, 15 m, and 15 m. b. What would be the difference(s) in their velocity-time graphs? Their velocity-time graphs would be identical. 14. Velocity Explain how you would use a velocity-time graph to find the time at which an object had a specified velocity. Draw or imagine a horizontal line at the specified velocity. Find the point where the graph intersects this line. Drop a line straight down to the t-axis. This would be the required time. 15. Velocity-Time Graph Sketch a velocity-time graph for a car that goes east at 25 m/s for 100 s, then west at 25 m/s for another 100 s. Velocity (m/s) Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 25 100 25 200 Time (s) 16. Average Velocity and Average Acceleration A canoeist paddles upstream at 2 m/s and then turns around and floats downstream at 4 m/s. The turnaround time is 8 s. a. What is the average velocity of the canoe? Choose a coordinate system with the positive direction upstream. v v i f v 2 2 m/s (4 m/s) 2 1 m/s b. What is the average acceleration of the canoe? v a t v v t f i (4 m/s) (2 m/s) 8s 0.8 m/s2 17. Critical Thinking A police officer clocked a driver going 32 km/h over the speed limit just as the driver passed a slower car. Both drivers were issued speeding tickets. The judge agreed with the officer that both were guilty. The judgement was issued based on the assumption that the cars must have been going the same speed because they were observed next to each other. Are the judge and the police officer correct? Explain with a sketch, a motion diagram, and a position-time graph. No, they had the same position, not velocity. To have the same velocity, they would have had to have the same relative position for a length of time. Physics: Principles and Problems Solutions Manual 31 Chapter 3 continued Position Sketch Position-Time Graph Passing 4 position 3 2 1 0 Motion diagram Faster car t1 t0 Slower car t2 t3 t4 t0 t1 t2 t3 t4 t1 t2 t3 t4 Time Practice Problems 3.2 Motion with Constant Acceleration pages 65–71 page 65 18. A golf ball rolls up a hill toward a miniature-golf hole. Assume that the direction toward the hole is positive. a. If the golf ball starts with a speed of 2.0 m/s and slows at a constant rate of 0.50 m/s2, what is its velocity after 2.0 s? vf vi at 2.0 m/s (0.50 m/s2)(2.0 s) 1.0 m/s vf vi at 2.0 m/s (0.50 m/s2)(6.0 s) 1.0 m/s c. Describe the motion of the golf ball in words and with a motion diagram. The ball’s velocity simply decreased in the first case. In the second case, the ball slowed to a stop and then began rolling back down the hill. Time interval Velocity Position Velocity Time interval 1 2 3 4 d 6 5 19. A bus that is traveling at 30.0 km/h speeds up at a constant rate of 3.5 m/s2. What velocity does it reach 6.8 s later? vf vi at 30.0 km/h (3.5 m/s2)(6.8 s) 1 km 3600 s 1000 m 1h 120 km/h 32 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. b. What is the golf ball’s velocity if the constant acceleration continues for 6.0 s? Chapter 3 continued 20. If a car accelerates from rest at a constant 5.5 m/s2, how long will it take for the car to reach a velocity of 28 m/s? vf vi at b. 2.0 s At 2.0 s, v 78 m/s c. 2.5 s At 2.5 s, v 80 m/s vf vi so t a 23. Use dimensional analysis to convert an airplane’s speed of 75 m/s to km/h. 28 m/s 0.0 m/s 2 5.5 m/s (75 m/s) 2.7102 km/h 1 km 3600 s 1000 m 1h 5.1 s 21. A car slows from 22 m/s to 3.0 m/s at a constant rate of 2.1 m/s2. How many seconds are required before the car is traveling at 3.0 m/s? 24. A position-time graph for a pony running in a field is shown in Figure 3-12. Draw the corresponding velocity-time graph using the same time scale. Displacement (m) vf vi at v v a f i so t 3.0 m/s 22 m/s 2 2.1 m/s 9.0 s y x Time (s) ■ 80 Velocity (m/s) Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 82 78 74 72 1.0 2.0 3.0 Time (s) ■ Figure 3-11 Graph B represents constant speed. So graph A should be used for the following calculations. a. 1.0 s At 1.0 s, v 74 m/s Physics: Principles and Problems Position-time graph Speeds up x Speeds up 76 70 0.0 y Velocity (m/s) Displacement (m) page 67 22. Use Figure 3-11 to determine the velocity of an airplane that is speeding up at each of the following times. Figure 3-12 Slows down Time (s) Stops and turns around 25. A car is driven at a constant velocity of 25 m/s for 10.0 min. The car runs out of gas and the driver walks in the same direction at 1.5 m/s for 20.0 min to the nearest gas station. The driver takes 2.0 min to fill a gasoline can, then walks back to the car at 1.2 m/s and eventually drives home at 25 m/s in the direction opposite that of the original trip. a. Draw a v-t graph using seconds as your time unit. Calculate the distance the driver walked to the gas station to find the time it took him to walk back to the car. Solutions Manual 33 Chapter 3 continued vf vi at 25 v v a 20 10 5 Time (s) 3500 4000 3000 2500 2000 1500 5 10 1000 0 500 Velocity (m/s) 15 (v v ) f i v 2 2 20 d vt 25 distance the driver walked to the gas station: (v v )t 2 f i (22 m/s 44 m/s)(11 s) 2 d vt (1.5 m/s)(20.0 min)(60 s/min) 1800 m 1.8 km time to walk back to the car: 1800 m 1.2 m/s t 1500 s 25 min b. Draw a position-time graph for the situation using the areas under the velocity-time graph. 1.2102 m 28. A car accelerates at a constant rate from 15 m/s to 25 m/s while it travels a distance of 125 m. How long does it take to achieve this speed? (v v ) v f i v 2 2 d vt (v v )t 2 f i 20,000 (vf vi) (2)(125 m) 25 m/s 15 m/s 10,000 25 s 5000 0 1000 2000 3000 4000 5000 Time (s) page 69 26. A skateboarder is moving at a constant velocity of 1.75 m/s when she starts up an incline that causes her to slow down with a constant acceleration of 0.20 m/s2. How much time passes from when she begins to slow down until she begins to move back down the incline? 29. A bike rider pedals with constant acceleration to reach a velocity of 7.5 m/s over a time of 4.5 s. During the period of acceleration, the bike’s displacement is 19 m. What was the initial velocity of the bike? v (v v ) f i v 2 2 (v v )t 2 f i d vt 2d t so vi vf (2)(19 m) 4.5 s 7.5 m/s 0.94 m/s 34 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 2d t 15,000 Position (m) 27. A race car travels on a racetrack at 44 m/s and slows at a constant rate to a velocity of 22 m/s over 11 s. How far does it move during this time? v 15 d v 0.0 m/s 1.75 m/s 0.20 m/s f i 8.8 s t 2 Chapter 3 continued page 71 30. A man runs at a velocity of 4.5 m/s for 15.0 min. When going up an increasingly steep hill, he slows down at a constant rate of 0.05 m/s2 for 90.0 s and comes to a stop. How far did he run? 1 2 d v1t1 (v2f v2i)t2 1 2 (4.5 m/s)(15.0 min)(60 s/min) (0.0 m/s 4.5 m/s)(90.0 s) 4.3103 m Velocity (m/s) 31. Sekazi is learning to ride a bike without training wheels. His father pushes him with a constant acceleration of 0.50 m/s2 for 6.0 s, and then Sekazi continues at 3.0 m/s for another 6.0 s before falling. What is Sekazi’s displacement? Solve this problem by constructing a velocity-time graph for Sekazi’s motion and computing the area underneath the graphed line. 3.0 1 2 6.0 12.0 Time (s) Part 1: Constant acceleration: 1 2 d1 (3.0 m/s)(6.0 s) 9.0 m Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Part 2: Constant velocity: d2 (3.0 m/s)(12.0 s 6.0 s) 18 m Thus d d1 d2 9.0 m 18 m 27 m 32. You start your bicycle ride at the top of a hill. You coast down the hill at a constant acceleration of 2.00 m/s2. When you get to the bottom of the hill, you are moving at 18.0 m/s, and you pedal to maintain that speed. If you continue at this speed for 1.00 min, how far will you have gone from the time you left the hilltop? Part 1: Constant acceleration: vf2 vi2 2a(df di) and di 0.00 m v2v2 2a f i so df since vi 0.00 m/s v2 2a df f (18.0 m/s)2 (2)(2.00 m/s ) 2 81.0 m Physics: Principles and Problems Solutions Manual 35 Chapter 3 continued Part 2: Constant velocity: d2 vt (18.0 m/s)(60.0 s) 1.08103 m Thus d d1 d2 81.0 m 1.08103 m 1.16103 m 33. Sunee is training for an upcoming 5.0-km race. She starts out her training run by moving at a constant pace of 4.3 m/s for 19 min. Then she accelerates at a constant rate until she crosses the finish line, 19.4 s later. What is her acceleration during the last portion of the training run? Part 1: Constant velocity: d vt (4.3 m/s)(19 min)(60 s/min) 4902 m Part 2: Constant acceleration: 1 2 df di vit at 2 2(df di vit) (2)(5.010 m 4902 m (4.3 m/s)(19.4 s)) a 2 2 t 3 (19.4 s) 0.077 m/s2 Section Review Motion with Constant Acceleration pages 65–71 page 71 34. Acceleration A woman driving at a speed of 23 m/s sees a deer on the road ahead and applies the brakes when she is 210 m from the deer. If the deer does not move and the car stops right before it hits the deer, what is the acceleration provided by the car’s brakes? vf2 vi2 2a(df di) v2v2 2(df di) f i a 0.0 m/s (23 m/s)2 (2)(210 m) 1.3 m/s2 35. Displacement If you were given initial and final velocities and the constant acceleration of an object, and you were asked to find the displacement, what equation would you use? vf2 vi2 2adf 36 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 3.2 Chapter 3 continued 36. Distance An in-line skater first accelerates from 0.0 m/s to 5.0 m/s in 4.5 s, then continues at this constant speed for another 4.5 s. What is the total distance traveled by the in-line skater? Accelerating v v 2 i f (t f) df vt f 0.0 m/s 5.0 m/s 2 (4.5 s) 11.25 m Constant speed df vft f (5.0 m/s)(4.5 s) 22.5 m total distance 11.25 m 22.5 m 34 m 37. Final Velocity A plane travels a distance of 5.0102 m while being accelerated uniformly from rest at the rate of 5.0 m/s2. What final velocity does it attain? vf2 vf2 2a(df di) and di 0, so vf2 vf2 2adf 2)( vf (0.0 m /s)2 2(5.0 m/s 5.01 02 m) Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 71 m/s 38. Final Velocity An airplane accelerated uniformly from rest at the rate of 5.0 m/s2 for 14 s. What final velocity did it attain? vf vi at f 0 (5.0 m/s2)(14 s) 7.0101 m/s 39. Distance An airplane starts from rest and accelerates at a constant 3.00 m/s2 for 30.0 s before leaving the ground. a. How far did it move? 1 2 df vit f at f2 (0.0 m/s)(30.0 s)2 (3.00 m/s2)(30.0 s)2 1 2 1.35103 m b. How fast was the airplane going when it took off? vf vi at f 0.0 m/s (3.00 m/s2)(30.0 s) 90.0 m/s Physics: Principles and Problems Solutions Manual 37 Chapter 3 continued Position 40. Graphs A sprinter walks up to the starting blocks at a constant speed and positions herself for the start of the race. She waits until she hears the starting pistol go off, and then accelerates rapidly until she attains a constant velocity. She maintains this velocity until she crosses the finish line, and then she slows down to a walk, taking more time to slow down than she did to speed up at the beginning of the race. Sketch a velocity-time and a position-time graph to represent her motion. Draw them one above the other on the same time scale. Indicate on your p-t graph where the starting blocks and finish line are. Velocity Time Time Starting blocks Finish line 41. Critical Thinking Describe how you could calculate the acceleration of an automobile. Specify the measuring instruments and the procedures that you would use. One person reads a stopwatch and calls out time intervals. Another person reads the speedometer at each time and records it. Plot speed versus time and find the slope. Practice Problems 3.3 page 74 42. A construction worker accidentally drops a brick from a high scaffold. a. What is the velocity of the brick after 4.0 s? Say upward is the positive direction. vf vi at, a g 9.80 m/s2 vf 0.0 m/s (9.80 m/s2)(4.0 s) 39 m/s when the upward direction is positive b. How far does the brick fall during this time? 1 2 d vit at 2 0 (9.80 m/s2)(4.0 s)2 1 2 78 m The brick falls 78 m. 43. Suppose for the previous problem you choose your coordinate system so that the opposite direction is positive. 38 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Free Fall pages 72–75 Chapter 3 continued a. What is the brick’s velocity after 4.0 s? Now the positive direction is downward. vf vi at, a g 9.80 m/s2 vf 0.0 m/s (9.80 m/s2)(4.0 s) 39 m/s when the downward direction is positive b. How far does the brick fall during this time? 1 2 d vit at 2, a g 9.80 m/s2 (0.0 m/s)(4.0 s) 12(9.80 m/s2)(4.0 s)2 78 m The brick still falls 78 m. 44. A student drops a ball from a window 3.5 m above the sidewalk. How fast is it moving when it hits the sidewalk? vf2 vi2 2ad, a g and vi 0 so vf 2gd (2)(9.8 0 m/s2 )(3.5 m) 8.3 m/s Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 45. A tennis ball is thrown straight up with an initial speed of 22.5 m/s. It is caught at the same distance above the ground. a. How high does the ball rise? a g, and at the maximum height, vf 0 vf2 vi2 2ad becomes vi2 2gd vi2 (22.5 m/s)2 (2)(9.80 m/s ) d 2 25.8 m 2g b. How long does the ball remain in the air? Hint: The time it takes the ball to rise equals the time it takes to fall. Calculate time to rise using vf vi at, with a g and vf 0 vi 22.5 m/s 9.80 m/s t 2 2.30 s g The time to fall equals the time to rise, so the time to remain in the air is tair 2trise (2)(2.30 s) 4.60 s 46. You decide to flip a coin to determine whether to do your physics or English homework first. The coin is flipped straight up. a. If the coin reaches a high point of 0.25 m above where you released it, what was its initial speed? Physics: Principles and Problems Solutions Manual 39 Chapter 3 continued vf2 vi2 2ad 2 vi v where a g f 2gd and vf 0 at the height of the toss, so 2)(0.2 (0.0 m /s)2 (2)(9. 80 m/s 5 m) vi 2.2 m/s b. If you catch it at the same height as you released it, how much time did it spend in the air? vf vi at where a g vi 2.2 m/s and vf 2.2 m/s vf vi t g 2.2 m/s 2.2 m/s 9.80 m/s2 0.45 s Section Review 3.3 Free Fall pages 72–75 a. How would the ball’s maximum height compare to that on Earth? At maximum height, vf 0, vi2 so df , or three times higher. 2g b. How would its flight time compare? 1 Time is found from df gt f2, or 2 tf g. Distance is multiplied by 3 and g is divided by 3, 2df so the flight time would be three times as long. 48. Velocity and Acceleration Suppose you throw a ball straight up into the air. Describe the changes in the velocity of the ball. Describe the changes in the acceleration of the ball. 40 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. page 75 47. Maximum Height and Flight Time Acceleration due to gravity on Mars is about one-third that on Earth. Suppose you throw a ball upward with the same velocity on Mars as on Earth. Chapter 3 continued Velocity is reduced at a constant rate as the ball travels upward. At its highest point, velocity is zero. As the ball begins to drop, the velocity begins to increase in the negative direction until it reaches the height from which it was initially released. At that point, the ball has the same speed it had upon release. The acceleration is constant throughout the ball’s flight. 49. Final Velocity Your sister drops your house keys down to you from the second floor window. If you catch them 4.3 m from where your sister dropped them, what is the velocity of the keys when you catch them? Upward is positive v 2 vi2 2ad where a g 2 v v i 2gd (0.0 m /s)2 (2)(9.80 m/ s2)(4.3 m) 9.2 m/s 50. Initial Velocity A student trying out for the football team kicks the football straight up in the air. The ball hits him on the way back down. If it took 3.0 s from the time when the student punted the ball until he gets hit by the ball, what was the football’s initial velocity? Choose a coordinate system with up as the positive direction and the origin at the punter. Choose the initial time at the punt and the final time at the top of the football’s flight. vf vi at f where a g Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. vi vf gt f 0.0 m/s (9.80 m/s2)(1.5 s) 15 m/s 51. Maximum Height When the student in the previous problem kicked the football, approximately how high did the football travel? vf2 vi2 2a(d) where a g vf2 vi2 d 2g (0.0 m/s) (15 m/s) 2 2 2 (2)(9.80 m/s ) 11 m 52. Critical Thinking When a ball is thrown vertically upward, it continues upward until it reaches a certain position, and then it falls downward. At that highest point, its velocity is instantaneously zero. Is the ball accelerating at the highest point? Devise an experiment to prove or disprove your answer. The ball is accelerating; its velocity is changing. Take a strobe photo to measure its position. From photos, calculate the ball’s velocity. Physics: Principles and Problems Solutions Manual 41 Chapter 3 continued Concept Mapping page 80 53. Complete the following concept map using the following symbols or terms: d, velocity, m/s2, v, m, acceleration. 30 Velocity (m/s) Chapter Assessment 25 20 15 10 5 0 5 10 15 20 25 30 35 Time (s) Quantities of motion ■ position velocity acceleration d v a m m/s m/s2 Mastering Concepts page 80 54. How are velocity and acceleration related? (3.1) Acceleration is the change in velocity divided by the time interval in which it occurs: it is the rate of change of velocity. a. an object that is slowing down, but has a positive acceleration if forward is the positive direction, a car moving backward at decreasing speed b. an object that is speeding up, but has a negative acceleration in the same coordinate system, a car moving backward at increasing speed 56. Figure 3-16 shows the velocity-time graph for an automobile on a test track. Describe how the velocity changes with time. (3.1) The car starts from rest and increases its speed. As the car’s speed increases, the driver shifts gears. 57. What does the slope of the tangent to the curve on a velocity-time graph measure? (3.1) instantaneous acceleration 58. Can a car traveling on an interstate highway have a negative velocity and a positive acceleration at the same time? Explain. Can the car’s velocity change signs while it is traveling with constant acceleration? Explain. (3.1) Yes, a car’s velocity is positive or negative with respect to its direction of motion from some point of reference. One direction of motion is defined as positive, and velocities in that direction are considered positive. The opposite direction of motion is considered negative; all velocities in that direction are negative. An object undergoing positive acceleration is either increasing its velocity in the positive direction or reducing its velocity in the negative direction. A car’s velocity can change signs when experiencing constant acceleration. For example, it can be traveling right, while the acceleration is to the left. The car slows down, stops, and then starts accelerating to the left. 59. Can the velocity of an object change when its acceleration is constant? If so, give an example. If not, explain. (3.1) Yes, the velocity of an object can change when its acceleration is constant. Example: dropping a book. The 42 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 55. Give an example of each of the following. (3.1) Figure 3-16 Chapter 3 continued longer it drops, the faster it goes, but the acceleration is constant at g. 60. If an object’s velocity-time graph is a straight line parallel to the t-axis, what can you conclude about the object’s acceleration? (3.1) When the velocity-time graph is a line parallel to the t-axis, the acceleration is zero. 61. What quantity is represented by the area under a velocity-time graph? (3.2) the change in displacement 62. Write a summary of the equations for position, velocity, and time for an object experiencing motion with uniform acceleration. (3.2) (vf vi) a vf vi v 2 2 d vt vf vi t 2 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. pages 80–81 66. Does a car that is slowing down always have a negative acceleration? Explain. No, if the positive axis points in the direction opposite the velocity, the acceleration will be positive. 67. Croquet A croquet ball, after being hit by a mallet, slows down and stops. Do the velocity and acceleration of the ball have the same signs? No, they have opposite signs. No, a 0 when velocity is constant. vf vi at f assuming ti 0, then t t f vf vi d tf 2 Applying Concepts 68. If an object has zero acceleration, does it mean its velocity is zero? Give an example. tf v Student answers will vary. Some examples are a steel ball, a rock, and a person falling through small distances. 63. Explain why an aluminum ball and a steel ball of similar size and shape, dropped from the same height, reach the ground at the same time. (3.3) All objects accelerate toward the ground at the same rate. 64. Give some examples of falling objects for which air resistance cannot be ignored. (3.3) Student answers will vary. Some examples are sheets of paper, parachutes, leaves, and feathers. 69. If an object has zero velocity at some instant, is its acceleration zero? Give an example. No, a ball rolling uphill has zero velocity at the instant it changes direction, but its acceleration is nonzero. 70. If you were given a table of velocities of an object at various times, how would you find out whether the acceleration was constant? Draw a velocity-time graph and see whether the curve is a straight line or v calculate accelerations using a t and compare the answers to see if they are the same. 71. The three notches in the graph in Figure 3-16 occur where the driver changed gears. Describe the changes in velocity and acceleration of the car while in first gear. Is the acceleration just before a gear change larger or smaller than the acceleration just after the change? Explain your answer. 65. Give some examples of falling objects for which air resistance can be ignored. (3.3) Physics: Principles and Problems Solutions Manual 43 Chapter 3 continued Velocity increases rapidly at first, then more slowly. Acceleration is greatest at the beginning but is reduced as velocity increases. Eventually, it is necessary for the driver to shift into second gear. The acceleration is smaller just before the gear change because the slope is less at that point on the graph. Once the driver shifts and the gears engage, acceleration and the slope of the curve increase. 72. Use the graph in Figure 3-16 and determine the time interval during which the acceleration is largest and the time interval during which the acceleration is smallest. The acceleration is largest during an interval starting at t 0 and lasting 1 about s. It is smallest beyond 33 s. 2 Displacement Displacement 73. Explain how you would walk to produce each of the position-time graphs in Figure 3-17. C D B A Time E F G H Time ■ Figure 3-17 Walk in the positive direction at a constant speed. Walk in the positive direction at an increasing speed for a short time; keep walking at a moderate speed for twice that amount of time; slow down over a short time and stop; remain stopped; and turn around and repeat the procedure until the original position is reached. Velocity Velocity Displacement ■ Time 44 Solutions Manual Time Figure 3-18 Time Velocity Displacement Displacement Time Time Time Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 74. Draw a velocity-time graph for each of the graphs in Figure 3-18. Chapter 3 continued 75. An object shot straight up rises for 7.0 s before it reaches its maximum height. A second object falling from rest takes 7.0 s to reach the ground. Compare the displacements of the two objects during this time interval. Both objects traveled the same distance. The object that is shot straight upward rises to the same height from which the other object fell. 76. The Moon The value of g on the Moon is one-sixth of its value on Earth. a. Would a ball that is dropped by an astronaut hit the surface of the Moon with a greater, equal, or lesser speed than that of a ball dropped from the same height to Earth? The ball will hit the Moon with a lesser speed because the acceleration due to gravity is less on the Moon. b. Would it take the ball more, less, or equal time to fall? Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. The ball will take more time to fall. 77. Jupiter The planet Jupiter has about three times the gravitational acceleration of Earth. Suppose a ball is thrown vertically upward with the same initial velocity on Earth and on Jupiter. Neglect the effects of Jupiter’s atmospheric resistance and assume that gravity is the only force on the ball. a. How does the maximum height reached by the ball on Jupiter compare to the maximum height reached on Earth? The relationship between d and g is (vf2 vi2) an inverse one: df . 2g If g increases by three times, or (vf2 vi2) 1 df , df changes by . 2(3g) 3 Therefore, a ball on Jupiter would 1 rise to a height of that on Earth. 3 Physics: Principles and Problems b. If the ball on Jupiter were thrown with an initial velocity that is three times greater, how would this affect your answer to part a? With vf 0, the value df is directly proportional to the square of initial (3vi)2 velocity, vi. That is, df vf2 . 2g On Earth, an initial velocity three times greater results in a ball rising nine times higher. On Jupiter, however, the height of nine times higher would be reduced to only three times higher because of df’s inverse relationship to a g that is three times greater. 78. Rock A is dropped from a cliff and rock B is thrown upward from the same position. a. When they reach the ground at the bottom of the cliff, which rock has a greater velocity? Rock B hits the ground with a greater velocity. b. Which has a greater acceleration? They have the same acceleration, the acceleration due to gravity. c. Which arrives first? rock A Mastering Problems 3.1 Acceleration pages 81–82 Level 1 79. A car is driven for 2.0 h at 40.0 km/h, then for another 2.0 h at 60.0 km/h in the same direction. a. What is the car’s average velocity? Total distance: 80.0 km 120.0 km 200.0 km total time is 4.0 hours, so, d 200.0 km v 5.0101 km/h t 4.0 h b. What is the car’s average velocity if it is driven 1.0102 km at each of the two speeds? Solutions Manual 45 Chapter 3 continued Total distance is 2.03102 km; 1.0102 km 40.0 km/h 1.0102 km 60.0 km/h total time 4.2 h d 2.0102 km so v t 4.2 h 48 km/h 80. Find the uniform acceleration that causes a car’s velocity to change from 32 m/s to 96 m/s in an 8.0-s period. v a t v2 v1 t 96 m/s 32 m/s 8.0 s 8.0 m/s2 81. A car with a velocity of 22 m/s is accelerated uniformly at the rate of 1.6 m/s2 for 6.8 s. What is its final velocity? vf vi atf 22 m/s (1.6 m/s2)(6.8 s) 33 m/s 30.0 20.0 10.0 0.0 5.0 10.0 15.0 20.0 25.0 30.0 Time (s) ■ Figure 3-19 a. during the first 5.0 s of travel v a t 30.0 m/s 0.0 m/s 5.0 s 6.0 m/s2 b. between 5.0 s and 10.0 s v a t 30.0 m/s 30.0 m/s 5.0 s 0.0 m/s2 46 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Velocity (m/s) 82. Refer to Figure 3-19 to find the acceleration of the moving object at each of the following times. Chapter 3 continued c. between 10.0 s and 15.0 s v a t 20.0 m/s 30.0 m/s 5.0 s 2.0 m/s2 d. between 20.0 s and 25.0 s v a t 0.0 m/s 20.0 m/s 5.0 s 4.0 m/s2 Velocity (m/s) Level 2 83. Plot a velocity-time graph using the information in Table 3-4, and answer the following questions. Velocity v. Time 16.0 Time (s) 12.0 0.00 4.00 8.00 1.00 8.00 4.00 0.00 Time (s) 2.00 4.00 6.00 8.00 10.0 12.0 4.00 8.00 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Table 3-4 a. During what time interval is the object speeding up? Slowing down? Velocity (m/s) 2.00 12.0 3.00 14.0 4.00 16.0 5.00 16.0 6.00 14.0 7.00 12.0 8.00 8.00 9.00 4.00 speeding up from 0.0 s to 4.0 s; slowing down from 5.0 s to 10.0 s 10.0 0.00 11.0 4.00 b. At what time does the object reverse direction? 12.0 8.00 at 10.0 s c. How does the average acceleration of the object in the interval between 0.0 s and 2.0 s differ from the average acceleration in the interval between 7.0 s and 12.0 s? v a t between 0.0 s and 2.0 s: 12.0 m/s 4.0 m/s a 4.0 m/s2 2.0 s 0.0 s between 7.0 s and 12.0 s: 8.0 m/s 12.0 m/s a 4.0 m/s2 12.0 s 7.0 s Physics: Principles and Problems Solutions Manual 47 Chapter 3 continued 84. Determine the final velocity of a proton that has an initial velocity of 2.35105 m/s and then is accelerated uniformly in an electric field at the rate of 1.101012 m/s2 for 1.50107 s. vf vi at f a. t 0.0 s and t 5.0 s 2.35105 m/s (1.101012 7.0104 3.2 Motion with Constant Acceleration page 82 Level 1 87. Refer to Figure 3-19 to find the distance traveled during the following time intervals. m/s2)(1.50107 s) 1 2 Area I bh (5.0 s)(30.0 m/s) 1 2 m/s Level 3 85. Sports Cars Marco is looking for a used sports car. He wants to buy the one with the greatest acceleration. Car A can go from 0 m/s to 17.9 m/s in 4.0 s; car B can accelerate from 0 m/s to 22.4 m/s in 3.5 s; and car C can go from 0 to 26.8 m/s in 6.0 s. Rank the three cars from greatest acceleration to least, specifically indicating any ties. 75 m b. t 5.0 s and t 10.0 s Area II bh (10.0 s 5.0 s)(30.0 m/s) 150 m c. t 10.0 s and t 15.0 s 1 2 Area III Area IV bh bh (15.0 s 10.0 s)(20.0 m/s) Car A: v 17.9 m/s 0 m/s a 4.5 m/s2 4.0 s 0.0 s t Car B: 12(15.0 s 10.0 s)(10.0 m/s) 125 m v 22.4 m/s 0 m/s a 6.4 m/s2 3.5 s 0.0 s t Car C: Car B has the greatest acceleration of 6.4 m/s2. Using significant digits, car A and car C tied at 4.5 m/s2. Area I Area II (Area III Area IV) Area V IV 75 m 150 m 125 m 1 2 bh bh 75 m 150 m 125 m (20.0 s 15.0 s)(20.0 m/s) 86. Supersonic Jet A supersonic jet flying at 145 m/s experiences uniform acceleration at the rate of 23.1 m/s2 for 20.0 s. a. What is its final velocity? vf vi at f 145 m/s (23.1 m/s2)(20.0 s) 607 m/s b. The speed of sound in air is 331 m/s. What is the plane’s speed in terms of the speed of sound? 12(25.0 s 20.0 s) 5.0102 m Level 2 88. A dragster starting from rest accelerates at 49 m/s2. How fast is it going when it has traveled 325 m? vf2 vi2 2a(df di) 2 di) vf v f i 2a(d 607 m/s 331 m/s N 1.83 times the speed of sound 48 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. v 26.8 m/s 0 m/s a 4.5 m/s2 6.0 s 0.0 s t d. t 0.0 s and t 25.0 s Chapter 3 continued 2 (0.0 m/s) (2)(49 m/s2) (325 m 0.0 m) 180 m/s 89. A car moves at 12 m/s and coasts up a hill with a uniform acceleration of 1.6 m/s2. a. What is its displacement after 6.0 s? 1 2 df vit f at f2 (12 m/s)(6.0 s) 12(1.6 m/s2)(6.0 s)2 43 m b. What is its displacement after 9.0 s? 1 2 df vit f at f2 (12 m/s)(9.0 s) 12(1.6 m/s2)(9.0 s)2 43 m The car is on the way back down the hill. The odometer will show that the car traveled 45 m up the hill 2 m back down 47 m. 90. Race Car A race car can be slowed with a constant acceleration of 11 m/s2. a. If the car is going 55 m/s, how many meters will it travel before it stops? Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. vf2 vi2 2adf vf2 vi2 df 2a (0.0 m/s)2 (55 m/s)2 (2)(11 m/s ) 2 1.4102 m b. How many meters will it take to stop a car going twice as fast? vf2 vi2 2adf vf2 vi2 df 2a (0.0 m/s)2 (110 m/s)2 (2)(11 m/s ) 550 m, 2 which is about 4 times longer than when going half the speed. 91. A car is traveling 20.0 m/s when the driver sees a child standing on the road. She takes 0.80 s to react, then steps on the brakes and slows at 7.0 m/s2. How far does the car go before it stops? reaction displacement dr (20.0 m/s)(0.80 s) 16 m Physics: Principles and Problems Solutions Manual 49 Chapter 3 continued therefore vf2 vi2 df 1 t 0 and apolicet vspeeder 0 2 2a braking displacement (0.0 m/s)2 (20.0 m/s)2 db (2)(7.0 m/s2) 2vspeeder t apolice (2)(30.0 m/s) 7.0 m/s 29 m 2 total displacement is dr db 16 m 29 m 45 m Level 3 92. Airplane Determine the displacement of a plane that experiences uniform acceleration from 66 m/s to 88 m/s in 12 s. (v v )t 2 8.6 s After t 8.6 s, the police car’s velocity was vf vi at 0.0 m/s (7.0 m/s2)(8.6 s) 6.0101 m/s f i df vt (88 m/s 66 m/s)(12 s) 2 9.2102 m 93. How far does a plane fly in 15 s while its velocity is changing from 145 m/s to 75 m/s at a uniform rate of acceleration? (v v )t 2 f i d vt a. Determine whether the car hits the barrier. The car will travel d vt (25.0 m/s)(0.75 s) 18.8 m (Round off at the end.) before the driver applies the brakes. Convert km/h to m/s. 1.6103 m 94. Police Car A speeding car is traveling at a constant speed of 30.0 m/s when it passes a stopped police car. The police car accelerates at 7.0 m/s2. How fast will it be going when it catches up with the speeding car? dspeeder vspeedert 1 dpolice vi policet apolicet 2 2 1 2 vspeedert vi policet apolicet 2 (90.0 km/h)(1000 m/km) 3600 s/h vi 25.0 m/s vf2 vi2 2a(df di) v2v2 f i df di 2a (0.0 m/s)2 (25.0 m/s)2 (2)(10.0 m/s ) 18.8 m 2 5.0101 m, yes it hits the barrier since vi police 0 then 1 2 vspeedert apolicet 2 1 2 0 apolicet 2 vspeedert 0 t apolicet vspeeder 1 2 50 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. (75 m/s 145 m/s)(15 m/s) 2 95. Road Barrier The driver of a car going 90.0 km/h suddenly sees the lights of a barrier 40.0 m ahead. It takes the driver 0.75 s to apply the brakes, and the average acceleration during braking is 10.0 m/s2. Chapter 3 continued b. What is the maximum speed at which the car could be moving and not hit the barrier 40.0 m ahead? Assume that the acceleration doesn’t change. dtotal dconstant ddecelerating dc vt (0.75 s)v dd 2 v2 v2 2a 2(10.0 m/s ) 2 v 2 2 v 40 m (0.75 s)v 2 20.0 m/s (15 m/s)v 800 m2/s2 0 Using the quadratic equation: v 22 m/s (The sense of the problem excludes the negative value.) 3.3 Free Fall page 82 Level 1 96. A student drops a penny from the top of a tower and decides that she will establish a coordinate system in which the direction of the penny’s motion is positive. What is the sign of the acceleration of the penny? Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 0.0 m/s (9.80 m/s2)(8.0 s) 78 m/s (downward) Choose the coordinate system to have the origin where the stone is at rest and positive to be upward. 1 2 1 2 vit gtf 2 df vit atf2 where a g 20.0 m/s v2 vi gtf b. What is the stone’s displacement during this time? 40.0 m 02 vf vi atf where a g 0.0 m (9.80 m/s2)(8.0 s)2 1 2 3.1102 m 99. A bag is dropped from a hovering helicopter. The bag has fallen for 2.0 s. What is the bag’s velocity? How far has the bag fallen? Velocity: vf vi atf where a g vi gtf 0.0 m/s (9.80 m/s2)(2.0 s) 2.0101 m/s The direction of the velocity is positive, and velocity is increasing. Therefore, the acceleration is also positive. Displacement: 97. Suppose an astronaut drops a feather from 1.2 m above the surface of the Moon. If the acceleration due to gravity on the Moon is 1.62 m/s2 downward, how long does it take the feather to hit the Moon’s surface? vitf gt2 df vitf atf2 (0 m/s)tf atf2 tf 1.2 s (1.62 m/s ) 2df a (2)(1.2 m) 2 98. A stone that starts at rest is in free fall for 8.0 s. a. Calculate the stone’s velocity after 8.0 s. Physics: Principles and Problems 1 2 df vitf atf2 where a g 1 2 0.0 m (9.80 m/s2)(2.0 s)2 1 2 2.0101 m Level 2 100. You throw a ball downward from a window at a speed of 2.0 m/s. How fast will it be moving when it hits the sidewalk 2.5 m below? Choose a coordinate system with the positive direction downward and the origin at the point where the ball leaves your hand. Solutions Manual 51 Chapter 3 continued vf2 vi2 2adf where a g 2 vf v f i 2gd (2.0 m /s)2 (2)(9. 80 m/ s2)(2.5 m) 7.3 m/s 101. If you throw the ball in the previous problem up instead of down, how fast will it be moving when it hits the sidewalk? Choose the same coordinate system. vf2 vi2 2adf where a g 2 vf v f i 2gd (2.0 m /s)2 (2)(9. 80 m/ s2)(2.5 m) 7.3 m/s (df is the displacement, not the total distance traveled.) Level 3 102. Beanbag You throw a beanbag in the air and catch it 2.2 s later. a. How high did it go? vf vi atf where a g vi vf gtf 0.0 m/s (9.80 m/s2)(1.1 s) 11 m/s Now you can use an equation that includes the displacement. 1 2 1 di vitf gtf2 2 df di vitf atf2 where a g 0.0 m (11 m/s)(1.1 s) 12(9.80 m/s2)(1.1 s)2 6.2 m b. What was its initial velocity? vi 11 m/s 52 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Choose a coordinate system with the upward direction positive and the origin at the point where the beanbag left your hand. Assume that you catch the beanbag at the same place where you threw it. Therefore, the time to reach the maximum height is half of the time in the air. Choose ti to be the time when the beanbag left your hand and tf to be the time at the maximum height. Each formula that you know includes vi, so you will have to calculate that first. Chapter 3 continued Mixed Review pages 82–84 Level 1 103. A spaceship far from any star or planet experiences uniform acceleration from 65.0 m/s to 162.0 m/s in 10.0 s. How far does it move? Choosing a coordinate system with the origin at the point where the speed is 65.0 m/s and given vi 65.0 m/s, vf 162.0 m/s, and tf 10.0 s and needing df, we use the formula with the average velocity. 1 2 df di (vi vf)tf 1 2 df 0 (65.0 m/s 162.0 m/s)(10.0 s) 1.14103 m 104. Figure 3-20 is a strobe photo of a horizontally moving ball. What information about the photo would you need and what measurements would you make to estimate the acceleration? v v 2 i f tf df vtf 0.0 m/s 4.0 m/s 2 (4.0 s) 8.0 m 106. A weather balloon is floating at a constant height above Earth when it releases a pack of instruments. a. If the pack hits the ground with a velocity of 73.5 m/s, how far did the pack fall? vf2 vi2 2adf vf2 vi2 df 2a (73.5 m/s)2 (0.00 m/s)2 (2)(9.80 m/s ) 2 276 m b. How long did it take for the pack to fall? vf vi atf where a g vf vi tf g 73.5 m/s 0.00 m/s 2 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 9.80 m/s 7.50 s ■ Figure 3-20 You need to know the time between flashes and the distance between the first two images and the distance between the last two. From these, you get two velocities. Between these two velocities, a time interval of t seconds occurred. Divide the difference between the two velocities by t. 105. Bicycle A bicycle accelerates from 0.0 m/s to 4.0 m/s in 4.0 s. What distance does it travel? Level 2 107. Baseball A baseball pitcher throws a fastball at a speed of 44 m/s. The acceleration occurs as the pitcher holds the ball in his hand and moves it through an almost straight-line distance of 3.5 m. Calculate the acceleration, assuming that it is constant and uniform. Compare this acceleration to the acceleration due to gravity. vf2 vi2 2adf vf2 vi2 a 2df (44 m/s)2 0 (2)(3.5 m) 2.8102 m/s2 2.8102 m/s2 29, or 29 times g 9.80 m/s2 Physics: Principles and Problems Solutions Manual 53 Chapter 3 continued 108. The total distance a steel ball rolls down an incline at various times is given in Table 3-5. vf2 Time (s) Distance (m) 0.0 0.0 1.0 2.0 2.0 8.0 3.0 18.0 4.0 32.0 5.0 50.0 b. Over what time interval does the acceleration take place? (vf vi)t d 2 2df (2)(0.020 m) 3.510 m/s 0.0 m/s t 3 vf vi 11106 s 50.0 11 microseconds 110. Sleds Rocket-powered sleds are used to test the responses of humans to acceleration. Starting from rest, one sled can reach a speed of 444 m/s in 1.80 s and can be brought to a stop again in 2.15 s. a. Calculate the acceleration of the sled when starting, and compare it to the magnitude of the acceleration due to gravity, 9.80 m/s2. 40.0 Position (m) 2 (2)(0.020 m) 3.1108 m/s2 a. Draw a position-time graph of the motion of the ball. When setting up the axes, use five divisions for each 10 m of travel on the d-axis. Use five divisions for 1 s of time on the t-axis. 30.0 20.0 v v2 v1 1.0 2.0 3.0 4.0 5.0 444 m/s 0.00 m/s 1.80 s Time (s) b. Calculate the distance the ball has rolled at the end of 2.2 s. After 2.2 seconds the ball has rolled approximately 10 m. 109. Engineers are developing new types of guns that might someday be used to launch satellites as if they were bullets. One such gun can give a small object a velocity of 3.5 km/s while moving it through a distance of only 2.0 cm. a. What acceleration does the gun give this object? Solutions Manual 247 m/s2 247 m/s2 25 times g 9.80 m/s2 b. Find the acceleration of the sled as it is braking and compare it to the magnitude of the acceleration due to gravity. v t v2 v1 a t 0.00 m/s 444 m/s 2.15 s 207 m/s2 207 m/s2 21 times g 9.80 m/s2 Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. a t t 10.0 54 3 2df Distance v. Time 0.0 or vf2 2adf (3.510 m/s) a Table 3-5 Top of incline vf2 vi2 2adf Chapter 3 continued 111. The velocity of a car changes over an 8.0-s time period, as shown in Table 3-6. 1 2 d bh bh Table 3-6 (5.0 s)(20.0 m/s 0.0 m/s) Velocity v. Time (8.0 s 5.0 s)(20.0 m/s) Time (s) 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 Velocity (m/s) 0.0 4.0 8.0 12.0 16.0 20.0 20.0 20.0 20.0 a. Plot the velocity-time graph of the motion. 1 2 110 m e. Find the slope of the line between t 0.0 s and t 4.0 s. What does this slope represent? v 16.0 m/s 0.00 m/s a t 4.0 s 0.0 s 4.0 m/s2, acceleration f. Find the slope of the line between t 5.0 s and t 7.0 s. What does this slope indicate? v 20.0 m/s 20.0 m/s a t Velocity (m/s) 20.0 0.0 m/s2, constant velocity 16.0 12.0 8.0 4.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 Time (s) Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 7.0 s 5.0 s b. Determine the displacement of the car during the first 2.0 s. Find the area under the v-t curve. Level 3 112. A truck is stopped at a stoplight. When the light turns green, the truck accelerates at 2.5 m/s2. At the same instant, a car passes the truck going 15 m/s. Where and when does the truck catch up with the car? Car: df di vtf dcar di vcartf vcartf 0 (15 m/s)tf 1 2 Truck: df di vitf atf2 d bh 1 (2.0 s)(8.0 m/s 0.0 m/s) 2 8.0 m c. What displacement does the car have during the first 4.0 s? Find the area under the v-t curve. 1 2 d bh (4.0 s)(16.0 m/s 0.0 m/s) 1 2 32 m d. What is the displacement of the car during the entire 8.0 s? Find the area under the v-t curve. Physics: Principles and Problems 1 2 1 2 dtruck atrucktf2 0 0 (2.5 m/s2)tf2 1 2 When the truck catches up, the displacements are equal. 1 2 vcartf atrucktf2 1 2 0 atrucktf2 vcartf 0 tfatrucktf vcar 1 2 Solutions Manual 55 therefore 1 tf 0 and atrucktf vcar 0 2 2vcar tf atruck (2)(15 m/s) 2.5 m/s Velocity (m/s) Fist Displacement (cm) Chapter 3 continued 12 s df (15 m/s)tf 10.0 5.0 0.0 5.0 0.0 5.0 10.0 15.0 5.0 10.0 15.0 20.0 25.0 30.0 (15 m/s)(12 s) Time (ms) 180 m ■ 113. Safety Barriers Highway safety engineers build soft barriers, such as the one shown in Figure 3-21, so that cars hitting them will slow down at a safe rate. A person wearing a safety belt can withstand an acceleration of 3.0102 m/s2. How thick should barriers be to safely stop a car that hits a barrier at 110 km/h? Figure 3-22 a. Use the velocity-time graph to describe the motion of George’s fist during the first 10 ms. The fist moves downward at about 13 m/s for about 4 ms. It then suddenly comes to a halt (accelerates). b. Estimate the slope of the velocity-time graph to determine the acceleration of his fist when it suddenly stops. 0 m/s (13 m/s) 7.5 ms 4.0 ms v a t 3.7103 m/s2 Figure 3-21 (110 km/h)(1000 m/km) 3600 s/h vi 31 m/s 3.7103 m/s2 3.8102 9.80 m/s2 vf2 vi2 2adf The acceleration is about 380g. with vf 0 m/s, vi2 2adf, or vi2 (31 m/s) df 2 2 2a 2 (2)(3.010 m/s ) 1.6 m thick 114. Karate The position-time and velocitytime graphs of George’s fist breaking a wooden board during karate practice are shown in Figure 3-22. 56 Solutions Manual d. Determine the area under the velocitytime curve to find the displacement of the fist in the first 6 ms. Compare this with the position-time graph. The area can be approximated by a rectangle: (13 m/s)(0.006 s) 8 cm This is in agreement with the position-time graph where the hand moves from 8 cm to 0 cm, for a net displacement of 8 cm. Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. ■ c. Express the acceleration as a multiple of the gravitational acceleration, g 9.80 m/s2. Chapter 3 continued 115. Cargo A helicopter is rising at 5.0 m/s when a bag of its cargo is dropped. The bag falls for 2.0 s. a. What is the bag’s velocity? vf vi atf where a g vi gtf 5.0 m/s (9.80 m/s2)(2.0 s) 15 m/s b. How far has the bag fallen? 1 df vitf atf2 where a g 2 1 2 vitf gtf2 (5.0 m/s)(2.0 s) 12(9.80 m/s2)(2.0 s)2 1.0101 m The bag has fallen 1.0101 m c. How far below the helicopter is the bag? The helicopter has risen df vitf (5.0 m/s2)(2.0 s) Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1.0101 m The bag is 1.0101 m below the origin and 2.0101 m below the helicopter. Thinking Critically page 84 116. Apply CBLs Design a lab to measure the distance an accelerated object moves over time. Use equal time intervals so that you can plot velocity over time as well as distance. A pulley at the edge of a table with a mass attached is a good way to achieve uniform acceleration. Suggested materials include a motion detector, CBL, lab cart, string, pulley, C-clamp, and masses. Generate distance-time and velocity-time graphs using different masses on the pulley. How does the change in mass affect your graphs? Students’ labs will vary. Students should find that a change in the mass over the edge of the table will not Physics: Principles and Problems change the distance the cart moves, because the acceleration is always the same: g. 117. Analyze and Conclude Which has the greater acceleration: a car that increases its speed from 50 km/h to 60 km/h, or a bike that goes from 0 km/h to 10 km/h in the same time? Explain. vf vi a t 60 km/h 50 km/h For car, a t 10 km/h t 10 km/h 0 km/h For bike, a t 10 km/h t The change in velocity is the same. 118. Analyze and Conclude An express train, traveling at 36.0 m/s, is accidentally sidetracked onto a local train track. The express engineer spots a local train exactly 1.00102 m ahead on the same track and traveling in the same direction. The local engineer is unaware of the situation. The express engineer jams on the brakes and slows the express train at a constant rate of 3.00 m/s2. If the speed of the local train is 11.0 m/s, will the express train be able to stop in time, or will there be a collision? To solve this problem, take the position of the express train when the engineer first sights the local train as a point of origin. Next, keeping in mind that the local train has exactly a 1.00102 m lead, calculate how far each train is from the origin at the end of the 12.0 s it would take the express train to stop (accelerate at 3.00 m/s2 from 36 m/s to 0 m/s). a. On the basis of your calculations, would you conclude that a collision will occur? Express: 1 2 df vitf atf2 (36.0 m/s)(12.0 s) Solutions Manual 57 Chapter 3 continued 12(3.00 m/s2)(12.0 s)2 432 m 216 m 216 m Local: df di vitf atf2 100 m (11.0 m/s)(12.0 s) 0 232 m On this basis, no collision will occur. b. The calculations that you made do not allow for the possibility that a collision might take place before the end of the 12 s required for the express train to come to a halt. To check this, take the position of the express train when the engineer first sights the local train as the point of origin and calculate the position of each train at the end of each second after the sighting. Make a table showing the distance of each train from the origin at the end of each second. Plot these positions on the same graph and draw two lines. Use your graph to check your answer to part a. d (Express) (m) (m) 1 111 35 2 122 66 3 133 95 4 144 120 5 155 145 6 166 162 7 177 179 8 188 192 9 199 203 10 210 210 11 221 215 12 232 216 d 250 200 150 Local 100 Express 50 2 4 6 Time (s) 8 10 12 t They collide between 6 and 7 s. Writing in Physics page 84 119. Research and describe Galileo’s contributions to physics. Student answers will vary. Answers should include Galileo’s experiments demonstrating how objects accelerate as they fall. Answers might include his use of a telescope to discover the moons of Jupiter and the rings of Saturn, and his reliance on experimental results rather than authority. 58 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. d (Local) Position (m) t (s) Chapter 3 continued Answers will vary. Because humans can experience negative effects, like blackouts, the designers of roller coasters need to structure the downward slopes in such a way that the coaster does not reach accelerations that cause blackouts. Likewise, engineers working on bullet trains, elevators, or airplanes need to design the system in such a way that allows the object to rapidly accelerate to high speeds, without causing the passengers to black out. Cumulative Review page 84 121. Solve the following problems. Express your answers in scientific notation. (Chapter 1) a. 6.2104 m 5.7103 m 6.3103 m b. 8.7108 km 3.4107 m Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 8.4108 km c. (9.21105 cm)(1.83108 cm) 1.69104 cm2 d. (2.63106 m)/(4.08106 s) 6.451013 m/s Position-Time Graph 400 300 Position (m) 120. Research the maximum acceleration a human body can withstand without blacking out. Discuss how this impacts the design of three common entertainment or transportation devices. 200 100 0 100 2 4 6 8 10 Time (s) Motion Diagram Challenge Problem page 75 You notice a water balloon fall past your classroom window. You estimate that it took the balloon about t seconds to fall the length of the window and that the window is about y meters high. Suppose the balloon started from rest, approximately how high above the top of the window was it released? Your answer should be in terms of t, y, g, and numerical constants. Down is positive. Work this problem in two stages. Stage 1 is falling the distance D to the top of the window. Stage 2 is falling the distance y from the top of the window to the bottom of the window. Stage 1: the origin is at the top of the fall. 122. The equation below describes the motion of an object. Create the corresponding position-time graph and motion diagram. Then write a physics problem that could be solved using that equation. Be creative. d (35.0 m/s) t 5.0 m (Chapter 2) vf12 vi12 2a(df1 di1) Graph and motion diagram indicate constant velocity motion with a velocity of 35.0 m/s and initial position of 5.0 m. Answers will vary for the create-a-problem part. df2 di1 vi1tf2 atf22 0 2g(D 0) vf1 2gD Stage 2: the origin is at the top of the window. 1 2 1 2 y 0 vf1t gt 2 0 2gD (t) gt 2 1 2 y gt 2gD t 2 1 y 2g t gt 2 2 D Physics: Principles and Problems Solutions Manual 59 CHAPTER 4 Forces in One Dimension 3. A cable pulls a crate at a constant speed across a horizontal surface. The surface provides a force that resists the crate’s motion. Practice Problems 4.1 Force and Motion pages 87–95 System page 89 For each of the following situations, specify the system and draw a motion diagram and a free-body diagram. Label all forces with their agents, and indicate the direction of the acceleration and of the net force. Draw vectors of appropriate lengths. x v 1. A flowerpot falls freely from a windowsill. (Ignore any forces due to air resistance.) on crate v Fnet 0 Fpull on crate 4. A rope lifts a bucket at a constant speed. (Ignore air resistance.) y v FEarth’s mass on flowerpot v a0 Ffriction System v a Fnet y Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. v 2. A sky diver falls downward through the air at constant velocity. (The air exerts an upward force on the person.) System Frope on bucket v v a0 Fnet 0 y v Fair resistance v on diver FEarth’s mass on bucket v a 0 Fnet 0 System v FEarth’s mass 5. A rope lowers a bucket at a constant speed. (Ignore air resistance.) on diver y v Frope on bucket System v v a0 Fnet 0 v FEarth’s mass on bucket Physics: Principles and Problems Solutions Manual 61 Chapter 4 continued page 93 6. Two horizontal forces, 225 N and 165 N, are exerted on a canoe. If these forces are applied in the same direction, find the net horizontal force on the canoe. Fnet 225 N 165 N 3.90102 N in the direction of the two forces 7. If the same two forces as in the previous problem are exerted on the canoe in opposite directions, what is the net horizontal force on the canoe? Be sure to indicate the direction of the net force. 10. Inertia Can you feel the inertia of a pencil? Of a book? If you can, describe how. Yes, you can feel the inertia of either object by using your hand to give either object an acceleration; that is, try to change the objects velocity. 11. Free-Body Diagram Draw a free-body diagram of a bag of sugar being lifted by your hand at a constant speed. Specifically identify the system. Label all forces with their agents and make the arrows the correct lengths. Fnet 225 N 165 N 6.0101 N F hand on bag in the direction of the larger force 8. Three confused sleigh dogs are trying to pull a sled across the Alaskan snow. Alutia pulls east with a force of 35 N, Seward also pulls east but with a force of 42 N, and big Kodiak pulls west with a force of 53 N. What is the net force on the sled? Identify east as positive and the sled as the system. Fnet FAlutia on sled FSeward on sled FKodiak on sled Section Review 4.1 Force and Motion pages 87–95 System a0 F Earth’s mass on bag 12. Direction of Velocity If you push a book in the forward direction, does this mean its velocity has to be forward? No, it could be moving backward and you would be reducing that velocity. 13. Free-Body Diagram Draw a free-body diagram of a water bucket being lifted by a rope at a decreasing speed. Specifically identify the system. Label all forces with their agents and make the arrows the correct lengths. page 95 9. Force Identify each of the following as either a, b, or c: weight, mass, inertia, the push of a hand, thrust, resistance, air resistance, spring force, and acceleration. F rope on bucket a. a contact force b. a field force a c. not a force weight (b), mass (c), inertia (c), push of a hand (a), thrust (a), resistance (a), air resistance (a), spring force (a), acceleration (c) 62 Solutions Manual System F Earth’s mass on bucket Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 35 N 42 N 53 N 24 N Fnet 24 N east Sugar Chapter 4 continued 14. Critical Thinking A force of 1 N is the only force exerted on a block, and the acceleration of the block is measured. When the same force is the only force exerted on a second block, the acceleration is three times as large. What can you conclude about the masses of the two blocks? 18. In Figure 4-8, the block has a mass of 1.2 kg and the sphere has a mass of 3.0 kg. What are the readings on the two scales? (Neglect the masses of the scales.) Because m F/a and the forces are the same, the mass of the second block is one-third the mass of the first block. Practice Problems 4.2 Using Newton’s Laws pages 96–101 page 97 15. You place a watermelon on a spring scale at the supermarket. If the mass of the watermelon is 4.0 kg, what is the reading on the scale? The scale reads the weight of the watermelon: Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Fg mg (4.0 kg)(9.80 m/s2) 39 N 16. Kamaria is learning how to ice-skate. She wants her mother to pull her along so that she has an acceleration of 0.80 m/s2. If Kamaria’s mass is 27.2 kg, with what force does her mother need to pull her? (Neglect any resistance between the ice and Kamaria’s skates.) Fnet ma (27.2 kg)(0.80 m/s2) 22 N 17. Taru and Reiko simultaneously grab a 0.75-kg piece of rope and begin tugging on it in opposite directions. If Taru pulls with a force of 16.0 N and the rope accelerates away from her at 1.25 m/s2, with what force is Reiko pulling? Identify Reiko’s direction as positive and the rope as the system. ■ Figure 4-8 Bottom scale: Identify the sphere as the system and up as positive. Fnet Fscale on sphere FEarth’s mass on sphere ma 0 Fscale on sphere FEarth’s mass on sphere msphereg (3.0 kg)(9.80 m/s2) 29 N Top scale: Identify the block as the system and up as positive. Fnet Ftop scale on block Fbottom scale on block FEarth’s mass on block ma 0 Ftop scale on block Fbottom scale on block FEarth’s mass on block Fbottom scale on block mblockg Fnet FRieko on rope FTaru on rope ma 29 N (1.2 kg) (9.80 m/s2) FRieko on rope ma FTaru on rope 41 N (0.75 kg)(1.25 m/s2) 16.0 N 17 N Physics: Principles and Problems Solutions Manual 63 Chapter 4 continued page 100 19. On Earth, a scale shows that you weigh 585 N. a. What is your mass? The scale reads 585 N. Since there is no acceleration, your weight equals the downward force of gravity: Fg mg c. It speeds up at 2.00 m/s2 while moving downward. Accelerating downward, so a 2.00 m/s2 Fscale Fnet Fg ma mg m(a g) Fg 585 N 9.80 m/s so m 2 59.7 kg g b. What would the scale read on the Moon (g 1.60 m/s2)? On the moon, g changes: Fg mgMoon (59.7 kg)(1.60 m/s2) 95.5 N 20. Use the results from Example Problem 2 to answer questions about a scale in an elevator on Earth. What force would be exerted by the scale on a person in the following situations? a. The elevator moves at constant speed. Fscale Fg mg (75.0 kg)(9.80 m/s2) 735 N b. It slows at 2.00 m/s2 while moving upward. 9.80 m/s2) 585 N d. It moves downward at constant speed. Constant speed, so a 0 and Fnet 0 Fscale Fg mg (75.0 kg)(9.80 m/s2) 735 N e. It slows to a stop at a constant magnitude of acceleration. Constant acceleration a, though the sign of a depends on the direction of the motion that is ending. Fscale Fnet Fg ma mg (75.0 kg)(a) (75.0 kg)(9.80 m/s2) (75.0 kg)(a) 735 N Slowing while moving upward, so a 2.00 m/s2 Fscale Fnet Fg ma mg m(a g) (75.0 kg)(2.00 m/s2 9.80 m/s2) 585 N 64 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Constant speed, so a 0 and Fnet 0. (75.0 kg)(2.00 m/s2 Chapter 4 continued Section Review 4.2 Constant Velocity Fscale Using Newton’s Laws pages 96–101 page 101 21. Lunar Gravity Compare the force holding a 10.0-kg rock on Earth and on the Moon. The acceleration due to gravity on the Moon is 1.62 m/s2. To hold the rock on Earth: Fnet FEarth on rock Fhold on rock 0 Fg apparent weight real weight Slowing While Rising/ Speeding Up While Descending Fscale Fhold on rock FEarth on rock mgEarth (10.0 kg)(9.80 m/s2) 98.0 N To hold the rock on the Moon: Fnet FMoon on rock Fhold on rock 0 Fhold on rock FMoon on rock mgMoon (10.0 kg)(1.62 m/s2) Fg apparent weight real weight Speeding Up While Rising/ Slowing While Descending Fscale Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 16.2 N 22. Real and Apparent Weight You take a ride in a fast elevator to the top of a tall building and ride back down while standing on a bathroom scale. During which parts of the ride will your apparent and real weights be the same? During which parts will your apparent weight be less than your real weight? More than your real weight? Sketch free-body diagrams to support your answers. Apparent weight and real weight are the same when you are traveling either up or down at a constant velocity. Apparent weight is less than real weight when the elevator is slowing while rising or speeding up while descending. Apparent weight is greater when speeding up while rising or slowing while going down. Fg apparent weight real weight 23. Acceleration Tecle, with a mass of 65.0 kg, is standing by the boards at the side of an ice-skating rink. He pushes off the boards with a force of 9.0 N. What is his resulting acceleration? Identify Tecle as the system and the direction away from the boards as positive. The ice can be treated as a resistance-free surface. Fnet Fboards on Tecle ma F boards on Tecle a m 9.0 N 65.0 kg 0.14 m/s2 away from the boards Physics: Principles and Problems Solutions Manual 65 Chapter 4 continued 24. Motion of an Elevator You are riding in an elevator holding a spring scale with a 1-kg mass suspended from it. You look at the scale and see that it reads 9.3 N. What, if anything, can you conclude about the elevator’s motion at that time? If the elevator is stationary or moving at a constant velocity, the scale should read 9.80 N. Because the scale reads a lighter weight, the elevator must be accelerating downward. To find the exact acceleration: identify up as positive and the 1-kg mass as the system. Fnet Fscale on 1 kg FEarth’s mass on 1 kg ma F m F scale on 1 kg Earth’s mass on 1 kg a 9.3 N 9.80 N 1 kg 0.5 m/s2 0.5 m/s2 downward Identify the toy as the system and the direction toward his cat as the positive direction. Fnet FMarcos on toy Fcat on toy ma FMarcos on toy Fcat on toy m a 22 N 19.5 N 6.25 m/s 2 27. Critical Thinking You have a job at a meat warehouse loading inventory onto trucks for shipment to grocery stores. Each truck has a weight limit of 10,000 N of cargo. You push each crate of meat along a low-resistance roller belt to a scale and weigh it before moving it onto the truck. However, right after you weigh a 1000-N crate, the scale breaks. Describe a way in which you could apply Newton’s laws to figure out the approximate masses of the remaining crates. Answers may vary. One possible answer is the following: You can neglect resistance if you do all your maneuvering on the roller belt. Because you know the weight of the 1000 N crate, you can use it as your standard. Pull on the 1000 N crate with a particular force for 1 s, estimate its velocity, and calculate the acceleration that your force gave to it. Next, pull on a crate of unknown mass with as close to the same force as you can for 1 s. Estimate the crate’s velocity and calculate the acceleration your force gave to it. The force you pulled with on each crate will be the net force in each case. Fnet 1000-N crate Fnet unknown crate (1000 N)(a1000-N crate) (munk)(aunk) (1000 N)(a aunk ) 1000-N crate munk 0.40 kg 26. Acceleration A sky diver falls at a constant speed in the spread-eagle position. After he opens his parachute, is the sky diver accelerating? If so, in which direction? Explain your answer using Newton’s laws. Yes, for a while the diver is accelerating upward because there is an additional 66 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 25. Mass Marcos is playing tug-of-war with his cat using a stuffed toy. At one instant during the game, Marcos pulls on the toy with a force of 22 N, the cat pulls in the opposite direction with a force of 19.5 N, and the toy experiences an acceleration of 6.25 m/s2. What is the mass of the toy? upward force due to air resistance on the parachute. The upward acceleration causes the driver’s downward velocity to decrease. Newton’s second law says that a net force in a certain direction will result in an acceleration in that direction (Fnet ma). Chapter 4 continued Practice Problems 4.3 Interaction Forces pages 102–107 31. A suitcase sits on a stationary airport luggage cart, as in Figure 4-13. Draw a free-body diagram for each object and specifically indicate any interaction pairs between the two. page 104 28. You lift a relatively light bowling ball with your hand, accelerating it upward. What are the forces on the ball? What forces does the ball exert? What objects are these forces exerted on? The forces on the ball are the force of your hand and the gravitational force of Earth’s mass. The ball exerts a force on your hand and a gravitational force on Earth. All these forces are exerted on your hand, on the ball, or on Earth. 29. A brick falls from a construction scaffold. Identify any forces acting on the brick. Also identify any forces that the brick exerts and the objects on which these forces are exerted. (Air resistance may be ignored.) Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. The only force acting on the brick is the gravitational attraction of Earth’s mass. The brick exerts an equal and opposite force on Earth. 30. You toss a ball up in the air. Draw a freebody diagram for the ball while it is still moving upward. Identify any forces acting on the ball. Also identify any forces that the ball exerts and the objects on which these forces are exerted. ■ Figure 4-13 Suitcase Cart Fsurface on cart Fcart on suitcase FEarth’s mass on cart Fsuitcase on cart FEarth’s mass on suitcase page 106 32. You are helping to repair a roof by loading equipment into a bucket that workers hoist to the rooftop. If the rope is guaranteed not to break as long as the tension does not exceed 450 N and you fill the bucket until it has a mass of 42 kg, what is the greatest acceleration that the workers can give the bucket as they pull it to the roof? Identify the bucket as the system and up as positive. Fnet Frope on bucket FEarth’s mass on bucket FEarth’s mass on ball The only force acting on the ball is the force of Earth’s mass on the ball, when ignoring air resistance. The ball exerts an equal and opposite force on Earth. ma Frope on bucket FEarth’s mass on bucket a m Frope on bucket mg m 450 N (42 kg)(9.80 m/s ) 2 42 kg 0.91 m/s2 Physics: Principles and Problems Solutions Manual 67 Chapter 4 continued 33. Diego and Mika are trying to fix a tire on Diego’s car, but they are having trouble getting the tire loose. When they pull together, Mika with a force of 23 N and Diego with a force of 31 N, they just barely get the tire to budge. What is the magnitude of the strength of the force between the tire and the wheel? Identify the tire as the system and the direction of pulling as positive. Fnet Fwheel on tire FMika on tire FDiego on tire ma 0 Fwheel on tire FMika on tire FDiego on tire 23 N 31 N 54 N For the bottom rope with the positive direction upward: Fnet Fbottom rope on bottom block FEarth’s mass on bottom block ma 0 Fbottom rope on bottom block FEarth’s mass on bottom block mg (5.0 kg)(9.80 m/s2) 49 N For the top rope, with the positive direction upward: Fnet Ftop rope on top block Fbottom rope on top block Section Review 4.3 Interaction Forces pages 102–107 page 107 34. Force Hold a book motionless in your hand in the air. Identify each force and its interaction pair on the book. 35. Force Lower the book from problem 34 at increasing speed. Do any of the forces or their interaction-pair partners change? Explain. Yes, the force of the hand on the book becomes smaller so there is a downward acceleration. The force of the book also becomes smaller; you can feel that. The interaction pair partners remain the same. ma 0 Ftop rope on top block FEarth’s mass on top block Fbottom rope on top block mg Fbottom rope on top block (5.0 kg)(9.80 m/s2) 49 N 98 N 37. Tension If the bottom block in problem 36 has a mass of 3.0 kg and the tension in the top rope is 63.0 N, calculate the tension in the bottom rope and the mass of the top block. For the bottom rope with the positive direction upward: Fnet Fbottom rope on bottom block FEarth’s mass on bottom block ma 0 Fbottom rope on bottom block 36. Tension A block hangs from the ceiling by a massless rope. A second block is attached to the first block and hangs below it on another piece of massless rope. If each of the two blocks has a mass of 5.0 kg, what is the tension in each rope? 68 Solutions Manual FEarth’s mass on bottom block (3.0 kg)(9.80 m/s2) 29 N Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. The forces on the book are downward force of gravity due to the mass of Earth and the upward force of the hand. The force of the book on Earth and the force of the book on the hand are the other halves of the interaction pairs. FEarth’s mass on top block Chapter 4 continued For the top mass with the positive direction upward: Chapter Assessment Fnet Ftop rope on top block Concept Mapping Fbottom rope on top block FEarth’s mass on top block ma 0 page 112 40. Complete the following concept map using the following term and symbols: normal, FT, Fg. FEarth’s mass on top block mg force Ftop rope on top block Fbottom rope on top block tension normal gravity Ftop rope on top block Fbottom rope on top block g FT FN Fg m 63.0 N 29 N 9.80 m/s 2 3.5 kg 38. Normal Force Poloma hands a 13-kg box to 61-kg Stephanie, who stands on a platform. What is the normal force exerted by the platform on Stephanie? Identify Stephanie as the system and positive to be upward. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Fnet Fplatform on Stephanie Fbox on Stephanie FEarth’s mass on Stephanie Fplatform on Stephanie Fbox on Stephanie FEarth’s mass on Stephanie mboxg mStephanieg (13 kg)(9.80 m/s2) (61 kg)(9.80 m/s2) 7.3102 N 39. Critical Thinking A curtain prevents two tug-of-war teams from seeing each other. One team ties its end of the rope to a tree. If the other team pulls with a 500-N force, what is the tension? Explain. The tension would be 500 N. The rope is in equilibrium, so the net force on it is zero. The team and the tree exert equal forces in opposite directions. Physics: Principles and Problems Mastering Concepts page 112 41. A physics book is motionless on the top of a table. If you give it a hard push with your hand, it slides across the table and slowly comes to a stop. Use Newton’s laws to answer the following questions. (4.1) a. Why does the book remain motionless before the force of your hand is applied? An object at rest tends to stay at rest if no outside force acts on it. b. Why does the book begin to move when your hand pushes hard enough on it? The force from your hand is greater than any opposing force such as friction. With a net force on it, the book slides in the direction of the net force. c. Under what conditions would the book remain in motion at a constant speed? The book would remain in motion if the net force acting on it is zero. 42. Cycling Why do you have to push harder on the pedals of a single-speed bicycle to start it moving than to keep it moving at a constant velocity? (4.1) A large force is required to accelerate the mass of the bicycle and rider. Once the desired constant velocity is reached, a much smaller force is sufficient to overcome the ever-present frictional forces. Solutions Manual 69 Chapter 4 continued 43. Suppose that the acceleration of an object is zero. Does this mean that there are no forces acting on it? Give an example supporting your answer. (4.2) No, it only means the forces acting on it are balanced and the net force is zero. For example, a book on a table is not moving but the force of gravity pulls down on it and the normal force of the table pushes up on it and these forces are balanced. 44. Basketball When a basketball player dribbles a ball, it falls to the floor and bounces up. Is a force required to make it bounce? Why? If a force is needed, what is the agent involved? (4.2) Yes, its velocity changed direction; thus, it was accelerated and a force is required to accelerate the basketball. The agent is the floor. 47. A rock is dropped from a bridge into a valley. Earth pulls on the rock and accelerates it downward. According to Newton’s third law, the rock must also be pulling on Earth, yet Earth does not seem to accelerate. Explain. (4.3) The rock does pull on Earth, but Earth’s enormous mass would undergo only a minute acceleration as a result of such a small force. This acceleration would go undetected. 48. Ramon pushes on a bed that has been pushed against a wall, as in Figure 4-17. Draw a free-body diagram for the bed and identify all the forces acting on it. Make a separate list of all the forces that the bed applies to other objects. (4.3) 45. Before a sky diver opens her parachute, she may be falling at a velocity higher than the terminal velocity that she will have after the parachute opens. (4.2) a. Describe what happens to her velocity as she opens the parachute. ■ The force of air resistance and the gravitational force are equal. Their sum is zero, so there is no longer any acceleration. The sky diver continues downward at a constant velocity. 46. If your textbook is in equilibrium, what can you say about the forces acting on it? (4.2) If the book is in equilibrium, the net force is zero. The forces acting on the book are balanced. 70 Solutions Manual Fwall on bed FRamon on bed Fg Forces that bed applies to other objects: Fbed on Ramon, Fbed on Earth, Fbed on floor, Fbed on wall Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Ffloor on bed Because the force of air resistance suddenly becomes larger, the velocity of the diver drops suddenly. b. Describe the sky diver’s velocity from when her parachute has been open for a time until she is about to land. Figure 4-17 Chapter 4 continued 49. Figure 4-18 shows a block in four different situations. Rank them according to the magnitude of the normal force between the block and the surface, greatest to least. Specifically indicate any ties. (4.3) 52. Baseball A slugger swings his bat and hits a baseball pitched to him. Draw free-body diagrams for the baseball and the bat at the moment of contact. Specifically indicate any interaction pairs between the two diagrams. (4.3) Fball on bat Fbatter on bat Fbat on ball Applying Concepts ■ pages 112–113 53. Whiplash If you are in a car that is struck from behind, you can receive a serious neck injury called whiplash. Figure 4-18 from left to right: second > fourth > third > first Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 50. Explain why the tension in a massless rope is constant throughout it. (4.3) If you draw a free-body diagram for any point on the rope, there will be two tension forces acting in opposite directions. Fnet Fup Fdown ma 0 (because it is massless). Therefore, Fup Fdown. According to Newton’s third law, the force that the adjoining piece of rope exerts on this point is equal and opposite to the force that this point exerts on it, so the force must be constant throughout. 51. A bird sits on top of a statue of Einstein. Draw free-body diagrams for the bird and the statue. Specifically indicate any interaction pairs between the two diagrams. (4.3) a. Using Newton’s laws, explain what happens to cause such an injury. The car is suddenly accelerated forward. The seat accelerates your body, but your neck has to accelerate your head. This can hurt your neck muscles. b. How does a headrest reduce whiplash? The headrest pushes on your head, accelerating it in the same direction as the car. 54. Space Should astronauts choose pencils with hard or soft lead for making notes in space? Explain. A soft lead pencil would work better because it would require less force to make a mark on the paper. The magnitude of the interaction force pair could push the astronaut away from the paper. FEarth on statue Fstatue on bird Fg, bird Fbird on statue Fg, statue Physics: Principles and Problems Solutions Manual 71 Chapter 4 continued 55. When you look at the label of the product in Figure 4-19 to get an idea of how much the box contains, does it tell you its mass, weight, or both? Would you need to make any changes to this label to make it correct for consumption on the Moon? 58. You toss a ball straight up into the air. a. Draw a free-body diagram for the ball at three points during its motion: on the way up, at the very top, and on the way down. Specifically identify the forces acting on the ball and their agents. On the way up FEarth’s mass on ball At the top FEarth’s mass on ball ■ Figure 4-19 The ounces tell you the weight in English units. The grams tell you the mass in metric units. The label would need to read “2 oz” to be correct on the Moon. The grams would remain unchanged. The lighter, air-filled table tennis ball reaches terminal velocity first. Its mass is less for the same shape and size, so the friction force of upward air resistance becomes equal to the downward force of mg sooner. Because the force of gravity on the water-filled table-tennis ball (more mass) is larger, its terminal velocity is larger, and it strikes the ground first. FEarth’s mass on ball b. What is the velocity of the ball at the very top of the motion? 0 m/s c. What is the acceleration of the ball at this same point? Because the only force acting on it is the gravitational attraction of Earth, a 9.80 m/s2. Mastering Problems 4.1 Force and Motion page 113 Level 1 59. What is the net force acting on a 1.0-kg ball in free-fall? Fnet Fg mg (1.0 kg)(9.80 m/s2) 57. It can be said that 1 kg equals 2.2 lb. What does this statement mean? What would be the proper way of making the comparison? 9.8 N It means that on Earth’s surface, the weight of 1 kg is equivalent to 2.2 lb. You should compare masses to masses and weights to weights. Thus 9.8 N equals 2.2 lb. 72 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 56. From the top of a tall building, you drop two table-tennis balls, one filled with air and the other with water. Both experience air resistance as they fall. Which ball reaches terminal velocity first? Do both hit the ground at the same time? On the way down Chapter 4 continued 60. Skating Joyce and Efua are skating. Joyce pushes Efua, whose mass is 40.0-kg, with a force of 5.0 N. What is Efua’s resulting acceleration? F ma F a m 65. Three objects are dropped simultaneously from the top of a tall building: a shot put, an air-filled balloon, and a basketball. a. Rank the objects in the order in which they will reach terminal velocity, from first to last. balloon, basketball, shot put 5.0 N 40.0 kg 0.12 m/s2 b. Rank the objects according to the order in which they will reach the ground, from first to last. shot put, basketball, balloon 61. A car of mass 2300 kg slows down at a rate of 3.0 m/s2 when approaching a stop sign. What is the magnitude of the net force causing it to slow down? F ma (2300 kg)(3.0 m/s2) 6.9103 N Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 62. Breaking the Wishbone After Thanksgiving, Kevin and Gamal use the turkey’s wishbone to make a wish. If Kevin pulls on it with a force 0.17 N larger than the force Gamal pulls with in the opposite direction, and the wishbone has a mass of 13 g, what is the wishbone’s initial acceleration? c. What is the relationship between your answers to parts a and b? They are inverses of each other. 66. What is the weight in pounds of a 100.0-N wooden shipping case? (100.0 N) 22 lb 1 kg 9.80 N 2.2 lb 1 kg 67. You place a 7.50-kg television on a spring scale. If the scale reads 78.4 N, what is the acceleration due to gravity at that location? Fg mg Fg g m 78.4 N F a 7.50 kg m 0.17 N 10.5 m/s2 0.013 kg 13 m/s2 4.2 Using Newton’s Laws pages 113–114 Level 1 63. What is your weight in newtons? Fg mg (9.80 m/s2)(m) Answers will vary. 64. Motorcycle Your new motorcycle weighs 2450 N. What is its mass in kilograms? Fg mg Fg 2450 N 9.80 m/s m 2 g 2.50102 N Physics: Principles and Problems Level 2 68. Drag Racing A 873-kg (1930-lb) dragster, starting from rest, attains a speed of 26.3 m/s (58.9 mph) in 0.59 s. a. Find the average acceleration of the dragster during this time interval. v a t (26.3 m/s 0.0 m/s) 0.59 s 45 m/s2 b. What is the magnitude of the average net force on the dragster during this time? F ma (873 kg)(45 m/s2) 3.9104 N Solutions Manual 73 Chapter 4 continued c. Assume that the driver has a mass of 68 kg. What horizontal force does the seat exert on the driver? F ma (68 kg)(45 m/s2) 3.1103 N e. It slows to a stop while moving downward with a constant acceleration. Depends on the magnitude of the acceleration. Fscale Fg Fslowing mg ma 69. Assume that a scale is in an elevator on Earth. What force would the scale exert on a 53-kg person standing on it during the following situations? a. The elevator moves up at a constant speed. Fscale Fg mg m(g a) (53 kg)(9.80 m/s2 a) 70. A grocery sack can withstand a maximum of 230 N before it rips. Will a bag holding 15 kg of groceries that is lifted from the checkout counter at an acceleration of 7.0 m/s2 hold? (53 kg)(9.80 m/s2) Use Newton’s second law Fnet ma. 5.2102 N If Fgroceries 230, then the bag rips. b. It slows at 2.0 m/s2 while moving upward. Fgroceries mgroceriesagroceries mgroceriesg Slows while moving up or speeds up while moving down, mgroceries(agroceries g) Fscale Fg Fslowing (15 kg)(7.0 m/s2 9.80 m/s2) mg ma m(g a) 4.1102 N c. It speeds up at 2.0 m/s2 while moving downward. Slows while moving up or speeds up while moving down, Fscale Fg Fspeeding mg ma m(g a) (53 kg)(9.80 m/s2 2.0 m/s2) 4.1102 N d. It moves downward at a constant speed. Fscale Fg mg (53 kg)(9.80 m/s2) 5.2102 N 74 Solutions Manual The bag does not hold. 71. A 0.50-kg guinea pig is lifted up from the ground. What is the smallest force needed to lift it? Describe its resulting motion. Flift Fg mg (0.50 kg)(9.80 m/s2) 4.9 N It would move at a constant speed. Level 3 72. Astronomy On the surface of Mercury, the gravitational acceleration is 0.38 times its value on Earth. a. What would a 6.0-kg mass weigh on Mercury? Fg mg(0.38) (6.0 kg)(9.80 m/s2)(0.38) 22 N Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. (53 kg)(9.80 m/s2 2.0 m/s2) 250 N Chapter 4 continued b. If the gravitational acceleration on the surface of Pluto is 0.08 times that of Mercury, what would a 7.0-kg mass weigh on Pluto? Fg mg(0.38)(0.08) (7.0 kg)(9.80 m/s2)(0.38)(0.08) 2.1 N 73. A 65-kg diver jumps off of a 10.0-m tower. a. Find the diver’s velocity when he hits the water. vf2 vi2 2gd Fnet N mg FN F7-kg block on 6-kg block mg 59 N; the direction is upward. so vf 2gd 2(9.80 m/s2)(10.0 m) 14.0 m/s b. The diver comes to a stop 2.0 m below the surface. Find the net force exerted by the water. vf2 vi2 2ad v 2 2d b. What is the force (magnitude and direction) exerted by the 6.0-kg block on the 7.0-kg block? equal and opposite to that in part a; therefore, 59 N downward 76. Rain A raindrop, with mass 2.45 mg, falls to the ground. As it is falling, what magnitude of force does it exert on Earth? Fraindrop on Earth Fg i vf 0, so a mg and F ma Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. a. What is the force (magnitude and direction) exerted by the 7.0-kg block on the 6.0-kg block? (6.0 kg)(9.80 m/s2) vi 0 m/s (0.00245 kg)(9.80 m/s2) mv 2 2d i (65 kg)(14.0 m/s)2 2(2.0 m) 4.3 Interaction Forces page 114 Level 1 75. A 6.0-kg block rests on top of a 7.0-kg block, which rests on a horizontal table. 3.2103 N 74. Car Racing A race car has a mass of 710 kg. It starts from rest and travels 40.0 m in 3.0 s. The car is uniformly accelerated during the entire time. What net force is exerted on it? d v0t at2 2 1 2.40102 N 77. A 90.0-kg man and a 55-kg man have a tug-of-war. The 90.0-kg man pulls on the rope such that the 55-kg man accelerates at 0.025 m/s2. What force does the rope exert on the 90.0-kg man? same in magnitude as the force the rope exerts on the 55-kg man: F ma (55 kg)(0.025 m/s2) 1.4 N Since v0 0, 2d t a 2 and F ma, so 2md t F 2 (2)(710 kg)(40.0 m) (3.0 s) 2 6.3103 N Physics: Principles and Problems Solutions Manual 75 Chapter 4 continued Level 2 78. Male lions and human sprinters can both accelerate at about 10.0 m/s2. If a typical lion weighs 170 kg and a typical sprinter weighs 75 kg, what is the difference in the force exerted on the ground during a race between these two species? Fnet Fmiddle block on top block FEarth’s mass on top block ma 0 Fmiddle block on top block FEarth’s mass on top block Use Newton’s second law, Fnet ma. The difference between Flion and Fhuman is (4.6 kg)(9.80 m/s2) Flion Fhuman 45 N mlionalion mhumanahuman (170 kg)(10.0 m/s2) (75 kg)(10.0 m/s2) The normal force is between the bottom and middle block; the middle block is the system; upward is positive. Fnet Fbottom block on middle block Ftop block on middle block 9.5102 N 79. A 4500-kg helicopter accelerates upward at 2.0 m/s2. What lift force is exerted by the air on the propellers? ma Fnet Fappl Fg Fappl mg so Fappl ma mg m(a g) (4500 kg)((2.0m/s2) (9.8 mtop blockg m/s2)) Level 3 80. Three blocks are stacked on top of one another, as in Figure 4-20. The top block has a mass of 4.6 kg, the middle one has a mass of 1.2 kg, and the bottom one has a mass of 3.7 kg. Identify and calculate any normal forces between the objects. ma 0 Ftop block on middle block Fmiddle block on top block Fbottom block on middle block Fmiddle block on top block FEarth’s mass on middle block Fmiddle block on top block mmiddle blockg 45 N (1.2 kg)(9.80 m/s2) 57 N The normal force is between the bottom block and the surface; the bottom block is the system; upward is positive. Fnet Fsurface on bottom block Fmiddle block on bottom block FEarth’s mass on bottom block ma 0 ■ Figure 4-20 The normal force is between the top and middle blocks; the top block is the system; upward is positive. 76 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 5.3104 N FEarth’s mass on middle block Chapter 4 continued Fsurface on bottom block F Fmiddle block on bottom block FEarth’s mass on bottom block Fmiddle block on bottom block m Fg mg Fg m g vf vi at and vi 0, so mbottom blockg 57 N (3.7 kg)(9.80 m/s2) 93 N v a t f Mixed Review pages 114–115 Level 1 81. The dragster in problem 68 completed a 402.3-m (0.2500-mi) run in 4.936 s. If the car had a constant acceleration, what was its acceleration and final velocity? 1 df di vit at 2 2 v f Fthrust m vm Fthrust f 1 df di vit at 2 2 di vi 0, so 1 df at 2 2 F vm F 1 thrust f di vi 0, so 2 m thrust 2 v 2m 2 Fthrust 2d t 1 f f a 2 (2)(402.3 m) 2 g Fthrust F (4.936 s) 1 2 33.02 m/s2 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. thrust Fthrust ma, so a 1 df di (vf vi)t 2 g v 2 f 2.75106 N 9.80 m/s (79.2 m/s)2 2 1 6 di vi 0, so 2 6.3510 N 139 m 2d t vf f (2)(402.3 m) 4.936 s 163.0 m/s Level 2 82. Jet A 2.75106-N catapult jet plane is ready for takeoff. If the jet’s engines supply a constant thrust of 6.35106 N, how much runway will it need to reach its minimum takeoff speed of 285 km/h? 83. The dragster in problem 68 crossed the finish line going 126.6 m/s. Does the assumption of constant acceleration hold true? What other piece of evidence could you use to determine if the acceleration was constant? 126.6 m/s is slower than found in problem 81, so the acceleration cannot be constant. Further, the acceleration in the first 0.59 s was 45 m/s2, not 33.02 m/s2. vf (285 km/h)(1000 m/km) 1h 3600s 79.2 m/s Physics: Principles and Problems Solutions Manual 77 Chapter 4 continued 84. Suppose a 65-kg boy and a 45-kg girl use a massless rope in a tug-of-war on an icy, resistance-free surface as in Figure 4-21. If the acceleration of the girl toward the boy is 3.0 m/s2, find the magnitude of the acceleration of the boy toward the girl. 86. Baseball As a baseball is being caught, its speed goes from 30.0 m/s to 0.0 m/s in about 0.0050 s. The mass of the baseball is 0.145 kg. a. What is the baseball’s acceleration? v v t f ti f i a 0.0 m/s 30.0 m/s 0.0050 s 0.0 s 6.0103 m/s2 ? 3.0 m/s2 b. What are the magnitude and direction of the force acting on it? F ma ■ (0.145 kg)(6.0103 m/s2) Figure 4-21 F1,2 F1,2, so m1a1 m2a2 m2a2 and a1 m1 (45 kg)(3.0 m/s2) (65 kg) 2.1 m/s2 8.7102 N (opposite direction of the velocity of the ball) c. What are the magnitude and direction of the force acting on the player who caught it? Same magnitude, opposite direction (in direction of velocity of ball) Use Newton’s second law to obtain Pratish’s mass, mPratish. Use Newton’s third law FA FB mAaA mBaB. Level 3 87. Air Hockey An air-hockey table works by pumping air through thousands of tiny holes in a table to support light pucks. This allows the pucks to move around on cushions of air with very little resistance. One of these pucks has a mass of 0.25 kg and is pushed along by a 12.0-N force for 9.0 s. a. What is the puck’s acceleration? Fg mPratish F ma g Fwall on Pratish FPratish on wall mPratishaPratish FgaPratish g 12.0 N 48 m/s2 (588 N)(3.00 m/s2) 9.80 m/s m 0.25 kg 2 1.80102 F a N b. What is the puck’s final velocity? vf vi at vi 0, so vf at (48 m/s2)(9.0 s) 4.3102 m/s 78 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 85. Space Station Pratish weighs 588 N and is weightless in a space station. If she pushes off the wall with a vertical acceleration of 3.00 m/s2, determine the force exerted by the wall during her push off. Chapter 4 continued 88. A student stands on a bathroom scale in an elevator at rest on the 64th floor of a building. The scale reads 836 N. a. As the elevator moves up, the scale reading increases to 936 N. Find the acceleration of the elevator. 89. Weather Balloon The instruments attached to a weather balloon in Figure 4-22 have a mass of 5.0 kg. The balloon is released and exerts an upward force of 98 N on the instruments. Fnet Fg Felevator Felevator Fnet Fg ma Fg m , so g 98 N Fnet Fg a F g g 5.0 kg g(Fnet Fg) Fg (9.80 m/s )(963 N 836 N) 2 836 N Figure 4-22 a. What is the acceleration of the balloon and instruments? Fnet Fappl Fg 1.17 m/s2 b. As the elevator approaches the 74th floor, the scale reading drops to 782 N. What is the acceleration of the elevator? Fnet Fg Felevator Felevator Fnet Fg ma Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. ■ Fg m , so Fappl mg 98 N (5.0 kg)(9.80 m/s2) 49 N (up) F m net a 49 N g 5.0 kg Fnet Fg a F 9.8 m/s2 g g g(Fnet Fg) Fg (9.80 m/s2)(782 N 836 N) 836 N 0.633 m/s2 c. Using your results from parts a and b, explain which change in velocity, starting or stopping, takes the longer time. Stopping, because the magnitude of the acceleration is less and b. After the balloon has accelerated for 10.0 s, the instruments are released. What is the velocity of the instruments at the moment of their release? v at (9.8 m/s2)(10.0 s) 98 m/s (up) c. What net force acts on the instruments after their release? just the instrument weight, 49 N (down) v t a Physics: Principles and Problems Solutions Manual 79 Chapter 4 continued d. When does the direction of the instruments’ velocity first become downward? The velocity becomes negative after it passes through zero. Thus, use 91. Two blocks, masses 4.3 kg and 5.4 kg, are pushed across a frictionless surface by a horizontal force of 22.5 N, as shown in Figure 4-23. vf vi gt, where vf 0, or v g t i (98 m/s) (9.80 m/s ) 2 1.0101 s after release 90. When a horizontal force of 4.5 N acts on a block on a resistance-free surface, it produces an acceleration of 2.5 m/s2. Suppose a second 4.0-kg block is dropped onto the first. What is the magnitude of the acceleration of the combination if the same force continues to act? Assume that the second block does not slide on the first block. F mfirst blockainitial ■ Figure 4-23 a. What is the acceleration of the blocks? Identify the two blocks together as the system, and right as positive. Fnet ma, and m m1 m2 F a m1 m2 F mfirst block 22.5 N F mboth blocksafinal 2.3 m/s2 to the right ainitial (mfirst block msecond block)afinal mfirst block msecond block F F msecond block ainitial 4.5 N 4.5 N 2 4.0 kg 2.5 m/s 0.78 m/s2 b. What is the force of the 4.3-kg block on the 5.4-kg block? Identify the 5.4-kg block as the system and right as positive. Fnet F4.3-kg block on 5.4-kg block ma (5.4 kg)(2.3 m/s2) 12 N to the right c. What is the force of the 5.4-kg block on the 4.3-kg block? According to Newton’s third law, this should be equal and opposite to the force found in part b, so the force is 12 N to the left. 80 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. F so, afinal 4.3 kg 5.4 kg Chapter 4 continued 92. Two blocks, one of mass 5.0 kg and the other of mass 3.0 kg, are tied together with a massless rope as in Figure 4-24. This rope is strung over a massless, resistance-free pulley. The blocks are released from rest. Find the following. a. the tension in the rope b. the acceleration of the blocks Hint: you will need to solve two simultaneous equations. 3.0 kg 5.0 kg ■ Figure 4-24 Equation 1 comes from a free-body diagram for the 5.0-kg block. Down is positive. Fnet FEarth’s mass on 5.0-kg block Frope on 5.0-kg block m5.0-kg blocka (1) Equation 2 comes from a free-body diagram for the 3.0-kg block. Up is positive. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Fnet Frope on 3.0-kg block FEarth’s mass on 3.0-kg block m3.0-kg blocka (2) The forces of the rope on each block will have the same magnitude, because the tension is constant throughout the rope. Call this force T. FEarth’s mass on 5.0-kg block T m5.0-kg blocka (1) T FEarth’s mass on 3.0-kg block m3.0-kg blocka (2) Solve equation 2 for T and plug into equation 1: m5.0-kg blocka FEarth’s mass on 5.0-kg block FEarth’s mass on 3.0-kg block m3.0-kg blocka FEarth’s mass on 5.0-kg block FEarth’s mass on 3.0-kg block a m5.0-kg block m3.0-kg block (m5.0-kg block m3.0-kg block)g m3.0-kg block m5.0-kg block (5.0 kg 3.0 kg)(9.80 m/s2) 3.0 kg 5.0 kg 2.4 m/s2 Physics: Principles and Problems Solutions Manual 81 Chapter 4 continued Solve equation 2 for T: For mA, mAa FT mAg T FEarth’s mass on 3.0-kg block m3.0-kg blocka For mB, mBa FT mBg m3.0-kg blockg m3.0-kg blocka FT mBg mBa mB(g a) m3.0-kg block(g a) Substituting into the equation for mA gives (3.0 kg)(9.80 m/s2 2.4 m/s2) mAa mBg mBa mAg 37 N or (mA mB)a (mB mA)g m m mA mB B A Therefore a g Thinking Critically pages 115–116 93. Formulate Models A 2.0-kg mass, mA, and a 3.0-kg mass, mB, are connected to a lightweight cord that passes over a frictionless pulley. The pulley only changes the direction of the force exerted by the rope. The hanging masses are free to move. Choose coordinate systems for the two masses with the positive direction being up for mA and down for mB. a. Create a pictorial model. 3.0 kg 2.0 kg 2.0 kg 3.0 kg (9.80 m/s2) 2.0 m/s2 upward 94. Use Models Suppose that the masses in problem 93 are now 1.00 kg and 4.00 kg. Find the acceleration of the larger mass. m m mA mB B A a g 4.00 kg 1.00 kg 1.00 kg 4.00 kg (9.80 m/s2) x 5.88 m/s2 downward mA b. Create a physical model with motion and free-body diagrams. a. Draw a graph of force versus time. What is the average force exerted on the ball by the bat? FT a 1.0104 mA a FT mAg mB Force (N) x 0.5104 mBg 0.0 0.5 1.0 1.5 2.0 Time (ms) c. What is the acceleration of the smaller mass? 82 1 Fave Fpeak 2 ma Fnet where m is the total mass 1 (1.0104 N) being accelerated. 5.0103 N Solutions Manual 2 Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. mB 95. Infer The force exerted on a 0.145-kg baseball by a bat changes from 0.0 N to 1.0104 N in 0.0010 s, then drops back to zero in the same amount of time. The baseball was going toward the bat at 25 m/s. Chapter 4 continued b. What is the acceleration of the ball? F m 5.0103 N net a FT2 m2a FT1 (4.0 kg)(3.0 m/s2) 6.0 N 0.145 kg 18 N 3.4104 m/s2 c. What is the final velocity of the ball, assuming that it reverses direction? vf vi at 25 m/s (3.4104 m/s2) (0.0020 s) 97. Critique Using the Example Problems in this chapter as models, write a solution to the following problem. A block of mass 3.46 kg is suspended from two vertical ropes attached to the ceiling. What is the tension in each rope? 43 m/s 96. Observe and Infer Three blocks that are connected by massless strings are pulled along a frictionless surface by a horizontal force, as shown in Figure 4-25. FT1 2.0 kg 4.0 kg m3 Fg F 36.0 N Figure 4-25 a. What is the acceleration of each block? Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Since they all move together, the acceleration is the same for all 3 blocks. F ma (m1 m2 m3)a F a m1 m2 m3 36 N 2.0 kg 4.0 kg 6.0 kg 3.0 m/s2 b. What are the tension forces in each of the strings? Hint: Draw a separate free-body diagram for each block. Fnet ma F FT2 m3a FT2 FT1 m2a T 3.46 kg 6.0 kg m2 m1 ■ FT2 T 1 Analyze and Sketch the Problem Draw free-body diagrams for the block and choose upward to be positive Solve for the Unknown Known: mblock 3.46 kg Unknown: Frope1 on block Frope2 on block ? 2 Solve for the Unknown Use Newton’s second law to find the tension in the ropes Fnet 2Frope1 on block FEarth’s mass on block ma 0 F Earth’s mass on block Frope1 on block 2 mg 2 Frope1 on block 2 (3.46 kg)(9.80 m/s ) 2 17.0 N FT1 m1a (2.0 kg)(3.0 m/s2) 6.0 N Physics: Principles and Problems Solutions Manual 83 Chapter 4 continued 3 Evaluate the Answer • Are the units correct? N is the correct unit for a tension, since it is a force. • Does the sign make sense? The positive sign indicates that the tension is pulling upwards. • Is the magnitude realistic? We would expect the magnitude to be on the same order as the block’s weight. Answers will vary. Here is one possible answer: She should take a known mass, say 5.00-kg, with her and place it on the scale. Since the gravitational force depends upon the local acceleration due to gravity, the scale will read a different number of newtons, depending on which planet she is on. The following analysis shows how to figure out what the scale would read on a given planet: 84 Solutions Manual Fnet Fscale on mass Fg ma 0 Fscale on mass Fg Fscale on mass mg Pluto: Fscale on mass (5.00 kg)(0.30 m/s2) 1.5 N Mercury: Fscale on mass (5.00 kg)(3.7 m/s2) 19 N 99. Apply Concepts Develop a CBL lab, using a motion detector, that graphs the distance a free-falling object moves over equal intervals of time. Also graph velocity versus time. Compare and contrast your graphs. Using your velocity graph, determine the acceleration. Does it equal g? Student labs will vary with equipment available and designs. p-t graphs and v-t graphs should reflect uniform acceleration. The acceleration should be close to g. Writing in Physics page 116 100. Research Newton’s contributions to physics and write a one-page summary. Do you think his three laws of motion were his greatest accomplishments? Explain why or why not. Answers will vary. Newton’s contributions should include his work on light and color, telescopes, astronomy, laws of motion, gravity, and perhaps calculus. One argument in favor of his three laws of motion being his greatest accomplishments is that mechanics is based on the foundation of these laws. His advances in the understanding of the concept of gravity may be suggested as his greatest accomplishment instead of his three laws of motion. Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 98. Think Critically Because of your physics knowledge, you are serving as a scientific consultant for a new science-fiction TV series about space exploration. In episode 3, the heroine, Misty Moonglow, has been asked to be the first person to ride in a new interplanetary transport for use in our solar system. She wants to be sure that the transport actually takes her to the planet she is supposed to be going to, so she needs to take a testing device along with her to measure the force of gravity when she arrives. The script writers don’t want her to just drop an object, because it will be hard to depict different accelerations of falling objects on TV. They think they’d like something involving a scale. It is your job to design a quick experiment Misty can conduct involving a scale to determine which planet in our solar system she has arrived on. Describe the experiment and include what the results would be for Pluto (g 0.30 m/s2), which is where she is supposed to go, and Mercury (g 3.70 m/s2), which is where she actually ends up. Identify the mass as the system and upward as positive. Chapter 4 continued Answers will vary. Newton’s first law of motion involves an object whose net forces are zero. If the object is at rest, it remains at rest; if it is in motion, it will continue to move in the same direction at a constant velocity. Only a force acting on an object at rest can cause it to move. Likewise, only a force acting on an object in motion can cause it to change its direction or speed. The two cases (object at rest, object in motion) could be viewed as two different frames of reference. This law can be demonstrated, but it cannot be proven. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 102. Physicists classify all forces into four fundamental categories: gravitational, electromagnetic, strong nuclear, and weak nuclear. Investigate these four forces and describe the situations in which they are found. Answers will vary. The strong nuclear force has a very short range and is what holds protons and neutrons together in the nucleus of an atom. The weak nuclear force is much less strong than the strong nuclear force and is involved in radioactive decay. The electromagnetic force is involved in holding atoms and molecules together and is based on the attraction of opposite charges. Gravity is a long-range force between two or more masses. Cumulative Review page 116 103. Cross-Country Skiing Your friend is training for a cross-country skiing race, and you and some other friends have agreed to provide him with food and water along his training route. It is a bitterly cold day, so none of you wants to wait outside longer than you have to. Taro, whose house is the stop before yours, calls you at 8:25 A.M. to tell you that the skier just passed his house and is planning to move at an average speed of 8.0 km/h. If it is 5.2 km from Physics: Principles and Problems Taro’s house to yours, when should you expect the skier to pass your house? (Chapter 2) d d vt, or t v d 5.2 km 5.2103 m v (8.0 km/h) 1000 m 1 km 1h 3600 s 2.2 m/s 3 5.210 m t 2.2 m/s 2.4103 s 39 min The skier should pass your house at 8:25 0:39 9:04 A.M. 104. Figure 4-26 is a position-time graph of the motion of two cars on a road. (Chapter 3) Distance (m) 101. Review, analyze, and critique Newton’s first law. Can we prove this law? Explain. Be sure to consider the role of resistance. Position of Two Cars 12 B 6 A 0 2 4 6 8 Time (s) ■ Figure 4-26 a. At what time(s) does one car pass the other? 3 s, 8 s b. Which car is moving faster at 7.0 s? car A c. At what time(s) do the cars have the same velocity? 5s d. Over what time interval is car B speeding up all the time? none e. Over what time interval is car B slowing down all the time? 3 s to 10 s Solutions Manual 85 Chapter 4 continued 105. Refer to Figure 4-26 to find the instantaneous speed for the following: (Chapter 3) a. car B at 2.0 s 0 m/s b. car B at 9.0 s 0 m/s c. car A at 2.0 s 1 m/s Challenge Problem page 100 An air-track glider passes through a photoelectric gate at an initial speed of 0.25 m/s. As it passes through the gate, a constant force of 0.40 N is applied to the glider in the same direction as its motion. The glider has a mass of 0.50 kg. 1. What is the acceleration of the glider? F ma F a m 0.40 N 0.50 kg 0.80 m/s2 2. It takes the glider 1.3 s to pass through a second gate. What is the distance between the two gates? 1 df di vit at 2 2 Let di position of first gate 0.0 m d 0.0 m (0.25 m/s)(1.3 s) 12 (0.80 m/s2)(1.3 s)2 1.0 m 3. The 0.40-N force is applied by means of a string attached to the glider. The other end of the string passes over a resistance-free pulley and is attached to a hanging mass, m. How big is m? Fg mmassg Fg mmass g 0.40 N 9.80 m/s 2 4.1102 kg T mg Ma 86 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 4. Derive an expression for the tension, T, in the string as a function of the mass, M, of the glider, the mass, m, of the hanging mass, and g. CHAPTER 5 Forces in Two Dimensions Practice Problems 5.1 Vectors pages 119–125 page 121 1. A car is driven 125.0 km due west, then 65.0 km due south. What is the magnitude of its displacement? Solve this problem both graphically and mathematically, and check your answers against each other. R 2 A2 B2 R A2 B2 (65.0 km)2 (12 5.0 km )2 125.0 km 65.0 km 141 km 141 km 2. Two shoppers walk from the door of the mall to their car, which is 250.0 m down a lane of cars, and then turn 90° to the right and walk an additional 60.0 m. What is the magnitude of the displacement of the shoppers’ car from the mall door? Solve this problem both graphically and mathematically, and check your answers against each other. R 2 A2 B2 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. R A2 B2 Mall door (250.0 m)2 (60.0 m)2 250.0 m 257 m 60.0 m Car 257 m 3. A hiker walks 4.5 km in one direction, then makes a 45° turn to the right and walks another 6.4 km. What is the magnitude of her displacement? R 2 A2 B 2 2AB cos 2 B 2 2AB R A cos (4.5 k m)2 (6.4 k m)2 2(4.5 km)(6 .4 km)(cos 135°) 1.0101 km 4. An ant is crawling on the sidewalk. At one moment, it is moving south a distance of 5.0 mm. It then turns southwest and crawls 4.0 mm. What is the magnitude of the ant’s displacement? R 2 A2 B 2 2AB cos R A2 B2 2AB c os (5.0 m m)2 (4.0 m m)2 2(5.0 mm)( 4.0 mm)(co s 135°) 8.3 mm Physics: Principles and Problems Solutions Manual 87 Chapter 5 continued page 125 Solve problems 5–10 algebraically. You may also choose to solve some of them graphically to check your answers. 5. Sudhir walks 0.40 km in a direction 60.0° west of north, then goes 0.50 km due west. What is his displacement? Identify north and west as the positive directions. d1W d1 sin (0.40 km)(sin 60.0°) 0.35 km d1N d1 cos (0.40 km)(cos 60.0°) 0.20 km d2W 0.50 km d2N 0.00 km RW d1W d2W 0.35 km 0.50 km 0.85 km RN d1N d2N 0.20 km 0.00 km 0.20 km RW2 RN2 R (0.85 km)2 (0.20 km)2 0.87 km R RW tan1 N tan1 0.85 km 0.20 km 77° R 0.87 km at 77° west of north Identify up and right as positive. FA on box,x FA on box cos A (20.4 N)(cos 120°) ■ Figure 5-6 10.2 N FA on box,y FA on box sin A (20.4 N)(sin 120°) 17.7 N FC on box,x FC on box cos A (17.7 N)(cos 55°) 88 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 6. Afua and Chrissy are going to sleep overnight in their tree house and are using some ropes to pull up a box containing their pillows and blankets, which have a total mass of 3.20 kg. The girls stand on different branches, as shown in Figure 5-6, and pull at the angles and with the forces indicated. Find the x- and y-components of the net force on the box. Hint: Draw a free-body diagram so that you do not leave out a force. Chapter 5 continued 10.2 N Fcombined Frope1 on swing cos FC on box,y FC on box sin A Frope2 on swing cos (17.7 N)(sin 55°) 2Frope2 on swing cos 14.5 N (2)(2.28 N)(cos 13.0°) Fg,x 0.0 N 4.44 N upward Fg,y mg (3.20 kg)(9.80 m/s2) 31.4 N 9. Could a vector ever be shorter than one of its components? Equal in length to one of its components? Explain. It could never be shorter than one of its components, but if it lies along either the x- or y-axis, then one of its components equals its length. Fnet on box,x FA on box,x FC on box,x Fg,x 10.2 N 10.2 N 0.0 N 0.0 N Fnet on box,y FA on box,y FC on box,y Fg,y 10. In a coordinate system in which the x-axis is east, for what range of angles is the x-component positive? For what range is it negative? The x-component is positive for angles less than 90° and for angles greater than 270°. It’s negative for angles greater than 90° but less than 270°. 17.7 N 14.5 N 31.4 N 0.8 N The net force is 0.8 N in the upward direction. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 7. You first walk 8.0 km north from home, then walk east until your displacement from home is 10.0 km. How far east did you walk? The resultant is 10.0 km. Using the Pythagorean Theorem, the distance east is R 2 A2 B2, so R2 A B 2 (10.0 km)2 (8.0 km)2 6.0 km 8. A child’s swing is held up by two ropes tied to a tree branch that hangs 13.0° from the vertical. If the tension in each rope is 2.28 N, what is the combined force (magnitude and direction) of the two ropes on the swing? Section Review 5.1 Vectors pages 119–125 page 125 11. Distance v. Displacement Is the distance that you walk equal to the magnitude of your displacement? Give an example that supports your conclusion. Not necessarily. For example, you could walk around the block (one km per side). Your displacement would be zero, but the distance that you walk would be 4 kilometers. 12. Vector Difference Subtract vector K from vector L, shown in Figure 5-7. The force will be straight up. Because the angles are equal, the horizontal forces will be equal and opposite and cancel out. The magnitude of this vertical force is Physics: Principles and Problems Solutions Manual 89 Chapter 5 continued 5.0 4.0 K 37.0° 6.0 L ■ M Figure 5-7 6.0 (4.0) 10.0 to the right 16. Critical Thinking A box is moved through one displacement and then through a second displacement. The magnitudes of the two displacements are unequal. Could the displacements have directions such that the resultant displacement is zero? Suppose the box was moved through three displacements of unequal magnitude. Could the resultant displacement be zero? Support your conclusion with a diagram. No, but if there are three displacements, the sum can be zero if the three vectors form a triangle when they are placed tip-to-tail. Also, the sum of three displacements can be zero without forming a triangle if the sum of two displacements in one direction equals the third in the opposite direction. 13. Components Find the components of vector M, shown in Figure 5-7. Mx m cos (5.0)(cos 37.0°) 4.0 to the right My m sin (5.0)(sin 37.0°) 3.0 upward 14. Vector Sum Find the sum of the three vectors shown in Figure 5-7. Rx Kx Lx Mx 4.0 6.0 4.0 6.0 Ry Ky Ly My 3.0 R Rx2 Ry2 6.02 3.02 6.7 R Ry tan1 x 6 3 tan1 27° R 6.7 at 27° 15. Commutative Operations The order in which vectors are added does not matter. Mathematicians say that vector addition is commutative. Which ordinary arithmetic operations are commutative? Which are not? Addition and multiplication are commutative. Subtraction and division are not. 90 Solutions Manual 5.2 Friction pages 126–130 page 128 17. A girl exerts a 36-N horizontal force as she pulls a 52-N sled across a cement sidewalk at constant speed. What is the coefficient of kinetic friction between the sidewalk and the metal sled runners? Ignore air resistance. FN mg 52 N Since the speed is constant, the friction force equals the force exerted by the girl, 36 N. Ff kFN F FN so k f 36 N 52 N 0.69 18. You need to move a 105-kg sofa to a different location in the room. It takes a force of 102 N to start it moving. What is the coefficient of static friction between the sofa and the carpet? Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 0.0 0.0 3.0 Practice Problems Chapter 5 continued Ff sFN 5.8 N F s f FN 1.0101 N 0.58 Ff, after k, afterFN Ff mg (0.06)(1.0101 N) 0.6 N 102 N 2 (105 kg)(9.80 m/s ) 0.0991 19. Mr. Ames is dragging a box full of books from his office to his car. The box and books together have a combined weight of 134 N. If the coefficient of static friction between the pavement and the box is 0.55, how hard must Mr. Ames push the box in order to start it moving? FAmes on box Ffriction sFN Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Fnet kFN ma kmg a k g 2 smg 1.25 m/s 2 (0.55)(134 N) 0.128 74 N 20. Suppose that the sled in problem 17 is resting on packed snow. The coefficient of kinetic friction is now only 0.12. If a person weighing 650 N sits on the sled, what force is needed to pull the sled across the snow at constant speed? At constant speed, applied force equals friction force, so Ff kFN (0.12)(52 N 650 N) 84 N 9.80 m/s 23. You help your mom move a 41-kg bookcase to a different place in the living room. If you push with a force of 65 N and the bookcase accelerates at 0.12 m/s2, what is the coefficient of kinetic friction between the bookcase and the carpet? Fnet F kFN F kmg ma F ma k mg 65 N (41 kg)(0.12 m/s2) (41 kg)(9.80 m/s ) 2 0.15 21. Suppose that a particular machine in a factory has two steel pieces that must rub against each other at a constant speed. Before either piece of steel has been treated to reduce friction, the force necessary to get them to perform properly is 5.8 N. After the pieces have been treated with oil, what will be the required force? Ff, before k, beforeFN Ff, page 130 22. A 1.4-kg block slides across a rough surface such that it slows down with an acceleration of 1.25 m/s2. What is the coefficient of kinetic friction between the block and the surface? before so FN k, before Physics: Principles and Problems 24. A shuffleboard disk is accelerated to a speed of 5.8 m/s and released. If the coefficient of kinetic friction between the disk and the concrete court is 0.31, how far does the disk go before it comes to a stop? The courts are 15.8 m long. Identify the direction of the disk’s motion as positive. Find the acceleration of the disk due to the force of friction. Fnet kFN kmg ma a kg Solutions Manual 91 Chapter 5 continued Then use the equation vf2 vi2 2a(df di ) to find the distance. (0.0 m/s) (23 m/s)2 (2)(0.41)(9.80 m/s ) 2 Let di 0 and solve for df . vf2 2a 66 m, so he hits the branch before he can stop. vi2 df vf2 vi2 Section Review (2)(kg) 5.2 (0.0 m/s)2 (5.8 m/s)2 (2)(0.31)(9.80 m/s ) 2 5.5 m 25. Consider the force pushing the box in Example Problem 4. How long would it take for the velocity of the box to double to 2.0 m/s? The initial velocity is 1.0 m/s, the final velocity is 2.0 m/s, and the acceleration is 2.0 m/s2, so v v f i ; let ti 0 and solve for tf. a t t f i v v a f i tf 2.0 m/s 1.0 m/s 2 2.0 m/s 0.50 s Choose positive direction as direction of car’s movement. Fnet kFN kmg ma Then use the equation vi2 2a(df di ) to find the distance. 28. Friction At a wedding reception, you notice a small boy who looks like his mass is about 25 kg, running part way across the dance floor, then sliding on his knees until he stops. If the kinetic coefficient of friction between the boy’s pants and the floor is 0.15, what is the frictional force acting on him as he slides? Ffriction kFN (0.15)(25 kg)(9.80 m/s2) 37 N (2)(kg) 29. Velocity Derek is playing cards with his friends, and it is his turn to deal. A card has a mass of 2.3 g, and it slides 0.35 m along the table before it stops. If the coefficient of kinetic friction between the card and the table is 0.24, what was the initial speed of the card as it left Derek’s hand? Solutions Manual Physics: Principles and Problems Let di 0 and solve for df. v2v2 2a f i df vf2 vi2 92 They are similar in that they both act in a direction opposite to the motion (or intended motion) and they both result from two surfaces rubbing against each other. Both are dependent on the normal force between these two surfaces. Static friction applies when there is no relative motion between the two surfaces. Kinetic friction is the type of friction when there is relative motion. The coefficient of static friction between two surfaces is greater than the coefficient of kinetic friction between those same two surfaces. kmg a kg vf2 page 130 27. Friction In this section, you learned about static and kinetic friction. How are these two types of friction similar? What are the differences between static and kinetic friction? Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 26. Ke Min is driving along on a rainy night at 23 m/s when he sees a tree branch lying across the road and slams on the brakes when the branch is 60.0 m in front of him. If the coefficient of kinetic friction between the car’s locked tires and the road is 0.41, will the car stop before hitting the branch? The car has a mass of 2400 kg. Friction pages 126–130 Chapter 5 continued Identify the direction of the card’s movement as positive kFN Fon table ma Fnet kFN kmg ma k Fon table ma mg a kg 25 N (13 kg)(0.26 m/s) (13 kg)(9.80 m/s ) 2 vf di 0 so 0.17 vi 2ad f All you can conclude about the coefficient of static friction is that it is between 2( kg)df 2( 9.80 m /s2)(0. 35 m) 0.24)( Fon table 1.3 m/s s mg 20 N 2 30. Force The coefficient of static friction between a 40.0-kg picnic table and the ground below it is 0.43 m. What is the greatest horizontal force that could be exerted on the table while it remains stationary? (13 kg)(9.80 m/s ) 0.16 Fon table and s mg 25 N 2 Ff sFN (13 kg)(9.80 m/s ) 0.20 smg (0.43)(40.0 kg)(9.80 m/s2) Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1.7102 N 31. Acceleration Ryan is moving to a new apartment and puts a dresser in the back of his pickup truck. When the truck accelerates forward, what force accelerates the dresser? Under what circumstances could the dresser slide? In which direction? Friction between the dresser and the truck accelerates the dresser forward. The dresser will slide backward if the force accelerating it is greater than smg. 32. Critical Thinking You push a 13-kg table in the cafeteria with a horizontal force of 20 N, but it does not move. You then push it with a horizontal force of 25 N, and it accelerates at 0.26 m/s2. What, if anything, can you conclude about the coefficients of static and kinetic friction? Practice Problems 5.3 Force and Motion in Two Dimensions pages 131–135 page 133 33. An ant climbs at a steady speed up the side of its anthill, which is inclined 30.0° from the vertical. Sketch a free-body diagram for the ant. x Ff y FN Fgy Fgx Fg 60.0° Fg From the sliding portion of your experiment you can determine that the coefficient of kinetic friction between the table and the floor is Ff Fon table F2 Physics: Principles and Problems Solutions Manual 93 Chapter 5 continued 34. Scott and Becca are moving a folding table out of the sunlight. A cup of lemonade, with a mass of 0.44 kg, is on the table. Scott lifts his end of the table before Becca does, and as a result, the table makes an angle of 15.0° with the horizontal. Find the components of the cup’s weight that are parallel and perpendicular to the plane of the table. Fg, parallel Fg sin 37. A suitcase is on an inclined plane. At what angle, relative to the vertical, will the component of the suitcase’s weight parallel to the plane be equal to half the perpendicular component of its weight? Fg, parallel Fg sin , when the angle is with respect to the horizontal Fg, perpendicular Fg cos , when the angle is with respect to the horizontal (0.44 kg)(9.80 m/s2)(sin 15.0°) 1.1 N Fg, perpendicular Fg cos (0.44 kg)(9.80 m/s2) (cos 15.0°) 4.2 N Fg, perpendicular 2Fg, parallel Fg, perpendicular 2 Fg, parallel Fg cos Fg sin 1 tan 35. Kohana, who has a mass of 50.0 kg, is at the dentist’s office having her teeth cleaned, as shown in Figure 5-14. If the component of her weight perpendicular to the plane of the seat of the chair is 449 N, at what angle is the chair tilted? tan1 1 2 26.6° relative to the horizontal, or 63.4° relative to the vertical page 135 38. Consider the crate on the incline in Example Problem 5. a. Calculate the magnitude of the acceleration. m Fg sin ■ m Figure 5-14 Fg, perpendicular Fg cos mg cos Fg, perpendicular cos1 mg (50.0 kg)(9.80 m/s ) 449 N cos1 2 23.6° 36. Fernando, who has a mass of 43.0 kg, slides down the banister at his grandparents’ house. If the banister makes an angle of 35.0° with the horizontal, what is the normal force between Fernando and the banister? FN mg cos (43.0 kg)(9.80 m/s2)(cos 35.0°) 345 N 94 Solutions Manual mg sin m g sin (9.80 m/s2)(sin 30.0°) 4.90 m/s2 b. After 4.00 s, how fast will the crate be moving? v v tf ti f i ; let vi ti 0. a Solve for vf. vf atf (4.90 m/s2)(4.00 s) 19.6 m/s Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. F a Chapter 5 continued 39. If the skier in Example Problem 6 were on a 31° downhill slope, what would be the magnitude of the acceleration? Since a g(sin cos ), a (9.80 m/s2)(sin 31° (0.15)(cos 31°)) 3.8 m/s2 40. Stacie, who has a mass of 45 kg, starts down a slide that is inclined at an angle of 45° with the horizontal. If the coefficient of kinetic friction between Stacie’s shorts and the slide is 0.25, what is her acceleration? FStacie’s weight parallel with slide Ff ma 5.3 Force and Motion in Two Dimensions pages 131–135 page 135 42. Forces One way to get a car unstuck is to tie one end of a strong rope to the car and the other end to a tree, then push the rope at its midpoint at right angles to the rope. Draw a free-body diagram and explain why even a small force on the rope can exert a large force on the car. g(sin k cos ) The vectors shown in the free body diagram indicate that even a small force perpendicular to the rope can increase the tension in the rope enough to overcome the friction force. Since F 2T sin (where is the angle between the rope’s original position and its displaced position), (9.80 m/s2)[sin 45° (0.25)(cos 45°)] T 5.2 m/s2 For smaller values of , the tension, T, will increase greatly. FStacie’s weight parallel with slide Ff a m mg sin F m k N mg sin mg cos m k Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Section Review 41. After the skier on the 37° hill in Example Problem 6 had been moving for 5.0 s, the friction of the snow suddenly increased and made the net force on the skier zero. What is the new coefficient of friction? a g(sin k cos ) F 2 sin Fe F car F friction a g sin gk cos If a 0, 0 g sin gk cos k cos sin sin k cos sin 37° k cos 37° 0.75 Physics: Principles and Problems Solutions Manual 95 Chapter 5 continued 43. Mass A large scoreboard is suspended from the ceiling of a sports arena by 10 strong cables. Six of the cables make an angle of 8.0° with the vertical while the other four make an angle of 10.0°. If the tension in each cable is 1300.0 N, what is the scoreboard’s mass? Fnet,y may 0 Fnet,y Fcables on board Fg 6Fcable cos 6 4Fcable cos 4 mg 0 6Fcable cos 6 4Fcable cos 4 m g 6(1300.0 N)(cos 8.0°) 4(1300.0 N)(cos 10.0°) 9.80 m/s 2 1.31103 kg 44. Acceleration A 63-kg water skier is pulled up a 14.0° incline by a rope parallel to the incline with a tension of 512 N. The coefficient of kinetic friction is 0.27. What are the magnitude and direction of the skier’s acceleration? FN mg cos Frope on skier Fg Ff ma Frope on skier mg sin kmg cos ma Frope on skier mg sin kmg cos a m 512 N (63 kg)(9.80 m/s2)(sin 14.0°) (0.27)(63 kg)(9.80 m/s2)(cos 14.0°) 63 kg 45. Equilibrium You are hanging a painting using two lengths of wire. The wires will break if the force is too great. Should you hang the painting as shown in Figures 5-15a or 5-15b? Explain. Fg Figure 5-15b; FT , so FT gets smaller as gets 2 sin ■ Figure 5-15a ■ Figure 5-15b larger, and is larger in 5-15b. 46. Critical Thinking Can the coefficient of friction ever have a value such that a skier would be able to slide uphill at a constant velocity? Explain why or why not. Assume there are no other forces acting on the skier. No, because both the frictional force opposing the motion of the skier and the component of Earth’s gravity parallel to the slope point downhill, not uphill. 96 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 3.2 m/s2, up the incline Chapter 5 continued 52. Explain the method that you would use to subtract two vectors graphically. (5.1) Chapter Assessment Concept Mapping page 140 47. Complete the concept map below by labeling the circles with sine, cosine, or tangent to indicate whether each function is positive or negative in each quadrant. Quadrant I sine cosine tangent II III 53. Explain the difference between these two symbols: A and A. (5.1) A is the symbol for the vector quantity. A is the signed magnitude (length) of the vector. 54. The Pythagorean theorem usually is written c 2 a2 b 2. If this relationship is used in vector addition, what do a, b, and c represent? (5.1) IV sine cosine tangent tangent cosine tangent cosine sine tangent Mastering Concepts page 140 48. Describe how you would add two vectors graphically. (5.1) Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Reverse the direction of the second vector and then add them. Make scale drawings of arrows representing the vector quantities. Place the arrows for the quantities to be added tipto-tail. Draw an arrow from the tail of the first to the tip of the last. Measure the length of that arrow and find its direction. 49. Which of the following actions is permissible when you graphically add one vector to another: moving the vector, rotating the vector, or changing the vector’s length? (5.1) allowed: moving the vector without changing length or direction 50. In your own words, write a clear definition of the resultant of two or more vectors. Do not explain how to find it; explain what it represents. (5.1) The resultant is the vector sum of two or more vectors. It represents the quantity that results from adding the vectors. 51. How is the resultant displacement affected when two displacement vectors are added in a different order? (5.1) a and b represent the lengths of two vectors that are at the right angles to one another. c represents the length of the sum of the two vectors. 55. When using a coordinate system, how is the angle or direction of a vector determined with respect to the axes of the coordinate system? (5.1) The angle is measured counterclockwise from the x-axis. 56. What is the meaning of a coefficient of friction that is greater than 1.0? How would you measure it? (5.2) The frictional force is greater than the normal force. You can pull the object along the surface, measuring the force needed to move it at constant speed. Also measure the weight of the object. 57. Cars Using the model of friction described in this textbook, would the friction between a tire and the road be increased by a wide rather than a narrow tire? Explain. (5.2) It would make no difference. Friction does not depend upon surface area. 58. Describe a coordinate system that would be suitable for dealing with a problem in which a ball is thrown up into the air. (5.3) One axis is vertical, with the positive direction either up or down. It is not affected. Physics: Principles and Problems Solutions Manual 97 Chapter 5 continued 59. If a coordinate system is set up such that the positive x-axis points in a direction 30° above the horizontal, what should be the angle between the x-axis and the y-axis? What should be the direction of the positive y-axis? (5.3) The two axes must be at right angles. The positive y-axis points 30° away from the vertical so that it is at right angles to the x-axis. 60. Explain how you would set up a coordinate system for motion on a hill. (5.3) For motion on a hill, the vertical (y) axis is usually set up perpendicular, or normal, to the surface of the hill. 61. If your textbook is in equilibrium, what can you say about the forces acting on it? (5.3) The net force acting on the book is zero. 62. Can an object that is in equilibrium be moving? Explain. (5.3) Yes, Newton’s first law permits motion as long as the object’s velocity is constant. It cannot accelerate. The vector sum of forces forming a closed triangle is zero. If these are the only forces acting on the object, the net force on the object is zero and the object is in equilibrium. 65. For a book on a sloping table, describe what happens to the component of the weight force parallel to the table and the force of friction on the book as you increase the angle that the table makes with the horizontal. (5.3) a. Which components of force(s) increase when the angle increases? As you increase the angle the table makes with the horizontal, the component of the book’s weight force along the table increases. b. Which components of force(s) decrease? When the angle increases, the component of the weight force normal to the table decreases and the friction force decreases. Applying Concepts pages 140–141 66. A vector that is 1 cm long represents a displacement of 5 km. How many kilometers are represented by a 3-cm vector drawn to the same scale? (3 cm) 15 km 5 km 1 cm 67. Mowing the Lawn If you are pushing a lawn mower across the grass, as shown in Figure 5-16, can you increase the horizontal component of the force that you exert on the mower without increasing the magnitude of the force? Explain. 64. You are asked to analyze the motion of a book placed on a sloping table. (5.3) a. Describe the best coordinate system for analyzing the motion. Set up the y-axis perpendicular to the surface of the table and the x-axis pointing uphill and parallel to the surface. F Fy Fx b. How are the components of the weight of the book related to the angle of the table? ■ 98 Solutions Manual Figure 5-16 Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 63. What is the sum of three vectors that, when placed tip to tail, form a triangle? If these vectors represent forces on an object, what does this imply about the object? (5.3) One component is parallel to the inclined surface and the other is perpendicular to it. Chapter 5 continued Yes, lower the handle to make the angle between the handle and the horizontal smaller. 68. A vector drawn 15 mm long represents a velocity of 30 m/s. How long should you draw a vector to represent a velocity of 20 m/s? (20 m/s) 10 mm 15 mm 30 m/s 69. What is the largest possible displacement resulting from two displacements with magnitudes 3 m and 4 m? What is the smallest possible resultant? Draw sketches to demonstrate your answers. The largest is 7 m; the smallest is 1 m. 3m 4m 7m 4m 73. Under what conditions can the Pythagorean theorem, rather than the law of cosines, be used to find the magnitude of a resultant vector? The Pythagorean theorem can be used only if the two vectors to be added are at right angles to one another. 74. A problem involves a car moving up a hill, so a coordinate system is chosen with the positive x-axis parallel to the surface of the hill. The problem also involves a stone that is dropped onto the car. Sketch the problem and show the components of the velocity vector of the stone. One component is in the negative x-direction, the other in the negative y-direction, assuming that the positive direction points upward, perpendicular to the hill. 3m vx 1m 70. How does the resultant displacement change as the angle between two vectors increases from 0° to 180°? vy v +x +y Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. The resultant increases. 71. A and B are two sides of a right triangle, where tan A/B. a. Which side of the triangle is longer if tan is greater than 1.0? A is longer. b. Which side is longer if tan is less than 1.0? B is longer. c. What does it mean if tan is equal to 1.0? A and B are equal in length. 72. Traveling by Car A car has a velocity of 50 km/h in a direction 60° north of east. A coordinate system with the positive x-axis pointing east and a positive y-axis pointing north is chosen. Which component of the velocity vector is larger, x or y? The northward component (y) is longer. Physics: Principles and Problems 75. Pulling a Cart According to legend, a horse learned Newton’s laws. When the horse was told to pull a cart, it refused, saying that if it pulled the cart forward, according to Newton’s third law, there would be an equal force backwards; thus, there would be balanced forces, and, according to Newton’s second law, the cart would not accelerate. How would you reason with this horse? The equal and opposite forces referred to in Newton’s third law are acting on different objects. The horse would pull on the cart, and the cart would pull on the horse. The cart would have an unbalanced net force on it (neglecting friction) and would thus accelerate. Solutions Manual 99 Chapter 5 continued 76. Tennis When stretching a tennis net between two posts, it is relatively easy to pull one end of the net hard enough to remove most of the slack, but you need a winch to take the last bit of slack out of the net to make the top almost completely horizontal. Why is this true? When stretching the net between the two posts, there is no perpendicular component upward to balance the weight of the net. All the force exerted on the net is horizontal. Stretching the net to remove the last bit of slack requires great force in order to reduce the flexibility of the net and to increase the internal forces that hold it together. 77. The weight of a book on an inclined plane can be resolved into two vector components, one along the plane, and the other perpendicular to it. a. At what angle are the components equal? 45° b. At what angle is the parallel component equal to zero? 0° 78. TV Towers The transmitting tower of a TV station is held upright by guy wires that extend from the top of the tower to the ground. The force along the guy wires can be resolved into two perpendicular components. Which one is larger? The component perpendicular to the ground is larger if the angle between the guy wire and horizontal is greater than 45°. 5.1 Vectors pages 141–142 Level 1 79. Cars A car moves 65 km due east, then 45 km due west. What is its total displacement? 65 km (45 km) 2.0101 km d 2.0101 km, east 80. Find the horizontal and vertical components of the following vectors, as shown in Figure 5-17. B(3.0) A(3.0) F(5.0) E(5.0) C(6.0) D(4.0) ■ Figure 5-17 a. E Ex E cos (5.0)(cos 45°) 3.5 Ey E sin (5.0)(sin 45°) 3.5 b. F Fx F cos (5.0)(cos 225°) 3.5 Fy F sin (5.0)(sin 225°) 3.5 c. A Ax A cos (3.0)(cos 180°) 3.0 Ay A sin (3.0)(sin 180°) 0.0 100 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. c. At what angle is the parallel component equal to the weight? 90° Mastering Problems Chapter 5 continued 81. Graphically find the sum of the following pairs of vectors, whose lengths and directions are shown in Figure 5-17. a. D and A 83. You walk 30 m south and 30 m east. Find the magnitude and direction of the resultant displacement both graphically and algebraically. R2 A2 B2 D R (30 m )2 (3 0 m)2 A 40 m R(1.0) 30 m tan 1 30 m b. C and D C D 45° R 40 m, 45° east of south R(10.0) 45 c. C and A R 42 30 m S C A 30 m E R(3.0) The difference in the answers is due to significant digits being considered in the calculation. d. E and F E F Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. R 0.0 Level 2 82. Graphically add the following sets of vectors, as shown in Figure 5-17. a. A, C, and D 84. Hiking A hiker’s trip consists of three segments. Path A is 8.0 km long heading 60.0° north of east. Path B is 7.0 km long in a direction due east. Path C is 4.0 km long heading 315° counterclockwise from east. a. Graphically add the hiker’s displacements in the order A, B, C. A B C D C A R R(7.0) b. Graphically add the hiker’s displacements in the order C, B, A. b. A, B, and E R E A R(6.5) C B A B c. What can you conclude about the resulting displacements? c. B, D, and F D B F You can add vectors in any order. The result is always the same. R Physics: Principles and Problems Solutions Manual 101 Chapter 5 continued 85. What is the net force acting on the ring in Figure 5-18? y (128 N)(cos 30.0°) 111 N By B sin B 400.0 N 500.0 N Bx B cos B (128 N)(sin 30.0°) 40.0° 64 N 50.0° x ■ 64 N 111 N Figure 5-18 47 N R 2 A2 B2 Ry Ay By R A2 B2 0 N 64 N 64 N (500.0 N)2 (400. 0 N)2 R Rx2 Ry2 640.3 N (47 N)2 (6 4 N)2 A tan B 79 N tan1 A B R Ry tan1 tan1 500.0 400.0 x tan1 64 47 51.34° from B 54° The net force is 640.3 N at 51.34° y 128 N 128 N 30.0° 64 N Figure 5-19 A 128 N 64 N 64 N Ax A cos A x Level 3 87. A Ship at Sea A ship at sea is due into a port 500.0 km due south in two days. However, a severe storm comes in and blows it 100.0 km due east from its original position. How far is the ship from its destination? In what direction must it travel to reach its destination? R2 A2 B2 R (100.0 km)2 (500.0 km)2 509.9 km R Ry tan1 x (64 N)(cos 180°) tan1 64 N 78.69° Ay A sin A (64 N)(sin 180°) 500.0 100.0 R 509.9 km, 78.69° south of west 0N 102 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 86. What is the net force acting on the ring in Figure 5-19? ■ Rx Ax Bx Chapter 5 continued 88. Space Exploration A descent vehicle landing on Mars has a vertical velocity toward the surface of Mars of 5.5 m/s. At the same time, it has a horizontal velocity of 3.5 m/s. a. At what speed does the vehicle move along its descent path? R2 A2 B2 R (5.5 m /s)2 (3.5 m /s)2 v R 6.5 m/s b. At what angle with the vertical is this path? R R y tan1 x tan1 5.5 3.5 58° from horizontal, which is 32° from vertical Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 89. Navigation Alfredo leaves camp and, using a compass, walks 4 km E, then 6 km S, 3 km E, 5 km N, 10 km W, 8 km N, and, finally, 3 km S. At the end of three days, he is lost. By drawing a diagram, compute how far Alfredo is from camp and which direction he should take to get back to camp. Take north and east to be positive directions. North: 6 km 5 km 8 km 3 km 4 km. East: 4 km 3 km 10 km 3 km. The hiker is 4 km north and 3 km west of camp. To return to camp, the hiker must go 3 km east and 4 km south. R2 A2 B2 R (3 km )2 (4 km)2 5 km R R y tan1 x tan1 4 km 3 km 53° R 5 km, 53° south of east 5.2 Friction page 142 Level 1 90. If you use a horizontal force of 30.0 N to slide a 12.0-kg wooden crate across a floor at a constant velocity, what is the coefficient of kinetic friction between the crate and the floor? Ff kFN kmg Fhorizontal F horizontal k mg 30.0 N 2 (12.0 kg)(9.80 m/s ) 0.255 91. A 225-kg crate is pushed horizontally with a force of 710 N. If the coefficient of friction is 0.20, calculate the acceleration of the crate. ma Fnet Fappl Ff where Ff kFN kmg Therefore Fappl kmg a m 710 N (0.20)(225 kg)(9.80 m/s2) 225 kg 1.2 m/s2 Level 2 92. A force of 40.0 N accelerates a 5.0-kg block at 6.0 m/s2 along a horizontal surface. a. How large is the frictional force? ma Fnet Fappl Ff so Ff Fappl ma 40.0 N (5.0 kg)(6.0 m/s2) 1.0101 N b. What is the coefficient of friction? Ff kFN kmg F mg f so k 1 1.010 N 2 (5.0 kg)(9.80 m/s ) 0.20 Physics: Principles and Problems Solutions Manual 103 Chapter 5 continued 93. Moving Appliances Your family just had a new refrigerator delivered. The delivery man has left and you realize that the refrigerator is not quite in the right position, so you plan to move it several centimeters. If the refrigerator has a mass of 180 kg, the coefficient of kinetic friction between the bottom of the refrigerator and the floor is 0.13, and the static coefficient of friction between these same surfaces is 0.21, how hard do you have to push horizontally to get the refrigerator to start moving? Fon fridge Ffriction F2 44.0 N, 60.0° smg m/s2) 370 N Ff kFN ma m(vf2 vi2) kmg where vf 0 2d (The minus sign indicates the force is acting opposite to the direction of motion.) v2 2dg (14.0 m/s)2 2(25.0 m)(9.80 m/s2) 0.400 F3 ? F1x F1 cos 1 (33.0 N)(cos 90.0°) 0.0 N F1y F1 sin 1 (33.0 N)(sin 90.0°) 33.0 N F2x F2 cos 2 (44.0 N)(cos 60.0°) 22.0 N F2y F2 sin 2 (44.0 N)(sin 60.0°) 38.1 N F3x F1x F2x 0.0 N 22.0 N 22.0 N F3y F1y F2y 33.0 N 38.1 N 71.1 N F3 F3x2 F3y2 (22.0 N)2 (71.1 N)2 74.4 N 104 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Level 3 94. Stopping at a Red Light You are driving a 2500.0-kg car at a constant speed of 14.0 m/s along a wet, but straight, level road. As you approach an intersection, the traffic light turns red. You slam on the brakes. The car’s wheels lock, the tires begin skidding, and the car slides to a halt in a distance of 25.0 m. What is the coefficient of kinetic friction between your tires and the wet road? k i First, find the magnitude of the sum of these two forces. The equilibrant will have the same magnitude but opposite direction. F1 33.0 N, 90.0° sFN (0.21)(180 kg)(9.80 5.3 Force and Motion in Two Dimensions pages 142–143 Level 1 95. An object in equilibrium has three forces exerted on it. A 33.0-N force acts at 90.0° from the x-axis and a 44.0-N force acts at 60.0° from the x-axis. What are the magnitude and direction of the third force? Chapter 5 continued For equilibrium, the sum of the components must equal zero, so 25.0 N F5y F5 sin 5 (50.0 N)(sin 60.0°) 22.0 N 180.0° F6x F1x F2x F3x F4x F5x F3y tan1 180.0° F3x F5x F5 cos 5 (50.0 N)(cos 60.0°) 71.1 N tan1 253° 43.3 N 0.0 N 40.0 N 0.0 N (40.0 N) 25.0 N F3 74.4 N, 253° Level 2 96. Five forces act on an object: (1) 60.0 N at 90.0°, (2) 40.0 N at 0.0°, (3) 80.0 N at 270.0°, (4) 40.0 N at 180.0°, and (5) 50.0 N at 60.0°. What are the magnitude and direction of a sixth force that would produce equilibrium? Solutions by components F1 60.0 N, 90.0° F2 40.0 N, 0.0° F3 80.0 N, 270.0° F6y F1y F2y F3y F4y F5y 60.0 N 0.0 N (80.0 N) 0.0 N 43.3 N 23.3 N 2 F6 F F6y2 6x (25.0 N)2 (23.3 N)2 34.2 N F F6y 6 tan1 180.0° 6x F4 40.0 N, 180.0° tan1 180.0° F5 50.0 N, 60.0° 223° F6 ? Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 25.0 N F1x F1 cos 1 (60.0 N)(cos 90.0°) 0.0 N F1y F1 sin 1 (60.0 N)(sin 90.0°) 60.0 N F2x F2 cos 2 (40.0 N)(cos 0.0°) 40.0 N F2y F2 sin 2 (40.0 N)(sin 0.0°) 23.3 N 25.0 N F6 34.2 N, 223° 97. Advertising Joe wishes to hang a sign weighing 7.50102 N so that cable A, attached to the store, makes a 30.0° angle, as shown in Figure 5-20. Cable B is horizontal and attached to an adjoining building. What is the tension in cable B? 30.0° A 0.0 N F3x F3 cos 3 (80.0 N)(cos 270.0°) B 0.0 N F3y F3 sin 3 (80.0 N)(sin 270.0°) 80.0 N ■ Figure 5-20 F4x F4 cos 4 (40.0 N)(cos 180.0°) 40.0 N F4y F4 sin 4 (40.0 N)(sin 180.0°) 0.0 N Physics: Principles and Problems Solutions Manual 105 Chapter 5 continued Solution by components. The sum of the components must equal zero, so FAy Fg 0 Fparallel Fg sin (215 N)(sin 35.0°) 123 N so FAy Fg 7.50102 N FAy FA sin 60.0° FAy so FA sin 60.0° 2 7.5010 N sin 60.0° 866 N Also, FB FA 0, so FB FA FA cos 60.0° (866 N)(cos 60.0°) 433 N, right 98. A street lamp weighs 150 N. It is supported by two wires that form an angle of 120.0° with each other. The tensions in the wires are equal. a. What is the tension in each wire supporting the street lamp? Level 3 100. Emergency Room You are shadowing a nurse in the emergency room of a local hospital. An orderly wheels in a patient who has been in a very serious accident and has had severe bleeding. The nurse quickly explains to you that in a case like this, the patient’s bed will be tilted with the head downward to make sure the brain gets enough blood. She tells you that, for most patients, the largest angle that the bed can be tilted without the patient beginning to slide off is 32.0° from the horizontal. a. On what factor or factors does this angle of tilting depend? The coefficient of static friction between the patient and the bed’s sheets. b. Find the coefficient of static friction between a typical patient and the bed’s sheets. Fg parallel to bed mg sin Ff Fg sFN so T 2 sin 150 N (2)(sin 30.0°) 1.5102 N b. If the angle between the wires supporting the street lamp is reduced to 90.0°, what is the tension in each wire? Fg T 2 sin smg cos mg sin mg cos so s sin cos tan tan 32.0° 0.625 150 N (2)(sin 45°) 1.1102 N 99. A 215-N box is placed on an inclined plane that makes a 35.0° angle with the horizontal. Find the component of the weight force parallel to the plane’s surface. 106 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Fg 2T sin Chapter 5 continued 101. Two blocks are connected by a string over a tionless, massless pulley such that one is resting on an inclined plane and the other is hanging over the top edge of the plane, as shown in Figure 5-21. The hanging block has a mass of 16.0 kg, and the one on the plane has a mass of 8.0 kg. The coefficient of kinetic friction between the block and the inclined plane is 0.23. The blocks are released from rest. fric- 37.0° ■ Figure 5-21 a. What is the acceleration of the blocks? F mbotha Fg hanging F plane Ff plane mhangingg Fg plane sin kFg plane cos so a mboth mhangingg mplaneg sin kmplaneg cos mboth g(mhanging mplanesin kmplane cos ) mhanging mplane (9.80 ms )(16.0 kg (8.0 kg)(sin 37.0°) (0.23)(8.0 kg)(cos 37.0°)) 2 (16.0 kg 80 kg) 4.0 m/s2 b. What is the tension in the string connecting the blocks? FT Fg Fa mg ma Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. m(g a) (16.0 kg)(9.80 m/s2 4.0 m/s2) 93 N 102. In Figure 5-22, a block of mass M is pushed with such a force, F, that the smaller block of mass m does not slide down the front of it. There is no friction between the larger block and the surface below it, but the coefficient of static friction between the two blocks is s. Find an expression for F in terms of M, m, s , and g. m F M Smaller block: Ff, M on m sFN, M on m mg ■ Figure 5-22 mg s FN, M on m ma g s a Physics: Principles and Problems Solutions Manual 107 Chapter 5 continued Larger block: b. What force will be needed to start the sled moving? F – FN, m on M Ma Ff sFN mg Mg F s s sFg g s (0.30)(4.90102 N) F (m M) 1.5102 N Mixed Review pages 143–144 Level 1 103. The scale in Figure 5-23 is being pulled on by three ropes. What net force does the scale read? c. What force is needed to keep the sled moving at a constant velocity? Ff sFN sFg (0.10)(4.90102 N) 49 N, kinetic friction d. Once moving, what total force must be applied to the sled to accelerate it at 3.0 m/s2? 27.0° 27.0° ma Fnet Fappl Ff so Fappl ma Ff (50.0 kg)(3.0 m/s2) 49 N 2.0102 N 75.0 N 75.0 N ■ Figure 5-7 Find the y-component of the two side ropes and then add them to the middle rope. Fy F cos (75.0 N)(cos 27.0°) 66.8 N Fy, total Fy, left Fy, middle Fy, right 66.8 N 150.0 N 66.8 N 283.6 N 104. Sledding A sled with a mass of 50.0 kg is pulled along flat, snow-covered ground. The static friction coefficient is 0.30, and the kinetic friction coefficient is 0.10. a. What does the sled weigh? Fg mg (50.0 kg)(9.80 m/s2) 4.90102 N 108 Solutions Manual a. If the coefficient of kinetic friction between the boulder and the mountainside is 0.40, the mass of the boulder is 20.0 kg, and the slope of the mountain is a constant 30.0°, what is the force that Sisyphus must exert on the boulder to move it up the mountain at a constant velocity? FS on rock Fg to slope Ff FS on rock mg sin kmg cos ma 0 FS on rock mg sin kmg cos Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 150.0 N Level 2 105. Mythology Sisyphus was a character in Greek mythology who was doomed in Hades to push a boulder to the top of a steep mountain. When he reached the top, the boulder would slide back down the mountain and he would have to start all over again. Assume that Sisyphus slides the boulder up the mountain without being able to roll it, even though in most versions of the myth, he rolled it. Chapter 5 continued mg(sin k cos ) (20.0 kg)(9.80 m/s2) (sin 30.0° (0.40)(cos 30.0°)) 166 N b. If Sisyphus pushes the boulder at a velocity of 0.25 m/s and it takes him 8.0 h to reach the top of the mountain, what is the mythical mountain’s vertical height? h d sin vt sin (0.25 m/s)(8.0 h)(3600 s/h)(sin 30.0°) 3.6103 m 3.6 km Level 3 106. Landscaping A tree is being transported on a flatbed trailer by a landscaper, as shown in Figure 5-24. If the base of the tree slides on the tree will the trailer, fall over and be damaged. If the coefficient of static friction between the tree and the trailer is 0.50, what is the minimum stopping distance of the truck, traveling at 55 km/h, if it is to accelerate uniformly and not have the tree slide forward and fall on the trailer? Ftruck Ff sFN smg ma ■ Figure 5-24 mg m a s sg Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. (0.50)(9.80 m/s2) 4.9 m/s2 vf2 vi2 2ad with vf 0, v2 2a so d i 1h 1000 m (55 km/h) 2 (2)(4.9 ms2) 1 km 3600 s 24 m Physics: Principles and Problems Solutions Manual 109 Chapter 5 continued Thinking Critically page 144 107. Use Models Using the Example Problems in this chapter as models, write an example problem to solve the following problem. Include the following sections: Analyze and Sketch the Problem, Solve for the Unknown (with a complete strategy), and Evaluate the Answer. A driver of a 975-kg car traveling 25 m/s puts on the brakes. What is the shortest distance it will take for the car to stop? Assume that the road is concrete, the force of friction of the road on the tires is constant, and the tires do not slip. df di 25 m/s FN +x v Ff a F net Fg Analyze and Sketch the Problem • Choose a coordinate system with a positive axis in the direction of motion. • Draw a motion diagram. • Label v and a. • Draw the free-body diagram. Known: Unknown: di 0 df ? vf 0 m 975 kg s 0.80 Solve for the Unknown Solve Newton’s second law for a. Fnet ma Ff ma Substitute Ff Fnet FN ma Substitute Ff FN mg ma Substitute FN mg a g Use the expression for acceleration to solve for distance. vf2 vi2 2a(df di) v2v2 2a f i df di v2 v2 f i di (2)(g) 110 Solutions Manual Substitute a g Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. vi 25 m/s Chapter 5 continued (0.0 m/s)2 (25 m/s)2 0.0 m (2)(0.65)(9.80 m/s)2 Substitute di 0.0 m, vf 0.0 m/s, vi 25 m/s, 0.65, g 9.80 m/s2 49 m 108. Analyze and Conclude Margaret Mary, Doug, and Kako are at a local amusement park and see an attraction called the Giant Slide, which is simply a very long and high inclined plane. Visitors at the amusement park climb a long flight of steps to the top of the 27° inclined plane and are given canvas sacks. They sit on the sacks and slide down the 70-m-long plane. At the time when the three friends walk past the slide, a 135-kg man and a 20-kg boy are each at the top preparing to slide down. “I wonder how much less time it will take the man to slide down than it will take the boy,” says Margaret Mary. “I think the boy will take less time,” says Doug. “You’re both wrong,” says Kako. “They will reach the bottom at the same time.” a. Perform the appropriate analysis to determine who is correct. Fnet Fg Ff Fg sin kFN mg sin kmg cos ma a g(sin k cos ), so the acceleration is independent of the mass. They will tie, so Kako is correct. b. If the man and the boy do not take the same amount of time to reach the of the slide, calculate how many seconds of difference there will be between the two times. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. They will reach the bottom at the same time. Writing in Physics page 144 109. Investigate some of the techniques used in industry to reduce the friction between various parts of machines. Describe two or three of these techniques and explain the physics of how they work. Answers will vary and may include lubricants and reduction of the normal force to reduce the force of friction. 110. Olympics In recent years, many Olympic athletes, such as sprinters, swimmers, skiers, and speed skaters, have used modified equipment to reduce the effects of friction and air or water drag. Research a piece of equipment used by one of these types of athletes and the way it has changed over the years. Explain how physics has impacted these changes. Answers will vary. Physics: Principles and Problems Solutions Manual 111 Chapter 5 continued Cumulative Review page 144 111. Add or subtract as indicated and state the answer with the correct number of significant digits. (Chapter 1) a. 85.26 g 4.7 g 90.0 g b. 1.07 km 0.608 km 1.68 km c. 186.4 kg 57.83 kg 128.6 kg d. 60.08 s 12.2 s Challenge Problem page 132 Find the equilibrant for the following forces. F1 61.0 N at 17.0° north of east F2 38.0 N at 64.0° north of east F3 54.0 N at 8.0° west of north F4 93.0 N at 53.0° west of north F5 65.0 N at 21.0° south of west F6 102.0 N at 15.0° west of south F7 26.0 N south F8 77.0 N at 22.0° east of south F9 51.0 N at 33.0° east of south F10 82.0 N at 5.0° south of east 47.9 s 112. You ride your bike for 1.5 h at an average velocity of 10 km/h, then for 30 min at 15 km/h. What is your average velocity? (Chapter 3) Average velocity is the total displacement divided by the total time. y 3 4 2 1 x 10 7 5 df di v tf ti v1t1 v2t2 di t1 t2 ti v1t1 v2t2 v t1 t2 (10 kmh)(1.5 h) (15 kmh)(0.5 h) 1.5 h 0.5 h 8 6 F1x (61.0 N)(cos 17.0°) 58.3 N F1y (61.0 N)(sin 17.0°) 17.8 N F2x (38.0 N)(cos 64.0°) 16.7 N 10 km/h F2y (38.0 N)(sin 64.0°) 34.2 N 113. A 45-N force is exerted in the upward direction on a 2.0-kg briefcase. What is the acceleration of the briefcase? (Chapter 4) Fnet Fapplied Fg Fapplied mg ma Fapplied mg so a m 45 N (2.0 kg)(9.80 ms2) 2.0 kg F3x (54.0 N)(sin 8.0°) 7.52 N F3y (54.0 N)(cos 8.0°) 53.5 N F4x (93.0 N)(sin 53.0°) 74.3 N F4y (93.0 N)(cos 53.0°) 56.0 N F5x (65.0 N)(cos 21.0°) 60.7 N F5y (65.0 N)(sin 21.0°) 23.3 N 13 m/s2 F6x (102 N)(sin 15.0°) 26.4 N F6y (102 N)(cos 15.0°) 98.5 N 112 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. di ti 0, so 9 Chapter 5 continued F7x 0.0 N F7y 26.0 N F8x (77.0 N)(sin 22.0°) 28.8 N F8y (77.0 N)(cos 22.0°) 71.4 N F9x (51.0 N)(sin 33.0°) 27.8 N F9y (51.0 N)(cos 33.0°) 42.8 N F10x (82.0 N)(cos 5.0°) 81.7 N F10y (82.0 N)(sin 5.0°) 7.15 N 10 Fx Fix i1 Fiy i1 107.65 N FR (Fx)2 (Fy)2 (44.38 N)2 (107.65 N)2 116 N F Fy R tan1 x 107.65 N 44.38 N tan1 67.6° Fequilibrant 116 N at 112.4° 116 N at 22.4° W of N Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 44.38 N 10 Fy Physics: Principles and Problems Solutions Manual 113 CHAPTER 6 Motion in Two Dimensions Practice Problems 6.1 Projectile Motion pages 147–152 page 150 1. A stone is thrown horizontally at a speed of 5.0 m/s from the top of a cliff that is 78.4 m high. a. How long does it take the stone to reach the bottom of the cliff? 1 2 Since vy 0, y vyt gt 2 1 becomes y gt 2 2 or t 2y g (2)(78.4 m) 9.80 m/s2 4.00 s Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. b. How far from the base of the cliff does the stone hit the ground? x vxt (5.0 m/s)(4.00 s) 2.0101 m c. What are the horizontal and vertical components of the stone’s velocity just before it hits the ground? vx 5.0 m/s. This is the same as the initial horizontal speed because the acceleration of gravity influences only the vertical motion. For the vertical component, use v vi gt with v vy and vi, the initial vertical component of velocity, zero. At t 4.00 s vy gt 2. Lucy and her friend are working at an assembly plant making wooden toy giraffes. At the end of the line, the giraffes go horizontally off the edge of the conveyor belt and fall into a box below. If the box is 0.6 m below the level of the conveyor belt and 0.4 m away from it, what must be the horizontal velocity of giraffes as they leave the conveyor belt? 2y g x vxt vx so vx x 2y g 0.4 m (2)(0.6 m) 9.80 m/s 2 1 m/s 3. You are visiting a friend from elementary school who now lives in a small town. One local amusement is the ice-cream parlor, where Stan, the short-order cook, slides his completed ice-cream sundaes down the counter at a constant speed of 2.0 m/s to the servers. (The counter is kept very well polished for this purpose.) If the servers catch the sundaes 7.0 cm from the edge of the counter, how far do they fall from the edge of the counter to the point at which the servers catch them? x vxt; x vx t 1 2 y gt 2 g 2 1 2 1 2 x vx 0.070 m 2 2.0 m/s (9.80 m/s2)(4.0 s) (9.80 m/s2) 39.2 m/s 0.0060 m or 0.60 cm Physics: Principles and Problems Solutions Manual 115 Chapter 6 continued page 152 4. A player kicks a football from ground level with an initial velocity of 27.0 m/s, 30.0° above the horizontal, as shown in Figure 6-4. Find each of the following. Assume that air resistance is negligible. 25 Trajectory y (m) 60.0° 30.0° 0 60 x (m) a. the ball’s hang time ■ vy vi sin Figure 6-4 1 When it lands, y vyt gt 2 0. 2 Therefore, 2vyt t 2 g 2vy t g 2v sin g i (2)(27.0 m/s)(sin 30.0°) 9.80 m/s 2 2.76 s b. the ball’s maximum height Maximum height occurs at half the “hang time,” or 1.38 s. Thus, 1 2 y vyt gt 2 2 1 (27.0 m/s)(sin 30.0°)(1.38 s) (9.80 m/s2)(1.38 s)2 2 9.30 m c. the ball’s range Distance: vx vi cos x vxt (vi cos )(t) (27.0 m/s)(cos 30.0°)(2.76 s) 64.5 m 5. The player in problem 4 then kicks the ball with the same speed, but at 60.0° from the horizontal. What is the ball’s hang time, range, and maximum height? Following the method of Practice Problem 4, Hangtime: 2v sin g i t (2)(27.0 m/s)(sin 60.0°) 2 9.80 m/s 4.77 s 116 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1 vi sin t gt 2 Chapter 6 continued Distance: x vi cos t (27.0 m/s)(cos 60.0°)(4.77 s) 64.4 m Maximum height: 1 2 at t (4.77 s) 2.38 s 1 2 y vi sin t gt 2 1 2 (27.0 m/s)(sin 60.0°)(2.38 s) (9.80 m/s2)(2.38 s)2 27.9 m 6. A rock is thrown from a 50.0-m-high cliff with an initial velocity of 7.0 m/s at an angle of 53.0° above the horizontal. Find the velocity vector for when it hits the ground below. vx vi cos vy vi sin gt 2y g vi sin g vi sin 2yg 2 v v vy2 x Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. (vi co s )2 (vi sin 2yg )2 2 2) ((7.0 m/s ) cos 53 .0°)2 ( 7.0 m/s) (sin 53.0 °) (2 )(50 .0 m )( 9.80 m /s 37 m/s vy tan1 vx v sin 2yg vi cos i i i tan1 (7.0 m/s)(sin 53.0°) (2)(50 .0 m)(9.80 m/s ) tan1 2 (7.0 m/s)(cos 53.0°) 83° from horizontal Section Review 6.1 Projectile Motion pages 147–152 page 152 7. Projectile Motion Two baseballs are pitched horizontally from the same height, but at different speeds. The faster ball crosses home plate within the strike zone, but the slower ball is below the batter’s knees. Why does the faster ball not fall as far as the slower one? Physics: Principles and Problems Solutions Manual 117 Chapter 6 continued The faster ball is in the air a shorter time, and thus gains a smaller vertical velocity. 8. Free-Body Diagram An ice cube slides without friction across a table at a constant velocity. It slides off the table and lands on the floor. Draw free-body and motion diagrams of the ice cube at two points on the table and at two points in the air. Free-Body Diagrams On the table In the air FN FN Fg Fg Motion Diagrams On the table In the air a0 a Fg Fg 9. Projectile Motion A softball is tossed into the air at an angle of 50.0° with the vertical at an initial velocity of 11.0 m/s. What is its maximum height? vf2 viy2 2a(df di); a g, di 0 At maximum height vf 0, so viy2 df 2g (v cos )2 2g i 2 ((11.0 m/s)(cos 50.0°)) 2 (2)(9.80 m/s ) 10. Projectile Motion A tennis ball is thrown out a window 28 m above the ground at an initial velocity of 15.0 m/s and 20.0° below the horizontal. How far does the ball move horizontally before it hits the ground? x v0xt, but need to find t First, determine vy f: vy f2 vy i2 2gy 2 2gy vy f v yi (vi sin )2 2gy ((15.0 m/s)( sin 20 .0°))2 (2)( 9.80 m /s2)(28 m) 24.0 m/s Now use vy f vy i gt to find t. vy f vy i t g vy f vi sin g 118 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 2.55 m Chapter 6 continued 2.40 m/s (15.0 m/s)(sin 20.0°) 2 9.80 m/s 1.92 s x vx it (vi cos )(t) (15.0 m/s)(cos 20.0°)(1.92 s) 27.1 m 11. Critical Thinking Suppose that an object is thrown with the same initial velocity and direction on Earth and on the Moon, where g is one-sixth that on Earth. How will the following quantities change? a. vx will not change b. the object’s time of flight 2v y will be larger; t g c. ymax will be larger d. R will be larger Practice Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 6.2 Circular Motion pages 153–156 page 156 12. A runner moving at a speed of 8.8 m/s rounds a bend with a radius of 25 m. What is the centripetal acceleration of the runner, and what agent exerts force on the runner? v2 r (8.8 m/s)2 25 m ac 3.1 m/s2, the frictional force of the track acting on the runner’s shoes exerts the force on the runner. 13. A car racing on a flat track travels at 22 m/s around a curve with a 56-m radius. Find the car’s centripetal acceleration. What minimum coefficient of static friction between the tires and road is necessary for the car to round the curve without slipping? v2 r (22 m/s)2 56 m ac 8.6 m/s2 Recall Ff FN. The friction force must supply the centripetal force so Ff mac. The normal force is FN mg. The coefficient of friction must be at least F FN ma mg a g 2 8.6 m/s c c f 2 0.88 9.80 m/s Physics: Principles and Problems Solutions Manual 119 Chapter 6 continued 14. An airplane traveling at 201 m/s makes a turn. What is the smallest radius of the circular path (in km) that the pilot can make and keep the centripetal acceleration under 5.0 m/s2? (201 m/s)2 v2 v2 8.1 km ac , so r r ac 5.0 m/s2 15. A 45-kg merry-go-round worker stands on the ride’s platform 6.3 m from the center. If her speed as she goes around the circle is 4.1 m/s, what is the force of friction necessary to keep her from falling off the platform? (45 kg)(4.1 m/s)2 mv 2 Ff Fc 120 N 6.3 m r b. What is the direction of the net force that is acting on you? The net force acting on your body is to the right c. What exerts this force? The force is exerted by the car’s seat. 18. Centripetal Force If a 40.0-g stone is whirled horizontally on the end of a 0.60-m string at a speed of 2.2 m/s, what is the tension in the string? FT mac mv 2 r 2 (0.0400 kg)(22 m/s) 0.60 m Section Review 6.2 Circular Motion pages 153–156 page 156 16. Uniform Circular Motion What is the direction of the force that acts on the clothes in the spin cycle of a washing machine? What exerts the force? 17. Free-Body Diagram You are sitting in the backseat of a car going around a curve to the right. Sketch motion and free-body diagrams to answer the following questions. a Fnet v a. What is the direction of your acceleration? Your body is accelerated to the right. 120 Solutions Manual 19. Centripetal Acceleration A newspaper article states that when turning a corner, a driver must be careful to balance the centripetal and centrifugal forces to keep from skidding. Write a letter to the editor that critiques this article. The letter should state that there is an acceleration because the direction of the velocity is changing; therefore, there must be a net force in the direction of the center of the circle. The road supplies that force and the friction between the road and the tires allows the force to be exerted on the tires. The car’s seat exerts the force on the driver that accelerates him or her toward the center of the circle. The note also should make it clear that centrifugal force is not a real force. 20. Centripetal Force A bowling ball has a mass of 7.3 kg. If you move it around a circle with a radius of 0.75 m at a speed of 2.5 m/s, what force would you have to exert on it? Fnet mac 2 mv r Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. The force is toward the center of the tub. The walls of the tub exert the force on the clothes. Of course, the whole point is that some of the water in the clothes goes out through holes in the wall of the tub rather than moving toward the center. 0.32 N Chapter 6 continued so, vw/g vb/g vb/w (7.3 kg)(2.5 m/s)2 0.75 m 0.5 m/s 2.5 m/s 61 N 21. Critical Thinking Because of Earth’s daily rotation, you always move with uniform circular motion. What is the agent that supplies the force that accelerates you? How does this motion affect your apparent weight? 2.0 m/s; against the boat 25. An airplane flies due north at 150 km/h relative to the air. There is a wind blowing at 75 km/h to the east relative to the ground. What is the plane’s speed relative to the ground? 2 v v vw2 p Earth’s gravity supplies the force that accelerates you in circular motion. Your uniform circular motion decreases your apparent weight. Practice Problems 6.3 Relative Velocity pages 157–159 page 159 22. You are riding in a bus moving slowly through heavy traffic at 2.0 m/s. You hurry to the front of the bus at 4.0 m/s relative to the bus. What is your speed relative to the street? Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. vy/g vb/g vy/b 2.0 m/s 4.0 m/s 6.0 m/s relative to street 23. Rafi is pulling a toy wagon through the neighborhood at a speed of 0.75 m/s. A caterpillar in the wagon is crawling toward the rear of the wagon at a rate of 2.0 cm/s. What is the caterpillar’s velocity relative to the ground? vc/g vw/g vc/w 0.75 m/s 0.02 m/s 0.73 m/s 24. A boat is rowed directly upriver at a speed of 2.5 m/s relative to the water. Viewers on the shore see that the boat is moving at only 0.5 m/s relative to the shore. What is the speed of the river? Is it moving with or against the boat? (150 k m/h)2 (75 km/h )2 1.7102 km/h Section Review 6.3 Relative Velocity pages 157–159 page 159 26. Relative Velocity A fishing boat with a maximum speed of 3 m/s relative to the water is in a river that is flowing at 2 m/s. What is the maximum speed the boat can obtain relative to the shore? The minimum speed? Give the direction of the boat, relative to the river’s current, for the maximum speed and the minimum speed relative to the shore. The maximum speed relative to the shore is when the boat moves at maximum speed in the same direction as the river’s flow: vb/s vb/w vw/s 3 m/s 2 m/s 5 m/s The minimum speed relative to the shore is when the boat moves in the opposite direction of the river’s flow with the same speed as the river: vb/s vb/w vw/s 3 m/s (2 m/s) 1 m/s vb/g vb/w vw/g; Physics: Principles and Problems Solutions Manual 121 Chapter 6 continued 27. Relative Velocity of a Boat A powerboat heads due northwest at 13 m/s relative to the water across a river that flows due north at 5.0 m/s. What is the velocity (both magnitude and direction) of the motorboat relative to the shore? 2 vR v vRW2 RN (vbN vrN)2 (vbW vrW )2 2 (vb sin v vb cos )2 r) ( ((13 m/s)(sin 45°) 5.0 m /s)2 ((13 m/s)(co s 45°) )2 17 m/s v vRN W tan1R v cos vb sin vr b tan1 (13 m/s)(cos 45°) tan1 (3 m/s)(sin 45°) 5.0 m/s 33° vR 17 m/s, 33° west of north 28. Relative Velocity An airplane flies due south at 175 km/h relative to the air. There is a wind blowing at 85 km/h to the east relative to the ground. What are the plane’s speed and direction relative to the ground? vR (175 k m/h)2 (85 km/h )2 190 km/h tan1 64° 175 km/h 85 km/h vR 190 km/h, 64° south of east vR vRE2 vRN2 2 (vpE vaE)2 (vp N vw N) (vw co s )2 (vp vw si n )2 ((65 k m/h)(c os 45 °))2 (235 km/h (65 km/h )(sin 4 5°))2 280 km/h v vRE N tan1R vp va sin tan1 va cos 235 km/h (65 km/h)(sin 45°) (65 km/h)(cos 45°) tan1 72° north of east 280 km/h, 72° north of east 122 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 29. A Plane’s Relative Velocity An airplane flies due north at 235 km/h relative to the air. There is a wind blowing at 65 km/h to the northeast relative to the ground. What are the plane’s speed and direction relative to the ground? Chapter 6 continued 30. Relative Velocity An airplane has a speed of 285 km/h relative to the air. There is a wind blowing at 95 km/h at 30.0° north of east relative to Earth. In which direction should the plane head to land at an airport due north of its present location? What is the plane’s speed relative to the ground? To travel north, the east components must be equal and opposite. vpW cos p , so vpR vpW p cos1 vpR vwE cos w cos1 vpR cos1 (95 km/h)(cos 30.0°) 285 km/h 73° north of west vpR vpN vwN vp sin p vw sin w (285 km/h)(sin 107°) (95 km/h)(sin 30.0°) 320 km/h 31. Critical Thinking You are piloting a boat across a fast-moving river. You want to reach a pier directly opposite your starting point. Describe how you would navigate the boat in terms of the components of your velocity relative to the water. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. You should choose the component of your velocity along the direction of the river to be equal and opposite to the velocity of the river. Chapter Assessment Concept Mapping page 164 32. Use the following terms to complete the concept map below: constant speed, horizontal part of projectile motion, constant acceleration, relative-velocity motion, uniform circular motion. Categories of Motion constant acceleration vertical part of projectile motion constant velocity relative-velocity motion Physics: Principles and Problems constant speed horizontal part of projectile motion uniform circular motion Solutions Manual 123 Chapter 6 continued Mastering Concepts page 164 33. Consider the trajectory of the cannonball shown in Figure 6-11. (6.1) B A C D E ■ Figure 6-11 Up is positive, down is negative. a. Where is the magnitude of the vertical-velocity component largest? The greatest vertical velocity occurs at point A. b. Where is the magnitude of the horizontal-velocity component largest? Neglecting air resistance, the horizontal velocity at all points is the same. Horizontal velocity is constant and independent of vertical velocity. c. Where is the vertical-velocity smallest? The least vertical velocity occurs at point E. d. Where is the magnitude of the acceleration smallest? The magnitude of the acceleration is the same everywhere. No, the horizontal component of motion does not affect the vertical component. 35. An airplane pilot flying at constant velocity and altitude drops a heavy crate. Ignoring air resistance, where will the plane be relative to the crate when the crate hits the ground? Draw the path of the crate as seen by an observer on the ground. (6.1) The plane will be directly over the crate when the crate hits the ground. Both have the same horizontal velocity. The crate will look like it is moving horizontally while falling vertically to an observer on the ground. 36. Can you go around a curve with the following accelerations? Explain. a. zero acceleration No, going around a curve causes a change in direction of velocity. Thus, the acceleration cannot be zero. b. constant acceleration (6.2) No, the magnitude of the acceleration may be constant, but the direction of the acceleration changes. 124 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 34. A student is playing with a radio-controlled race car on the balcony of a sixthfloor apartment. An accidental turn sends the car through the railing and over the edge of the balcony. Does the time it takes the car to fall depend upon the speed it had when it left the balcony? (6.1) Chapter 6 continued 37. To obtain uniform circular motion, how must the net force depend on the speed of the moving object? (6.2) Circular motion results when the direction of the force is constantly perpendicular to the instantaneous velocity of the object. 38. If you whirl a yo-yo about your head in a horizontal circle, in what direction must a force act on the yo-yo? What exerts the force? (6.2) The force is along the string toward the center of the circle that the yo-yo follows. The string exerts the force. 39. Why is it that a car traveling in the opposite direction as the car in which you are riding on the freeway often looks like it is moving faster than the speed limit? (6.3) Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. The magnitude of the relative velocity of that car to your car can be found by adding the magnitudes of the two cars’ velocities together. Since each car probably is moving at close to the speed limit, the resulting relative velocity will be larger than the posted speed limit. Applying Concepts pages 164–165 40. Projectile Motion Analyze how horizontal motion can be uniform while vertical motion is accelerated. How will projectile motion be affected when drag due to air resistance is taken into consideration? The horizontal motion is uniform because there are no forces acting in that direction (ignoring friction). The vertical motion is accelerated due to the force of gravity. The projectile motion equations in this book do not hold when friction is taken into account. Projectile motion in both directions will be impacted when drag due to air resistance is taken into consideration. There will be a friction force opposing the motion. Physics: Principles and Problems 41. Baseball A batter hits a pop-up straight up over home plate at an initial velocity of 20 m/s. The ball is caught by the catcher at the same height that it was hit. At what velocity does the ball land in the catcher’s mitt? Neglect air resistance. 20 m/s, where the negative sign indicates down 42. Fastball In baseball, a fastball takes about 1 s to reach the plate. Assuming that such a 2 pitch is thrown horizontally, compare the 1 4 1 the distance it falls in the second s. 4 distance the ball falls in the first s with Because of the acceleration due to gravity, the baseball falls a greater distance 1 4 during the second s than during the 1 4 first s. 43. You throw a rock horizontally. In a second horizontal throw, you throw the rock harder and give it even more speed. a. How will the time it takes the rock to hit the ground be affected? Ignore air resistance. The time does not change—the time it takes to hit the ground depends only on vertical velocities and acceleration. b. How will the increased speed affect the distance from where the rock left your hand to where the rock hits the ground? A higher horizontal speed produces a longer horizontal distance. 44. Field Biology A zoologist standing on a cliff aims a tranquilizer gun at a monkey hanging from a distant tree branch. The barrel of the gun is horizontal. Just as the zoologist pulls the trigger, the monkey lets go and begins to fall. Will the dart hit the monkey? Ignore air resistance. Yes, in fact, the monkey would be safe if it did not let go of the branch. The vertical acceleration of the dart is the same as that of the monkey. Therefore, the dart is at the same vertical height when Solutions Manual 125 Chapter 6 continued it reaches the monkey. 45. Football A quarterback throws a football at 24 m/s at a 45° angle. If it takes the ball 3.0 s to reach the top of its path and the ball is caught at the same height at which it is thrown, how long is it in the air? Ignore air resistance. 6.0 s: 3.0 s up and 3.0 s down 46. Track and Field You are working on improving your performance in the long jump and believe that the information in this chapter can help. Does the height that you reach make any difference to your jump? What influences the length of your jump? Both speed and angle of launch matter, so height does make a difference. Maximum range is achieved when the resultant velocity has equal vertical and horizontal components—in other words, a launch angle of 45°. For this reason, height and speed affect the range. 47. Imagine that you are sitting in a car tossing a ball straight up into the air. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. a. If the car is moving at a constant velocity, will the ball land in front of, behind, or in your hand? The ball will land in your hand because you, the ball, and the car all are moving forward with the same speed. b. If the car rounds a curve at a constant speed, where will the ball land? The ball will land beside you, toward the outside of the curve. A top view would show the ball moving straight while you and the car moved out from under the ball. 48. You swing one yo-yo around your head in a horizontal circle. Then you swing another yo-yo with twice the mass of the first one, but you don’t change the length of the string or the period. How do the tensions in the strings differ? The tension in the string is doubled since FT mac. 126 Solutions Manual Physics: Principles and Problems Chapter 6 continued 49. Car Racing The curves on a race track are banked to make it easier for cars to go around the curves at high speeds. Draw a free-body diagram of a car on a banked curve. From the motion diagram, find the direction of the acceleration. Front View v2 v1 v1 v v2 1 y vyt gt 2 2 Since initial vertical velocity is zero, t g 9.80 m/s (2)(64 m) 2y 2 3.6 s Top View FN should you look for the keys? a Fg x vxt (8.0 m/s)(3.6) 28.8 m 29 m 52. The toy car in Figure 6-12 runs off the edge of a table that is 1.225-m high. The car lands 0.400 m from the base of the table. v The acceleration is directed toward the center of the track. a. What exerts the force in the direction of the acceleration? Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. The component of the normal force acting toward the center of the curve, and depending on the car’s speed, the component of the friction force acting toward the center, both contribute to the net force in the direction of acceleration. 1.225 m b. Can you have such a force without friction? Yes, the centripetal acceleration need only be due to the normal force. 0.400 m ■ a. How long did it take the car to fall? 1 y vy 0t gt 2 2 50. Driving on the Highway Explain why it is that when you pass a car going in the same direction as you on the freeway, it takes a longer time than when you pass a car going in the opposite direction. The relative speed of two cars going in the same direction is less than the relative speed of two cars going in the opposite direction. Passing with the lesser relative speed will take longer. Mastering Problems 6.1 Projectile Motion page 165 Level 1 51. You accidentally throw your car keys horizontally at 8.0 m/s from a cliff 64-m high. How far from the base of the cliff Physics: Principles and Problems Figure 6-12 Since initial vertical velocity is zero, t g 9.80 m/s 2y (2)(1.225 m) 2 0.500 s b. How fast was the car going on the table? x 0.400 m vx 0.800 m/s t 0.500 s 53. A dart player throws a dart horizontally at 12.4 m/s. The dart hits the board 0.32 m below the height from which it was thrown. How far away is the player from the board? 1 y vy 0t gt 2 2 and because initial velocity is zero, t 2y g Solutions Manual 127 Chapter 6 continued 55. Swimming You took a running leap off a high-diving platform. You were running at 2.8 m/s and hit the water 2.6 s later. How high was the platform, and how far from the edge of the platform did you hit the water? Ignore air resistance. (2)(0.32 m) 9.80 m/s2 0.26 s Now x vxt (12.4 m/s)(0.26 s) 3.2 m 1 y vy it gt 2 2 54. The two baseballs in Figure 6-13 were hit with the same speed, 25 m/s. Draw separate graphs of y versus t and x versus t for each ball. 1 2 0(2.6 s) (9.80 m/s2)(2.6 s)2 33 m, so the platform is 33 m high x vxt (2.8 m/s)(2.6 s) 7.3 m A yA B 60° yB 30° a. What is the maximum height the arrow will attain? xA xB ■ Figure 6-13 (vy 0)2 d 60 2g (v sin )2 2g 15 i 10 30 5 2 ((49 m/s)(sin 30.0°)) 2 (2)(9.80 m/s ) 0 1 2 3 Time (s) 4 5 b. The target is at the height from which the arrow was shot. How far away is it? Horizontal v. Time 80 Horizontal (m) 31 m 1 y vy 0t gt 2 2 60 but the arrow lands at the same height, so 30 40 60 20 1 y 0 and 0 vy i gt 2 0 0 1 2 3 Time (s) 4 5 so t 0 or 2vy g t i 2v sin g i (2)(49 m/s)(sin 30.0°) 2 9.80 m/s 5.0 s 128 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Vertical (m) 25 0 vy2 vyi2 2gd At the high point vy 0, so Vertical v. Time 20 Level 2 56. Archery An arrow is shot at 30.0° above the horizontal. Its velocity is 49 m/s, and it hits the target. Chapter 6 continued and x vxt but vy i 0, so (vi cos )(t) (49 m/s)(cos 30.0°)(5.0 s) 2.1102 m 57. Hitting a Home Run A pitched ball is hit by a batter at a 45° angle and just clears the outfield fence, 98 m away. If the fence is at the same height as the pitch, find the velocity of the ball when it left the bat. Ignore air resistance. 1 x vi sin i g 0 vi cos i Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Multiplying by vi cos i gives x vx t (125 km/h) 1h 3600 s 1 497 m 59. Diving Divers in Acapulco dive from a cliff that is 61 m high. If the rocks below the cliff extend outward for 23 m, what is the minimum horizontal velocity a diver must have to clear the rocks? 1 y vy it gt 2 and since vy i 0, t 2y g (2)(61 m) 9.80 m/s gx so vi2 (2)(sin i)(cos i) 3.53 s 1000 m 1 km (14.3 s) vi2 sin i cos i gx 0 2 thus, vi 2 2 From above 2 (2)(1001 m) b. What is the horizontal distance between the plane and the victims when the box is dropped? Now x vxt (vi cos i)t, so x t 1 And y vy it gt 2, but y 0, so 2 1 0 vy i gt t 2 1 so t 0 or vy i gt 0 2 9.80 m/s 14.3 s The components of the initial velocity are vx vi cos i and vy i vi sin i vi cos i 2y g t gx (2)(sin i)(cos i) (9.80 m/s )(98 m) (2)(sin 45°)(cos 45°) 2 31 m/s at 45° Level 3 58. At-Sea Rescue An airplane traveling 1001 m above the ocean at 125 km/h is going to drop a box of supplies to shipwrecked victims below. a. How many seconds before the plane is directly overhead should the box be dropped? 1 y vy it gt 2 2 Physics: Principles and Problems 2 x vxt x t 23 m 3.53 s vx 6.5 m/s 60. Jump Shot A basketball player is trying to make a half-court jump shot and releases the ball at the height of the basket. Assuming that the ball is launched at 51.0°, 14.0 m from the basket, what speed must the player give the ball? The components of the initial velocity are vx i vi cos i and vy i vi sin i Now x vx it (vi cos i)t, so Solutions Manual 129 Chapter 6 continued x vi cos i t 1 And y vy it gt 2, but y 0, so 2 1 0 vy i gt t 2 1 so t 0 or vy i gt 2 0 2 From above 1 x v0 sin i g 0 vi cos i 2 Multiplying by vi cos i gives 1 vi2(sin i)(cos i) gx 0 2 so vi (2)(sin )(cos ) gx i i (9.80 m/s )(14.0 m) (2)(sin 51.0°)(cos 51.0°) 2 11.8 m/s a. What is the acceleration of the car? 2 v ac r 2 4 r 2 T 4 2(50.0 m) (14.3 s) 2 9.59 m/s2 b. What force must the track exert on the tires to produce this acceleration? Fc mac (615 kg)(9.59 m/s2) 5.90103 N 130 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 6.2 Circular Motion page 166 Level 1 61. Car Racing A 615-kg racing car completes one lap in 14.3 s around a circular track with a radius of 50.0 m. The car moves at a constant speed. Chapter 6 continued 62. Hammer Throw An athlete whirls a 7.00-kg hammer 1.8 m from the axis of rotation in a horizontal circle, as shown in Figure 6-14. If the hammer makes one revolution in 1.0 s, what is the centripetal acceleration of the hammer? What is the tension in the chain? vtang (4 2)(0.050 m) (1.80 s) 0.61 m/s2 2 r 10.0 cm: (4 2)(0.100 m) (1.80 s) 4 2r T ac 2 2 1.22 m/s2 r 15.0 cm: 4 2r T (4 2)(0.150 m) (1.80 s) ac 2 2 1.83 m/s2 c. What force accelerates the coin? frictional force between coin and record ■ d. At which of the three radii in part b would the coin be most likely to fly off the turntable? Why? Figure 6-14 15.0 cm, the largest radius; the friction force needed to hold it is the greatest. 2 4 r ac 2 T (4 2)(1.8 m) (1 . 0 s) 2 71 m/s2 Fc mac Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. (7.00 kg)(71 m/s2) 5.0102 N Level 2 63. A coin is placed on a vinyl stereo record 1 that is making 33 revolutions per minute 3 on a turntable. a. In what direction is the acceleration of the coin? The acceleration is toward the center of the record. b. Find the magnitude of the acceleration when the coin is placed 5.0, 10.0, and 15.0 cm from the center of the record. T 1 f 1 1 33 revmin 3 (0.0300 min) 1.80 s 60 s 1 min r 5.0 cm: 4 2r T ac 2 Physics: Principles and Problems 64. A rotating rod that is 15.3 cm long is spun with its axis through one end of the rod so that the other end of the rod has a speed of 2010 m/s (4500 mph). a. What is the centripetal acceleration of the end of the rod? v2 r (2010 m/s)2 0.153 m ac 2.64107 m/s2 b. If you were to attach a 1.0-g object to the end of the rod, what force would be needed to hold it on the rod? Fc mac (0.0010 kg)(2.64107 m/s2) 2.6104 N 65. Friction provides the force needed for a car to travel around a flat, circular race track. What is the maximum speed at which a car can safely travel if the radius of the track is 80.0 m and the coefficient of friction is 0.40? Fc Ff FN mg mv 2 mv 2 But Fc , thus mg. r r Solutions Manual 131 Chapter 6 continued The mass of the car divides out to give v 2 gr, so v gr (0.40) m/s2)(80.0 (9.80 m) 18 m/s Level 3 66. A carnival clown rides a motorcycle down a ramp and around a vertical loop. If the loop has a radius of 18 m, what is the slowest speed the rider can have at the top of the loop to avoid falling? Hint: At this slowest speed, the track exerts no force on the motorcycle at the top of the loop. Fc mac Fg mg, so ac g v2 g, so r v gr (9.80 m/s2) (18 m) 13 m/s vtang 120 m/s r v2 so r g (120 m/s)2 9.80 m/s 2 1.5103 m 6.3 Relative Velocity pages 166–167 Level 1 68. Navigating an Airplane An airplane flies at 200.0 km/h relative to the air. What is the velocity of the plane relative to the ground if it flies during the following wind conditions? a. a 50.0-km/h tailwind Tailwind is in the same direction as the airplane 200.0 km/h 50.0 km/h 250.0 km/h b. a 50.0-km/h headwind Head wind is in the opposite direction of the airplane 200.0 km/h 50.0 km/h 150.0 km/h 69. Odina and LaToya are sitting by a river and decide to have a race. Odina will run down the shore to a dock, 1.5 km away, then turn around and run back. LaToya will also race to the dock and back, but she will row a boat in the river, which has a current of 2.0 m/s. If Odina’s running speed is equal to LaToya’s rowing speed in still water, which is 4.0 m/s, who will win the race? Assume that they both turn instantaneously. x x vt, so t v for Odina, 3 3.010 m t 4.0 m/s ■ 132 Figure 6-15 7.5102 s Because the net force is equal to the weight of the pilot, For LaToya (assume against current on the way to the dock), Fc mac Fg mg, so 1 2 t Solutions Manual x v1 x v2 Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 67. A 75-kg pilot flies a plane in a loop as shown in Figure 6-15. At the top of the loop, when the plane is completely upside-down for an instant, the pilot hangs freely in the seat and does not push against the seat belt. The airspeed indicator reads 120 m/s. What is the radius of the plane’s loop? 2 v ac g or g Chapter 6 continued 3 3 1.510 m 1.510 m 4.0 m/s 2.0 m/s 4.0 m/s 2.0 m/s 1.0103 s v vair b/air tan1 tan1 15 m/s 6.5 m/s Odina wins. 67° from the horizon toward the west Level 2 70. Crossing a River You row a boat, such as the one in Figure 6-16, perpendicular to the shore of a river that flows at 3.0 m/s. The velocity of your boat is 4.0 m/s relative to the water. Level 3 72. Boating You are boating on a river that flows toward the east. Because of your knowledge of physics, you head your boat 53° west of north and have a velocity of 6.0 m/s due north relative to the shore. vb a. What is the velocity of the current? v vb/s w/s tan , so vw vw/s (tan )(vb/s) (tan 53°)(6.0 m/s) ■ a. What is the velocity of your boat relative to the shore? 2 (v 2 vb/s (v b/w) w/s) (4.0 m /s)2 (3.0 m /s)2 5.0 m/s Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 8.0 m/s east Figure 6-16 b. What is the speed of your boat relative to the water? v vb/w /s , so cos b v cos /s vb/w b 6.0 m/s cos 53° vb/w tan1 vw/s 3.0 m/s 4.0 m/s tan1 53° from shore b. What is the component of your velocity parallel to the shore? Perpendicular to it? 3.0 m/s; 4.0 m/s 71. Studying the Weather A weather station releases a balloon to measure cloud conditions that rises at a constant 15 m/s relative to the air, but there is also a wind blowing at 6.5 m/s toward the west. What are the magnitude and direction of the velocity of the balloon? vb (vb/air )2 (vair )2 (15 m /s)2 (6.5 m /s)2 16 m/s Physics: Principles and Problems 1.0101 m/s 73. Air Travel You are piloting a small plane, and you want to reach an airport 450 km due south in 3.0 h. A wind is blowing from the west at 50.0 km/h. What heading and airspeed should you choose to reach your destination in time? d t 450 km s vs 150 km/h 3.0 h (150 k m/h)2 (50 .0 km /h)2 vp 1.6102 km/h v wind tan1 vs tan1 50.0 km/h 150 km/h 18° west of south Mixed Review Solutions Manual 133 Chapter 6 continued page 167 Level 1 74. Early skeptics of the idea of a rotating Earth said that the fast spin of Earth would throw people at the equator into space. The radius of Earth is about 6.38103 km. Show why this idea is wrong by calculating the following. a. the speed of a 97-kg person at the equator d t 2r T v 2(6.38106 m) 3600 s (24 h) 1h 464 m/s b. the force needed to accelerate the person in the circle Fc mac mv 2 r 2 (97 kg)(464 m/s) 6 6.3810 m 3.3 N c. the weight of the person Fg mg (97)(9.80 m/s2) 9.5102 N d. the normal force of Earth on the person, that is, the person’s apparent weight FN 9.5102 N 3.3 N 75. Firing a Missile An airplane, moving at 375 m/s relative to the ground, fires a missile forward at a speed of 782 m/s relative to the plane. What is the speed of the missile relative to the ground? vm/g vp/g vm/p 375 m/s 782 m/s 1157 m/s 76. Rocketry A rocket in outer space that is moving at a speed of 1.25 km/s relative to an observer fires its motor. Hot gases are expelled out the back at 2.75 km/s relative to the rocket. What is the speed of the gases relative to the observer? vg/o vr/o vg/r 1.25 km/s (2.75 km/s) 1.50 km/s Level 2 77. Two dogs, initially separated by 500.0 m, are running towards each other, each moving with a constant speed of 2.5 m/s. A dragonfly, moving with a constant speed of 3.0 m/s, flies from the nose of one dog to the other, then turns around 134 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 9.5102 N Chapter 6 continued instantaneously and flies back to the other dog. It continues to fly back and forth until the dogs run into each other. What distance does the dragonfly fly during this time? The dogs will meet in 500.0 m 1.0102 s 5.0 m/s The dragonfly flies (3.0 m/s)(1.0102 s) 3.0102 m. 78. A 1.13-kg ball is swung vertically from a 0.50-m cord in uniform circular motion at a speed of 2.4 m/s. What is the tension in the cord at the bottom of the ball’s motion? FT Fg Fc Fc Fg mv 2 cos mg sin r v2 sin g g tan r cos v2 gr tan1 (15.7 m/s)2 (9.80 m/s )(56.0 m) tan1 2 34.9° For 45 mph: v2 gr tan1 (20.1 m/s)2 (9.80 m/s )(36.0 m) mv mg tan1 2 (1.13 kg)(9.80 m/s2) 48.9° 2 r (1.13 kg)(2.4 m/s)2 0.50 m 24 N Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. For 35 mph: 79. Banked Roads Curves on roads often are banked to help prevent cars from slipping off the road. If the posted speed limit for a particular curve of radius 36.0 m is 15.7 m/s (35 mph), at what angle should the road be banked so that cars will stay on a circular path even if there were no friction between the road and the tires? If the speed limit was increased to 20.1 m/s (45 mph), at what angle should the road be banked? Level 3 80. The 1.45-kg ball in Figure 6-17 is suspended 14.0° from a 0.80m string and swung in a ■ Figure 6-17 horizontal circle at a constant speed such that the string makes an angle of 14.0° with the vertical. a. What is the tension in the string? FT cos mg mg cos so FT 2 (1.45 m/s)(9.80 m/s ) cos 14.0° 14.6 N b. What is the speed of the ball? 2 mv Fc FT sin Fg FT cos r mg FT sin mv 2 so FT cos rmg 2 v or tan rg Physics: Principles and Problems Solutions Manual 135 Chapter 6 continued so v rg tan (0.80 0 m/s2 )(tan 14.0°) m)(9.8 1.4 m/s 81. A baseball is hit directly in line with an outfielder at an angle of 35.0° above the horizontal with an initial velocity of 22.0 m/s. The outfielder starts running as soon as the ball is hit at a constant velocity of 2.5 m/s and barely catches the ball. Assuming that the ball is caught at the same height at which it was hit, what was the initial separation between the hitter and outfielder? Hint: There are two possible answers. x vxit vpt t(vx i vp) To get t, 1 y vyit gt 2, y 0 2 1 so vy it gt 2, t 0 or 2 1 vy i gt 2 2vy i t g 2v sin g i 2v sin i (vx i vp) so x g 2v sin 2 (2)(22.0 m/s)(sin 35.0°) 9.80 m/s ((22.0 m/s)(cos 35.0°) 2.5 m/s) 53 m or 4.0101 m 82. A Jewel Heist You are serving as a technical consultant for a locally produced cartoon. In one episode, two criminals, Shifty and Lefty, have stolen some jewels. Lefty has the jewels when the police start to chase him, and he runs to the top of a 60.0-m tall building in his attempt to escape. Meanwhile, Shifty runs to the convenient hot-air balloon 20.0 m from the base of the building and untethers it, so it begins to rise at a constant speed. Lefty tosses the bag of jewels horizontally with a speed of 7.3 m/s just as the balloon 136 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. i (vi cos vp) g Chapter 6 continued begins its ascent. What must the velocity of the balloon be for Shifty to easily catch the bag? x vx i x vx it, so t 1 ybag vy it gt 2, but vy i 0 2 1 2 so ybag gt 2 y balloon vballoon t 60.0 m ybag t 1 2 60.0 m gt 2 t 60.0 m 1 gt t 2 1 x 60.0 m g x 2 v xi vxi (60.0 m)v x gx xi 2vx i (60.0 m)(7.3 m/s) (20.0 m) (9.80 m/s)(20.0 m) 2(7.3 m/s) Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 8.5 m/s Thinking Critically page 168 83. Apply Concepts Consider a roller-coaster loop like the one in Figure 6-18. Are the cars traveling through the loop in uniform circular motion? Explain. ■ Figure 6-18 The vertical gravitational force changes the speed of the cars, so the motion is not uniform circular motion. 84. Use Numbers A 3-point jump shot is released 2.2 m above the ground and 6.02 m from the basket. The basket is 3.05 m above the floor. For launch angles of 30.0° and 60.0°, find the speed the ball needs to be thrown to make the basket. Physics: Principles and Problems Solutions Manual 137 Chapter 6 continued x vix a. Is the hit a home run? x vi cos x vixt, so t Yes, the hit is a home run; the ball clears the wall by 2.1 m. 1 y viyt gt 2 2 1 x x vi sin g 2 vi cos 2 vi cos b. What is the minimum speed at which the ball could be hit and clear the wall? g cos x tan 2 2 2vi (cos ) x g so vi cos 2((tan )x y) 96.0 cos 29.80 m/s2 (2)((tan 26°)(96.0 m) 13 m) m 6° For 30.0° vi 6.02 cos 3 m 9.80 m/s2 0.0° .02 m) (3.05 m 2.2 m)) 9.5 m/s For 60.0° vi 6.02 co s6 9.80 m/s2 m 2((tan 60°(6.02 m) (3.05 m 2.2 m)) 0.0° 8.6 m/s Varying speed by 5 percent at 30.0° changes R by about 0.90 m in either direction. At 60.0° it changes R by only about 0.65 m. Thus, the high launch angle is less sensitive to speed variations. 86. Apply Computers and Calculators A baseball player hits a belt-high (1.0 m) fastball down the left-field line. The ball is hit with an initial velocity of 42.0 m/s at 26°. The left-field wall is 96.0 m from home plate at the foul pole and is 14-m high. Write the equation for the height of the ball, y, as a function of its distance from home plate, x. Use a computer or graphing calculator to plot the path of the ball. Trace along the path to find how high above the ground the ball is when it is at the wall. 138 Solutions Manual 41 m/s c. If the initial velocity of the ball is 42.0 m/s, for what range of angles will the ball go over the wall? For the ball to go over the wall, the range of angles needs to be 25° 70°. 87. Analyze Albert Einstein showed that the rule you learned for the addition of velocities does not work for objects moving near the speed of light. For example, if a rocket moving at velocity vA releases a missile that has velocity vB relative to the rocket, then the velocity of the missile relative to an observer that is at rest is given by v (vA vB)/(1 vAvB/c2), where c is the speed of light, 3.00108 m/s. This formula gives the correct values for objects moving at slow speeds as well. Suppose a rocket moving at 11 km/s shoots a laser beam out in front of it. What speed would an unmoving observer find for the laser light? Suppose that a rocket moves at a speed c/2, half the speed of light, and shoots a missile forward at a speed of c/2 relative to the rocket. How fast would the missile be moving relative to a fixed observer? vl/o (v v ) r/o l/r v v r/o 1 c 2 1/r 1.1104 m/s 3.00108 m/s (1.1104 m/s)(3.00108 m/s) 1 (3.00108 m/s)2 3.0108 m/s vm/o vr/o vm/r v v r/o m/r 1 2 c Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 85. Analyze For which angle in problem 84 is it more important that the player get the speed right? To explore this question, vary the speed at each angle by 5 percent and find the change in the range of the attempted shot. 2((tan ) x y) x vi gx 2 Chapter 6 continued c c 2 2 c2c2 1 c2 4 c 5 88. Analyze and Conclude A ball on a light string moves in a vertical circle. Analyze and describe the motion of this system. Be sure to consider the effects of gravity and tension. Is this system in uniform circular motion? Explain your answer. Writing in Physics page 168 89. Roller Coasters If you take a look at vertical loops on roller coasters, you will notice that most of them are not circular in shape. Research why this is so and explain the physics behind this decision by the coaster engineers. mv 2 Answers will vary. Since Fc , as v r decreases due to gravity when going uphill, r is reduced to keep the force constant. 90. Many amusement-park rides utilize centripetal acceleration to create thrills for the park’s customers. Choose two rides other than roller coasters that involve circular motion and explain how the physics of circular motion creates the sensations for the riders. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. It is not uniform circular motion. Gravity increases the speed of the ball when it moves downward and reduces the speed when it is moving upward. Therefore, the centripetal acceleration needed to keep it moving in a circle will be larger at the bottom and smaller at the top of the circle. At the top, tension and gravity are in the same direction, so the tension needed will be even smaller. At the bottom, gravity is outward while the tension is inward. Thus, the tension exerted by the string must be even larger. Physics: Principles and Problems Solutions Manual 139 CHAPTER 7 Gravitation Practice Problems Planetary Motion and Gravitation pages 171–178 7.1 page 174 1. If Ganymede, one of Jupiter’s moons, has a period of 32 days, how many units are there in its orbital radius? Use the information given in Example Problem 1. 2 T TGI rG r 3 rI G 32 days nits) (4.2 u 1.8 days 3 2 3 23.4 103 un its3 3 29 units 2. An asteroid revolves around the Sun with a mean orbital radius twice that of Earth’s. Predict the period of the asteroid in Earth years. T 2 TEa r 3 a with ra 2rE r E T y) (1.0 ra 3 2 E rE Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Ta 2rE 3 rE 2 2.8 y 3. From Table 7-1, on page 173, you can find that, on average, Mars is 1.52 times as far from the Sun as Earth is. Predict the time required for Mars to orbit the Sun in Earth days. T 2 TME 3 r M with rM 1.52rE r E Thus, TM T (365 days) r r rM 3 E 2 1.52rE 3 E 2 E 4.68 105 da ys2 684 days 4. The Moon has a period of 27.3 days and a mean distance of 3.90105 km from the center of Earth. a. Use Kepler’s laws to find the period of a satellite in orbit 6.70103 km from the center of Earth. T 2 TMs r 3 rM s Physics: Principles and Problems Solutions Manual 141 Chapter 7 continued Ts T rs 3 2 M rM 3 day s) (27. 6.70103 km 3 3.90105 km 2 3.78 103 days2 6.15102 days 88.6 min b. How far above Earth’s surface is this satellite? h rs rE 6.70106 m 6.38106 m 3.2105 m 3.2102 km 5. Using the data in the previous problem for the period and radius of revolution of the Moon, predict what the mean distance from Earth's center would be for an artificial satellite that has a period of exactly 1.00 day. 2 T TMs rs r 3 rM s T 1.00 days (3.90 10 km) r T 27. 3 days 3 s 3 M 2 3 5 3 2 M 7.96 1013 k m3 3 4.30104 km 7.1 Planetary Motion and Gravitation pages 171–178 page 178 6. Neptune’s Orbital Period Neptune orbits the Sun with an orbital radius of 4.4951012 m, which allows gases, such as methane, to condense and form an atmosphere, as shown in Figure 7-8. If the mass of the Sun is 1.991030 kg, calculate the period of Neptune’s orbit. Gm 3 r T 2 S 2 (4.4951012 m)3 11 (6.6710 Nm2/kg2)(1.991030 kg) ■ Figure 7-8 5.20109 s 6.02105 days 7. Gravity If Earth began to shrink, but its mass remained the same, what would happen to the value of g on Earth’s surface? The value of g would increase. 142 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Section Review Chapter 7 continued 8. Gravitational Force What is the gravitational force between two 15-kg packages that are 35 cm apart? What fraction is this of the weight of one package? m m r Fg G E 2 (6.671011 Nm2kg2)(15 kg)2 (0.35 m) 2 1.2107 N Because the weight is mg 147 N, the gravitational force is 8.21010 or 0.82 parts per billion of the weight. 9. Universal Gravitational Constant Cavendish did his experiment using lead spheres. Suppose he had replaced the lead spheres with copper spheres of equal mass. Would his value of G be the same or different? Explain. It would be the same, because the same value of G has been used successfully to describe the attraction of bodies having diverse chemical compositions: the Sun (a star), the planets, and satellites. 10. Laws or Theories? Kepler’s three statements and Newton’s equation for gravitational attraction are called “laws.” Were they ever theories? Will they ever become theories? No. A scientific law is a statement of what has been observed to happen many times. A theory explains scientific results. None of these statements offers explanations for why the motion of planets are as they are or for why gravitational attraction acts as it does. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 11. Critical Thinking Picking up a rock requires less effort on the Moon than on Earth. a. How will the weaker gravitational force on the Moon’s surface affect the path of the rock if it is thrown horizontally? Horizontal throwing requires the same effort because the inertial character, F ma, of the rock is involved. The mass of the rock depends only on the amount of matter in the rock, not on its location in the universe. The path would still be a parabola, but it could be much wider because the rock would go farther before it hits the ground, given the smaller acceleration rate and longer time of flight. b. If the thrower accidentally drops the rock on her toe, will it hurt more or less than it would on Earth? Explain. Assume the rocks would be dropped from the same height on Earth and on the Moon. It will hurt less because the smaller value of g on the Moon means that the rock strikes the toe with a smaller velocity than on Earth. Physics: Principles and Problems Solutions Manual 143 Chapter 7 continued Practice Problems 7.2 Using the Law of Universal of Gravitation pages 179–185 page 181 For the following problems, assume a circular orbit for all calculations. 12. Suppose that the satellite in Example Problem 2 is moved to an orbit that is 24 km larger in radius than its previous orbit. What would its speed be? Is this faster or slower than its previous speed? r (h 2.40104 m) rE (2.25105 m 2.40104 m) 6.38106 m 6.63106 m v r GmE (6.671011 Nm2/kg2)(5.971024 kg) 6.63106 m 7.75103 m/s, slower 13. Use Newton’s thought experiment on the motion of satellites to solve the following. a. Calculate the speed that a satellite shot from a cannon must have to orbit Earth 150 km above its surface. v GmE r (6.671011 Nm2/kg2)(5.971024 kg) (6.38106 m 1.5105 m) 7.8103 m/s b. How long, in seconds and minutes, would it take for the satellite to complete one orbit and return to the cannon? (6.3810 m 1.510 m) r 2 Gm (6.6710 Nm /kg )(5.9710 kg) 6 3 E 11 5 2 2 3 24 5.3103 s 88 min 14. Use the data for Mercury in Table 7-1 on page 173 to find the following. a. the speed of a satellite that is in orbit 260 km above Mercury’s surface v r GmM r rM 260 km 2.44106 m 0.26106 m 2.70106 m v (6.671011 Nm2/kg2)(3.301023 kg) 2.70106 m 2.86103 m/s b. the period of the satellite T 2 (2.7010 m) r 2 Gm (6.6710 Nm /kg )(3.3010 kg) 6 3 M 11 2 2 3 23 5.94103 s 1.65 h 144 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. T 2 Chapter 7 continued Section Review 7.2 Using the Law of Universal of Gravitation pages 179–185 page 185 15. Gravitational Fields The Moon is 3.9105 km from Earth’s center and 1.5108 km from the Sun’s center. The masses of Earth and the Sun are 6.01024 kg and 2.01030 kg, respectively. a. Find the ratio of the gravitational fields due to Earth and the Sun at the center of the Moon. m S Gravitational field due to the Sun: gS G r 2 S mE Gravitational field due to Earth: gE G r E2 gS mS rE2 gE mE rS2 (2.01030 kg)(3.9105 km)2 (6.010 kg)(1.510 km) 24 8 2 2.3 b. When the Moon is in its third quarter phase, as shown in Figure 7-17, its direction from Earth is at right angles to the Sun’s direction. What is the net gravitational field due to the Sun and Earth at the center of the Moon? Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Moon Sun Earth ■ Figure 7-17 Because the directions are at right angles, the net field is the square root of the sum of the squares of the two fields. Gm r S gS 2 (6.671011 Nm2kg2)(2.01030 kg) (1.510 m) 11 2 5.9103 N/kg Similarly, gE 2.6103 N/kg 3 N/kg 103 N/kg)2 (2.6 10 )2 gnet (5.9 6.4103 N/kg 16. Gravitational Field The mass of the Moon is 7.31022 kg and its radius is 1785 km. What is the strength of the gravitational field on the surface of the Moon? (6.671011 Nm2kg2)(7.31022 kg) (1.78510 m) GM g 2 3 2 r 1.5 N/kg, about one-sixth that on Earth Physics: Principles and Problems Solutions Manual 145 Chapter 7 continued 17. A Satellite’s Mass When the first artificial satellite was launched into orbit by the former Soviet Union in 1957, U.S. president Dwight D. Eisenhower asked his scientific advisors to calculate the mass of the satellite. Would they have been able to make this calculation? Explain. No. Because the speed and period of the orbit don’t depend at all on the mass of the satellite, the scientific advisors would not have been able to calculate the mass of the satellite. 18. Orbital Period and Speed Two satellites are in circular orbits about Earth. One is 150 km above the surface, the other 160 km. a. Which satellite has the larger orbital period? When the orbital radius is large, the period also will be large. Thus, the one at 160 km will have the larger period. b. Which one has the greater speed? The one at 150 km, because the smaller the orbital radius, the greater the speed. 19. Theories and Laws Why is Einstein’s description of gravity called a “theory,” while Newton’s is a “law?” Newton’s law describes how to calculate the force between two massive objects. Einstein’s theory explains how an object, such as Earth, attracts the Moon. 20. Weightlessness Chairs in an orbiting spacecraft are weightless. If you were on board such a spacecraft and you were barefoot, would you stub your toe if you kicked a chair? Explain. 21. Critical Thinking It is easier to launch a satellite from Earth into an orbit that circles eastward than it is to launch one that circles westward. Explain. Earth rotates toward the east, and its velocity adds to the velocity given to the satellite by the rocket, thereby reducing the velocity that the rocket must supply. Chapter Assessment Concept Mapping page 190 22. Create a concept map using these terms: planets, stars, Newton’s law of universal gravitation, Kepler’s first law, Kepler’s second law, Kepler’s third law, Einstein’s general theory of relativity. 146 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Yes. The chairs are weightless but not massless. They still have inertia and can exert contact forces on your toe. Chapter 7 continued Einstein’s general theory of relativity Explains Newton’s law of universal gravitation Supports Kepler’s first law Kepler’s second law Describes motion of Kepler’s third law Relates periods and radii of planets Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. moving around stars Kepler’s first and second laws describe the motion of a single planet. Kepler’s third law describes the periods versus the orbital radii of all planets around a star. Newton’s law of universal gravitation supports Kepler’s laws. Einstein’s theory explains Newton’s and Kepler’s laws. Mastering Concepts page 190 23. In 1609, Galileo looked through his telescope at Jupiter and saw four moons. The name of one of the moons that he saw is Io. Restate Kepler’s first law for Io and Jupiter. (7.1) The path of Io is an ellipse, with Jupiter at one focus. 24. Earth moves more slowly in its orbit during summer in the northern hemisphere than it does during winter. Is it closer to the Sun in summer or in winter? (7.1) Because Earth moves more slowly in its orbit during summer, by Kepler’s second law, it must be farther from the Sun. Therefore, Earth is closer to the Sun in the winter months. Physics: Principles and Problems Solutions Manual 147 Chapter 7 continued 25. Is the area swept out per unit of time by Earth moving around the Sun equal to the area swept out per unit of time by Mars moving around the Sun? (7.1) No. The equality of the area swept out per unit of time applies to each planet individually. 26. Why did Newton think that a force must act on the Moon? (7.1) Newton knew that the Moon followed a curved path; therefore, it was accelerated. He also knew that a force is required for acceleration. 27. How did Cavendish demonstrate that a gravitational force of attraction exists between two small objects? (7.1) He carefully measured the masses, the distance between the masses, and the force of attraction. He then calculated G using Newton’s law of universal gravitation. 28. What happens to the gravitational force between two masses when the distance between the masses is doubled? (7.1) 29. According to Newton’s version of Kepler’s third law, how would the ratio T 2/r 3 change if the mass of the Sun were doubled? (7.1) Because T 2/r 3 4 2/GmS, if the mass of the Sun, mS, is doubled, the ratio will be halved. 30. How do you answer the question, “What keeps a satellite up?” (7.2) gravitational attraction to the central body 33. During space flight, astronauts often refer to forces as multiples of the force of gravity on Earth’s surface. What does a force of 5g mean to an astronaut? (7.2) A force of 5g means that an astronaut’s weight is five times heavier than it is on Earth. The force exerted on the astronaut is five times the force of Earth’s gravitational force. 34. Newton assumed that a gravitational force acts directly between Earth and the Moon. How does Einstein’s view of the attractive force between the two bodies differ from Newton’s view? (7.2) Einstein’s view is that gravity is an effect of the curvature of space as a result of the presence of mass, whereas Newton’s view of gravity is that it is a force acting directly between objects. Thus, according to Einstein, the attraction between Earth and the Moon is the effect of curvature of space caused by their combined masses. 35. Show that the dimensions of g in the equation g F/m are in m/s2. (7.2) kgm/s2 kg N F The units of are m/s2 m kg 36. If Earth were twice as massive but remained the same size, what would happen to the value of g? (7.2) The value of g would double. Its speed; it is falling all the time. 31. A satellite is orbiting Earth. On which of the following does its speed depend? (7.2) a. mass of the satellite b. distance from Earth c. mass of Earth Speed depends only on b, the distance from the Earth, and c, the mass of Earth. 148 Solutions Manual Applying Concepts pages 190–191 37. Golf Ball The force of gravity acting on an object near Earth’s surface is proportional to the mass of the object. Figure 7-18 shows a tennis ball and golf ball in free fall. Why does a tennis ball not fall faster than a golf ball? Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. According to Newton, F 1/r 2. If the distance is doubled, the force is cut to one-fourth. 32. What provides the force that causes the centripetal acceleration of a satellite in orbit? (7.2) Chapter 7 continued 40. Decide whether each of the orbits shown in Figure 7-19 is a possible orbit for a planet. c c Sun Sun c ■ Figure 7-18 c Sun Sun m m r 2 F G 1 2 m1 Earth’s mass F a m2 m2 object’s mass Gm r 1 Thus, a 2 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. The acceleration is independent of the object’s mass. This is because more massive objects require more force to accelerate at the same rate. 38. What information do you need to find the mass of Jupiter using Newton’s version of Kepler’s third law? You must know the period and orbital radius of at least one of its satellites. 39. The mass of Pluto was not known until a satellite of the planet was discovered. Why? Orbital motion of a planet does not depend on its mass and cannot be used to find the mass. A satellite orbiting the planet is necessary to find the planet’s mass. ■ Figure 7-19 Only d (lower right) is possible. a (top left) and b (top right) do not have the Sun at a focus, and in c (lower left), the planet is not in orbit around the Sun. 41. The Moon and Earth are attracted to each other by gravitational force. Does the moremassive Earth attract the Moon with a greater force than the Moon attracts Earth? Explain. No. The forces constitute an actionreaction pair, so under Newton’s third law, they are equal and opposite. 42. What would happen to the value of G if Earth were twice as massive, but remained the same size? Nothing. G is a universal constant, and it is independent of Earth’s mass. However, the force of attraction would double. 43. Figure 7-20 shows a satellite orbiting Earth. Examine the equation v , relating r GmE the speed of an orbiting satellite and its distance from the center of Earth. Does a satellite with a large or small orbital radius have the greater velocity? Physics: Principles and Problems Solutions Manual 149 Chapter 7 continued 300 g 2 3. Thus, g on Jupiter is three 10 Satellite times that on Earth. 47. A satellite is one Earth radius above the surface of Earth. How does the acceleration due to gravity at that location compare to acceleration at the surface of Earth? r Earth d rE rE 2rE r 2 2rE E so, a g 1 g ■ Figure 7-20 (Not to scale) A satellite with a small radius has the greater velocity. 44. Space Shuttle If a space shuttle goes into a higher orbit, what happens to the shuttle’s period? Because T 2 , if the orbital Gm r3 radius increases, so will the period. 45. Mars has about one-ninth the mass of Earth. Figure 7-21 shows satellite M, which orbits Mars with the same orbital radius as satellite E, which orbits Earth. Which satellite has a smaller period? Earth rM E Figure 7-21 (Not to scale) Because T 2 r3 , as the mass of Gm the planet increases, the period of the satellite decreases. Because Earth has a larger mass than Mars, Earth’s satellite will have a smaller period. 46. Jupiter has about 300 times the mass of Earth and about ten times Earth’s radius. Estimate the size of g on the surface of Jupiter. m rE E g 2 . If Jupiter has 300 times the mass and ten times the radius of Earth, 150 Solutions Manual It also will double. 49. Weight Suppose that yesterday your body had a mass of 50.0 kg. This morning you stepped on a scale and found that you had gained weight. a. What happened, if anything, to your mass? Your mass increased. b. What happened, if anything, to the ratio of your weight to your mass? The ratio remained constant because it is equal to the gravitational field at the location, a constant g. rE M ■ 48. If a mass in Earth’s gravitational field is doubled, what will happen to the force exerted by the field upon the mass? 50. As an astronaut in an orbiting space shuttle, how would you go about “dropping” an object down to Earth? To “drop” an object down to Earth, you would have to launch it backward at the same speed at which you are traveling in orbit. With respect to Earth, the object’s speed perpendicular to Earth’s gravity would be zero, and it could then “drop” down to Earth. However, the object is likely to burn up as a result of friction with Earth’s atmosphere on the way down. 51. Weather Satellites The weather pictures that you see every day on TV come from a spacecraft in a stationary position relative to Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Mars 4 Chapter 7 continued the surface of Earth, 35,700 km above Earth’s equator. Explain how it can stay in exactly the same position day after day. What would happen if it were closer? Farther out? Hint: Draw a pictorial model. The satellite is positioned as close to the equator as possible so it doesn’t appear to have much north-south movement. Because it is placed at that distance, the satellite has a period of 24.0 h. If it were positioned any closer, its period would be less than 24.0 h and it would appear to move toward the east. If it were positioned any farther, its period would be longer than 24.0 h. Mastering Problems 7.1 Planetary Motion and Gravitation pages 191–192 Level 1 52. Jupiter is 5.2 times farther from the Sun than Earth is. Find Jupiter’s orbital period in Earth years. 2 T J TE Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. TJ r 3 rE J (5.9 kg)(4.7102 kg) (5.5102 m)2 6.1109 N 54. Use Table 7-1 on p. 173 to compute the gravitational force that the Sun exerts on Jupiter. m m r 1 2 F G 2 (6.671011 Nm2/kg2) (1.991030 kg)(1.901027 kg) (7.781011 m)2 4.171023 N 55. Tom has a mass of 70.0 kg and Sally has a mass of 50.0 kg. Tom and Sally are standing 20.0 m apart on the dance floor. Sally looks up and sees Tom. She feels an attraction. If the attraction is gravitational, find its size. Assume that both Tom and Sally can be replaced by spherical masses. m m r (6.671011 Nm2/kg2) (1.0 y) 1.0 (6.671011 Nm2/kg2) T S F G 2 rJ 3 2 TE rE 5.2 3 m m r S J F G 2 2 (70.0 kg)(50.0 kg) (20.0 m)2 141 y2 5.841010 N 12 y 53. Figure 7-22 shows a Cavendish apparatus like the one used to find G. It has a large lead sphere that is 5.9 kg in mass and a small one with a mass of 0.047 kg. Their centers are separated by 0.055 m. Find the force of attraction between them. 56. Two balls have their centers 2.0 m apart, as shown in Figure 7-23. One ball has a mass of 8.0 kg. The other has a mass of 6.0 kg. What is the gravitational force between them? 2.0 m 8.0 kg 6.0 kg 5.9 kg 0.047 kg 0.055 m 0.055 m 5.9 kg Figure 7-22 Physics: Principles and Problems Figure 7-23 (6.671011 Nm2/kg2) (8.0 kg)(6.0 kg) (2.0 m)2 0.047 kg ■ ■ m1m2 F G r2 8.01010 N Solutions Manual 151 Chapter 7 continued 57. Two bowling balls each have a mass of 6.8 kg. They are located next to each other with their centers 21.8 cm apart. What gravitational force do they exert on each other? 2 Fr mE Gm 6 6.01024 kg b. Calculate the average density of Earth. (6.671011 (6.8 kg)(6.8 kg) Nm2/kg2) 2 (4)(6.4106 m)3 4 V r 3 (0.218 m) 6.5108 3 N a. What is the force of gravitational attraction between you and Earth? Sample answer: m m r E F G S 2 24 6.010 kg m D 21 3 V 3 3 b. What is your weight? TU 2 TE 84 y 2 1.0 y 19 Sample answer: 4.90102 N 59. The gravitational force between two electrons that are 1.00 m apart is 5.541071 N. Find the mass of an electron. m m r 1 2 F G 2 , where m1 m2 me (5.5410 N)(1.00 m) 6.6710 Nm /kg Fr 2 G 71 So rU 19rE 62. Venus Venus has a period of revolution of 225 Earth days. Find the distance between the Sun and Venus as a multiple of Earth’s orbital radius. T 60. A 1.0-kg mass weighs 9.8 N on Earth’s surface, and the radius of Earth is roughly 6.4106 m. a. Calculate the mass of Earth. 3 3 2 9.111031 kg r 3 rE V r V rE 2 2 2 TVE TV 2 TE 225 days 2 365 days 0.724 So rV 0.724rE Level 2 63. If a small planet, D, were located 8.0 times as far from the Sun as Earth is, how many years would it take the planet to orbit the Sun? T 2 TDE r 3 rE D Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Fg mg (50.0 kg)(9.80 m/s2) Solutions Manual r 3 rE U 489 N 152 2 r U rE (50.0 kg)(5.971024 kg) (6.38106 m)2 m m r m 61. Uranus Uranus requires 84 years to circle the Sun. Find Uranus’s orbital radius as a multiple of Earth’s orbital radius. T 11 1.110 5.5103 kg/m3 TUE (6.671011 Nm2/kg2) 1 2 F G 2 3 1.11021 m3 58. Assume that you have a mass of 50.0 kg. Earth has a mass of 5.971024 kg and a radius of 6.38106 m. So me Nm kg )(1.0 kg) (6.6710 m1m2 F G r2 2 (9.8 N)(6.410 m) 11 2 2 Chapter 7 continued TD T (1.0 y) 23 years r 3 2 D E rE 8.0 3 1.0 2 64. Two spheres are placed so that their centers are 2.6 m apart. The force between the two spheres is 2.751012 N. What is the mass of each sphere if one sphere is twice the mass of the other sphere? m m r 2 F G 1 2 , where m2 2m1 (m )(2m ) r 1 1 F G 2 m1 Fr 2G 2 (2.7510 N)(2.6 m) (2)(6.6710 Nm kg ) 12 2 11 2 2 0.3733 kg or 0.37 kg to two significant digits m2 2m1 (2)(0.3733 kg) 0.7466 kg or 0.75 to two significant digits 65. The Moon is 3.9105 km from Earth’s center and 1.5108 km from the Sun’s center. If the masses of the Moon, Earth, and the Sun are 7.31022 kg, 6.01024 kg, and 2.01030 kg, respectively, find the ratio of the gravitational forces exerted by Earth and the Sun on the Moon. m m r 2 F G 1 2 (6.01024 kg)(7.31022 kg) (3.910 m) Earth on Moon: FE (6.671011 Nm2/kg2) 8 2 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1.91020 N (2.01030 kg)(7.31022 kg) Sun on Moon: FE (6.671011 Nm2/kg2) (1.51011 m)2 4.31020 N FE 1.0 1.91020 N Ratio is 20 FS 4.310 N 2.3 The Sun pulls more than twice as hard on the Moon as does Earth. 66. Toy Boat A force of 40.0 N is required to pull a 10.0-kg wooden toy boat at a constant velocity across a smooth glass surface on Earth. What force would be required to pull the same wooden toy boat across the same glass surface on the planet Jupiter? F FN F mbg f f , where mb is the mass of the toy boat. On Jupiter, the normal force is equal to the gravitational attraction between the toy boat and Jupiter, or m m rJ J FN G b 2 F FN m m rJ J Now f , so FfJ FN G b 2 Physics: Principles and Problems Solutions Manual 153 Chapter 7 continued F mbg f But , so FfJ m m mbgrJ m grJ (40.0 N)(6.671011 Nm2/kg2)(1.901027 kg) (9.80 m/s )(7.1510 m) J J FfG b 2 FfG 2 2 7 2 101 N 67. Mimas, one of Saturn’s moons, has an orbital radius of 1.87108 m and an orbital period of about 23.0 h. Use Newton’s version of Kepler’s third law to find Saturn’s mass. 42 Gm T 2 r 3 4 2r 3 GT m 2 4 2(1.87108 m)3 11 2 2 4 2 (6.6710 Nm /kg )(8.2810 s) 5.61026 kg Level 3 68. The Moon is 3.9108 m away from Earth and has a period of 27.33 days. Use Newton’s version of Kepler’s third law to find the mass of Earth. Compare this mass to the mass found in problem 60. 4 2 Gm T 2 r 3 4 2 r 3 G T m 2 4 2 (3.9108 m)3 (6.6710 Nm /kg ) (2.36110 s) 11 2 2 6 2 The mass is considerably close to that found in problem 60. 69. Halley’s Comet Every 74 years, comet Halley is visible from Earth. Find the average distance of the comet from the Sun in astronomical units (AU). For Earth, r 1.0 AU and T 1.0 y r 3 T 2 Tb rba a ra 3 T 2 Tb a rb3 3 74 y 2 1.0 y (1.0 AU)3 18 AU 70. Area is measured in m2, so the rate at which area is swept out by a planet or satellite is measured in m2/s. a. How quickly is an area swept out by Earth in its orbit about the Sun? r 1.501011 m and T 3.156107 s, in 365.25 days 1.00 y (1.501011 m)2 r 2 2.241015 m2/s T 3.156107 s 154 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 6.31024 kg Chapter 7 continued b. How quickly is an area swept out by the Moon in its orbit about Earth? Use 3.9108 m as the average distance between Earth and the Moon, and 27.33 days as the period of the Moon. (3.9108 m)2 2.01011 m2/s 2.36106 s 7.2 Using the Law of Universal Gravitation pages 192–193 Level 1 71. Satellite A geosynchronous satellite is one that appears to remain over one spot on Earth, as shown in Figure 7-24. Assume that a geosynchronous satellite has an orbital radius of 4.23107 m. Satellite r Earth ■ Figure 7-24 (Not to scale) a. Calculate its speed in orbit. r (6.6710 Nm /kg )(5.97x10 kg) 4.2310 m v GmE 11 2 2 24 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 7 3.07103 m/s or 3.07 km/s b. Calculate its period. T 2 r3 GmE (6.6710 Nm /kg )(5.9710 kg) (4.23107 m)3 2 11 2 2 24 8 s2 21.90 10 8.66104 s or 24.1 h 72. Asteroid The asteroid Ceres has a mass of 71020 kg and a radius of 500 km. a. What is g on the surface of Ceres? Gm r g 2 (6.671011 Nm2/kg2)(71020 kg) 3 2 (50010 m) 0.2 m/s2 b. How much would a 90-kg astronaut weigh on Ceres? Fg mg (90 kg)(0.2 m/s2) 20 N Physics: Principles and Problems Solutions Manual 155 Chapter 7 continued 73. Book A 1.25-kg book in space has a weight of 8.35 N. What is the value of the gravitational field at that location? F m 8.35 N 1.25 kg g 6.68 N/kg 74. The Moon’s mass is 7.341022 kg, and it is 3.8108 m away from Earth. Earth’s mass is 5.971024 kg. a. Calculate the gravitational force of attraction between Earth and the Moon. m m r E M F G 2 (5.981024 kg)(7.341022 kg) (3.810 m) (6.671011 Nm2/kg2) 8 2 2.01020 N b. Find Earth’s gravitational field at the Moon. F 2.031020 N g 0.0028 N/kg 22 m 7.3410 kg Note that 2.031020 N instead of 2.01020 N is used to prevent roundoff error. 75. Two 1.00-kg masses have their centers 1.00 m apart. What is the force of attraction between them? m m r 2 Fg G 1 2 (6.671011 Nm2/kg2)(1.00 kg)(1.00 kg) (1.00 m) 2 6.671011 N a. 6.38103 km d rE rE 2rE Therefore, Fg original weight (7.20103 N) 1.80103 N 1 4 1 4 b. 1.28104 km d rE 2rE 3rE Fg (7.20103 N) 8.00102 N 1 9 77. Rocket How high does a rocket have to go above Earth’s surface before its weight is half of what it is on Earth? 1 r Fg 2 1 F So r g 156 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Level 2 76. The radius of Earth is about 6.38103 km. A 7.20103-N spacecraft travels away from Earth. What is the weight of the spacecraft at the following distances from Earth’s surface? Chapter 7 continued 1 If the weight is , the distance is 2 (rE) or 2 r 2 (6.38106 m) 9.02106 m 9.02106 m 6.38106 m 2.64106 m 2.64103 km 78. Two satellites of equal mass are put into orbit 30.0 m apart. The gravitational force between them is 2.0107 N. a. What is the mass of each satellite? m m r 2 F G 1 2 , m1 m2 m m 7 Fr (2.010 N)(30.0 m) G 6.6710 Nm kg 2 2 11 2 2 1.6103 kg b. What is the initial acceleration given to each satellite by gravitational force? Fnet ma F m 2.0107 N 1.610 kg 1.31010 m/s2 a net 3 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 79. Two large spheres are suspended close to each other. Their centers are 4.0 m apart, as shown in Figure 7-25. One sphere weighs 9.8102 N. The other sphere has a weight of 1.96102 N. What is the gravitational force between them? 4.0 m 9.8102 N ■ Fg g 1.96102 N Figure 7-25 9.810 N 2 m1 2 1.010 kg Fg g 2 9.80 m/s 1.9610 N 1 m2 2 2.0010 kg 2 9.80 m/s m m r 2 F G 1 2 (6.671011 Nm2/kg2)(1.0101 kg)(2.00101 kg) 2 (4.0 m) 8.3109 N Physics: Principles and Problems Solutions Manual 157 Chapter 7 continued 80. Suppose the centers of Earth and the Moon are 3.9108 m apart, and the gravitational force between them is about 1.91020 N. What is the approximate mass of the Moon? m m r E M F G 2 Fr 2 GmE mM (1.91020 N)(3.9108 m)2 11 2 24 (6.6710 7.31022 Nm/kg )(5.9710 kg) kg 81. On the surface of the Moon, a 91.0-kg physics teacher weighs only 145.6 N. What is the value of the Moon’s gravitational field at its surface? Fg mg, Fg 145.6 N 91.0 kg So g 1.60 N/kg m Level 3 82. The mass of an electron is 9.11031 kg. The mass of a proton is 1.71027 kg. An electron and a proton are about 0.591010 m apart in a hydrogen atom. What gravitational force exists between the proton and the electron of a hydrogen atom? memp F G 2 r (6.671011 Nm2/kg2)(9.11031 kg)(1.71027 kg) (1.010 m) 10 2 83. Consider two spherical 8.0-kg objects that are 5.0 m apart. a. What is the gravitational force between the two objects? m m r 2 F G 1 2 (6.671011 Nm2/kg2)(8.0 kg)(8.0 kg) (5.0 m) 2 1.71010 N b. What is the gravitational force between them when they are 5.0101 m apart? m m r 2 F G 1 2 (6.671011 Nm2/kg2)(8.0 kg)(8.0 kg) (5.010 m) 1 2 1.71012 N 158 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1.01047 N Chapter 7 continued 84. If you weigh 637 N on Earth’s surface, how much would you weigh on the planet Mars? Mars has a mass of 6.421023 kg and a radius of 3.40106 m. Fg 637 N 9.80 m/s m 2 65.0 kg g m m r 2 F G 1 2 (6.671011 Nm2/kg2)(65.0 kg)(6.371023 kg) (3.4310 m) 6 2 235 N 85. Using Newton’s version of Kepler’s third law and information from Table 7-1 on page 173, calculate the period of Earth’s Moon if the orbital radius were twice the actual value of 3.9108 m. TM (r ) 4 2 GmE 3 4 (7.8 10 m ) (6.6710 Nm /kg )(5.97910 kg) 2 11 2 2 24 8 3 6.85106 s or 79 days 86. Find the value of g, acceleration due to gravity, in the following situations. a. Earth’s mass is triple its actual value, but its radius remains the same. Gm (rE) E 2 g 2 9.80 m/s Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 2mE → 2g 2(9.80 m/s2) 19.6 m/s2 b. Earth’s radius is tripled, but its mass remains the same. Gm (rE) E 2 g 2 9.80 m/s g 9.80 m/s2 2rE → 2.45 m/s2 4 4 c. Both the mass and radius of Earth are doubled. Gm (rE) E g 2 2g 4 2(9.80 m/s2) 4 2mE and 2rE → 4.90 m/s2 87. Astronaut What would be the strength of Earth’s gravitational field at a point where an 80.0-kg astronaut would experience a 25.0 percent reduction in weight? Fg mg (80.0 kg)(9.80 m/s2) 784 N Fg, reduced (784 N)(0.750) 588 N Fg, reduced 588 N 80.0 kg greduced 7.35 m/s2 m Physics: Principles and Problems Solutions Manual 159 Chapter 7 continued Mixed Review pages 193–194 Level 1 88. Use the information for Earth in Table 7-1 on page 173 to calculate the mass of the Sun, using Newton’s version of Kepler’s third law. 4 2 Gm T 2 r 3, 4 2 G so mT 2 r 3 and 4 2 r 3 G T m 2 4 2 6.6710 Nm /kg (1.501011 m)3 (3.15610 s) 11 2 2 7 2 2.011030 kg 89. Earth’s gravitational field is 7.83 N/kg at the altitude of the space shuttle. At this altitude, what is the size of the force of attraction between a student with a mass of 45.0 kg and Earth? F m g F mg (45.0 kg)(7.83 N/kg) 352 N 90. Use the data from Table 7-1 on page 173 to find the speed and period of a satellite that orbits Mars 175 km above its surface. r rM 175 km 3.40106 m 0.175106 m v Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 3.58106 m GM r M 11 (6.6710 Nm /kg )(6.4210 kg) 3.5810 m 2 2 23 6 3.46103 m/s T 2 r3 GMM (6.6710 Nm /kg )(6.4210 kg) 6 3 (3.5810 m) 2 11 2 2 23 6.45103 s or 1.79 h 160 Solutions Manual Physics: Principles and Problems Chapter 7 continued Level 2 91. Satellite A satellite is placed in orbit, as shown in Figure 7-26, with a radius that is half the radius of the Moon’s orbit. Find the period of the satellite in units of the period of the Moon. Moon Earth rM 1 rM 2 Satellite ■ 2 T TMs r rM Figure 7-26 3 s So, Ts T T r r rs 3 0.50rM 3 M M M 2 M 0.125 TM2 0.35TM 92. Cannonball The Moon’s mass is 7.31022 kg and its radius is 1785 km. If Newton’s thought experiment of firing a cannonball from a high mountain were attempted on the Moon, how fast would the cannonball have to be fired? How long would it take the cannonball to return to the cannon? Gm r Nm /kg )(7.310 kg) (6.6710 1.78510 m Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. v 11 2 2 22 6 1.7103 m/s T 2 r Gm 3 (6.6710 Nm /kg )(7.310 kg) 6 3 (1.78510 m) 2 11 2 2 22 6.8103 s 93. The period of the Moon is one month. Answer the following questions assuming that the mass of Earth is doubled. a. What would the period of the Moon be? Express your results in months. T 2 r3 Gm So, if Earth’s mass were doubled, but the radius remained the same, 1 then the period would be reduced by a factor of , or 0.707 months. 2 Physics: Principles and Problems Solutions Manual 161 Chapter 7 continued b. Where would a satellite with an orbital period of one month be located? 2 T 2 3 r Gm T 2 2 r 3 (Gm) 1 Thus, r 3 would be doubled, or r would be increased by 2 3 1.26 times the present radius of the Moon. c. How would the length of a year on Earth be affected? The length of a year on Earth would not be affected. It does not depend on Earth’s mass. Level 3 94. How fast would a planet of Earth’s mass and size have to spin so that an object at the equator would be weightless? Give the period of rotation of the planet in minutes. The centripetal acceleration must equal the acceleration due to gravity so that the surface of the planet would not have to supply any force (otherwise known as weight). m m mv 2 G E r r2 v r GmE 2r 2r But, v , so T T v 2r r GmE Gm 3 E 2 (6.38106 m)3 11 (6.6710 Nm2/kg2)(5.971024 kg) 5.07103 s 84.5 min 95. Car Races Suppose that a Martian base has been established and car races are being considered. A flat, circular race track has been built for the race. If a car can achieve speeds of up to 12 m/s, what is the smallest radius of a track for which the coefficient of friction is 0.50? The force that causes the centripetal acceleration is the static friction force: smg Fstatic 2 v The centripetal acceleration is ac r 2 mv So, r Thus, r smg v2 sg Gm (rplanet) Note that g 2 162 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. r 2 Chapter 7 continued To find g on Mars, the following calculation is used. mMars 6.371023 kg, and RMars 3.43106 m So, gMars 3.61 m/s2 (12 m/s)2 (0.50)(3.61 m/s2) Therefore, R R 8.0101 m 96. Apollo 11 On July 19, 1969, Apollo 11’s revolution around the Moon was adjusted to an average orbit of 111 km. The radius of the Moon is 1785 km, and the mass of the Moon is 7.31022 kg. a. How many minutes did Apollo 11 take to orbit the Moon once? rorbit 1785103 m 111103 m T 2 r Gm 3 (6.6710 Nm /kg )(7.310 kg) 3 3 (189610 m) 2 11 2 2 22 7.4103 s 1.2102 min b. At what velocity did Apollo 11 orbit the Moon? v Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. r Gm orbit 11 (6.6710 Nm /kg )(7.310 kg) 189610 m 2 2 22 3 1.6103 m/s Thinking Critically page 194 97. Analyze and Conclude Some people say that the tides on Earth are caused by the pull of the Moon. Is this statement true? a. Determine the forces that the Moon and the Sun exert on a mass, m, of water on Earth. Your answer will be in terms of m with units of N. (1.991030 kg)(m) FS, m 6.671011 Nm2/kg2 (1.501011 m)2 (5.90103 N)m (7.361022 kg)(m) FM, m 6.671011 Nm2/kg2 (3.80108 m)2 (3.40105 N)m b. Which celestial body, the Sun or the Moon, has a greater pull on the waters of Earth? The Sun pulls approximately 100 times stronger on the waters of Earth. Physics: Principles and Problems Solutions Manual 163 Chapter 7 continued c. Determine the difference in force exerted by the Moon on the water at the near surface and the water at the far surface (on the opposite side) of Earth, as illustrated in Figure 7-27. Again, your answer will be in terms of m with units of N. Far tidal bulge Near tidal bulge Earth ■ Moon Figure 7-27 (Not to scale) Fm, mA Fm, mB (6.671011 Nm2kg2)(7.361022 kg)(m) 1 1 (3.80108 m 6.37106 m)2 (3.80108 m 6.37106 m)2 (2.28106 N)m d. Determine the difference in force exerted by the Sun on water at the near surface and on water at the far surface (on the opposite side) of Earth. FS, mA FS, mB (6.671011 Nm2kg2)(1.991030 kg)(m) 1 1 (1.501011 m 6.37106 m)2 (1.501011 m 6.37106 m)2 (1.00106 N)m the Moon f. Why is the statement that the tides result from the pull of the Moon misleading? Make a correct statement to explain how the Moon causes tides on Earth. The tides are primarily due to the difference between the pull of the Moon on Earth’s near side and Earth’s far side. 98. Make and Use Graphs Use Newton’s law of universal gravitation to find an equation where x is equal to an object’s distance from Earth’s center, and y is its acceleration due to gravity. Use a graphing calculator to graph this equation, using 6400–6600 km as the range for x and 9–10 m/s2 as the range for y. The equation should be of the form y c(1/x 2). Trace along this graph and find y for the following locations. a. at sea level, 6400 km c (GmE)(106 m2/km2) 4.0108 units acc 9.77 m/s2 b. on top of Mt. Everest, 6410 km 9.74 m/s2 164 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. e. Which celestial body has a greater difference in pull from one side of Earth to the other? Chapter 7 continued c. in a typical satellite orbit, 6500 km 9.47 m/s2 d. in a much higher orbit, 6600 km 9.18 m/s2 Writing in Physics page 194 99. Research and describe the historical development of the measurement of the distance between the Sun and Earth. One of the earliest crude measurements was made by James Bradley in 1732. The answers also should discuss measurements of the transits of Venus done in the 1690s. 100. Explore the discovery of planets around other stars. What methods did the astronomers use? What measurements did they take? How did they use Kepler’s third law? Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Astronomers measure the star’s tiny velocity due to the gravitational force exerted on it by a massive planet. The velocity is calculated by measuring the Doppler shift of the star’s light that results from that motion. The velocity oscillates back and forth as the planets orbit the star, allowing calculation of the planet’s period. From the size of the velocity they can estimate the planet’s distance and mass. By comparing the distances and periods of planets in solar systems with multiple planets and using Kepler’s third law, astronomers can better separate the distances and masses of stars and planets. Cumulative Review page 194 101. Airplanes A jet airplane took off from Pittsburgh at 2:20 P.M. and landed in Washington, DC, at 3:15 P.M. on the same day. If the jet’s average speed while in the air was 441.0 km/h, what is the distance between the cities? (Chapter 2) t 55 min 0.917 h d v t d v t (441.0 km/h)(0.917 h) 404 km 102. Carolyn wants to know how much her brother Jared weighs. He agrees to stand on a scale for her, but only if they are riding in an elevator. If he steps on the scale while the elevator is accelerating upward at 1.75 m/s2 and the scale reads 716 N, what is Jared’s usual weight on Earth? (Chapter 4) Identify Jared as the system and upward as positive. Fnet Fscale on Jared FEarth’s mass on Jared ma Physics: Principles and Problems Solutions Manual 165 Chapter 7 continued FEarth’s mass on Jared Fscale on Jared FEarth’s mass on Jared a g FEarth’s mass on Jared1 a g FEarth’s mass on Jared Fscale on Jared 1 ga 716 N 1.75 m/s 1 9.80 m/s2 2 608 N 103. Potato Bug A 1.0-g potato bug is walking around the outer rim of an upsidedown flying disk. If the disk has a diameter of 17.2 cm and the bug moves at a rate of 0.63 cm/s, what is the centripetal force acting on the bug? What agent provides this force? (Chapter 6) 2 (0.0010 kg)(0.0063 cm)2 0.086 m mv Fc 5.0107 N, r provided by the frictional force between the bug and the flying disk Challenge Problem D B C rC rD rB Upsilon Andromedae 1. Do these planets obey Kepler’s third law? 3 r Test by calculating the following ratio 2. T For planet B, 166 Solutions Manual r 3 B TB2 (0.059 AU)3 (4.6170 d ays) 6 AU3/days2 2 9.610 Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. page 176 Astronomers have detected three planets that orbit the star Upsilon Andromedae. Planet B has an average orbital radius of 0.059 AU and a period of 4.6170 days. Planet C has an average orbital radius of 0.829 AU and a period of 241.5 days. Planet D has an average orbital radius of 2.53 AU and a period of 1284 days. (Distances are given in astronomical units (AU)—Earth’s average distance from the Sun. The distance from Earth to the Sun is 1.00 AU.) Chapter 7 continued r 3 TC (0.829 AU)3 (241.5 d ays) r 3 TD (2.53 AU)3 (1284 d ays) 6 AU3/days2 For planet C, C 2 2 9.7710 6 AU3/days2 For planet D, D 2 2 9.8210 These values are quite close, so Kepler’s third law is obeyed. 2. Find the mass of the star Upsilon Andromedae in units of the Sun’s mass. Gmcentral body r3 2 4 2 T r3 T (1.000 AU)3 (1.000 y) 3 2 For the Earth-Sun system, 2 2 1.000 AU /y For the planet C-Upsilon system, r3 9.77106 AU3/days2 T2 (9.77106 AU3/days2)(365 days/y)2 1.30 AU3/y2 Gm 4 star 2 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. The ratio of the two shows the star’s slightly heavier mass to be 1.30 that of the Sun. Physics: Principles and Problems Solutions Manual 167 CHAPTER 8 Rotational Motion Because the radius of the wheel is reduced from 35.4 cm to 24 cm, the angular acceleration will be increased. Practice Problems 8.1 Describing Rotational Motion pages 197–200 1 5.23 rad/s2 a2 page 200 1. What is the angular displacement of each of the following hands of a clock in 1 h? State your answer in three significant digits. a. the second hand (60)(2 rad) 120 rad or 377 rad b. the minute hand 2 rad or 6.28 rad 7.7 rad/s2 4. You want to replace the tires on your car with tires that have a larger diameter. After you change the tires, for trips at the same speed and over the same distance, how will the angular velocity and number of revolutions change? c. the hour hand v Because , if r is increased, will r Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1 (2 rad) 12 rad or 0.524 rad 6 2. If a truck has a linear acceleration of 1.85 m/s2 and the wheels have an angular acceleration of 5.23 rad/s2, what is the diameter of the truck’s wheels? a r 1.85 m/s2 5.23 rad/s2 0.354 m Thus, the diameter is 0.707 m. 3. The truck in the previous problem is towing a trailer with wheels that have a diameter of 48 cm. a. How does the linear acceleration of the trailer compare with that of the truck? The changes in velocity are the same, so the linear accelerations are the same. 2 1.85 m/s 2 r 0.24 m decrease. The number of revolutions will also decrease. Section Review 8.1 Describing Rotational Motion pages 197–200 page 200 5. Angular Displacement A movie lasts 2 h. During that time, what is the angular displacement of each of the following? a. the hour hand 1 (2 rad) rad 6 3 b. the minute hand (2)(2 rad) 4 rad 6. Angular Velocity The Moon rotates once on its axis in 27.3 days. Its radius is 1.74106 m. b. How do the angular accelerations of the wheels of the trailer and the wheels of the truck compare? Physics: Principles and Problems Solutions Manual 169 Chapter 8 continued a. What is the period of the Moon’s rotation in seconds? period (27.3 day)(24 h/day) (3600 s/h) 2.36106 s b. What is the frequency of the Moon’s rotation in rad/s? 1 period 1 rev/s, or 2.36106 2.66106 rad/s c. What is the linear speed of a rock on the Moon’s equator due only to the Moon’s rotation v r (1.74106 m)(2.66106 rad/s) 4.63 m/s d. Compare this speed with the speed of a person on Earth’s equator due to Earth’s rotation. The speed on Earth’s equator is 464 m/s, or about 100 times faster. d r d 12 cm so 12 rad r 1.0 cm 8. Angular Displacement Do all parts of the minute hand on a watch have the same angular displacement? Do they move the same linear distance? Explain. angular displacement—yes; linear distance—no, because linear distance is a function of the radius 170 Solutions Manual i 635 rpm 66.53 rad/s f 0.0, so 66.5 rad/s 66.5 rad/s and 8.3 rad/s2 t 8.0 s 10. Critical Thinking A CD-ROM has a spiral track that starts 2.7 cm from the center of the disk and ends 5.5 cm from the center. The disk drive must turn the disk so that the linear velocity of the track is a constant 1.4 m/s. Find the following. a. the angular velocity of the disk (in rad/s and rev/min) for the start of the track v r 1.4 m/s 0.027 m 52 rad/s or 5.0102 rev/min b. the disk’s angular velocity at the end of the track v r 1.4 m/s 0.055 m 25 rad/s or 2.4102 rev/min c. the disk’s angular acceleration if the disk is played for 76 min t f i t 25 rad/s 52 rad/s (76 min)(60 s/min) 5.9103 rad/s2 Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 7. Angular Displacement The ball in a computer mouse is 2.0 cm in diameter. If you move the mouse 12 cm, what is the angular displacement of the ball? 9. Angular Acceleration In the spin cycle of a clothes washer, the drum turns at 635 rev/min. If the lid of the washer is opened, the motor is turned off. If the drum requires 8.0 s to slow to a stop, what is the angular acceleration of the drum? Chapter 8 continued Practice Problems 8.2 Rotational Dynamics pages 201–210 page 203 11. Consider the wrench in Example Problem 1. What force is needed if it is applied to the wrench at a point perpendicular to the wrench? Fr sin so F r sin 35 Nm (0.25 m)(sin 90.0°) 1.4102 N 12. If a torque of 55.0 Nm is required and the largest force that can be exerted by you is 135 N, what is the length of the lever arm that must be used? Fr sin mgr sin (65 kg)(9.80 m/s2)(0.18 m)(sin 55°) 94 Nm 15. If the pedal in problem 14 is horizontal, how much torque would you exert? How much torque would you exert when the pedal is vertical? Horizontal 90.0° so Fr sin mgr sin (65 kg)(9.80 m/s2)(0.18 m) (sin 90.0°) 1.1102 Nm Vertical = 0.0° So Fr sin mgr sin For the shortest possible lever arm, 90.0°. (65 kg)(9.80 m/s2)(0.18 m) (sin 0.0°) Fr sin 0.0 Nm so r F sin Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 55.0 Nm (135 N)(sin 90.0°) 0.407 m 13. You have a 0.234-m-long wrench. A job requires a torque of 32.4 Nm, and you can exert a force of 232 N. What is the smallest angle, with respect to the vertical, at which the force can be exerted? Fr sin Fr so sin1 sin1 32.4 Nm (232 N)(0.234 m) 36.6° 14. You stand on the pedal of a bicycle. If you have a mass of 65 kg, the pedal makes an angle of 35° above the horizontal, and the pedal is 18 cm from the center of the chain ring, how much torque would you exert? The angle between the force and the radius is 90° 35° 55°. Physics: Principles and Problems page 205 16. Ashok, whose mass is 43 kg, sits 1.8 m from the center of a seesaw. Steve, whose mass is 52 kg, wants to balance Ashok. How far from the center of the seesaw should Steve sit? FArA FSrS F r FS A A so rS m gr mSg A A m r mS A A (43 kg)(1.8 m) 52 kg 1.5 m 17. A bicycle-chain wheel has a radius of 7.70 cm. If the chain exerts a 35.0-N force on the wheel in the clockwise direction, what torque is needed to keep the wheel from turning? Solutions Manual 171 Chapter 8 continued chain Fgr (35.0 N)(0.0770 m) 2.70 Nm Thus, a torque of 2.70 Nm must be exerted to balance this torque. 18. Two baskets of fruit hang from strings going around pulleys of different diameters, as shown in Figure 8-6. What is the mass of basket A? 1 2 4.5 cm 1.1 cm F1r1 F2r2 m1gr1 m2gr2 m r r1 2 2 m1 A (0.23 kg)(1.1 cm) 4.5 cm 0.23 kg 0.056 kg ■ Figure 8-6 19. Suppose the radius of the larger pulley in problem 18 was increased to 6.0 cm. What is the mass of basket A now? m r r1 2 2 m1 (0.23 kg)(1.1 cm) 6.0 cm 0.042 kg 45.0° 9.70 cm cr ch 0.170 m ■ Figure 8-7 Fcrrcr sin Fchrch F r sin rch cr cr so Fch mgr sin rch cr (65.0 kg)(9.80 m/s )(0.170 m)(sin 45.0°) 2 0.097 m 789 N 172 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 20. A bicyclist, of mass 65.0 kg, stands on the pedal of a bicycle. The crank, which is 0.170 m long, makes a 45.0° angle with the vertical, as shown in Figure 8-7. The crank is attached to the chain wheel, which has a radius of 9.70 cm. What force must the chain exert to keep the wheel from turning? Chapter 8 continued page 208 21. Two children of equal masses sit 0.3 m from the center of a seesaw. Assuming that their masses are much greater than that of the seesaw, by how much is the moment of inertia increased when they sit 0.6 m from the center? For the mass of the two children, I mr 2 mr 2 2mr 2. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. When r is doubled, I is multiplied by a factor of 4. 24. Each sphere in the previous problem has a mass of 0.10 kg. The distance between spheres A and C is 0.20 m. Find the moment of inertia in the following instances: rotation about sphere A, rotation about sphere C. About sphere A: I mr 2 m(2r)2 5mr 2 (5)(0.10 kg)(0.20 m)2 0.020 kgm2 22. Suppose there are two balls with equal diameters and masses. One is solid, and the other is hollow, with all its mass distributed at its surface. Are the moments of inertia of the balls equal? If not, which is greater? About sphere C: The more of the mass that is located far from the center, the greater the moment of inertia. Thus, the hollow ball has a greater value of I. 0.0080 kgm2 23. Figure 8-9 shows three massive spheres on a rod of very small mass. Consider the moment of inertia of the system, first when it is rotated about sphere A, and then when it is rotated about sphere C. Are the moments of inertia the same or different? Explain. If the moments of inertia are different, in which case is the moment of inertia greater? A ■ C Figure 8-9 The moments of inertia are different. If the spacing between spheres is r and each sphere has mass m, then rotation about sphere A is I mr 2 m(2r)2 5mr 2. Rotation about sphere C is I mr 2 mr 2 2mr 2. I mr 2 mr 2 2mr 2 (2)(0.10 kg)(0.20 m)2 page 210 25. Consider the wheel in Example Problem 4. If the force on the strap were twice as great, what would be the speed of rotation of the wheel after 15 s? Torque is now twice as great. The angular acceleration is also twice as great, so the change in angular velocity is twice as great. Thus, the final angular velocity is 32 rad/s, or 16 rev/s. 26. A solid wheel accelerates at 3.25 rad/s2 when a force of 4.5 N exerts a torque on it. If the wheel is replaced by a wheel with all of its mass on the rim, the moment of inertia is given by I mr 2. If the same angular velocity were desired, what force would have to be exerted on the strap? The angular acceleration has not changed, but the moment of inertia is twice as great. Therefore, the torque must be twice as great. The radius of the wheel is the same, so the force must be twice as great, or 9.0 N. The moment of inertia is greater when rotating around sphere A. Physics: Principles and Problems Solutions Manual 173 Chapter 8 continued 27. A bicycle wheel can be accelerated either by pulling on the chain that is on the gear or by pulling on a string wrapped around the tire. The wheel’s radius is 0.38 m, while the radius of the gear is 0.14 m. If you obtained the needed acceleration with a force of 15 N on the chain, what force would you need to exert on the string? The torque on the wheek comes from either the chain or the string. chain Iwheelwheel wheel Iwheelwheel Thus, chain wheel Fchainrgear Fstringrwheel Fchainrgear Fstring rwheel (15 N)(0.14 m) 0.38 m 5.5 N 28. The bicycle wheel in problem 27 is used with a smaller gear whose radius is 0.11 m. The wheel can be accelerated either by pulling on the chain that is on thegear or by pulling string that is wrapped around the tire. If you obtained the needed acceleration with a force of 15 N on the chain, what force would you need to exert on the string? Fchainrgear Fstring rwheel (15 N)(0.11 m) 4.3 N 29. A disk with a moment of inertia of 0.26 kgm2 is attached to a smaller disk mounted on the same axle. The smaller disk has a diameter of 0.180 m and a mass of 2.5 kg. A strap is wrapped around the smaller disk, as shown in Figure 8-10. Find the force needed to give this system an angular acceleration of 2.57 rad/s2. F Fr I I I so F r (Ismall Ilarge) ■ Figure 8-10 r 12msmallrsmall2 Ilarge r 12(2.5 kg)(0.090 m)2 0.26 kgm2(2.57 rad/s2) 0.090 m 7.7 N 174 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 0.38 m Chapter 8 continued Section Review 8.2 Rotational Dynamics pages 201–210 page 210 30. Torque Vijesh enters a revolving door that is not moving. Explain where and how Vijesh should push to produce a torque with the least amount of force. To produce a torque with the least force, you should push as close to the edge as possible and at right angles to the door. 31. Lever Arm You try to open a door, but you are unable to push at a right angle to the door. So, you push the door at an angle of 55° from the perpendicular. How much harder would you have to push to open the door just as fast as if you were to push it at 90°? The angle between the force and the radius is 35°. Torque is Fr sin . Because sin 90° 1, and sin 35° 0.57, you would have to increase the Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1 force by a ratio of 1.75 to obtain 0.57 the same torque. 32. Net Torque Two people are pulling on ropes wrapped around the edge of a large wheel. The wheel has a mass of 12 kg and a diameter of 2.4 m. One person pulls in a clockwise direction with a 43-N force, while the other pulls in a counterclockwise direction with a 67-N force. What is the net torque on the wheel? net 1 2 F1r F2r 33. Moment of Inertia Refer to Table 8-2 on page 206 and rank the moments of inertia from least to greatest of the following objects: a sphere, a wheel with almost all of its mass at the rim, and a solid disk. All have equal masses and diameters. Explain the advantage of using the one with the least moment of inertia. 2 from least to greatest: sphere mr 2, 5 1 solid disk mr 2, wheel (mr 2) 2 The less the moment of inertia, the less torque needed to give an object the same angular acceleration. 34. Newton’s Second Law for Rotational Motion A rope is wrapped around a pulley and pulled with a force of 13.0 N. The pulley’s radius is 0.150 m. The pulley’s rotational speed goes from 0.0 to 14.0 rev/min in 4.50 s. What is the moment of inertia of the pulley? I Fr t Frt f i (13.0 N)(0.150 m)(4.50 s) rev rev 2 rad min 0.0 14.0 min min 60 s rev 5.99 kgm2 35. Critical Thinking A ball on an extremely low-friction, tilted surface, will slide downhill without rotating. If the surface is rough, however, the ball will roll. Explain why, using a free-body diagram. (F1 F2) d 2 1 (43 N 67 N) (2.4 m) 1 2 29 Nm Physics: Principles and Problems Solutions Manual 175 Chapter 8 continued Torque: Fr sin . The force is due to friction, and the torque causes the ball to rotate clockwise. If the surface is friction-free, then there is no force parallel to the surface, no torque, and thus no rotation. Remember, forces acting on the pivot point (the center of the ball) are ignored. Extremely low-friction surface Fnormal Ffriction r Ffriction Surface Fg Equilibrium pages 211–217 B ■ Figure 8-15 clockwise: A FArA (0.66 m)FA counterclockwise: B FBrB FB(0.96 m 0.45 m) (0.51 m)FB b. Write the equation for rotational equilibrium. page 215 36. What would be the forces exerted by the two sawhorses if the ladder in Example Problem 5 had a mass of 11.4 kg? Because no distances have changed, the equations are still valid: net A B 0 so B A (0.51 m)FB ((0.66 m)FA) (0.51 m)FB (0.66 m)FA Fg FA FB rB m)(11.4 kg)(9.80 m/s2) 32 N FB (0.30 1.05 m (0.66 m)F 0.51 m rg A Fg FA mg 1 rB 0.30 m (11.4 kg)(9.80 m/s2)1 1.05 m 8.0101 N 37. A 7.3-kg ladder, 1.92 m long, rests on two sawhorses, as shown in Figure 8-15. Sawhorse A, on the left, is located 0.30 m from the end, and sawhorse B, on the right, is located 0.45 m from the other end. Choose the axis of rotation to be the center of mass of the ladder. Solutions Manual thus, FA Fg FB Fg or FA 0.66 m 1 0.51 m ma 0.66 m 1 0.51 m 2 (7.3 kg)(9.80 m/s ) 0.66 m 1 0.51 m 31 N d. How would the forces exerted by the two sawhorses change if A were moved very close to, but not directly under, the center of mass? Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. c. Solve the equation for FA in terms of Fg. rgmg FB 176 0.45 m FA(0.96 m – 0.30 m) Practice Problems 8.3 1.92 m 0.51 m A r Surface Fg 0.66 m a. What are the torques acting on the ladder? Rough surface Fnormal 0.30 m Chapter 8 continued g(2mdiverg mboard) FA would become greater, and FB would be less. 38. A 4.5-m-long wooden plank with a 24-kg mass is supported in two places. One support is directly under the center of the board, and the other is at one end. What are the forces exerted by the two supports? Pick the center of mass of the board as the pivot. The unsupported end exerts no torque, so the supported end does not have to exert any torque. Therefore, all the force is exerted by the center support. That force is equal to the weight of the board: Fcenter Fg (24 kg)(9.80 m/s2) 2.4102 N Fend 0 N Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 39. A 85-kg diver walks to the end of a diving board. The board, which is 3.5 m long with a mass of 14 kg, is supported at the center of mass of the board and at one end. What are the forces on the two supports? Choose the center of mass of the board as the pivot. The force of Earth’s gravity on the board is exerted totally on the support under the center of mass. end diver Fendrend Fdiverrdiver F r rend diver diver Thus, Fend m gr diver diver rend (85 kg)(9.80 m) m/s2)(1.75 1.75 m 8.3102 N To find the force on the center support, notice that because the board is not moving, Fend Fcenter Fdiver Fg Thus, Fcenter Fdiver Fg Fend 2Fdiver Fg 2mdiverg mboardg Physics: Principles and Problems (9.80 m/s2) (2(85 kg) 14 kg) 1.8103 N Section Review 8.3 Equilibrium pages 211–217 page 217 40. Center of Mass Can the center of mass of an object be located in an area where the object has no mass? Explain. Yes, an object moves as if all its mass is concentrated at the center of mass. There is nothing in the definition that requires any or all of the object’s mass to be at that location. 41. Stability of an Object Why is a modified vehicle with its body raised high on risers less stable than a similar vehicle with its body at normal height? The center of mass of the vehicle will be raised, but the size of its base will not be increased. Therefore, it needs to be tipped at a smaller angle to get the center of mass outside the base of the vehicle. 42. Conditions for Equilibrium Give an example of an object for each of the following conditions. a. rotational equilibrium, but not translational equilibrium a book that is dropped so it falls without rotating b. translational equilibrium, but not rotational equilibrium a seesaw that is not balanced and rotates until one person’s feet hit the ground 43. Center of Mass Where is the center of mass of a roll of masking tape? It is in the middle of the roll, in the open space. Solutions Manual 177 Chapter 8 continued 44. Locating the Center of Mass Describe how you would find the center of mass of this textbook. Obtain a piece of string and attach a small weight to it. Suspend the string and the weight from one corner of the book. Draw a line along the string. Suspend the two from another corner of the book. Again, draw a line along the string. The point where the lines cross is the center of mass. Chapter Assessment Concept Mapping page 222 47. Complete the following concept map using the following terms: angular acceleration, radius, tangential acceleration, centripetal acceleration. Angular acceleration Angular velocity Radius 45. Rotating Frames of Reference A penny is placed on a rotating, old-fashioned record turntable. At the highest speed, the penny starts sliding outward. What are the forces acting on the penny? Earth’s mass exerts a downward force. The turntable’s surface exerts both an upward force to balance gravity and an inward force due to friction that gives the penny its centripetal acceleration. There is no outward force. If it were not for friction, the penny would move in a straight line. They would move in the opposite direction. Winds from the north move from the equator, where the linear speed due to rotation is highest, to mid-latitudes, where it is lower. Thus, the winds bend to the east. Winds from the south blow from regions where the linear speed is low to regions where it is higher; thus, they bend to the west. These two factors result in a clockwise rotation around a low-pressure area. Tangential acceleration Mastering Concepts page 222 48. A bicycle wheel rotates at a constant 25 rev/min. Is its angular velocity decreasing, increasing, or constant? (8.1) It is constant. 49. A toy rotates at a constant 5 rev/min. Is its angular acceleration positive, negative, or zero? (8.1) It is zero. 50. Do all parts of Earth rotate at the same rate? Explain. (8.1) Yes, because all parts of a rigid body rotate at the same rate. 51. A unicycle wheel rotates at a constant 14 rev/min. Is the total acceleration of a point on the tire inward, outward, tangential, or zero? (8.1) It is inward (centripetal). 52. Think about some possible rotations of your textbook. Are the moments of inertia about these three axes the same or different? Explain. (8.2) They are all different. The one with the most mass, farthest from the axis, has the greatest moment of inertia. 178 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 46. Critical Thinking You have learned why the winds around a low-pressure area move in a counterclockwise direction. Would the winds move in the same or opposite direction in the southern hemisphere? Explain. Centripetal acceleration Chapter 8 continued 53. Torque is important when tightening bolts. Why is force not important? (8.2) An angular acceleration must be produced to tighten a bolt. Different torques can be exerted with wrenches of different lengths. 54. Rank the torques on the five doors shown in Figure 8-18 from least to greatest. Note that the magnitude of all the forces is the same. (8.2) B A C D E ■ Figure 8-18 0EDCBA Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 55. Explain how you can change an object’s angular frequency. (8.2) Change the amount of torque applied to the object or change the moment of inertia. 56. To balance a car’s wheel, it is placed on a vertical shaft and weights are added to make the wheel horizontal. Why is this equivalent to moving the center of mass until it is at the center of the wheel? (8.3) When the wheel is balanced so it does not tilt (rotate) in any direction, then there is no net torque on it. This means that the center of mass is at the pivot point. 58. Suppose you stand flat-footed, then you rise and balance on tiptoe. If you stand with your toes touching a wall, you cannot balance on tiptoe. Explain. (8.3) Your center of mass must be above the point of support. But your center of mass is roughly in the center of your body. Thus, while you are on your toes, about half of your body must be in front of your toes, and half must be behind. If your toes are against the wall, no part of your body can be in front of your toes. 59. Why does a gymnast appear to be floating on air when she raises her arms above her head in a leap? (8.3) She moves her center of mass closer to her head. 60. Why is a vehicle with wheels that have a large diameter more likely to roll over than a vehicle with wheels that have a smaller diameter? (8.3) The center of mass of the vehicle with the larger wheels is located at a higher point. Thus it does not have to be tilted very far before it rolls over. Applying Concepts pages 222–223 61. Two gears are in contact and rotating. One is larger than the other, as shown in Figure 8-19. Compare their angular velocities. Also compare the linear velocities of two teeth that are in contact. 57. A stunt driver maneuvers a monster truck so that it is traveling on only two wheels. Where is the center of mass of the truck? (8.3) It is directly above the line between the points where the two wheels are touching the ground. There is no net torque on the truck, so it is momentarily stable. Physics: Principles and Problems ■ Figure 8-19 Solutions Manual 179 Chapter 8 continued The teeth have identical linear velocities. Because the radii are different v and , the angular velocities are r different. 62. Videotape When a videotape is rewound, why does it wind up fastest towards the end? The machine turns the spool at a constant angular velocity. Towards the end, the spool has the greatest radius. Because v r, the velocity of the tape is fastest when the radius is greatest. 63. Spin Cycle What does a spin cycle of a washing machine do? Explain in terms of the forces on the clothes and water. In the spin cycle, the water and clothes undergo great centripetal accelerations. The drum can exert forces on the clothes, but when the water reaches the holes in the drum, no inward force can be exerted on it and it therefore moves in a straight line, out of the drum. 64. How can you experimentally find the moment of inertia of an object? 65. Bicycle Wheels Three bicycle wheels have masses that are distributed in three different ways: mostly at the rim, uniformly, and mostly at the hub. The wheels all have the same mass. If equal torques are applied to them, which one will have the greatest angular acceleration? Which one will have the least? The more mass there is far from the axis, the greater the moment of inertia. If torque is fixed, the greater the moment of inertia, the less the angular acceleration. Thus, the wheel with mass mostly at the hub has the least moment of inertia and the greatest angular acceleration. The wheel with mass mostly near the rim has the greatest moment of inertia and the least angular acceleration. 180 Solutions Manual Its rotation rate can be increased only if a torque is applied to it. The frictional force of the alley on the ball provides this force. Once the ball is rolling so that there is no velocity difference between the surface of the ball and the alley, then there is no more frictional force and thus no more torque. 67. Flat Tire Suppose your car has a flat tire. You get out your tools and find a lug wrench to remove the nuts off the bolt studs. You find it impossible to turn the nuts. Your friend suggests ways you might produce enough torque to turn them. What three ways might your friend suggest? Put an extension pipe on the end of the wrench to increase the lever arm, exert your force at right angles to the wrench, or exert a greater force, perhaps by standing on the end of the wrench. 68. Tightrope Walkers Tightrope walkers often carry long poles that sag so that the ends are lower than the center as shown in Figure 8-20. How does such a pole increase the tightrope walker’s stability? Hint: Consider both center of mass and moment of inertia. ■ Figure 8-20 Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. You can apply a known torque and measure the resulting angular acceleration. 66. Bowling Ball When a bowling ball leaves a bowler’s hand, it does not spin. After it has gone about half the length of the lane, however, it does spin. Explain how its rotation rate increased and why it does not continue to increase. Chapter 8 continued Mastering Problems 8.1 Describing Rotational Motion pages 223–224 Level 1 72. A wheel is rotated so that a point on the edge moves through 1.50 m. The radius of the wheel is 2.50 m, as shown in Figure 8-21. Through what angle (in radians) is the wheel rotated? 2. 50 m The pole increases moment of inertia because of its mass and length. The drooping ends of the pole bring the center of mass closer to the wire, thus reducing the torque on the walker. The increased moment of inertia and decreased torque both reduce the angular acceleration if the walker becomes unbalanced. The walker can also use the pole to easily shift the center of mass over the wire to compensate for instability. 69. Merry-Go-Round While riding a merry-goround, you toss a key to a friend standing on the ground. For your friend to be able to catch the key, should you toss it a second or two before you reach the spot where your friend is standing or wait until your friend is directly behind you? Explain. You have forward tangential velocity, so the key will leave your hand with that velocity. Therefore, you should toss it early. 1.50 m ■ Figure 8-21 d r d so r 1.50 m 2.50 m Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 70. Why can you ignore forces that act on the axis of rotation of an object in static equilibrium when determining the net torque? The torque caused by these forces is zero because the lever arm is zero. 71. In solving problems about static equilibrium, why is the axis of rotation often placed at a point where one or more forces are acting on the object? That makes the torque caused by that force equal to zero, reducing the number of torques that must be calculated. 0.600 rad 73. The outer edge of a truck tire that has a radius of 45 cm has a velocity of 23 m/s. What is the angular velocity of the tire in rad/s? v r, v r 23 m/s 51 rad/s 0.45 m 74. A steering wheel is rotated through 128°, as shown in Figure 8-22. Its radius is 22 cm. How far would a point on the steering wheel’s edge move? 22 cm 128° ■ Physics: Principles and Problems Figure 8-22 Solutions Manual 181 Chapter 8 continued d r (0.22 m)(128°) 0.49 m 2 rad 360° 75. Propeller A propeller spins at 1880 rev/min. a. What is its angular velocity in rad/s? 2 rad rev a. What is the ratio of the centripetal accelerations for the fast and slow spin cycles? v2 r 1880 rev min Level 2 78. Washing Machine A washing machine’s two spin cycles are 328 rev/min and 542 rev/min. The diameter of the drum is 0.43 m. min 60 s Recall that ac and v rw. 197 rad/s b. What is the angular displacement of the propeller in 2.50 s? a fast rfast2 aslow rslow2 2 (542 rev/min) 2 t (328 rev/min) 2.73 (197 rad/s)(2.50 s) 492 rad 76. The propeller in the previous problem slows from 475 rev/min to 187 rev/min in 4.00 s. What is its angular acceleration? b. What is the ratio of the linear velocity of an object at the surface of the drum for the fast and slow spin cycles? vfast fastr vslow slowr slow fast t f i t 542 rev/min (187 rev/min 475 rev/min) 2 rad 4.00 s rev 328 rev/min 1 min 60 s 7.54 rad/s2 79. Find the maximum centripetal acceleration in terms of g for the washing machine in problem 78. ac 2r 2 1g 9.80 m/s 2 542 rev/min 2 rad rev 1 min 60 s 1g 0.43 m 2 9.80 m/s2 71g Level 3 80. A laboratory ultracentrifuge is designed to produce a centripetal acceleration of 0.35106 g at a distance of 2.50 cm from the axis. What angular velocity in rev/min is required? ■ Figure 8-23 v r (7.00 cm)(2.50 rad/s) 17.5 cm/s 182 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 77. An automobile wheel with a 9.00 cm radius, as shown in Figure 8-23, rotates at 2.50 rad/s. How fast does a point 7.00 cm from the center travel? 1.65 Chapter 8 continued ac 2r so a r c (0.3510 )(9.80 m/s ) rev 0.025 m 2 rad 6 2 60 s 1 min 83. A toy consisting of two balls, each 0.45 kg, at the ends of a 0.46-m-long, thin, lightweight rod is shown in Figure 8-25. Find the moment of inertia of the toy. The moment of inertia is to be found about the center of the rod. 0.45 kg 0.45 kg 1.1105 rev/min 8.2 Rotational Dynamics page 224 Level 1 81. Wrench A bolt is to be tightened with a torque of 8.0 Nm. If you have a wrench that is 0.35 m long, what is the least amount of force you must exert? Fr sin so F r sin For the least possible force, the angle is 90.0°, then 8.0 Nm F (0.35 m)(sin 90.0°) ■ Figure 8-25 I mr 2 (0.45 kg)(0.23 m)2 (0.45 kg)(0.23 m)2 0.048 kgm2 Level 2 84. A bicycle wheel with a radius of 38 cm is given an angular acceleration of 2.67 rad/s2 by applying a force of 0.35 N on the edge of the wheel. What is the wheel’s moment of inertia? 23 N Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 0.46 m I 82. What is the torque on a bolt produced by a 15-N force exerted perpendicular to a wrench that is 25 cm long, as shown in Figure 8-24? F 90° 25 cm ■ Figure 8-24 Fr sin I Fr sin (0.35 N)(0.38 m)(sin 90.0°) 2 2.67 rad/s 0.050 kgm2 Level 3 85. Toy Top A toy top consists of a rod with a diameter of 8.0-mm and a disk of mass 0.0125 kg and a diameter of 3.5 cm. The moment of inertia of the rod can be neglected. The top is spun by wrapping a string around the rod and pulling it with a velocity that increases from zero to 3.0 m/s over 0.50 s. (15 N)(0.25 m)(sin 90.0°) 3.8 Nm Physics: Principles and Problems Solutions Manual 183 Chapter 8 continued a. What is the resulting angular velocity of the top? v f f rrod 3.0 m/s 12(0.0080 m) 7.5102 rad/s b. What force was exerted on the string? Frrod sin and I Thus, Frrod sin I I F rrod sin t 12mrdisk2 r sin rod 2 mr 2trrod sin disk ( )mr 2trrod sin 2 f i disk 0.035 m 2 2 (7.5102 rad/s 0.0 rad/s)(0.0125 kg) 0.0080 m (2)(0.50 s)(sin 90.0°) 2 0.72 N a. What is the least force that Judi could exert to lift the board to the horizontal position? What part of the board should she lift to exert this force? At the opposite end, she will only lift half the mass. Fleast mg 1 (12.5 kg)(9.80 m/s2) 2 61.2 N b. What is the greatest force that Judi could exert to lift the board to the horizontal position? What part of the board should she lift to exert this force? At the board’s center of mass (middle), she will lift the entire mass. Fgreatest mg (12.5 kg)(9.80 m/s2) 184 Solutions Manual 122 N Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 8.3 Equilibrium page 224 Level 1 86. A 12.5-kg board, 4.00 m long, is being held up on one end by Ahmed. He calls for help, and Judi responds. Chapter 8 continued Level 2 87. Two people are holding up the ends of a 4.25-kg wooden board that is 1.75 m long. A 6.00-kg box sits on the board, 0.50 m from one end, as shown in Figure 8-26. What forces do the two people exert? In equilibrium, the sum of all forces is zero and the sum of the torques about an axis of rotation is zero. 6.00 kg 1.25 m 0.50 m Fleft Fright Fboard Fbox 0 ■ Figure 8-26 left right board box 0 We can choose the axis of rotation at the location of one of the unknown forces (Fleft) so that torque is eliminated, thus simplifying the calculations. Fleftrleft Frightrright Fboardrboard Fboxrbox 0 Fleftrleft Frightrright mboardgrboard mboxgrbox 0 1.25 m 0.50 m 2 Fleft(0) Fright(1.25 m 0.50 m) (4.25 kg)(9.80 m/s2) (6.00 kg)(9.80 m/s2)(1.25 m) 0 Fright 63 N Substituting this into the force equation, Fleft Fright Fboard Fbox 0 Fleft Fright Fboard Fbox Fright mboardg mboxg (63 N) (4.25 kg)(9.80 m/s2) Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. (6.00 kg)(9.80 m/s2) 37 N Level 3 88. A car’s specifications state that its weight distribution is 53 percent on the front tires and 47 percent on the rear tires. The wheel base is 2.46 m. Where is the car’s center of mass? Let the center of mass be x from the front of the car. Let the weight of the car be Fg. front rear Ffrontrfront Frearrrear (0.53 Fg)x (0.47 Fg)(2.46 m x) x 1.16 m Mixed Review pages 224–225 Level 1 89. A wooden door of mass, m, and length, l, is held horizontally by Dan and Ajit. Dan suddenly drops his end. Physics: Principles and Problems Solutions Manual 185 Chapter 8 continued a. What is the angular acceleration of the door just after Dan lets go? The torque is due to the gravitational force. The force at the center of mass is mg. Thus, Fleftrleft Fbagsrbags Fbagsrbags Fleft rleft I Then, left bags Fr sin 13ml 2 mg l(sin 90.0°) 1 2 13ml 2 3 gl 2 b. Is the acceleration constant? Explain. No; the angle between the door and the weight is changing, and therefore the torque is also changing. Thus, the acceleration changes. 2.43 m 3.6102 N Substitute this into the force equation. Fleft Fright Fbags 0 Fright Fleft Fbags 3.6102 N 10(175 N) 1.4102 N 91. Basketball A basketball is rolled down the court. A regulation basketball has a diameter of 24.1 cm, a mass of 0.60 kg, and a moment of inertia of 5.8103 kgm2. The basketball’s initial velocity is 2.5 m/s. a. What is its initial angular velocity? v r 2.5 m/s 1 (0.241 m) 2 21 rad/s b. The ball rolls a total of 12 m. How many revolutions does it make? d r d so r 16 rev 1.93 m 0.50 m ■ Figure 8-27 In equilibrium, the sum of the forces is zero and the sum of the torques is zero. Fleft Fright Fbags 0 left right bags 0 Choose the location of Fright as the axis of rotation to make that torque zero. 186 Solutions Manual 12 m rev 2 rad 1 2(0.241 m) c. What is its total angular displacement? d r d so r 12 m 12(0.241 m) 1.0102 rad Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 90. Topsoil Ten bags of topsoil, each weighing 175 N, are placed on a 2.43-m-long sheet of wood. They are stacked 0.50 m from one end of the sheet of wood, as shown in Figure 8-27. Two people lift the sheet of wood, one at each end. Ignoring the weight of the wood, how much force must each person exert? (10)(175 N)(0.50 m) Chapter 8 continued 92. The basketball in the previous problem stops rolling after traveling 12 m. a. If its acceleration was constant, what was its angular acceleration? v f2 vi2 2ad v 2 2rd a i Thus, (2.50 m) (1.25 s) 2.00 m/s vi2 so a 2d r d v t (2.5 m/s)2 1 2(0.241 m)(12 m) 2 2.2 rad/s2 b. What torque was acting on it as it was slowing down? I (5.8103 kgm2)(2.2 rad/s2) 1.3102 Nm c. What is the angular velocity of the cylinder? v r 2.00 m/s 12(50 m) 8102 rad/s 94. Hard Drive A hard drive on a modern computer spins at 7200 rpm (revolutions per minute). If the drive is designed to start from rest and reach operating speed in 1.5 s, what is the angular acceleration of the disk? Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 93. A cylinder with a 50 m diameter, as shown in Figure 8-28, is at rest on a surface. A rope is wrapped around the cylinder and pulled. The cylinder rolls without slipping. 50 m f i t t (7200 rpm 0 rpm) 2 rad 1 min 1.5 s rev 60 s 5.0102 rad/s2 95. Speedometers Most speedometers in automobiles measure the angular velocity of the transmission and convert it to speed. How will increasing the diameter of the tires affect the reading of the speedometer? Because increasing the diameter decreases the angular velocity, it will also decrease the reading of the speedometer. ■ Figure 8-28 a. After the rope has been pulled a distance of 2.50 m at a constant speed, how far has the center of mass of the cylinder moved? The center of mass is always over the point of contact with the surface for a uniform cylinder. Therefore the center of mass has moved 2.50 m. b. If the rope was pulled a distance of 2.50 m in 1.25 s, how fast was the center of mass of the cylinder moving? Physics: Principles and Problems 96. A box is dragged across the floor using a rope that is a distance h above the floor. The coefficient of friction is 0.35. The box is 0.50 m high and 0.25 m wide. Find the force that just tips the box. Let M equal the mass of the box. The center of mass of the box is 0.25 m above the floor. The box just tips when the torques on it are equal. rope friction Froperrope Ffrictionrfriction Solutions Manual 187 Chapter 8 continued F The longer piece of lumber would be easier to keep from rotating because it has a greater moment of inertia. r rrope friction friction Frope Mgr rrope friction 99. Surfboard Harris and Paul carry a surfboard that is 2.43 m long and weighs 143 N. Paul lifts one end with a force of 57 N. m/s2)(0.25 (0.35)M(9.80 m) h 0.25 m 2 2 a. What force must Harris exert? (0.86 m /s )M h 0.25 m Note that when you pull the box at the height of its center of mass, the denominator becomes zero. That is, you can pull with any amount of force and not tip the box. 97. The second hand on a watch is 12 mm long. What is the velocity of its tip? v r min (0.012 m)(2 rad/min) 60 s 2.44 m 1.22 m 1.22 m Figure 8-29 a. Which load is easier to lift? Why? The masses are the same, so the weights are the same. Thus, the same upward force is required to lift each load. b. Both you and your friend apply a torque with your hands to keep the lumber from rotating. Which load is easier to keep from rotating? Why? Solutions Manual 86 N b. What part of the board should Harris lift? Choose the point of rotation at the end where Paul lifts. H g FHrH Fgrg Fgrg rH (143 N) 2.43 m 2 86 N 2.0 m Thus, Harris has to lift 2.0 m from Paul’s end of the board. 100. A steel beam that is 6.50 m long weighs 325 N. It rests on two supports, 3.00 m apart, with equal amounts of the beam extending from each end. Suki, who weighs 575 N, stands on the beam in the center and then walks toward one end. How close to the end can she come before the beam begins to tip? Each support is 1.75 m from the end of the beam. Choose the point of rotation to be the support at the end closer to Suki. The center of mass of the beam is 1.50 m from that support. The beam will just begin to tip when Suki’s torque (S) equals the torque of the beam’s center of mass (cm) and the entire weight is on the support closest to Suki. S cm FSrS Fcmrcm Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Level 2 98. Lumber You buy a 2.44-m-long piece of 10 cm 10 cm lumber. Your friend buys a piece of the same size and cuts it into two lengths, each 1.22 m long, as shown in Figure 8-29. You each carry your lumber on your shoulders. 188 143 N 57 N FH 1.3103 m/s ■ FH Fg FP Chapter 8 continued F r FS cm cm rS (325 N) 3.00 m 2 575 N Cable 0.848 m That is Suki can move 0.848 m from the support or 1.75 0.848 0.90 m from the end. Axis of rotation Center of mass 25.0° 1.05 m 1.80 m Thinking Critically 2.10 m pages 225–226 101. Apply Concepts Consider a point on the edge of a rotating wheel. a. Under what conditions can the centripetal acceleration be zero? when 0.0 b. Under what conditions can the tangential (linear) acceleration be zero? when 0.0 c. Can the tangential acceleration be nonzero while the centripetal acceleration is zero? Explain. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. When 0.0 instantaneously, but is not zero, will keep changing. d. Can the centripetal acceleration be nonzero while the tangential acceleration is zero? Explain. Yes, as long as is constant but not zero. 102. Apply Concepts When you apply the brakes in a car, the front end dips. Why? The road exerts a force on the tires that brings the car to rest. The center of mass is above the road. Therefore, there is a net torque on the car, causing it to rotate in the direction that forces the front down. ■ Figure 8-30 We can use torques to find the vertical component (F Ty) of the tension. The counterclockwise torques are in equilibrium with the clockwise torques. ccw cw cable pole banner FTyrcable Fpolerpole Fbannerrbanner Fpolerpole Fbannerrbanner FTy rcable The total tension, then, is FTy Fpolerpole Fbannerrbanner sin 25° rcable sin 25° FT (175 N)(1.05 m) (105 N)(1.80 m) (2.10 m) sin 25° 420 N 103. Analyze and Conclude A banner is suspended from a horizontal, pivoted pole, as shown in Figure 8-30. The pole is 2.10 m long and weighs 175 N. The banner, which weighs 105 N, is suspended 1.80 m from the pivot point or axis of rotation. What is the tension in the cable supporting the pole? Physics: Principles and Problems Solutions Manual 189 Chapter 8 continued 104. Analyze and Conclude A pivoted lamp pole is shown in Figure 8-31. The pole weighs 27 N, and the lamp weighs 64 N. The total tension, then, is FTy FT sin 105° Fpolerpole Flamprlamp rrope sin 105° Rope y 105.0° 0.33 m 0.44 m Lamp ■ 0.44 m 2 (0.44 m)(sin 105°) 64 N x Axis of rotation (27 N) (64 N)(0.33 m) Figure 8-31 a. What is the torque caused by each force? g Fgr sin 105. Analyze and Conclude Gerald and Evelyn carry the following objects up a flight of stairs: a large mirror, a dresser, and a television. Evelyn is at the front end, and Gerald is at the bottom end. Assume that both Evelyn and Gerald exert only upward forces. a. Draw a free-body diagram showing Gerald and Evelyn exerting the same force on the mirror. Mirror (27 N)(0.22 m)(sin 90.0°) 5.9 Nm FE (64 N)(0.33 m)(sin 90.0°) FG rE lamp Flampr sin cm rG rE rG 21 Nm E G Use the torques to find the vertical component (F Ty ) of the tension. The counterclockwise torques are in equilibrium with the clockwise torques. ccw cw rope pole lamp Fg b. Draw a free-body diagram showing Gerald exerting more force on the bottom of the dresser. Dresser FG rE FE FTy rrope Fpolerpole Flamprlamp Fpolerpole Flamprlamp cm rE rG rG E G FTy rrope Fg 190 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. b. Determine the tension in the rope supporting the lamp pole. Chapter 8 continued c. Where would the center of mass of the television have to be so that Gerald carries all the weight? directly above where Gerald is lifting TV cm FG Cumulative Review page 226 108. Two blocks, one of mass 2.0 kg and the other of mass 3.0 kg, are tied together with a massless rope. This rope is strung over a massless, resistance-free pulley. The blocks are released from rest. Find the following. (Chapter 4) a. the tension in the rope The tension on the rope is Fg T m2g m2a T m3g m3a Thus, Writing in Physics Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. page 226 106. Astronomers know that if a satellite is too close to a planet, it will be torn apart by tidal forces. That is, the difference in the gravitational force on the part of the satellite nearest the planet and the part farthest from the planet is stronger than the forces holding the satellite together. Do research on the Roche limit and determine how close the Moon would have to orbit Earth to be at the Roche limit. For a planet and a moon with identical densities, the Roche limit is 2.446 times the radius of the planet. Earth’s Roche limit is 18,470 km. m m m3 m2 3 2 a g Substitute into the first equation. 2m m m2 m3 2 3 T g 2(2 kg)(3 kg) (9.80 m/s2) 2 kg 3 kg 24 N b. the acceleration of the blocks. Substitute into the equation for a. m m m3 m2 3 2 a g 3 kg 2 kg (9.80 m/s2) 3 kg 2 kg 1.96 m/s2 107. Automobile engines are rated by the torque that they produce. Research and explain why torque is an important quantity to measure. The force exerted by the ground on the tire accelerates the car. This force is produced by the engine. It creates the force by rotating the axle. The torque is equal to the force on the edge of the tire multiplied by the radius of the tire. Gears in the transmission may cause the force to change, but they do not change the torque. Therefore, the amount of torque created by the engine is delivered to the wheels. 109. Eric sits on a see-saw. At what angle, relative to the vertical, will the component of his weight parallel to the plane be equal to one-third the perpendicular component of his weight? (Chapter 5) Fg, parallel Fg sin Fg, perpendicular Fg cos Fg, perpendicular 3Fg, parallel Fg, perpendicular 3 Fg, parallel Fg cos 1 3 Physics: Principles and Problems Fg sin tan tan1(3) 71.6° Solutions Manual 191 Chapter 8 continued 110. The pilot of a plane wants to reach an airport 325 km due north in 2.75 hours. A wind is blowing from the west at 30.0 km/h. What heading and airspeed should be chosen to reach the destination on time? (Chapter 6) Let g be the distance north to the airport, x be the westward deflection, and h be the actual distance traveled. First, find the heading as the angle traveled eastward from the northward path. v vy x tan 111. A 60.0-kg speed skater with a velocity of 18.0 m/s comes into a curve of 20.0-m radius. How much friction must be exerted between the skates and ice to negotiate the curve? (Chapter 6) 2 mv Ff Fnet r 2 (60.0 kg)(18.0 m/s) 972 N 20.0 m Challenge Problem page 208 Rank the objects shown in the diagram according to their moments of inertia about the indicated axes. All spheres have equal masses and all separations are the same. v vy x tan1 vxty tan1 dy b tan1 (30.0 km/h)(2.75 h) 325 km c d a A B 14.3° west of north The airspeed, then, should be vh2 vx2 vy2 C 1 2 vh (vx2 vy2) dy2 1 2 D Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. vx2 2 ty 2 1 (325 km) 2 (30.0 km/h)2 2 (2.75 h) 122 km/h 192 Solutions Manual Physics: Principles and Problems Momentum and Its Conservation CHAPTER 9 impulse Ft Practice Problems 9.1 (5.0103 N)(2.0 s) Impulse and Momentum pages 229–235 page 233 1. A compact car, with mass 725 kg, is moving at 115 km/h toward the east. Sketch the moving car. a. Find the magnitude and direction of its momentum. Draw an arrow on your sketch showing the momentum. 1.0104 Ns The impulse is directed westward and has a magnitude of 1.0104 Ns. b. Complete the “before” and “after” sketches, and determine the momentum and the velocity of the car now. Before After E pi pf v E p pi 2.32104 kgm/s eastward p mv Ft p pf pi (725 kg)(115 km/h) 1h 3600 s pf Ft pi 1.0104 kgm/s 2.32104 kgm/s b. A second car, with a mass of 2175 kg, has the same momentum. What is its velocity? 1.3104 kgm/s eastward pf mvf p m p 1.3104 kgm/s vf f v m 2175 kg 1 km 3600 s (2.32104 kgm/s) 1000 m 1h 65 km/h eastward 38.4 km/h eastward 2. The driver of the compact car in the previous problem suddenly applies the brakes hard for 2.0 s. As a result, an average force of 5.0103 N is exerted on the car to slow it down. t 2.0 s F 5.0103 N a. What is the change in momentum; that is, the magnitude and direction of the impulse, on the car? Physics: Principles and Problems 725 kg 18 m/s 3. A 7.0-kg bowling ball is rolling down the alley with a velocity of 2.0 m/s. For each impulse, shown in Figures 9-3a and 9-3b, find the resulting speed and direction of motion of the bowling ball. a. Force (N) Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 2.32104 kgm/s eastward b. 5 0 5 1 2 Force (N) 1000 m 1 km 5 0 5 Time (s) ■ 1 2 Time (s) Figure 9-3 Solutions Manual 193 Chapter 9 continued a. Ft pf pi mvf mvi 141 km Ft mv vf f m (5.0 N)(1.0 s) (7.0 kg)(2.0 m/s) 125.0 km 7.0 kg a. What is the average force exerted on the person? 2.7 m/s in the same direction as the original velocity Ft p pf pi Ft mv b. vf f m p p t f i F (5.0 N)(1.0 s) (7.0 kg)(2.0 m/s) p mv t f i F 7.0 kg 1.3 m/s in the same direction as the original velocity (0.0 kgm/s) (60.0 kg)(94 km/h) 0.20 s 1000 m 1h 1 km 3600 s 4. The driver accelerates a 240.0-kg snowmobile, which results in a force being exerted that speeds up the snowmobile from 6.00 m/s to 28.0 m/s over a time interval of 60.0 s. 7.8103 N opposite to the direction of motion b. Some people think that they can stop their bodies from lurching forward in a vehicle that is suddenly braking by putting their hands on the dashboard. Find the mass of an object that has a weight equal to the force you just calculated. Could you lift such a mass? Are you strong enough to stop your body with your arms? a. Sketch the event, showing the initial and final situations. Before After vi vf x Fg 7.8103 N 2 m 2 8.010 kg g p Ft Such a mass is too heavy to lift. You cannot safely stop yourself with your arms. m(vf vi) (240.0 kg)(28.0 m/s 6.00 m/s) 5.28103 kgm/s c. What is the magnitude of the average force that is exerted on the snowmobile? p 5.28103 kgm/s F t 60.0 s 88.0 N 5. Suppose a 60.0-kg person was in the vehicle that hit the concrete wall in Example Problem 1. The velocity of the person equals that of the car both before and after the crash, and the velocity changes in 0.20 s. Sketch the problem. Solutions Manual 9.80 m/s Section Review 9.1 Impulse and Momentum pages 229–235 page 235 6. Momentum Is the momentum of a car traveling south different from that of the same car when it travels north at the same speed? Draw the momentum vectors to support your answer. Yes, momentum is a vector quantity, and the momenta of the two cars are in opposite directions. Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Fg mg b. What is the snowmobile’s change in momentum? What is the impulse on the snowmobile? 194 65.0 km Chapter 9 continued p 11.1 kgm/s N W 11.1 Ns E d. If the bat and softball are in contact for 0.80 ms, what is the average force that the bat exerts on the ball? S Ft m(vf vi) p 7. Impulse and Momentum When you jump from a height to the ground, you let your legs bend at the knees as your feet hit the floor. Explain why you do this in terms of the physics concepts introduced in this chapter. You reduce the force by increasing the length of time it takes to stop the motion of your body. 8. Momentum Which has more momentum, a supertanker tied to a dock or a falling raindrop? Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. The raindrop has more momentum, because a supertanker at rest has zero momentum. 9. Impulse and Momentum A 0.174-kg softball is pitched horizontally at 26.0 m/s. The ball moves in the opposite direction at 38.0 m/s after it is hit by the bat. m(v v ) t f i F (0.174 kg)(38.0 m/s (26.0 m/s)) 1s (0.80 ms) 1000 ms 1.4104 N 10. Momentum The speed of a basketball as it is dribbled is the same when the ball is going toward the floor as it is when the ball rises from the floor. Is the basketball’s change in momentum equal to zero when it hits the floor? If not, in which direction is the change in momentum? Draw the basketball’s momentum vectors before and after it hits the floor. No, the change in momentum is upward. Before the ball hits the floor, its momentum vector is downward. After the ball hits the floor, its momentum vector is upward. Before After a. Draw arrows showing the ball’s momentum before and after the bat hits it. Before vf p After p vi vi y vf x b. What is the change in momentum of the ball? p m(vf vi) (0.174 kg) (38.0 m/s (26.0 m/s)) 11.1 kgm/s c. What is the impulse delivered by the bat? Ft pf pi p Physics: Principles and Problems 11. Angular Momentum An ice-skater spins with his arms outstretched. When he pulls his arms in and raises them above his head, he spins much faster than before. Did a torque act on the ice-skater? If not, how could his angular velocity have increased? No torque acted on him. By drawing his arms in and keeping them close to the axis of rotation, he decreased his moment of inertia. Because the angular momentum did not change, the skater’s angular velocity increased. Solutions Manual 195 Chapter 9 continued 12. Critical Thinking An archer shoots arrows at a target. Some of the arrows stick in the target, while others bounce off. Assuming that the masses of the arrows and the velocities of the arrows are the same, which arrows produce a bigger impulse on the target? Hint: Draw a diagram to show the momentum of the arrows before and after hitting the target for the two instances. The ones that bounce off give more impulse because they end up with some momentum in the reverse direction, meaning they have a larger change in momentum. m v (mP mG) P Pi vf (0.105 kg)(24 m/s) 0.034 m/s (0.105 kg 75 kg) 15. A 35.0-g bullet strikes a 5.0-kg stationary piece of lumber and embeds itself in the wood. The piece of lumber and bullet fly off together at 8.6 m/s. What was the original speed of the bullet? mbvbi mwvwi (mb mw)vf where vf is the common final speed of the bullet and piece of lumber. Because vwi 0.0 m/s, Practice Problems 9.2 Conservation of Momentum pages 236–245 page 238 13. Two freight cars, each with a mass of 3.0105 kg, collide and stick together. One was initially moving at 2.2 m/s, and the other was at rest. What is their final speed? pi pf (m m )v mb b w f vbi (0.0350 kg 5.0 kg)(8.6 m/s) 0.0350 kg 1.2103 m/s 16. A 35.0-g bullet moving at 475 m/s strikes a 2.5-kg bag of flour that is on ice, at rest. The bullet passes through the bag, as shown in Figure 9-7, and exits it at 275 m/s. How fast is the bag moving when the bullet exits? v v 2 Ai Bi vf 275 m/s 2.2 m/s 0.0 m/s 2 1.1 m/s 14. A 0.105-kg hockey puck moving at 24 m/s is caught and held by a 75-kg goalie at rest. With what speed does the goalie slide on the ice? pPi pGi pPf pGf mPvPi mGvGi mPvPf mGvGf Because vGi 0.0 kgm/s, 196 ■ mBvBi mFvFi mBvBf mFvFf where vFi 0.0 m/s (m v m v ) mF mPvPi (mP mG)vf B Bi B Bf vFf where vf vPf vGf is the common final speed of the goalie and the puck. B Bi Bf vFf Solutions Manual Figure 9-7 m (v v ) mF Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. mvAi mvBi 2mvf Chapter 9 continued kg)(475 m/s 275 m/s) (0.0350 2.5 kg 2.8 m/s 17. The bullet in the previous problem strikes a 2.5-kg steel ball that is at rest. The bullet bounces backward after its collision at a speed of 5.0 m/s. How fast is the ball moving when the bullet bounces backward? The system is the bullet and the ball. mbulletvbullet, i mballvball, i mbulletvbullet, f mballvball, f vball, i 0.0 m/s and vbullet, f 5.0 m/s mbullet(vbullet, i vbullet, f) (0.0350 kg)(475 m/s (5.0 m/s)) so vball, f mball 2.5 kg 6.7 m/s 18. A 0.50-kg ball that is traveling at 6.0 m/s collides head-on with a 1.00-kg ball moving in the opposite direction at a speed of 12.0 m/s. The 0.50-kg ball bounces backward at 14 m/s after the collision. Find the speed of the second ball after the collision. Say that the first ball (ball C) is initially moving in the positive (forward) direction. mCvCi mDvDi mCvCf mDvDf m v m v m v mD C Ci D Di C Cf so vDf (0.50 kg)(6.0 m/s) (1.00 kg)(12.0 m/s) (0.50 kg)(14 m/s) Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1.00 kg 2.0 m/s, or 2.0 m/s in the opposite direction page 240 19. A 4.00-kg model rocket is launched, expelling 50.0 g of burned fuel from its exhaust at a speed of 625 m/s. What is the velocity of the rocket after the fuel has burned? Hint: Ignore the external forces of gravity and air resistance. pri pfuel, i prf pfuel, f where prf pfuel, f 0.0 kgm/s If the initial mass of the rocket (including fuel) is mr 4.00 kg, then the final mass of the rocket is mrf 4.00 kg 0.0500 kg 3.95 kg 0.0 kgm/s mrfvrf mfuelvfuel, f m v fuel fuel, f vrf mrf (0.0500 kg)(625 m/s) 3.95 kg 7.91 m/s Physics: Principles and Problems Solutions Manual 197 Chapter 9 continued 20. A thread holds a 1.5-kg cart and a 4.5-kg cart together. After the thread is burned, a compressed spring pushes the carts apart, giving the 1.5-kg cart a speed of 27 cm/s to the left. What is the velocity of the 4.5-kg cart? Let the 1.5-kg cart be represented by “C” and the 4.5-kg cart be represented by “D”. pCi pDi pCf pDf where pCi pDi 0.0 kgm/s mDvDf mCvCf m v mD C Cf so vDf (1.5 kg)(27 cm/s) 4.5 kg 9.0 cm/s to the right 21. Carmen and Judi dock a canoe. 80.0-kg Carmen moves forward at 4.0 m/s as she leaves the canoe. At what speed and in what direction do the canoe and Judi move if their combined mass is 115 kg? pCi pJi pCf pJf where pCi pJi 0.0 kgm/s mCvCf mJvJf m v mJ C Cf so vJf (80.0 kg)(4.0 m/s) 115 kg page 243 22. A 925-kg car moving north at 20.1 m/s collides with a 1865-kg car moving west at 13.4 m/s. The two cars are stuck together. In what direction and at what speed do they move after the collision? Before: pi, y myvi, y (925 kg)(20.1 m/s) 1.86104 kgm/s pi, x mxvi, x (1865 kg)(13.4 m/s) 2.50104 kgm/s pf, y pi, y pf, x pi, x pf pi 2 (pf, x)2 (pf , y) 198 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 2.8 m/s in the opposite direction Chapter 9 continued (2.5 kgm /s)2 (1.86 104 kgm/ s)2 010 4 3.12104 kgm/s p m1 m2 f vf 4 3.1210 kgm/s 11.2 m/s (925 kg 1865 kg) pf, y tan1 pf, x 1.86104 kgm/s 2.5010 kgm/s tan1 4 36.6° north of west 23. A 1383-kg car moving south at 11.2 m/s is struck by a 1732-kg car moving east at 31.3 m/s. The cars are stuck together. How fast and in what direction do they move immediately after the collision? Before: pi, x p1, x p2, x 0 m2v2i pi, y p1, y p2, y m1v1i 0 pf pi Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. p1, x2 pi, y2 2 (m2v2 m1v1i )2 i) ( p m1 m2 f vf v )2 (m v )2 (m 2 2i 1 1i m1 m2 2 kg)(31.3 m/s)) ((138 3 kg)(11.2 m/s))2 ((1732 1383 kg 1782 kg 18.1 m/s pi, y m v m2v2i 1 1i tan1 tan1 tan1 15.9° pi, x (1383 kg)(11.2 m/s) (1732 kg)(31.3 m/s) south of east 24. A stationary billiard ball, with a mass of 0.17 kg, is struck by an identical ball moving at 4.0 m/s. After the collision, the second ball moves 60.0° to the left of its original direction. The stationary ball moves 30.0° to the right of the moving ball’s original direction. What is the velocity of each ball after the collision? pCi pDi pCf pDf where pCi 0.0 kgm/s mC mD m 0.17 kg Physics: Principles and Problems Solutions Manual 199 Chapter 9 continued vDf Df Di vDi Ci 60.0 30.0 Cf vCf Vector Diagram pDi 30.0 60.0 pCf pDf The vector diagram provides final momentum equations for the ball that is initially stationary, C, and the ball that is initially moving, D. pCf pDi sin 60.0° pDf pDi cos 60.0° We can use the momentum equation for the stationary ball to find its final velocity. pCf pDi sin 60.0° mvCf mvDi sin 60.0° vCf vDi sin 60.0° (4.0 m/s)(sin 60.0°) 3.5 m/s, 30.0° to the right We can use the momentum equation for the moving ball to find its velocity. pDf pDi cos 60.0° mvDf mvDi cos 60.0° vDf vDi cos 60.0° 2.0 m/s, 60.0° to the left 25. A 1345-kg car moving east at 15.7 m/s is struck by a 1923-kg car moving north. They are stuck together and move with an initial velocity of 14.5 m/s at 63.5°. Was the north-moving car exceeding the 20.1 m/s speed limit? Before: pi, x m1v1, i (1345 kg)(15.7 m/s) 2.11104 kgm/s pf pi (m1 m2)vf (1345 kg 1923 kg)(14.5 m/s) 4.74104 kgm/s pf, y pf sin (4.74104 kgm/s)(sin 63.5°) 4.24104 kgm/s 200 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. (4.0 m/s)(cos 60.0°) Chapter 9 continued pf, y pi, y m2v2, i where pf, y 4.24104 kgm/s v2, i a and m m2 1923 kg F Fg m g Fg sin 22.1 m/s so, a g sin Fg/g Yes, it was exceeding the speed limit. The velocity and acceleration of the cart are related by the motion equation, v 2 vi2 2a(d di) with vi 0 and di 0. Thus, Section Review 9.2 Conservation of Momentum pages 236–245 v 2 2ad page 245 26. Angular Momentum The outer rim of a plastic disk is thick and heavy. Besides making it easier to catch, how does this affect the rotational properties of the plastic disk? Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Most of the mass of the disk is located at the rim, thereby increasing its moment of inertia. Therefore, when the disk is spinning, its angular momentum is larger than it would be if more mass were near the center. With a larger angular momentum, the disk flies through the air with more stability. 27. Speed A cart, weighing 24.5 N, is released from rest on a 1.00-m ramp, inclined at an angle of 30.0° as shown in Figure 9-14. The cart rolls down the incline and strikes a second cart weighing 36.8 N. 24 .5 N 1.0 0 m 36.8 N 30.0° ■ Figure 9-14 a. Calculate the speed of the first cart at the bottom of the incline. The force parallel to the surface of the ramp is F Fg sin Physics: Principles and Problems v 2ad (2)(g s in )(d) (2)(9.8 0 m/s2 )(sin 30.0°) (1.00 m) 3.13 m/s b. If the two carts stick together, with what initial speed will they move along? mCvCi (mC mC)vf m v mC mD C Ci so, vf F gC vCi FC FD g g F v FC FD C Ci (24.5 N)(3.13 m/s) 24.5 N 36.8 N 1.25 m/s 28. Conservation of Momentum During a tennis serve, the racket of a tennis player continues forward after it hits the ball. Is momentum conserved in the collision? Explain, making sure that you define the system. No, because the mass of the racket is much larger than that of the ball, only a small change in its velocity is required. In addition, it is being held by a massive, moving arm that is attached to a body in contact with Earth. Thus, the racket and ball do not comprise an isolated system. 29. Momentum A pole-vaulter runs toward the Solutions Manual 201 Chapter 9 continued launch point with horizontal momentum. Where does the vertical momentum come from as the athlete vaults over the crossbar? Product is the The vertical momentum comes from the impulsive force of Earth against the pole. Earth acquires an equal and opposite vertical momentum. 30. Initial Momentum During a soccer game, two players come from opposite directions and collide when trying to head the ball. They come to rest in midair and fall to the ground. Describe their initial momenta. Because their final momenta are zero, their initial momenta were equal and opposite. 31. Critical Thinking You catch a heavy ball while you are standing on a skateboard, and then you roll backward. If you were standing on the ground, however, you would be able to avoid moving while catching the ball. Explain both situations using the law of conservation of momentum. Explain which system you use in each case. Chapter Assessment Concept Mapping page 250 32. Complete the following concept map using the following terms: mass, momentum, average force, time over which the force is exerted. impulse Produces a change in momentum Product is the mass velocity Mastering Concepts page 250 33. Can a bullet have the same momentum as a truck? Explain. (9.1) Yes, for a bullet to have the same momentum as a truck, it must have a higher velocity because the two masses are not the same. mbulletvbullet mtruckvtruck 34. A pitcher throws a curve ball to the catcher. Assume that the speed of the ball doesn’t change in flight. (9.1) a. Which player exerts the larger impulse on the ball? The pitcher and the catcher exert the same amount of impulse on the ball, but the two impulses are in opposite directions. b. Which player exerts the larger force on the ball? The catcher exerts the larger force on the ball because the time interval over which the force is exerted is smaller. 35. Newton’s second law of motion states that if no net force is exerted on a system, no acceleration is possible. Does it follow that no change in momentum can occur? (9.1) No net force on the system means no net impulse on the system and no net change in momentum. However, individual parts of the system may have a 202 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. In the case of the skateboard, the ball, the skateboard, and you are an isolated system, and the momentum of the ball is shared. In the second case, unless Earth is included, there is an external force, so momentum is not conserved. If Earth’s large mass is included in the system, the change in its velocity is negligible. time over which the force is exerted average force Chapter 9 continued change in momentum as long as the net change in momentum is zero. 36. Why are cars made with bumpers that can be pushed in during a crash? (9.1) Cars are made with bumpers that compress during a crash to increase the time of a collision, thereby reducing the force. 37. An ice-skater is doing a spin. (9.1) a. How can the skater’s angular momentum be changed? by applying an external torque b. How can the skater’s angular velocity be changed without changing the angular momentum? by changing the moment of inertia 38. What is meant by “an isolated system?” (9.2) Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. An isolated system has no external forces acting on it. 39. A spacecraft in outer space increases its velocity by firing its rockets. How can hot gases escaping from its rocket engine change the velocity of the craft when there is nothing in space for the gases to push against? (9.2) Momentum is conserved. The change in momentum of gases in one direction must be balanced by an equal change in momentum of the spacecraft in the opposite direction. 40. A cue ball travels across a pool table and collides with the stationary eight ball. The two balls have equal masses. After the collision, the cue ball is at rest. What must be true regarding the speed of the eight ball? (9.2) The eight ball must be moving with the same velocity that the cue ball had just before the collision. 41. Consider a ball falling toward Earth. (9.2) a. Why is the momentum of the ball not conserved? Physics: Principles and Problems The momentum of a falling ball is not conserved because a net external force, gravity, is acting on it. b. In what system that includes the falling ball is the momentum conserved? One such system in which total momentum is conserved includes the ball plus Earth. 42. A falling basketball hits the floor. Just before it hits, the momentum is in the downward direction, and after it hits the floor, the momentum is in the upward direction. (9.2) a. Why isn’t the momentum of the basketball conserved even though the bounce is a collision? The floor is outside the system, so it exerts an external force, and therefore, an impulse on the ball. b. In what system is the momentum conserved? Momentum is conserved in the system of ball plus Earth. 43. Only an external force can change the momentum of a system. Explain how the internal force of a car’s brakes brings the car to a stop. (9.2) The external force of a car’s brakes can bring the car to a stop by stopping the wheels and allowing the external frictional force of the road against the tires to stop the car. If there is no friction— for example, if the road is icy—then there is no external force and the car does not stop. 44. Children’s playgrounds often have circularmotion rides. How could a child change the angular momentum of such a ride as it is turning? (9.2) The child would have to exert a torque on it. He or she could stand next to it and exert a force tangential to the circle on the handles as they go past. He or she also could run at the ride and jump onboard. Solutions Manual 203 Chapter 9 continued Applying Concepts pages 250–251 45. Explain the concept of impulse using physical ideas rather than mathematics. A force, F, exerted on an object over a time, t, causes the momentum of the object to change by the quantity Ft. 46. Is it possible for an object to obtain a larger impulse from a smaller force than it does from a larger force? Explain. Yes, if the smaller force acts for a long enough time, it can provide a larger impulse. 47. Foul Ball You are sitting at a baseball game when a foul ball comes in your direction. You prepare to catch it bare-handed. To catch it safely, should you move your hands toward the ball, hold them still, or move them in the same direction as the moving ball? Explain. After time A, the object moves with a constant, positive velocity. After time B, the object is at rest. After time C, the object moves with a constant, negative velocity. 50. During a space walk, the tether connecting an astronaut to the spaceship breaks. Using a gas pistol, the astronaut manages to get back to the ship. Use the language of the impulsemomentum theorem and a diagram to explain why this method was effective. When the gas pistol is fired in the opposite direction, it provides the impulse needed to move the astronaut toward the spaceship. p You should move your hands in the same direction the ball is traveling to increase the time of the collision, thereby reducing the force. The bullet is in the rifle a longer time, so the momentum it gains is larger. 49. An object initially at rest experiences the impulses described by the graph in Figure 9-15. Describe the object’s motion after impulses A, B, and C. Force (N) 2 A 0 2 6 4 8 B 2 Time (s) ■ 204 Figure 9-15 Solutions Manual 10 11 C 51. Tennis Ball As a tennis ball bounces off a wall, its momentum is reversed. Explain this action in terms of the law of conservation of momentum. Define the system and draw a diagram as a part of your explanation. Before After p p vi vf pEarth and wall +x Consider the system to be the ball, the wall, and Earth. The wall and Earth gain some momentum in the collision. 52. Imagine that you command spaceship Zeldon, which is moving through interplanetary space at high speed. How could you Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 48. A 0.11-g bullet leaves a pistol at 323 m/s, while a similar bullet leaves a rifle at 396 m/s. Explain the difference in exit speeds of the two bullets, assuming that the forces exerted on the bullets by the expanding gases have the same magnitude. p Chapter 9 continued slow your ship by applying the law of conservation of momentum? By shooting mass in the form of exhaust gas, at high velocity in the same direction in which you are moving, its momentum would cause the ship’s momentum to decrease. 53. Two trucks that appear to be identical collide on an icy road. One was originally at rest. The trucks are stuck together and move at more than half the original speed of the moving truck. What can you conclude about the contents of the two trucks? If the two trucks had equal masses, they would have moved off at half the speed of the moving truck. Thus, the moving truck must have had a more massive load. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 54. Explain, in terms of impulse and momentum, why it is advisable to place the butt of a rifle against your shoulder when first learning to shoot. When held loosely, the recoil momentum of the rifle works against only the mass of the rifle, thereby producing a larger velocity and striking your shoulder. The recoil momentum must work against the mass of the rifle and you, resulting in a smaller velocity. 55. Bullets Two bullets of equal mass are shot at equal speeds at blocks of wood on a smooth ice rink. One bullet, made of rubber, bounces off of the wood. The other bullet, made of aluminum, burrows into the wood. In which case does the block of wood move faster? Explain. Momentum is conserved, so the momentum of the block and bullet after the collision equals the momentum of the bullet before the collision. The rubber bullet has a negative momentum after impact, with respect to the block, so the block’s momentum must be greater in this case. Mastering Problems 9.1 Impulse and Momentum pages 251–252 Level 1 56. Golf Rocío strikes a 0.058-kg golf ball with a force of 272 N and gives it a velocity of 62.0 m/s. How long was Rocío’s club in contact with the ball? mv (0.058 kg)(62.0 m/s) t F 272 N 0.013 s 57. A 0.145-kg baseball is pitched at 42 m/s. The batter hits it horizontally to the pitcher at 58 m/s. a. Find the change in momentum of the ball. Take the direction of the pitch to be positive. p mvf mvi m(vf vi) (0.145 kg)(58 m/s (42 m/s)) 14 kgm/s b. If the ball and bat are in contact for 4.6104 s, what is the average force during contact? Ft p p t F m(v v ) t i f (0.145 kg)(58 m/s (42 m/s)) 4.6104 s 3.2104 N 58. Bowling A force of 186 N acts on a 7.3-kg bowling ball for 0.40 s. What is the bowling ball’s change in momentum? What is its change in velocity? p Ft (186 N)(0.40 s) 74 Ns 74 kgm/s p m v Ft m Physics: Principles and Problems Solutions Manual 205 Chapter 9 continued 62. Hockey A hockey player makes a slap shot, exerting a constant force of 30.0 N on the hockey puck for 0.16 s. What is the magnitude of the impulse given to the puck? (186 N)(0.40 s) 7.3 kg 1.0101 m/s 59. A 5500-kg freight truck accelerates from 4.2 m/s to 7.8 m/s in 15.0 s by the application of a constant force. a. What change in momentum occurs? p mv m(vf vi) (5500 kg)(7.8 m/s 4.2 m/s) 2.0104 kgm/s b. How large of a force is exerted? p t F Ft (30.0 N)(0.16 s) 4.8 Ns 63. Skateboarding Your brother’s mass is 35.6 kg, and he has a 1.3-kg skateboard. What is the combined momentum of your brother and his skateboard if they are moving at 9.50 m/s? p mv (mboy mboard)v (35.6 kg 1.3 kg)(9.50 m/s) m(vf vi) t 3.5102 kgm/s (5500 kg)(7.8 m/s 4.2 m/s) 15.0 s 1.3103 N 60. In a ballistics test at the police department, Officer Rios fires a 6.0-g bullet at 350 m/s into a container that stops it in 1.8 ms. What is the average force that stops the bullet? p t 64. A hockey puck has a mass of 0.115 kg and is at rest. A hockey player makes a shot, exerting a constant force of 30.0 N on the puck for 0.16 s. With what speed does it head toward the goal? Ft mv m(vf vi) where vi 0 Ft Thus vf F m (30.0 N)(0.16 s) (0.0060 kg)(0.0 m/s 350 m/s) 3 42 m/s 0.115 kg 1.810 1.2103 s N 61. Volleyball A 0.24-kg volleyball approaches Tina with a velocity of 3.8 m/s. Tina bumps the ball, giving it a speed of 2.4 m/s but in the opposite direction. What average force did she apply if the interaction time between her hands and the ball was 0.025 s? mv F t (0.24 kg)(2.4 m/s 3.8 m/s) 0.025 s 6.0101 N 65. Before a collision, a 25-kg object was moving at 12 m/s. Find the impulse that acted on the object if, after the collision, it moved at the following velocities. a. 8.0 m/s Ft mv m(vf vi) (25 kg)(8.0 m/s 12 m/s) 1.0102 kgm/s b. 8.0 m/s Ft mv m(vf vi) (25 kg)(8.0 m/s 12 m/s) 5.0102 kgm/s 206 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. m(v v ) t f i Chapter 9 continued Ft mv 2 Force (N) Level 2 66. A 0.150-kg ball, moving in the positive direction at 12 m/s, is acted on by the impulse shown in the graph in Figure 9-16. What is the ball’s speed at 4.0 s? 0 1 2 3 4 2 Area of graph mv Time (s) 1 (2.0 N)(2.0 s) m(vf vi) 2 ■ Figure 9-16 2.0 Ns (0.150 kg)(vf 12 m/s) 2.0 kgm/s vf 12 m/s 0.150 kg 25 m/s 67. Baseball A 0.145-kg baseball is moving at 35 m/s when it is caught by a player. a. Find the change in momentum of the ball. p m(vf vi) (0.145 kg)(0.0 m/s 35 m/s) 5.1 kgm/s b. If the ball is caught with the mitt held in a stationary position so that the ball stops in 0.050 s, what is the average force exerted on the ball? p Faveraget p t m(v v ) t Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. f i so, Faverage (0.145 kg)(0.0 m/s 35 m/s) 0.500 s 1.0102 N c. If, instead, the mitt is moving backward so that the ball takes 0.500 s to stop, what is the average force exerted by the mitt on the ball? p Faveraget p t m(v v ) t f i so, Faverage (0.145 kg)(0.0 m/s 35 m/s) 0.500 s 1.0101 N 68. Hockey A hockey puck has a mass of 0.115 kg and strikes the pole of the net at 37 m/s. It bounces off in the opposite direction at 25 m/s, as shown in Figure 9-17. a. What is the impulse on the puck? Ft m(vf vi) 0.115 kg (0.115 kg)(25 m/s 37 m/s) 7.1 kgm/s 25 m/s ■ Physics: Principles and Problems Figure 9-17 Solutions Manual 207 Chapter 9 continued b. If the collision takes 5.0104 s, what is the average force on the puck? Ft m(vf vi) m(v v ) t f i F (0.115 kg)(25 m/s 37 m/s) 4 5.010 s 1.4104 N 69. A nitrogen molecule with a mass of 4.71026 kg, moving at 550 m/s, strikes the wall of a container and bounces back at the same speed. a. What is the impulse the molecule delivers to the wall? Ft m(vf vi) (4.71026 kg)(550 m/s 550 m/s) 5.21023 kgm/s The impulse the wall delivers to the molecule is 5.21023 kgm/s. The impulse the molecule delivers to the wall is 5.21023 kgm/s. b. If there are 1.51023 collisions each second, what is the average force on the wall? Ft m(vf vi) m(v v ) t f i F For all the collisions, the force is m(v v ) t f i Ftotal (1.51023) 7.8 N Level 3 70. Rockets Small rockets are used to make tiny adjustments in the speeds of satellites. One such rocket has a thrust of 35 N. If it is fired to change the velocity of a 72,000-kg spacecraft by 63 cm/s, how long should it be fired? Ft mv mt so, t F (72,000 kg)(0.63 m/s) 35 N 1.3103 s, or 22 min 208 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. (4.71026 kg)(550 m/s 550 m/s) 1.0 s (1.51023) Chapter 9 continued 71. An animal rescue plane flying due east at 36.0 m/s drops a bale of hay from an altitude of 60.0 m, as shown in Figure 9-18. If the bale of hay weighs 175 N, what is the momentum of the bale the moment before it strikes the ground? Give both magnitude and direction. 36.0 m/s 175 N First use projectile motion to find the velocity of the bale. 60.0 m p mv To find v, consider the horizontal and vertical components. vx 36.0 m/s vy2 viy2 2dg 2dg ■ Figure 9-18 Thus, 2 2 2 v v x v y v x 2dg The momentum, then, is F v 2 Fgv x 2dg g g g p (175 N)(36.0 (2)(60.0 m)(9.80 m/s ) m/s) 2 2 2 9.80 m/s 888 kgm/s The angle from the horizontal is vy tan vx Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 2dg v x 2) .0 m)(9.80 m/s (2)(60 36.0 m/s 43.6° 72. Accident A car moving at 10.0 m/s crashes into a barrier and stops in 0.050 s. There is a 20.0-kg child in the car. Assume that the child’s velocity is changed by the same amount as that of the car, and in the same time period. a. What is the impulse needed to stop the child? Ft mv m(vf vi) (20.0 kg)(0.0 m/s 10.0 m/s) 2.00102 kgm/s b. What is the average force on the child? Ft mv m(vf vi) m(v v ) t f i so, F (20.0 kg)(0.0 m/s 10.0 m/s) 0.050 s 4.0103 N Physics: Principles and Problems Solutions Manual 209 Chapter 9 continued c. What is the approximate mass of an object whose weight equals the force in part b? Before vFB vDT Fg mg Fg 4.0103 N so, m 2 g 9.80 m/s 4.1102 kg d. Could you lift such a weight with your arm? +x No. After vFB vDT 0 e. Why is it advisable to use a proper restraining seat rather than hold a child on your lap? You would not be able to protect a child on your lap in the event of a collision. 9.2 Conservation of Momentum pages 252–253 Level 1 73. Football A 95-kg fullback, running at 8.2 m/s, collides in midair with a 128-kg defensive tackle moving in the opposite direction. Both players end up with zero speed. +x b. What was the fullback’s momentum before the collision? pFB mFBvFB (95 kg)(8.2 m/s) 7.8102 kgm/s c. What was the change in the fullback’s momentum? pFB pf pFB 0 pFB 7.8102 kgm/s Before: mFB 95 kg vFB 8.2 m/s d. What was the change in the defensive tackle’s momentum? mDT 128 kg vDT ? After: m 223 kg 7.8102 kgm/s e. What was the defensive tackle’s original momentum? 7.8102 kgm/s vf 0 m/s f. How fast was the defensive tackle moving originally? mDTvDT 7.8102 kgm/s 7.810 kgm/s so, vDT 2 128 kg 6.1 m/s 210 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. a. Identify the “before” and “after” situations and draw a diagram of both. Chapter 9 continued 74. Marble C, with mass 5.0 g, moves at a speed of 20.0 cm/s. It collides with a second marble, D, with mass 10.0 g, moving at 10.0 cm/s in the same direction. After the collision, marble C continues with a speed of 8.0 cm/s in the same direction. a. Sketch the situation and identify the system. Identify the “before” and “after” situations and set up a coordinate system. Before: mC 5.0 g pDf pCi pDi pCf 1.00103 kgm/s 1.00103 kgm/s 4.0104 kgm/s 1.6103 kgm/s e. What is the speed of marble D after the collision? pDf mDvDf mD 10.0 g vCi 20.0 cm/s vDi 10.0 cm/s p mD Df so, vDf After: mC 5.0 g 3 mD 10.0 g 1.610 kgm/s 2 vCf 8.0 cm/s 1.6101 m/s 0.16 m/s vDf ? 16 cm/s 1.0010 Before vCi C Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. d. Calculate the momentum of marble D after the collision. pCi pDi pCf pDf D vDi +x C 75. Two lab carts are pushed together with a spring mechanism compressed between them. Upon release, the 5.0-kg cart repels one way with a velocity of 0.12 m/s, while the 2.0-kg cart goes in the opposite direction. What is the velocity of the 2.0-kg cart? m1vi m2vf m v m2 vf 1i After vCf kg D vDf (5.0 kg)(0.12 m/s) (2.0 kg) 0.30 m/s +x b. Calculate the marbles’ momenta before the collision. mCvCi (5.0103 kg)(0.200 m/s) 1.0103 kgm/s mDvDi (1.00102 kg)(0.100 m/s) 1.0103 kgm/s c. Calculate the momentum of marble C after the collision. mCvCf (5.0103 kg)(0.080 m/s) 76. A 50.0-g projectile is launched with a horizontal velocity of 647 m/s from a 4.65-kg launcher moving in the same direction at 2.00 m/s. What is the launcher’s velocity after the launch? pCi pDi pCf pDf mCvCi mDvDi mCvCf mDvDf m v m v m v mD C Ci D Di C Cf so, vDf Assuming that the projectile, C, is launched in the direction of the launcher, D, motion, 4.0104 kgm/s Physics: Principles and Problems Solutions Manual 211 Chapter 9 continued (0.0500 kg)(2.00 m/s) (4.65 kg)(2.00 m/s) (0.0500 kg)(647 m/s) vDf 4.65 kg 4.94 m/s, or 4.94 m/s backwards Level 2 77. A 12.0-g rubber bullet travels at a velocity of 150 m/s, hits a stationary 8.5-kg concrete block resting on a frictionless surface, and ricochets in the opposite direction with a velocity of 1.0102 m/s, as shown in Figure 9-19. How fast will the concrete block be moving? mCvCi mDvDi mCvCf mDvDf 1.0102 m/s mCvCi mDvDi mCvCf vDf mD 8.5 kg 12.0 g ■ Figure 9-19 since the block is initially at rest, this becomes m (v v ) mD C Ci Cf vDf (0.0120 kg)(150 m/s (1.010 m/s)) 2 8.5 kg 0.35 m/s 78. Skateboarding Kofi, with mass 42.00 kg, is riding a skateboard with a mass of 2.00 kg and traveling at 1.20 m/s. Kofi jumps off and the skateboard stops dead in its tracks. In what direction and with what velocity did he jump? where vsf 0 and vLi vsi vi (m m )v mL L s i Thus vLf (42.00 kg 2.00 kg)(1.20 m/s) 42.00 kg 1.26 m/s in the same direction as she was riding 79. Billiards A cue ball, with mass 0.16 kg, rolling at 4.0 m/s, hits a stationary eight ball of similar mass. If the cue ball travels 45° above its original path and the eight ball travels 45° below the horizontal, as shown in Figure 9-20, what is the velocity of each ball after the collision? 212 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. (mLvLi msvsi)vi mLvLf msvsf Chapter 9 continued 80. A 2575-kg van runs into the back of an 825-kg compact car at rest. They move off together at 8.5 m/s. Assuming that the friction with the road is negligible, calculate the initial speed of the van. 45° pCi pDi pCf pDf mCvCi (mC mD)vf 45° C D so, vCi m m mC (2575 kg 825 kg)(8.5 m/s) vf 2575 kg 11 m/s ■ Level 3 81. In-line Skating Diego and Keshia are on in-line skates and stand face-to-face, then push each other away with their hands. Diego has a mass of 90.0 kg and Keshia has a mass of 60.0 kg. Figure 9-20 pCi 45 p8f pCf Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. We can get momentum equations from the vector diagram. a. Sketch the event, identifying the “before” and “after” situations, and set up a coordinate axis. Before: mK 60.0 kg pCf pCi sin 45° mD 90.0 kg mCvCf mCvCi sin 45° vi 0.0 m/s vCf vCi sin 45° (4.0 m/s)(sin 45°) 2.8 m/s For the eight ball, p8f pci cos 45° After: mK 60.0 kg mD 90.0 kg vKf ? vDf ? m8v8f mCvCi(cos 45°) where m8 mC. Thus, v8f vCi cos 45° (4.0 m/s)(cos 45°) 2.8 m/s Physics: Principles and Problems Solutions Manual 213 Chapter 9 continued Before vKi vDi 0 +x After vDf vKf +x b. Find the ratio of the skaters’ velocities just after their hands lose contact. pKi pDi 0.0 kgm/s pKf pDf so, mKvKf mDvDf 0.0 kgm/s Thus, the ratios of the velocities are mD vKf 90.0 kg 1.50 vDf mK 60.0 kg The negative sign shows that the velocities are in opposite directions. c. Which skater has the greater speed? Keshia, who has the smaller mass, has the greater speed. d. Which skater pushed harder? The forces were equal and opposite. 82. A 0.200-kg plastic ball moves with a velocity of 0.30 m/s. It collides with a second plastic ball of mass 0.100 kg, which is moving along the same line at a speed of 0.10 m/s. After the collision, both balls continue moving in the same, original direction. The speed of the 0.100-kg ball is 0.26 m/s. What is the new velocity of the 0.200-kg ball? mCvCi mDvDi mCvCf mDvDf m v m v m v mC C Ci D Di D Df so, vCf 214 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. and mKvKf mDvDf Chapter 9 continued (0.200 kg)(0.30 m/s) (0.100 kg)(0.10 m/s) (0.100 kg)(0.26 m/s) 0.200 kg 0.22 m/s in the original direction Mixed Review pages 253–254 Level 1 83. A constant force of 6.00 N acts on a 3.00-kg object for 10.0 s. What are the changes in the object’s momentum and velocity? The change in momentum is p Ft (6.00 N)(10.0 s) 60.0 Ns 60.0 kgm/s The change in velocity is found from the impulse. Ft mv Ft v m (6.00 N)(10.0 s) 3.00 kg 20.0 m/s 84. The velocity of a 625-kg car is changed from 10.0 m/s to 44.0 m/s in 68.0 s by an external, constant force. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. a. What is the resulting change in momentum of the car? p mv m(vf vi) (625 kg)(44.0 m/s 10.0 m/s) 2.12104 kgm/s b. What is the magnitude of the force? Ft mv mv so, F t m(v v ) t f i (625 kg)(44.0 m/s 10.0 m/s) 68.0 s 313 N 85. Dragster An 845-kg dragster accelerates on a race track from rest to 100.0 km/h in 0.90 s. a. What is the change in momentum of the dragster? p m(vf vi) (845 kg)(100.0 km/h 0.0 km/h) 1000 m 1 km 1h 3600 s 2.35104 kgm/s Physics: Principles and Problems Solutions Manual 215 Chapter 9 continued b. What is the average force exerted on the dragster? m(v v ) t f i F (845 kg)(100.0 km/h 0.0 km/h) 1000 m 1h 0.90 s 1 km 3600 s 2.6104 N c. What exerts that force? The force is exerted by the track through friction. Level 2 86. Ice Hockey A 0.115-kg hockey puck, moving at 35.0 m/s, strikes a 0.365-kg jacket that is thrown onto the ice by a fan of a certain hockey team. The puck and jacket slide off together. Find their velocity. mPvPi (mP mJ)vf m m mP mJ P Pi vf (0.115 kg)(35.0 m/s) (0.115 kg 0.365 kg) 8.39 m/s 87. A 50.0-kg woman, riding on a 10.0-kg cart, is moving east at 5.0 m/s. The woman jumps off the front of the cart and lands on the ground at 7.0 m/s eastward, relative to the ground. a. Sketch the “before” and “after” situations and assign a coordinate axis to them. mc 10.0 kg vi 5.0 m/s After: mw 50.0 kg mc 10.0 kg vwf 7.0 m/s vcf ? Before After vi vwf vCf +x 216 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Before: mw 50.0 kg Chapter 9 continued b. Find the cart’s velocity after the woman jumps off. (mw mc)vi mwvwf mcvcf (m m )v m v mc w c i w wf so, vcf (50.0 kg 10.0 kg)(5.0 m/s) (50.0 kg)(7.0 m/s) 10.0 kg 5.0 m/s, or 5.0 m/s west 88. Gymnastics Figure 9-21 shows a gymnast performing a routine. First, she does giant swings on the high bar, holding her body straight and pivoting around her hands. Then, she lets go of the high bar and grabs her knees with her hands in the tuck position. Finally, she straightens up and lands on her feet. a. In the second and final parts of the gymnast’s routine, around what axis does she spin? She spins around the center of mass of her body, first in the tuck position and then also as she straightens out. b. Rank in order, from greatest to least, her moments of inertia for the three positions. giant swing (greatest), straight, tuck (least) c. Rank in order, from greatest to least, her angular velocities in the three positions. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. tuck (greatest), straight, giant swing (least) ■ Figure 9-21 Level 3 89. A 60.0-kg male dancer leaps 0.32 m high. a. With what momentum does he reach the ground? v v02 2dg Thus, the velocity of the dancer is v 2dg Physics: Principles and Problems Solutions Manual 217 Chapter 9 continued and his momentum is p mv m2dg (60.0 kg)(2)(0.3 2 m)(9.80 m /s2) 1.5102 kgm/s down b. What impulse is needed to stop the dancer? Ft mv m(vf vi) To stop the dancer, vf 0. Thus, Ft mvf p 1.5102 kgm/s up c. As the dancer lands, his knees bend, lengthening the stopping time to 0.050 s. Find the average force exerted on the dancer’s body. Ft mv m2dg m2dg so, F t (60.0 kg)(2)(0.3 2 m)(9.80 m /s2) 0.050 s 3.0103 N d. Compare the stopping force with his weight. Fg mg (60.0 kg)(9.80 m/s2) 5.98102 N The force is about five times the weight. Thinking Critically a. Sketch the event, identifying the “before” and “after” situations. Before: mA 92 kg mB 75 kg mC 75 kg vAi 5.0 m/s vBi 2.0 m/s vCi 4.0 m/s After: mA 92 kg mB 75 kg mC 75 kg vf ? 218 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. page 254 90. Apply Concepts A 92-kg fullback, running at 5.0 m/s, attempts to dive directly across the goal line for a touchdown. Just as he reaches the line, he is met head-on in midair by two 75-kg linebackers, both moving in the direction opposite the fullback. One is moving at 2.0 m/s and the other at 4.0 m/s. They all become entangled as one mass. Chapter 9 continued Before vAi vBi After vCi vf +x b. What is the velocity of the football players after the collision? pAi pBi pCi pAf pBf pCf mAvAi mBvBi mCvCi mAvAf mBvBf mCvCf (mA mB mC)vf m v m v m v mA mB mC A Ai B Bi Ci Ci vf (92 kg)(5.0 m/s) (75 kg)(2.0 m/s) (75 kg)(4.0 m/s) 92 kg 75 kg 75 kg 0.041 m/s c. Does the fullback score a touchdown? Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Yes. The velocity is positive, so the football crosses the goal line for a touchdown. 91. Analyze and Conclude A student, holding a bicycle wheel with its axis vertical, sits on a stool that can rotate without friction. She uses her hand to get the wheel spinning. Would you expect the student and stool to turn? If so, in which direction? Explain. The student and the stool would spin slowly in the direction opposite to that of the wheel. Without friction there are no external torques. Thus, the angular momentum of the system is not changed. The angular momentum of the student and stool must be equal and opposite to the angular momentum of the spinning wheel. 92. Analyze and Conclude Two balls during a collision are shown in Figure 9-22, which is drawn to scale. The balls enter from the left, collide, and then bounce away. The heavier ball, at the bottom of the diagram, has a mass of 0.600 kg, and the other has a mass of 0.400 kg. Using a vector diagram, determine whether momentum is conserved in this collision. Explain any difference in the momentum of the system before and after the collision. ■ Figure 9-22 Dotted lines show that the changes of momentum for each ball are equal and opposite: (mAvA) (mBvB). Because the masses have a 3:2 ratio, a 2:3 ratio of velocity changes will compensate. Physics: Principles and Problems Solutions Manual 219 Chapter 9 continued vB1 mB 0.400 kg vB vB2 vB1 vA2 vA1 vA mA 0.600 kg vA1 Writing in Physics page 254 93. How can highway barriers be designed to be more effective in saving people’s lives? Research this issue and describe how impulse and change in momentum can be used to analyze barrier designs. 94. While air bags save many lives, they also have caused injuries and even death. Research the arguments and responses of automobile makers to this statement. Determine whether the problems involve impulse and momentum or other issues. There are two ways an air bag reduces injury. First, an air bag extends the time over which the impulse acts, thereby reducing the force. Second, an air bag spreads the force over a larger area, thereby reducing the pressure. Thus, the injuries due to forces from small objects are reduced. The dangers of air bags mostly center on the fact that air bags must be inflated very rapidly. The surface of an air bag can approach the passenger at speeds of up to 322 km/h (200 mph). Injuries can occur when the moving bag collides with the person. 220 Solutions Manual Cumulative Review page 254 95. A 0.72-kg ball is swung vertically from a 0.60-m string in uniform circular motion at a speed of 3.3 m/s. What is the tension in the cord at the top of the ball’s motion? (Chapter 6) The tension is the gravitational force minus the centripetal force. FT Fg Fc v2 mv 2 mg m g r r 2 (3.3 m/s) (0.72 kg)(9.80 m/s2) 0.60 m 6.0 N 96. You wish to launch a satellite that will remain above the same spot on Earth’s surface. This means the satellite must have a period of exactly one day. Calculate the radius of the circular orbit this satellite must have. Hint: The Moon also circles Earth and both the Moon and the satellite will obey Kepler’s third law. The Moon is 3.9108 m from Earth and its period is 27.33 days. (Chapter 7) T 2 s Tm r 3 rm s T 2 Tm 1 s so rs rm3 3 2 1 1.000 day (3.9108 m)3 3 27.33 days 4.3107 m 97. A rope is wrapped around a drum that is 0.600 m in diameter. A machine pulls with a constant 40.0 N force for a total of 2.00 s. In that time, 5.00 m of rope is unwound. Find , at 2.00 s, and I. (Chapter 8) The angular acceleration is the ratio of the linear acceleration of the drum’s edge and drum’s radius. Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. The change in a car’s momentum does not depend on how it is brought to a stop. Thus, the impulse also does not change. To reduce the force, the time over which a car is stopped must be increased. Using barriers that can extend the time it takes to stop a car will reduce the force. Flexible, plastic containers filled with sand often are used. Systems are being developed that will adjust the rate at which gases fill the air bags to match the size of the person. Chapter 9 continued a r The linear acceleration is found from the equation of motion. 1 x at 2 2 2x a 2 t Thus, the angular acceleration is a 2x 2 r rt (2)(5.00 m) 0.60 0m (2.00 s)2 2 8.33 rad/s2 At the end of 2.00 s, the angular velocity is t 2xt 2 rt 2x rt (2)(5.00 m) 0.60 0 m (2.00 s) 2 16.7 rad/s The moment of inertia is Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. I Fr sin r2tx2 Fr t sin 2 2 2x (40.0 N) (2.00 s)2(sin 90.0°) 0.600 m 2 2 (2)(5.00 m) 1.44 kgm2 Challenge Problem page 244 Your friend was driving her 1265-kg car north on Oak Street when she was hit by a 925-kg compact car going west on Maple Street. The cars stuck together and slid 23.1 m at 42° north of west. The speed limit on both streets is 22 m/s (50 mph). Assume that momentum was conserved during the collision and that acceleration was constant during the skid. The coefficient of kinetic friction between the tires and the pavement is 0.65. Physics: Principles and Problems Solutions Manual 221 Chapter 9 continued y x 925 kg 42° 1265 kg 1. Your friend claims that she wasn’t speeding, but that the driver of other car was. How fast was your friend driving before the crash? The vector diagram provides a momentum equation for the friend’s car. pCi pf sin 42° The friend’s initial velocity, then, is (m m )v sin 42° mC p mC Ci C D f vCi We can find vf first by finding the acceleration and time of the skid. The acceleration is Fg (m m )g mC mD F C D a g m m The time can be derived from the distance equation. 2 t 2d 2d g a The final velocity, then, is 2d vf at g 2dg g Using this, we now can find the friend’s initial velocity. (m m )v sin 42° mC C D f vCi (m m )2dg sin 42° C D m C (1265 kg 925 kg)(2)(23 .1 m)(0.65)( 9.80 m /s2)(sin 42°) 1265 kg 2.0101 m/s 222 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1 d at 2 Chapter 9 continued 2. How fast was the other car moving before the crash? Can you support your friend’s case in court? From the vector diagram, the momentum equation for the other car is pDi pf cos 42° (mC mD)vf cos 42° (mC mD)2dg (cos 42°) The other car’s initial velocity, then, is, p mD Di vDi (m m )2dg (sin 42°) mD C D (1265 kg 925 kg)(2)(23 m)(0.65)( m /s2) (cos 42°) .1 9.80 925 kg 3.0101 m/s Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. The friend was not exceeding the 22 m/s speed limit. The other car was exceeding the speed limit. Physics: Principles and Problems Solutions Manual 223 Energy, Work, and Simple Machines CHAPTER 10 Practice Problems 10.1 Energy and Work pages 257–265 page 261 1. Refer to Example Problem 1 to solve the following problem. a. If the hockey player exerted twice as much force, 9.00 N, on the puck, how would the puck’s change in kinetic energy be affected? Because W Fd and KE W, doubling the force would double the work, which would double the change in kinetic energy to 1.35 J. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. b. If the player exerted a 9.00 N-force, but the stick was in contact with the puck for only half the distance, 0.075 m, what would be the change in kinetic energy? Because W Fd, halving the distance would cut the work in half, which also would cut the change in kinetic energy in half, to 0.68 J. 2. Together, two students exert a force of 825 N in pushing a car a distance of 35 m. a. How much work do the students do on the car? W Fd (825 N)(35 m) 2.9104 J b. If the force was doubled, how much work would they do pushing the car the same distance? W Fd (2)(825 N)(35 m) 5.8104 J which is twice as much work 3. A rock climber wears a 7.5-kg backpack while scaling a cliff. After 30.0 min, the climber is 8.2 m above the starting point. a. How much work does the climber do on the backpack? W Fd mgd (7.5 kg)(9.80 m/s2)(8.2 m) 6.0102 J b. If the climber weighs 645 N, how much work does she do lifting herself and the backpack? W Fd 6.0102 J (645 N)(8.2 m) 6.0102 J 5.9103 J c. What is the average power developed by the climber? 5.9103 J 30.0 min P W t 1 min 60 s 3.3 W page 262 4. If the sailor in Example Problem 2 pulled with the same force, and along the same distance, but at an angle of 50.0°, how much work would he do? W Fd cos (255 N)(30.0 m)(cos 50.0°) 4.92103 J 5. Two people lift a heavy box a distance of 15 m. They use ropes, each of which makes an angle of 15° with the vertical. Each person exerts a force of 225 N. How much work do they do? W Fd cos (2)(225 N)(15 m)(cos 15°) 6.5103 J Physics: Principles and Problems Solutions Manual 225 Chapter 10 continued 6. An airplane passenger carries a 215-N suitcase up the stairs, a displacement of 4.20 m vertically, and 4.60 m horizontally. a. How much work does the passenger do? Since gravity acts vertically, only the vertical displacement needs to be considered. W Fd (215 N)(4.20 m) 903 J b. The same passenger carries the same suitcase back down the same set of stairs. How much work does the passenger do now? W Fd (25 N)(275 m) 6.9103 J b. How much work is done by the force of gravity on the bike? The force is downward (90°), and the displacement is 25° above the horizontal or 115° from the force. W Fd cos mgd cos (13 kg)(9.80 m/s2)(275 m) (cos 115°) Force is upward, but vertical displacement is downward, so W Fd cos (215 N)(4.20 m)(cos 180.0°) 903 J 7. A rope is used to pull a metal box a distance of 15.0 m across the floor. The rope is held at an angle of 46.0° with the floor, and a force of 628 N is applied to the rope. How much work does the force on the rope do? W Fd cos (628 N)(15.0 m)(cos 46.0°) 6.54103 J 1.5104 J page 264 9. A box that weighs 575 N is lifted a distance of 20.0 m straight up by a cable attached to a motor. The job is done in 10.0 s. What power is developed by the motor in W and kW? W t (575 N)(20.0 m) 10.0 s Fd t P 1.15103 W 1.15 kW 10. You push a wheelbarrow a distance of 60.0 m at a constant speed for 25.0 s, by exerting a 145-N force horizontally. W t Fd t (145 N)(60.0 m) 25.0 s P 348 W b. If you move the wheelbarrow twice as fast, how much power is developed? t is halved, so P is doubled to 696 W. m 275 11. What power does a pump develop to lift 35 L of water per minute from a depth of 110 m? (1 L of water has a mass of 1.00 kg.) mgd t W t P mt gd N 25 ■ 25° Figure 10-4 (Not to scale) a. How much work does the rider do on the bike? Force and displacement are in the same direction. m where (35 L /min)(1.00 kg/L) t Thus, P mt gd (35 L /min)(1.00 kg/L)(9.80 m/s2) (110 m)(1 min/60s) 0.63 kW 226 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. a. What power do you develop? 8. A bicycle rider pushes a bicycle that has a mass of 13 kg up a steep hill. The incline is 25° and the road is 275 m long, as shown in Figure 10-4. The rider pushes the bike parallel to the road with a force of 25 N. Chapter 10 continued 12. An electric motor develops 65 kW of power as it lifts a loaded elevator 17.5 m in 35 s. How much force does the motor exert? W t 1 2 (15 m)(210.0 N 40.0 N) 1.9103 J Fd t P d 1.3105 N 13. A winch designed to be mounted on a truck, as shown in Figure 10-7, is advertised as being able to exert a 6.8103-N force and to develop a power of 0.30 kW. How long would it take the truck and the winch to pull an object 15 m? 210.0 Force (N) (65103 W)(35 s) 17.5 m Pt F 40.0 0 15 Displacement (m) Section Review 10.1 Energy and Work pages 257–265 page 265 15. Work Murimi pushes a 20-kg mass 10 m across a floor with a horizontal force of 80 N. Calculate the amount of work done by Murimi. W Fd (80 N)(10 m) 8102 J The mass is not important to this problem. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. ■ Figure 10-7 W Fd P t t Fd P t (6.8103 N)(15 m) (0.3010 W) 340 s 3 5.7 min 14. Your car has stalled and you need to push it. You notice as the car gets going that you need less and less force to keep it going. Suppose that for the first 15 m, your force decreased at a constant rate from 210.0 N to 40.0 N. How much work did you do on the car? Draw a force-displacement graph to represent the work done during this period. The work done is the area of the trapezoid under the solid line: 1 2 W d(F1 F2) Physics: Principles and Problems 16. Work A mover loads a 185-kg refrigerator into a moving van by pushing it up a 10.0-m, friction-free ramp at an angle of inclination of 11.0°. How much work is done by the mover? y (10.0 m)(sin 11.0°) 1.91 m W Fd mgd sin (185 kg)(9.80 m/s2)(10.0 m)(sin 11.0°) 3.46103 J 17. Work and Power Does the work required to lift a book to a high shelf depend on how fast you raise it? Does the power required to lift the book depend on how fast you raise it? Explain. No, work is not a function of time. However, power is a function of time, so the power required to lift the book does depend on how fast you raise it. Solutions Manual 227 Chapter 10 continued 18. Power An elevator lifts a total mass of 1.1103 kg a distance of 40.0 m in 12.5 s. How much power does the elevator generate? W t mgd t Fd t P (1.1103 kg)(9.80 m/s2)(40.0 m) 12.5 s 3.4104 W 19. Work A 0.180-kg ball falls 2.5 m. How much work does the force of gravity do on the ball? W Fgd mgd (0.180 kg)(9.80 m/s2)(2.5 m) 4.4 J 20. Mass A forklift raises a box 1.2 m and does 7.0 kJ of work on it. What is the mass of the box? W Fd mgd 7.0103 J (9.80 m/s )(1.2 m) W so m 2 gd 6.0102 kg Both do the same amount of work. Only the height lifted and the vertical force exerted count. Since work is the change in kinetic energy, calculate the work done by each force. The work can be positive, negative, or zero, depending on the relative angles of the force and displacement of the object. The sum of the three works is the change in energy of the system. Practice Problems 10.2 Machines pages 266–273 page 272 24. If the gear radius in the bicycle in Example Problem 4 is doubled, while the force exerted on the chain and the distance the wheel rim moves remain the same, what quantities change, and by how much? re 8.00 cm 35.6 cm IMA 0.225 (doubled) rr MA IMA (0.225) e 100 95.0 100 0.214 (doubled) Fr MA so Fr (MA)(Fe) Fe (0.214)(155 N) 33.2 N de IMA dr so de (IMA)(dr) (0.225)(14.0 cm) 22. Work and Kinetic Energy If the work done on an object doubles its kinetic energy, does it double its velocity? If not, by what ratio does it change the velocity? Kinetic energy is proportional to the square of the velocity, so doubling the energy doubles the square of the velocity. The velocity increases by a factor of the square root of 2, or 1.4. 3.15 cm 25. A sledgehammer is used to drive a wedge into a log to split it. When the wedge is driven 0.20 m into the log, the log is separated a distance of 5.0 cm. A force of 1.7104 N is needed to split the log, and the sledgehammer exerts a force of 1.1104 N. a. What is the IMA of the wedge? de (0.20 m) (0.050 m) IMA 4.0 dr 228 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 21. Work You and a friend each carry identical boxes from the first floor of a building to a room located on the second floor, farther down the hall. You choose to carry the box first up the stairs, and then down the hall to the room. Your friend carries it down the hall on the first floor, then up a different stairwell to the second floor. Who does more work? 23. Critical Thinking Explain how to find the change in energy of a system if three agents exert forces on the system at once. Chapter 10 continued b. What is the MA of the wedge? Fr MA Fe (1.7104 N) (1.110 N) 1.5 4 27. You exert a force of 225 N on a lever to raise a 1.25103-N rock a distance of 13 cm. If the efficiency of the lever is 88.7 percent, how far did you move your end of the lever? Wo c. Calculate the efficiency of the wedge as a machine. efficiency 100 Wi Frdr 100 MA IMA e 100 Fede 1.5 4.0 100 38% 26. A worker uses a pulley system to raise a 24.0-kg carton 16.5 m, as shown in Figure 10-14. A force of 129 N is exerted, and the rope is pulled 33.0 m. Frdr(100) So de Fe(efficiency) (1.25103 N)(0.13 m)(100) (225 N)(88.7) 0.81 m 28. A winch has a crank with a 45-cm radius. A rope is wrapped around a drum with a 7.5-cm radius. One revolution of the crank turns the drum one revolution. 33.0 m 24.0 kg a. What is the ideal mechanical advantage of this machine? Compare effort and resistance distances for 1 rev: 16.5 m dr Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 129 N ■ Figure 10-14 a. What is the MA of the pulley system? Fr mg Fe MA Fe de (24.0 kg)(9.80 m/s2) 129 N (2)45 cm (2)7.5 cm IMA 6.0 b. If, due to friction, the machine is only 75 percent efficient, how much force would have to be exerted on the handle of the crank to exert 750 N of force on the rope? efficiency 100 MA IMA Fr 100 1.82 b. What is the efficiency of the system? MA efficiency 100 IMA (MA)(100) d e dr (Fe)(IMA) (Fr)(100) so Fe (efficiency)(IMA) (750 N)(100) (75)(6.0) 1.7102 N (MA)(d )(100) de r (1.82)(16.5 m)(100) 33.0 m 91.0% Physics: Principles and Problems Solutions Manual 229 Chapter 10 continued Section Review 10.2 Machines pages 266–273 page 273 29. Simple Machines Classify the tools below as a lever, a wheel and axle, an inclined plane, a wedge, or a pulley. a. screwdriver wheel and axle b. pliers lever c. chisel wedge de (3)(2r) 2r IMA dr (3)(2)(45 cm) 2)(7.5 cm) ( 18 32. Efficiency Suppose you increase the efficiency of a simple machine. Do the MA and IMA increase, decrease, or remain the same? Either MA increases while IMA remains the same, or IMA decreases while MA remains the same, or MA increases while IMA decreases. 33. Critical Thinking The mechanical advantage of a multi-gear bicycle is changed by moving the chain to a suitable rear gear. d. nail puller lever 30. IMA A worker is testing a multiple pulley system to estimate the heaviest object that he could lift. The largest downward force he could exert is equal to his weight, 875 N. When the worker moves the rope 1.5 m, the object moves 0.25 m. What is the heaviest object that he could lift? Fr MA Fe Assuming the efficiency is 100%, de MA IMA (Fe) dr (1.5 m) (0.25 m) (875 N) 5.2103 N 31. Compound Machines A winch has a crank on a 45-cm arm that turns a drum with a 7.5-cm radius through a set of gears. It takes three revolutions of the crank to rotate the drum through one revolution. What is the IMA of this compound machine? rgear large, to increase IMA rwheel b. As you reach your traveling speed, you want to rotate the pedals as few times as possible. Should you choose a small or large gear? Small, because less chain travel, hence few pedal revolutions, will be required per wheel revolution. c. Many bicycles also let you choose the size of the front gear. If you want even more force to accelerate while climbing a hill, would you move to a larger or smaller front gear? smaller, to increase pedal-front gear IMA because rpedal IMA rfront gear The IMA of the system is the product of the IMA of each machine. For the crank and drum, the ratio of distances is 2(45 cm) 6.0. 2(7.5 cm) 230 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. so Fr (MA)(Fe) a. To start out, you must accelerate the bicycle, so you want to have the bicycle exert the greatest possible force. Should you choose a small or large gear? Chapter 10 continued Chapter Assessment Concept Mapping W J/s page 278 34. Create a concept map using the following terms: force, displacement, direction of motion, work, change in kinetic energy. Force Displacement 39. What is a watt equivalent to in terms of kilograms, meters, and seconds? (10.1) Direction of motion Nm/s (kgm/s2)m/s kgm2/s3 40. Is it possible to get more work out of a machine than you put into it? (10.2) no, e 100% Work Change in kinetic energy Mastering Concepts page 278 35. In what units is work measured? (10.1) joules Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 36. Suppose a satellite revolves around Earth in a circular orbit. Does Earth’s gravity do any work on the satellite? (10.1) No, the force of gravity is directed toward Earth and is perpendicular to the direction of displacement of the satellite. 37. An object slides at constant speed on a frictionless surface. What forces act on the object? What work is done by each force? (10.1) Only gravity and an upward, normal force act on the object. No work is done because the displacement is perpendicular to these forces. There is no force in the direction of displacement because the object is sliding at a constant speed. 38. Define work and power. (10.1) Work is the product of force and the distance over which an object is moved in the direction of the force. Power is the time rate at which work is done. Physics: Principles and Problems 41. Explain how the pedals of a bicycle are a simple machine. (10.2) Pedals transfer force from the rider to the bike through a wheel and axle. Applying Concepts page 278 42. Which requires more work, carrying a 420-N backpack up a 200-m-high hill or carrying a 210-N backpack up a 400-m-high hill? Why? Each requires the same amount of work because force times distance is the same. 43. Lifting You slowly lift a box of books from the floor and put it on a table. Earth’s gravity exerts a force, magnitude mg, downward, and you exert a force, magnitude mg, upward. The two forces have equal magnitudes and opposite directions. It appears that no work is done, but you know that you did work. Explain what work was done. You do positive work on the box because the force and motion are in the same direction. Gravity does negative work on the box because the force of gravity is opposite to the direction of motion. The work done by you and by gravity are separate and do not cancel each other. 44. You have an after-school job carrying cartons of new copy paper up a flight of stairs, and then carrying recycled paper back down the stairs. The mass of the paper does not Solutions Manual 231 Chapter 10 continued change. Your physics teacher says that you do not work all day, so you should not be paid. In what sense is the physics teacher correct? What arrangement of payments might you make to ensure that you are properly compensated? The net work is zero. Carrying the carton upstairs requires positive work; carrying it back down is negative work. The work done in both cases is equal and opposite because the distances are equal and opposite. The student might arrange the payments on the basis of the time it takes to carry paper, whether up or down, not on the basis of work done. 45. You carry the cartons of copy paper down the stairs, and then along a 15-m-long hallway. Are you working now? Explain. No, the force on the box is up and the displacement is down the hall. They are perpendicular and no work is done. 46. Climbing Stairs Two people of the same mass climb the same flight of stairs. The first person climbs the stairs in 25 s; the second person does so in 35 s. 48. How can you increase the ideal mechanical advantage of a machine? Increase the ratio of de/dr to increase the IMA of a machine. 49. Wedge How can you increase the mechanical advantage of a wedge without changing its ideal mechanical advantage? Reduce friction as much as possible to reduce the resistance force. 50. Orbits Explain why a planet orbiting the Sun does not violate the work-energy theorem. Assuming a circular orbit, the force due to gravity is perpendicular to the direction of motion. This means the work done is zero. Hence, there is no change in kinetic energy of the planet, so it does not speed up or slow down. This is true for a circular orbit. 51. Claw Hammer A claw hammer is used to pull a nail from a piece of wood, as shown in Figure 10-16. Where should you place your hand on the handle and where should the nail be located in the claw to make the effort force as small as possible? Both people are doing the same amount of work because they both are climbing the same flight of stairs and they have the same mass. b. Which person produces more power? Explain your answer. The person who climbs in 25 s expends more power, as less time is needed to cover the distance. 47. Show that power delivered can be written as P Fv cos . ■ Figure 10-16 Your hand should be as far from the head as possible to make de as large as possible. The nail should be as close to the head as possible to make dr as small as possible. W t P , but W Fd cos Fd cos t so, P d t because v , P Fv cos 232 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. a. Which person does more work? Explain your answer. Chapter 10 continued Mastering Problems 10.1 Energy and Work pages 278–280 Level 1 52. The third floor of a house is 8 m above street level. How much work is needed to move a 150-kg refrigerator to the third floor? W Fd mgd (150 kg)(9.80 m/s2)(8 m) 1104 J 53. Haloke does 176 J of work lifting himself 0.300 m. What is Haloke’s mass? Fd t (15.0 N)(2.51 m) 30.0 s 126 W 58. A student librarian lifts a 2.2-kg book from the floor to a height of 1.25 m. He carries the book 8.0 m to the stacks and places the book on a shelf that is 0.35 m above the floor. How much work does he do on the book? Only the net vertical displacement counts. W Fd mgd W Fd mgd; therefore, (2.2 kg)(9.80 m/s2)(0.35 m) 176 J W m 2 7.5 J gd (9.80 m/s )(0.300 m) 59.9 kg 54. Football After scoring a touchdown, an 84.0-kg wide receiver celebrates by leaping 1.20 m off the ground. How much work was done by the wide receiver in the celebration? W Fd mgd (84.0 kg)(9.80 m/s2)(1.20 m) Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. W t P 988 J 55. Tug-of-War During a tug-of-war, team A does 2.20105 J of work in pulling team B 8.00 m. What force was team A exerting? W Fd, so W d 2.20105 J 8.00 m F 2.75104 N 56. To keep a car traveling at a constant velocity, a 551-N force is needed to balance frictional forces. How much work is done against friction by the car as it travels from Columbus to Cincinnati, a distance of 161 km? W Fd (551 N)(1.61105 m) 8.87107 J 57. Cycling A cyclist exerts a force of 15.0 N as he rides a bike 251 m in 30.0 s. How much power does the cyclist develop? Physics: Principles and Problems 59. A force of 300.0 N is used to push a 145-kg mass 30.0 m horizontally in 3.00 s. a. Calculate the work done on the mass. W Fd (300.0 N)(30.0 m) 9.00103 J 9.00 kJ b. Calculate the power developed. W t 9.00103 J 3.00 s P 3.00103 W 3.00 kW Level 2 60. Wagon A wagon is pulled by a force of 38.0 N exerted on the handle at an angle of 42.0° with the horizontal. If the wagon is pulled in a circle of radius 25.0 m, how much work is done? W Fd cos (F)(2r) cos (38.0 N)(2)(25.0 m)(cos 42.0°) 4.44103 J 61. Lawn Mower Shani is pushing a lawn mower with a force of 88.0 N along a handle that makes an angle of 41.0° with the horizontal. How much work is done by Solutions Manual 233 Chapter 10 continued Shani in moving the lawn mower 1.2 km to mow the yard? W Fd cos 518 J (88.0 N)(1.2103 m)(cos 41.0°) 8.0104 J 62. A 17.0-kg crate is to be pulled a distance of 20.0 m, requiring 1210 J of work to be done. If the job is done by attaching a rope and pulling with a force of 75.0 N, at what angle is the rope held? W Fd cos (225 N)(1.15 m) sin 30.0° 65. Piano A 4.2103-N piano is to be slid up a 3.5-m frictionless plank at a constant speed. The plank makes an angle of 30.0° with the horizontal. Calculate the work done by the person sliding the piano up the plank. The force parallel to the plane is given by F F sin so W F d Fd sin W (4200 N)(3.5 m)(sin 30.0°) cos1 W Fd 7.4103 J cos1 1210 J (75.0 N)(20.0 m) 36.2° 63. Lawn Tractor A 120-kg lawn tractor, shown in Figure 10-17, goes up a 21° incline that is 12.0 m long in 2.5 s. Calculate the power that is developed by the tractor. 66. Sled Diego pulls a 4.5-kg sled across level snow with a force of 225 N on a rope that is 35.0° above the horizontal, as shown in Figure 10-18. If the sled moves a distance of 65.3 m, how much work does Diego do? 22 35.0° 4.5 kg 0 kg 120. 5N ■ W Fd cos 21° ■ (225 N)(65.3 m)(cos 35.0°) Figure 10-17 W t Fd sin t mgd sin t P (120 kg)(9.80 m/s2)(12.0 m)(sin 21°) 2.5 s 2.0103 W 2.0 kW 64. You slide a crate up a ramp at an angle of 30.0° by exerting a 225-N force parallel to the ramp. The crate moves at a constant speed. The coefficient of friction is 0.28. How much work did you do on the crate as it was raised a vertical distance of 1.15 m? F and d are parallel so h sin W Fd F 234 Solutions Manual Figure 10-18 1.20104 J 67. Escalator Sau-Lan has a mass of 52 kg. She rides up the escalator at Ocean Park in Hong Kong. This is the world’s longest escalator, with a length of 227 m and an average inclination of 31°. How much work does the escalator do on Sau-Lan? W Fd sin mgd sin (52 kg)(9.80 m/s2)(227 m)(sin 31°) 6.0104 J 68. Lawn Roller A lawn roller is pushed across a lawn by a force of 115 N along the direction of the handle, which is 22.5° Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 0m 12. Chapter 10 continued above the horizontal. If 64.6 W of power is developed for 90.0 s, what distance is the roller pushed? Fd cos t W t k .0 60 P so, Pt d F cos g m 2.0 N 0.0 40 1.0 m (64.6 W)(90.0 s) (115 N)(cos 22.5°) ■ 54.7 m 69. John pushes a crate across the floor of a factory with a horizontal force. The roughness of the floor changes, and John must exert a force of 20 N for 5 m, then 35 N for 12 m, and then 10 N for 8 m. a. How much work does Maricruz do in sliding the crate up the ramp? W Fd (400.0 N)(2.0 m) 8.0102 J b. How much work would be done if Maricruz simply lifted the crate straight up from the floor to the platform? W Fd mgd a. Draw a graph of force as a function of distance. (60.0 kg)(9.80 m/s2)(1.0 m) 5.9102 J 30 20 10 5 10 15 20 25 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Displacement (m) b. Find the work John does pushing the crate. W F1d1 F2d2 F3d3 (20 N)(5 m) (35 N)(12 m) (10 N)(8 m) 600 J 70. Maricruz slides a 60.0-kg crate up an inclined ramp that is 2.0-m long and attached to a platform 1.0 m above floor level, as shown in Figure 10-19. A 400.0-N force, parallel to the ramp, is needed to slide the crate up the ramp at a constant speed. 71. Boat Engine An engine moves a boat through the water at a constant speed of 15 m/s. The engine must exert a force of 6.0 kN to balance the force that the water exerts against the hull. What power does the engine develop? W t Fd t P Fv (6.0103 N)(15 m/s) 9.0104 W 9.0101 kW Level 3 72. In Figure 10-20, the magnitude of the force necessary to stretch a spring is plotted against the distance the spring is stretched. 8.00 Force (N) Force (N) 40 0 Figure 10-19 6.00 4.00 2.00 0.00 0.10 ■ Physics: Principles and Problems 0.20 0.30 Figure 10-20 Solutions Manual 235 Chapter 10 continued a. Calculate the slope of the graph, k, and show that F kd, where k 25 N/m. y x 5.00 N 0.00 N 0.20 m 0.00 m k 74. A worker pushes a crate weighing 93 N up an inclined plane. The worker pushes the crate horizontally, parallel to the ground, as illustrated in Figure 10-21. F1 kd1 Let d1 0.20 m From the graph, F1 is 5.00 N. 85 N 0 5. 3.0 m m F1 So k d1 4.0 m 5.00 N 0.20 m 25 N/m 93 N b. Find the amount of work done in stretching the spring from 0.00 m to 0.20 m by calculating the area under the graph from 0.00 m to 0.20 m. 1 2 A (base)(height) (0.20 m)(5.00 N) 1 2 0.50 J c. Show that the answer to part b can be calculated using the formula 1 W kd2, where W is the work, 2 W kd 2 (25 N/m)(0.20 m)2 1 2 1 2 0.50 J 73. Use the graph in Figure 10-20 to find the work needed to stretch the spring from 0.12 m to 0.28 m. Add the areas of the triangle and rectangle. The area of the triangle is: 1 1 bh (0.28 m 0.12 m)(7.00 N 3.00 N) 2 2 0.32 J The area of the rectangle is: bh (0.28 m 0.12 m)(3.00 N 0.00 N) 0.48 J Total work is 0.32 J 0.48 J 0.80 J 236 Solutions Manual Figure 10-21 a. The worker exerts a force of 85 N. How much work does he do? Displacement in direction of force is 4.0 m, so W Fd (85 N)(4.0 m) 3.4102 J b. How much work is done by gravity? (Be careful with the signs you use.) Displacement in direction of force is 3.0 m, so W Fd (93 N)(3.0 m) 2.8102 J c. The coefficient of friction is 0.20. How much work is done by friction? (Be careful with the signs you use.) W FNd (Fyou, Fg, )d 0.20(85 N)(sin ) (93 N)(cos )(5.0 m) 0.20(85 N) 3.0 5.0 (93 N)( 5.0 m) 4.0 5.0 1.3102 J (work done against friction) 75. Oil Pump In 35.0 s, a pump delivers 0.550 m3 of oil into barrels on a platform 25.0 m above the intake pipe. The oil’s density is 0.820 g/cm3. Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. k 25 N/m (the slope of the graph), and d is the distance the spring is stretched (0.20 m). ■ Chapter 10 continued a. Calculate the work done by the pump. The work done is W Fgd mgh 78. The graph in Figure 10-22 shows the force and displacement of an object being pulled. 1 kg 1000 g (0.550 m3)(0.820 g/cm3) (1.00106 cm3/m3)(9.80 m/s2) Force (N) (volume)(density)gh 40.0 20.0 (25.0 m) 1.10105 J b. Calculate the power produced by the pump. W t 1.10105 0.0 2.0 6.0 Displacement (m) J ■ P Figure 10-22 35.0 s 3.14103 W 3.14 kW 76. Conveyor Belt A 12.0-m-long conveyor belt, inclined at 30.0°, is used to transport bundles of newspapers from the mail room up to the cargo bay to be loaded onto delivery trucks. Each newspaper has a mass of 1.0 kg, and there are 25 newspapers per bundle. Determine the power that the conveyor develops if it delivers 15 bundles per minute. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 4.0 mgd W Fd P t t t (25 newspapers)(15 bundles/min) a. Calculate the work done to pull the object 7.0 m. Find the area under the curve (see graph): 0.0 to 2.0 m: 1 (20.0 N)(2.0 m) 2.0101 J 2 2.0 m to 3.0 m: 1 (30.0 N)(1.0 m) (20 N)(1.0 m) 35 J 2 3.0 m to 7.0 m: (50.0 N)(4.0 m) 2.0102 J Total work: (1.0 kg/newspaper)(9.80 m/s2) 2.0101 J 35 J 2.0102 J (12.0 m)(sin 30.0°)(1 min/60 s) 2.6102 J 3.7102 W 77. A car is driven at a constant speed of 76 km/h down a road. The car’s engine delivers 48 kW of power. Calculate the average force that is resisting the motion of the car. W t b. Calculate the power that would be developed if the work was done in 2.0 s. W t 2.6102 J 2.0 s P 1.3102 W Fd t P Fv P v so F 48,000 W 1 h 76 k m 1000 m 1 h 1 km 3600 s 2.3103 N Physics: Principles and Problems Solutions Manual 237 Chapter 10 continued 10.2 Machines pages 280–281 Level 1 79. Piano Takeshi raises a 1200-N piano a distance of 5.00 m using a set of pulleys. He pulls in 20.0 m of rope. a. How much effort force would Takeshi apply if this were an ideal machine? Fr de Fe dr 81. A pulley system lifts a 1345-N weight a distance of 0.975 m. Paul pulls the rope a distance of 3.90 m, exerting a force of 375 N. a. What is the ideal mechanical advantage of the system? de 3.90 m 0.975 m IMA 4.00 dr b. What is the mechanical advantage? Fr 1345 N MA 3.59 Frdr (1200 N)(5.00 m) so Fe 20.0 m de 3.0102 N Fe c. How efficient is the system? MA IMA efficiency 100 3.59 4.00 b. What force is used to balance the friction force if the actual effort is 340 N? 100 89.8% Fe Ff Fe, ideal Ff Fe Fe, ideal 340 N 3.0102 N 40 N c. What is the output work? Wo Frdr (1200 N)(5.00 m) 6.0103 J d. What is the input work? Wi Fede (340 N)(20.0 m) e. What is the mechanical advantage? Fr 1200 N MA 3.5 Fe 340 N 80. Lever Because there is very little friction, the lever is an extremely efficient simple machine. Using a 90.0-percent-efficient lever, what input work is required to lift an 18.0-kg mass through a distance of 0.50 m? 82. A force of 1.4 N is exerted through a distance of 40.0 cm on a rope in a pulley system to lift a 0.50-kg mass 10.0 cm. Calculate the following. a. the MA Fr mg Fe MA Fe (0.50 kg)(9.80 m/s2) 1.4 N 3.5 b. the IMA de 40.0 cm 10.0 cm IMA 4.00 dr c. the efficiency MA IMA efficiency 100 3.5 4.00 100 88% Wo efficiency 100 Wi (Wo)(100) (mgd)(100) 90.0 Wi efficiency (18.0 kg)(9.80 m/s2)(0.50 m)(100) 90.0 83. A student exerts a force of 250 N on a lever, through a distance of 1.6 m, as he lifts a 150-kg crate. If the efficiency of the lever is 90.0 percent, how far is the crate lifted? 98 J MA IMA e 90 100 Frdr 100 Fr Fe d 100 e dr Fede 238 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 6.8103 J 375 N Chapter 10 continued eFede eFede 100Fr 100mg b. What is the IMA of the ramp? so, dr de 18 m 4.5 m IMA 4.0 df (90.0)(250 N)(1.6 m) (100)(150 kg)(9.80 m/s2) c. What are the real MA and the efficiency of the ramp if a parallel force of 75 N is actually required? 0.24 m Level 2 84. What work is required to lift a 215-kg mass a distance of 5.65 m, using a machine that is 72.5 percent efficient? Wo Fr MA Fe mg (25 kg)(9.80 m/s2) 75 N Fe 3.3 MA IMA efficiency 100 e 100 Wi 3.3 4.0 100 82% Frdr 100 Wi 86. Bicycle Luisa pedals a bicycle with a gear radius of 5.00 cm and a wheel radius of 38.6 cm, as shown in Figure 10-24. If the wheel revolves once, what is the length of the chain that was used? mgdr 100 Wi mgdr Wi 100 e (215 kg)(9.80 m/s2)(5.65 m)(100) 72.5 1.64104 J 38.6 cm Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 85. The ramp in Figure 10-23 is 18 m long and 4.5 m high. 18 m 5.00 cm FA 4.5 m ■ Fg ■ Figure 10-24 d 2r 2(5.00 cm) 31.4 cm Figure 10-23 a. What force, parallel to the ramp (FA), is required to slide a 25-kg box at constant speed to the top of the ramp if friction is disregarded? W Fgd mgh mgh d so F Fg (25 kg)(9.80 m/s2)(4.5 m) 18 m 61 N Physics: Principles and Problems Level 3 87. Crane A motor with an efficiency of 88 percent operates a crane with an efficiency of 42 percent. If the power supplied to the motor is 5.5 kW, with what constant speed does the crane lift a 410-kg crate of machine parts? Total efficiency (88%)(42%) 37% Useful Power (5.5 kW)(37%) 2.0 kW 2.0103 W Solutions Manual 239 Chapter 10 continued c. If you move the effort side of the lever 12.0 cm, how far is the box lifted? P F Fv W t Fd t d t 3 2.010 W P P v 2 Fg mg (410 kg)(9.80 m/s ) 0.50 m/s de1 a. Show that the IMA of this compound machine is 6.0. Wi1 Wo1 Wi2 Wo2 Wi1 Wo2 Fe1de1 Fr2dr2 For the compound machine de1 IMAc dr2 de1 de2 IMA1 and IMA2 dr1 dr2 dr1 de2 de1 (IMA1)(IMA2)(dr2) de1 IMAc (IMA1)(IMA2) dr2 (3.0)(2.0) 6.0 b. If the compound machine is 60.0 percent efficient, how much effort must be applied to the lever to lift a 540-N box? Fr Fe 100 IMA (Fr)(100) (Fe)(IMA) (Fr)(100) so Fe (e)(IMA) (540 N)(100) (60.0)(6.0) 150 N Solutions Manual IMAc Mixed Review pages 281–282 Level 1 89. Ramps Isra has to get a piano onto a 2.0-m-high platform. She can use a 3.0-m-long frictionless ramp or a 4.0-m-long frictionless ramp. Which ramp should Isra use if she wants to do the least amount of work? Either ramp: only the vertical distance is important. If Isra used a longer ramp, she would require less force. The work done would be the same. 90. Brutus, a champion weightlifter, raises 240 kg of weights a distance of 2.35 m. a. How much work is done by Brutus lifting the weights? W Fd mgd (240 kg)(9.80 m/s2)(2.35 m) 5.5103 J b. How much work is done by Brutus holding the weights above his head? d 0, so no work c. How much work is done by Brutus lowering them back to the ground? d is opposite of motion in part a, so W is also the opposite, 5.5103 J. d. Does Brutus do work if he lets go of the weights and they fall back to the ground? No. He exerts no force, so he does no work, positive or negative. e. If Brutus completes the lift in 2.5 s, how much power is developed? W t 5.5103 J 2.5 s P 2.2 kW Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. de1 dr1 de2 (IMA2)(dr2) IMA1 240 12.0 cm 6.0 dr2 2.0 cm 88. A compound machine is constructed by attaching a lever to a pulley system. Consider an ideal compound machine consisting of a lever with an IMA of 3.0 and a pulley system with an IMA of 2.0. MA e 100 IMA de1 IMAc dr2 Chapter 10 continued Level 2 91. A horizontal force of 805 N is needed to drag a crate across a horizontal floor with a constant speed. You drag the crate using a rope held at an angle of 32°. a. What force do you exert on the rope? a. How much force does the rope exert on the crate? Fx F cos W Fd cos Fx 805 N so F cos cos 32° 11,400 J W so F 9.5102 N b. How much work do you do on the crate if you move it 22 m? W Fxd (805 N)(22 m) 1.8104 J c. If you complete the job in 8.0 s, what power is developed? W t 1.8104 J P 2.2 kW 8.0 s 92. Dolly and Ramp A mover’s dolly is used to transport a refrigerator up a ramp into a house. The refrigerator has a mass of 115 kg. The ramp is 2.10 m long and rises 0.850 m. The mover pulls the dolly with a force of 496 N up the ramp. The dolly and ramp constitute a machine. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 93. Sally does 11.4 kJ of work dragging a wooden crate 25.0 m across a floor at a constant speed. The rope makes an angle of 48.0° with the horizontal. a. What work does the mover do? Wi Fd (496 N)(2.10 m) 1.04103 J b. What is the work done on the refrigerator by the machine? d height raised 0.850 m Wo Fgd mgd (115 kg)(9.80 m/s2)(0.850 m) 958 J c. What is the efficiency of the machine? Wo efficiency 100 Wi 958 J 1.0410 J 100 3 d cos (25.0 m)(cos 48.0°) 681 N b. What is the force of friction acting on the crate? The crate moves with constant speed, so the force of friction equals the horizontal component of the force of the rope. Ff Fx F cos (681 N)(cos 48.0°) 456 N, opposite to the direction of motion c. What work is done by the floor through the force of friction between the floor and the crate? Force and displacement are in opposite directions, so W Fd (456 N)(25.0 m) 1.14104 J (Because no net forces act on the crate, the work done on the crate must be equal in magnitude but opposite in sign to the energy Sally expands: 1.14104 J) 94. Sledding An 845-N sled is pulled a distance of 185 m. The task requires 1.20104 J of work and is done by pulling on a rope with a force of 125 N. At what angle is the rope held? W Fd cos , so 1.20104 J (125 N)(185 m) cos1 cos1 W Fd 58.7° 92.1% Physics: Principles and Problems Solutions Manual 241 Chapter 10 continued Level 3 95. An electric winch pulls a 875-N crate up a 15° incline at 0.25 m/s. The coefficient of friction between the crate and incline is 0.45. a. What power does the winch develop? Work is done on the crate by the winch, gravity, and friction. Because the kinetic energy of the crate does not change, the sum of the three works is equal to zero. Therefore, Wwinch Wfriction Wgravity only one box at a time, most of the energy will go into raising your own body. The power (in watts) that your body can develop over a long time depends on the mass that you carry, as shown in Figure 10-25. This is an example of a power curve that applies to machines as well as to people. Find the number of boxes to carry on each trip that would minimize the time required. What time would you spend doing the job? Ignore the time needed to go back down the stairs and to lift and lower each box. Power v. Mass or, Pwinch Pfriction Pgravity FN Fg d t d t Power (W) 25 FNd Fgd t t 20 15 10 FNv Fgv 5 (Fg )(cos )(v) Fgv 0 5 10 (0.45)(875 N)(cos 15°) 3.1102 W Wo Po e Wi Pi Po so, Pi e 3.1102 W 0.85 W Fgd mgd (150 kg)(9.80 m/s2)(12 m) 1.76104 J. From the graph, the maximum power is 25 W at 15 kg. Since the mass per box is 150 kg 5 kg, this represents three 30 boxes boxes. W t W t P so t 3.6102 W 1.76104 J 25 W 7.0102 s page 282 96. Analyze and Conclude You work at a store, carrying boxes to a storage loft that is 12 m above the ground. You have 30 boxes with a total mass of 150 kg that must be moved as quickly as possible, so you consider carrying more than one up at a time. If you try to move too many at once, you know that you will go very slowly, resting often. If you carry 12 min Solutions Manual 30 The work has to be done the same, Thinking Critically 242 25 97. Apply Concepts A sprinter of mass 75 kg runs the 50.0-m dash in 8.50 s. Assume that the sprinter’s acceleration is constant throughout the race. a. What is the average power of the sprinter over the 50.0 m? Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. b. If the winch is 85 percent efficient, what is the electrical power that must be delivered to the winch? 20 Figure 10-25 ■ (0.25 m/s) (875 N)(0.25 m/s) 15 Mass (kg) Chapter 10 continued Assume constant acceleration, therefore constant force Final velocity: 1 d di vit at 2 2 vi 0 so but di vi 0 W t Fd t mad t P m2d 2d t t (2)(75 kg)(50.0 m) 3 (8.50 s)3 t 2md 2 6.1102 W b. What is the maximum power generated by the sprinter? vf at a(t1) Therefore, 1 2 df at12 at1t2 a t12 t1t2 1 2 df a 1 t12 t1t2 2 Power increases linearly from zero, since the velocity increases linearly as shown by P F Fv. 6.25 m/s2 W t Fd t d t 50.0 m 1 2(1.00 s)2 (1.00 s)(7.50 s) Therefore For the first second: Pmax 2Pave 1.2103 W d at 2 (6.25 m/s2)(1.00 s)2 1 2 Power (W) c. Make a quantitative graph of power versus time for the entire race. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. vf vi at 1 2 3.12 m From Problem 97, mad t P 1.2x103 (75 kg)(6.25 m/s2)(3.12 m) 1.00 s Pave 0 8.50 Time (s) 98. Apply Concepts The sprinter in the previous problem runs the 50.0-m dash in the same time, 8.50 s. However, this time the sprinter accelerates in the first second and runs the rest of the race at a constant velocity. a. Calculate the average power produced for that first second. Distance first second Distance rest of race 50.0 m 1 2 df di vit at 2 di vi 0 so 1 2 df a(t1)2 vf (t2) 50.0 m Physics: Principles and Problems 1.5103 W b. What is the maximum power that the sprinter now generates? Pmax 2Pave 3.0103 W Writing in Physics page 282 99. Just as a bicycle is a compound machine, so is an automobile. Find the efficiencies of the component parts of the power train (engine, transmission, wheels, and tires). Explore possible improvements in each of these efficiencies. The overall efficiency is 15–30 percent. The transmission’s efficiency is about 90 percent. Rolling friction in the tires is about 1 percent (ratio of pushing force to weight moved). The largest gain is possible in the engine. Solutions Manual 243 Chapter 10 continued 100. The terms force, work, power, and energy often mean the same thing in everyday use. Obtain examples from advertisements, print media, radio, and television that illustrate meanings for these terms that differ from those used in physics. Answers will vary. Some examples include, the company Consumers’ Power changed its name to Consumers’ Energy without changing its product, natural gas. “It’s not just energy, it’s power!” has appeared in the popular press. Cumulative Review diy viy 0 1 2 so dfy gt 2 (9.80 m/s2)(0.457 s)2 1 2 1.02 m 103. People sometimes say that the Moon stays in its orbit because the “centrifugal force just balances the centripetal force, giving no net force.” Explain why this idea is wrong. (Chapter 8) There is only one force on the moon, the gravitational force of Earth’s mass on it. This net force gives it an acceleration which is its centripetal acceleration toward Earth’s center. Challenge Problem page 268 An electric pump pulls water at a rate of 0.25 m3/s from a well that is 25 m deep. The water leaves the pump at a speed of 8.5 m/s. 8.5 m/s Fyou on can Ffriction sFN smg (0.35)(24 kg)(9.80 m/s2) 82 N 102. Baseball If a major league pitcher throws a fastball horizontally at a speed of 40.3 m/s (90 mph) and it travels 18.4 m (60 ft, 6 in), how far has it dropped by the time it crosses home plate? (Chapter 6) 25 m dfx dix vxt dfx dix so t vx 18.4 m 0.0 m 40.3 m/s 0.457 s 244 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. page 282 101. You are helping your grandmother with some gardening and have filled a garbage can with weeds and soil. Now you have to move the garbage can across the yard and realize it is so heavy that you will need to push it, rather than lift it. If the can has a mass of 24 kg, the coefficient of kinetic friction between the can’s bottom and the muddy grass is 0.27, and the static coefficient of friction between those same surfaces is 0.35, how hard do you have to push horizontally to get the can to just start moving? (Chapter 5) 1 2 dfy diy viyt gt2 Chapter 10 continued 1. What power is needed to lift the water to the surface? The work done in lifting is Fgd mgd. Therefore, the power is W t Fgd mgd t Plift t (0.25 m3)(1.00103 kg/m3)(9.80 m/s2)(25 m) 1.0 s 6.1104 W 61 kW 2. What power is needed to increase the pump’s kinetic energy? 1 The work done in increasing the pump’s kinetic energy is mv 2. 2 W t KE t Therefore, P 1 mv 2 2 t 2 mv 2t (0.25 m3)(1.00103 kg/m3)(8.5 m/s)2 (2)(1.0 s) 9.0103 W 9.0 kW 3. If the pump’s efficiency is 80 percent, how much power must be delivered to the pump? Wo e 100 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Wi Wo t Po 100 100 so, W Pi i t Po 3 9.010 W Pi 100 100 e 80 1.1104 W 11 kW Physics: Principles and Problems Solutions Manual 245 CHAPTER 11 Energy and Its Conservation Practice Problems 11.1 The Many Forms of Energy pages 285–292 page 287 1. A skater with a mass of 52.0 kg moving at 2.5 m/s glides to a stop over a distance of 24.0 m. How much work did the friction of the ice do to bring the skater to a stop? How much work would the skater have to do to speed up to 2.5 m/s again? To bring the skater to a stop: W KEf KEi 1 1 mvf2 mvi2 2 2 1 (52.0 kg)(0.00 m/s)2 2 1 (52.0 kg)(2.5 m/s)2 2 160 J This is the reverse of the first question. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1 1 KEf mv 2 (875.0 kg)(44.0 m/s)2 2 2 8.47105 J The work done is KEf KEi 8.47105 J 2.12105 J 6.35105 J 3. A comet with a mass of 7.851011 kg strikes Earth at a speed of 25.0 km/s. Find the kinetic energy of the comet in joules, and compare the work that is done by Earth in stopping the comet to the 4.21015 J of energy that was released by the largest nuclear weapon ever built. 1 KE mv 2 2 1 (7.851011 kg)(2.50104 m/s)2 2 2.451020 J To speed up again: W KEf KEi 1 1 mvf2 mvi2 2 2 1 (52.0 kg)(2.5 m/s)2 2 1 (52.0 kg)(0.00 m/s)2 2 160 J 2. An 875.0-kg compact car speeds up from 22.0 m/s to 44.0 m/s while passing another car. What are its initial and final energies, and how much work is done on the car to increase its speed? The initial kinetic energy of the car is 1 1 KEi mv 2 (875.0 kg)(22.0 m/s)2 2 The final kinetic energy is 2 2.12105 J KE met 2.451020 J co 5.8104 KEbomb 4.21015 J 5.8104 bombs would be required to produce the same amount of energy used by Earth in stopping the comet. page 291 4. In Example Problem 1, what is the potential energy of the bowling ball relative to the rack when it is on the floor? PE mgh (7.30 kg)(9.80 m/s2)(0.610 m) 43.6 J 5. If you slowly lower a 20.0-kg bag of sand 1.20 m from the trunk of a car to the driveway, how much work do you do? W Fd mg(hf hi) (20.0 kg)(9.80 m/s2)(0.00 m 1.20 m) 2.35102 J Physics: Principles and Problems Solutions Manual 247 Chapter 11 continued 6. A boy lifts a 2.2-kg book from his desk, which is 0.80 m high, to a bookshelf that is 2.10 m high. What is the potential energy of the book relative to the desk? PE mg(hf hi) (2.2 kg)(9.80 m/s2)(2.10 m 0.80 m) 28 J 7. If a 1.8-kg brick falls to the ground from a chimney that is 6.7 m high, what is the change in its potential energy? Choose the ground as the reference level. PE mg(hf hi) (1.8 kg)(9.80 m/s2)(0.0 m 6.7 m) Section Review 11.1 The Many Forms of Energy pages 285–292 page 292 9. Elastic Potential Energy You get a springloaded toy pistol ready to fire by compressing the spring. The elastic potential energy of the spring pushes the rubber dart out of the pistol. You use the toy pistol to shoot the dart straight up. Draw bar graphs that describe the forms of energy present in the following instances. a. The dart is pushed into the gun barrel, thereby compressing the spring. 1.2102 J 8. A warehouse worker picks up a 10.1-kg box from the floor and sets it on a long, 1.1-m-high table. He slides the box 5.0 m along the table and then lowers it back to the floor. What were the changes in the energy of the box, and how did the total energy of the box change? (Ignore friction.) To lift the box to the table: W Fd PE (10.1 kg)(9.80 m/s2)(1.1 m 0.0 m) Dart kinetic energy There should be three bars: one for the spring’s potential energy, one for gravitational potential energy, and one for kinetic energy. The spring’s potential energy is at the maximum level, and the other two are zero. b. The spring expands and the dart leaves the gun barrel after the trigger is pulled. 1.1102 J To slide the box across the table, W 0.0 because the height did not change and we ignored friction. To lower the box to the floor: W Fd mg(hf hi) PE (10.1 kg)(9.80 m/s2)(0.0 m 1.1 m) Dart Spring gravitational elastic potential potential energy energy Dart kinetic energy The kinetic energy is at the maximum level, and the other two are zero. 1.1102 J The sum of the three energy changes is 1.1102 J 0.0 J (1.1102 J) 0.0 J 248 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. mg(hf hi) Dart Spring gravitational elastic potential potential energy energy Chapter 11 continued c. The dart reaches the top of its flight. Dart Spring gravitational elastic potential potential energy energy Dart kinetic energy The gravitational potential energy is at the maximum level, and the other two are zero. 10. Potential Energy A 25.0-kg shell is shot from a cannon at Earth’s surface. The reference level is Earth’s surface. What is the gravitational potential energy of the system when the shell is at 425 m? What is the change in potential energy when the shell falls to a height of 225 m? The bowling ball has zero kinetic energy when it is resting on the rack or when it is held near your shoulder. Therefore, the total work done on the ball by you and by gravity must equal zero. 13. Potential Energy A 90.0-kg rock climber first climbs 45.0 m up to the top of a quarry, then descends 85.0 m from the top to the bottom of the quarry. If the initial height is the reference level, find the potential energy of the system (the climber and Earth) at the top and at the bottom. Draw bar graphs for both situations. 3.97104 J At the bottom At the top 3.53104 J a. PE mgh (25.0 kg)(9.80 m/s2)(425 m) PE mgh 1.04105 J At the top, b. PE mgh (25.0 kg)(9.80 m/s2)(225 m) Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 5.51104 J The change in energy is (1.04105 J) (5.51104 J) 4.89104 J 11. Rotational Kinetic Energy Suppose some children push a merry-go-round so that it turns twice as fast as it did before they pushed it. What are the relative changes in angular momentum and rotational kinetic energy? The angular momentum is doubled because it is proportional to the angular velocity. The rotational kinetic energy is quadrupled because it is proportional to the square of the angular velocity. The children did work in rotating the merry-go-round. 12. Work-Energy Theorem How can you apply the work-energy theorem to lifting a bowling ball from a storage rack to your shoulder? Physics: Principles and Problems PE (90.0 kg)(9.80 m/s2)(45.0 m) 3.97104 J At the bottom, PE (90.0 kg)(9.80 m/s2) (45.0 m 85.0 m) 3.53104 J 14. Critical Thinking Karl uses an air hose to exert a constant horizontal force on a puck, which is on a frictionless air table. He keeps the hose aimed at the puck, thereby creating a constant force as the puck moves a fixed distance. a. Explain what happens in terms of work and energy. Draw bar graphs. KEinitial W KEfinal Karl exerted a constant force F over a distance d and did an amount of work W Fd on the puck. This work changed the kinetic energy of the puck. Solutions Manual 249 Chapter 11 continued W (KEf KEi) 1 1 mvf2 mvi2 2 2 1 mvf2 2 b. Suppose Karl uses a different puck with half the mass of the first one. All other conditions remain the same. How will the kinetic energy and work differ from those in the first situation? If the puck has half the mass, it still receives the same amount of work and has the same change in kinetic energy. However, the smaller mass will move faster by a factor of 1.414. c. Explain what happened in parts a and b in terms of impulse and momentum. The two pucks do not have the same final momentum. Momentum of the first puck: p1 m1v1 Momentum of the second puck: p2 m2v2 1 m1(1.414v1) 2 Thus, the second puck has less momentum than the first puck does. Because the change in momentum is equal to the impulse provided by the air hose, the second puck receives a smaller impulse. Practice Problems 11.2 Conservation of Energy pages 293–301 page 297 15. A bike rider approaches a hill at a speed of 8.5 m/s. The combined mass of the bike and the rider is 85.0 kg. Choose a suitable system. Find the initial kinetic energy of the system. The rider coasts up the hill. Assuming there is no friction, at what height will the bike come to rest? 250 Solutions Manual 1 KE mv 2 2 1 (85.0 kg)(8.5 m/s)2 2 3.1103 J KEi PEi KEf PEf 1 mv 2 0 0 mgh 2 2 (8.5 m/s)2 (2)(9.80 m/s ) v h 2 2g 3.7 m 16. Suppose that the bike rider in problem 15 pedaled up the hill and never came to a stop. In what system is energy conserved? From what form of energy did the bike gain mechanical energy? The system of Earth, bike, and rider remains the same, but now the energy involved is not mechanical energy alone. The rider must be considered as having stored energy, some of which is converted to mechanical energy. Energy came from the chemical potential energy stored in the rider’s body. 17. A skier starts from rest at the top of a 45.0-m-high hill, skis down a 30° incline into a valley, and continues up a 40.0-mhigh hill. The heights of both hills are measured from the valley floor. Assume that you can neglect friction and the effect of the ski poles. How fast is the skier moving at the bottom of the valley? What is the skier’s speed at the top of the next hill? Do the angles of the hills affect your answers? Bottom of valley: KEi PEi KEf PEf 1 0 mgh mv 2 0 2 v2 2gh v 2gh (2)(9.8 0 m/s2 )(45.0 m) Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 0.707 p1 The system is the bike rider Earth. There are no external forces, so total energy is conserved. Chapter 11 continued 29.7 m/s Conservation of momentum: Top of next hill: mv (m M)V, or KEi PEi KEf PEf (m M)V v m 1 0 mghi mv 2 mghf 2 v 2 2g(hi hf) 2g(hi hf) 2)(45.0 (2)(9.8 0 ms m 40.0 m) 9.90 m/s No, the angles do not have any impact. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 18. In a belly-flop diving contest, the winner is the diver who makes the biggest splash upon hitting the water. The size of the splash depends not only on the diver’s style, but also on the amount of kinetic energy that the diver has. Consider a contest in which each diver jumps from a 3.00-m platform. One diver has a mass of 136 kg and simply steps off the platform. Another diver has a mass of 102 kg and leaps upward from the platform. How high would the second diver have to leap to make a competitive splash? Using the water as a reference level, the kinetic energy on entry is equal to the potential energy of the diver at the top of his flight. The large diver has PE mgh (136 kg)(9.80 m/s2)(3.00 m) 4.00103 J To equal this, the smaller diver would have to jump to 3 4.0010 J h 2 4.00 m (102 kg)(9.80 m/s ) Thus, the smaller diver would have to leap 1.00 m above the platform. (0.00800 kg 9.00 kg)(0.100 m/s) 0.00800 kg 1.13102 m/s 20. A 0.73-kg magnetic target is suspended on a string. A 0.025-kg magnetic dart, shot horizontally, strikes the target head-on. The dart and the target together, acting like a pendulum, swing 12.0 cm above the initial level before instantaneously coming to rest. a. Sketch the situation and choose a system. System 12 cm PE 0 The system includes the suspended target and the dart. b. Decide what is conserved in each part and explain your decision. Only momentum is conserved in the inelastic dart-target collision, so mvi MVi (m M)Vf where Vi 0 since the target is initially at rest and Vf is the common velocity just after impact. As the dart-target combination swings upward, energy is conserved, so PE KE or, at the top of the swing, 1 page 300 19. An 8.00-g bullet is fired horizontally into a 9.00-kg block of wood on an air table and is embedded in it. After the collision, the block and bullet slide along the frictionless surface together with a speed of 10.0 cm/s. What was the initial speed of the bullet? Physics: Principles and Problems (m M)ghf (m M)(Vf)2 2 Solutions Manual 251 Chapter 11 continued c. What was the initial velocity of the dart? Solve for Vf. 2ghf Vf Substitute vf into the momentum equation and solve for vi. mM vi 2gh f m 0.025 kg 0.73 kg (2)(9.8 0 m/s2 )(0.12 0 m) 0.025 kg 46 m/s 21. A 91.0-kg hockey player is skating on ice at 5.50 m/s. Another hockey player of equal mass, moving at 8.1 m/s in the same direction, hits him from behind. They slide off together. a. What are the total energy and momentum in the system before the collision? 1 1 KEi m1v12 m2v22 2 2 1 1 (91.0 kg)(5.50 m/s)2 (91.0 kg)(8.1 m/s)2 2 2 4.4103 J pi m1v1 m2v2 (91.0 kg)(5.5 m/s) (91.0 kg)(8.1 m/s) 1.2103 kg m/s b. What is the velocity of the two hockey players after the collision? After the collision: Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. pi pf m1v1 m2v2 (m1 m2)vf m v m v m1 m2 1 1 2 2 vf (91.0 kg)(5.50 ms) (91.0 kg)(8.1 ms) 91.0 kg 91.0 kg 6.8 m/s c. How much energy was lost in the collision? The final kinetic energy is 1 KEf (mi mf)vf2 2 1 (91.0 kg 91.0 kg)(6.8 m/s)2 2 4.2103 J Thus, the energy lost in the collision is KEi KEf 4.4103 J 4.2103 J 2102 J 252 Solutions Manual Physics: Principles and Problems Chapter 11 continued Section Review 11.2 Conservation of Energy pages 293–301 page 301 22. Closed Systems Is Earth a closed, isolated system? Support your answer. To simplify problems that take place over a short time, Earth is considered a closed system. It is not actually isolated, however, because it is acted upon by the gravitational forces from the planets, the Sun, and other stars. In addition, Earth is the recipient of continuous electromagnetic energy, primarily from the Sun. 23. Energy A child jumps on a trampoline. Draw bar graphs to show the forms of energy present in the following situations. a. The child is at the highest point. 25. Kinetic Energy In table tennis, a very light but hard ball is hit with a hard rubber or wooden paddle. In tennis, a much softer ball is hit with a racket. Why are the two sets of equipment designed in this way? Can you think of other ball-paddle pairs in sports? How are they designed? The balls and the paddle and racket are designed to match so that the maximum amount of kinetic energy is passed from the paddle or racket to the ball. A softer ball receives energy with less loss from a softer paddle or racket. Other combinations are a golf ball and club (both hard) and a baseball and bat (also both hard). 26. Potential Energy A rubber ball is dropped from a height of 8.0 m onto a hard concrete floor. It hits the floor and bounces repeat1 5 edly. Each time it hits the floor, it loses of its total energy. How many times will it bounce before it bounces back up to a height of only about 4 m? Etotal mgh Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Child Trampoline gravitational elastic potential potential energy energy Child kinetic energy b. The child is at the lowest point. Since the rebound height is proportional to energy, each bounce will rebound to 4 the height of the previous bounce. 5 After one bounce: h (8 m) 6.4 m 4 5 After two bounces: h (6.4 m) 5 5.12 m 4 Child Trampoline gravitational elastic potential potential energy energy Child kinetic energy After three bounces: h (5.12 m) 5 4.1 m 4 24. Kinetic Energy Suppose a glob of chewing gum and a small, rubber ball collide head-on in midair and then rebound apart. Would you expect kinetic energy to be conserved? If not, what happens to the energy? Even though the rubber ball rebounds with little energy loss, kinetic energy would not be conserved in this case because the glob of chewing gum probably was deformed in the collision. Physics: Principles and Problems Solutions Manual 253 Chapter 11 continued 27. Energy As shown in Figure 11-15, a 36.0-kg child slides down a playground slide that is 2.5 m high. At the bottom of the slide, she is moving at 3.0 m/s. How much energy was lost as she slid down the slide? Chapter Assessment Concept Mapping page 306 29. Complete the concept map using the following terms: gravitational potential energy, elastic potential energy, kinetic energy. 36.0 kg Energy potential kinetic 2.5 m linear rotational gravitational elastic Mastering Concepts ■ Figure 11-15 Ei mgh (36.0 kg)(9.80 m/s2)(2.5 m) 880 J 1 Ef mv 2 2 1 (36.0 kg)(3.0 m/s)2 2 160 J 720 J 28. Critical Thinking A ball drops 20 m. When it has fallen half the distance, or 10 m, half of its energy is potential and half is kinetic. When the ball has fallen for half the amount of time it takes to fall, will more, less, or exactly half of its energy be potential energy? The ball falls more slowly during the beginning part of its drop. Therefore, in the first half of the time that it falls, it will not have traveled half of the distance that it will fall. Therefore, the ball will have more potential energy than kinetic energy. 30. Explain how work and a change in energy are related. (11.1) The work done on an object causes a change in the object’s energy. This is the work-energy theorem. 31. What form of energy does a wound-up watch spring have? What form of energy does a functioning mechanical watch have? When a watch runs down, what has happened to the energy? (11.1) The wound-up watch spring has elastic potential energy. The functioning watch has elastic potential energy and rotational kinetic energy. The watch runs down when all of the energy has been converted to heat by friction in the gears and bearings. 32. Explain how energy change and force are related. (11.1) A force exerted over a distance does work, which produces a change in energy. 33. A ball is dropped from the top of a building. You choose the top of the building to be the reference level, while your friend chooses the bottom. Explain whether the 254 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Energy loss 880 J – 160 J page 306 Unless otherwise directed, assume that air resistance is negligible. Chapter 11 continued energy calculated using these two reference levels is the same or different for the following situations. (11.1) a. the ball’s potential energy at any point The potential energies are different due to the different reference levels. b. the change in the ball’s potential energy as a result of the fall The changes in the potential energies as a result of the fall are equal because the change in h is the same for both reference levels. 38. The sport of pole-vaulting was drastically changed when the stiff, wooden poles were replaced by flexible, fiberglass poles. Explain why. (11.2) A flexible, fiberglass pole can store elastic potential energy because it can be bent easily. This energy can be released to push the pole-vaulter higher vertically. By contrast, the wooden pole does not store elastic potential energy, and the pole-vaulter’s maximum height is limited by the direct conversion of kinetic energy to gravitational potential energy. c. the kinetic energy of the ball at any point The kinetic energies of the ball at any point are equal because the velocities are the same. 34. Can the kinetic energy of a baseball ever be negative? (11.1) The kinetic energy of a baseball can never be negative because the kinetic energy depends on the square of the velocity, which is always positive. 39. You throw a clay ball at a hockey puck on ice. The smashed clay ball and the hockey puck stick together and move slowly. (11.2) a. Is momentum conserved in the collision? Explain. The total momentum of the ball and the puck together is conserved in the collision because there are no unbalanced forces on this system. b. Is kinetic energy conserved? Explain. The total kinetic energy is not conserved. Part of it is lost in the smashing of the clay ball and the adhesion of the ball to the puck. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 35. Can the gravitational potential energy of a baseball ever be negative? Explain without using a formula. (11.1) The gravitational potential energy of a baseball can be negative if the height of the ball is lower than the reference level. 36. If a sprinter’s velocity increases to three times the original velocity, by what factor does the kinetic energy increase? (11.1) 40. Draw energy bar graphs for the following processes. (11.2) a. An ice cube, initially at rest, slides down a frictionless slope. The sprinter’s kinetic energy increases by a factor of 9, because the velocity is squared. PEi 37. What energy transformations take place when an athlete is pole-vaulting? (11.2) The pole-vaulter runs (kinetic energy) and bends the pole, thereby adding elastic potential energy to the pole. As he/she lifts his/her body, that kinetic and elastic potential energy is transferred into kinetic and gravitational potential energy. When he/she releases the pole, all of his/her energy is kinetic and gravitational potential energy. Physics: Principles and Problems KEi PEf KEf b. An ice cube, initially moving, slides up a frictionless slope and instantaneously comes to rest. PEi KEi PEf KEf Solutions Manual 255 Chapter 11 continued 41. Describe the transformations from kinetic energy to potential energy and vice versa for a roller-coaster ride. (11.2) On a roller-coaster ride, the car has mostly potential energy at the tops of the hills and mostly kinetic energy at the bottoms of the hills. 42. Describe how the kinetic energy and elastic potential energy are lost in a bouncing rubber ball. Describe what happens to the motion of the ball. (11.2) On each bounce, some, but not all, of the ball’s kinetic energy is stored as elastic potential energy; the ball’s deformation dissipates the rest of the energy as thermal energy and sound. After the bounce, the stored elastic potential energy is released as kinetic energy. Due to the energy losses in the deformation, each subsequent bounce begins with a smaller amount of kinetic energy, and results in the ball reaching a lower height. Eventually, all of the ball’s energy is dissipated, and the ball comes to rest. Applying Concepts a. The car’s wheels do not skid. If the car wheels do not skid, the brake surfaces rub against each other and do work that stops the car. The work that the brakes do is equal to the change in kinetic energy of the car. The brake surfaces heat up because the kinetic energy is transformed to thermal energy. b. The brakes lock and the car’s wheels skid. If the brakes lock and the car wheels skid, the wheels rubbing on the road are doing the work that stops the car. The tire surfaces heat up, not the brakes. This is not an efficient way to stop a car, and it ruins the tires. 256 Solutions Manual The trailer truck has more kinetic energy, 1 KE mv 2, because it has greater mass 2 than the compact car. Thus, according to the work-energy theorem, the truck’s engine must have done more work. 45. Catapults Medieval warriors used catapults to assault castles. Some catapults worked by using a tightly wound rope to turn the catapult arm. What forms of energy are involved in catapulting a rock to the castle wall? Elastic potential energy is stored in the wound rope, which does work on the rock. The rock has kinetic and potential energy as it flies through the air. When it hits the wall, the inelastic collision causes most of the mechanical energy to be converted to thermal and sound energy and to do work breaking apart the wall structure. Some of the mechanical energy appears in the fragments thrown from the collision. 46. Two cars collide and come to a complete stop. Where did all of their energy go? The energy went into bending sheet metal on the cars. Energy also was lost due to frictional forces between the cars and the tires, and in the form of thermal energy and sound. 47. During a process, positive work is done on a system, and the potential energy decreases. Can you determine anything about the change in kinetic energy of the system? Explain. The work equals the change in the total mechanical energy, W (KE PE ). If W is positive and PE is negative, then KE must be positive and greater than W. 48. During a process, positive work is done on a system, and the potential energy increases. Can you tell whether the kinetic energy increased, decreased, or remained the same? Explain. Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. pages 306–307 43. The driver of a speeding car applies the brakes and the car comes to a stop. The system includes the car but not the road. Apply the work-energy theorem to the following situations. 44. A compact car and a trailer truck are both traveling at the same velocity. Did the car engine or the truck engine do more work in accelerating its vehicle? Chapter 11 continued The work equals the change in the total mechanical energy, W (KE PE ). If W is positive and PE is positive, then you cannot say anything conclusive about KE. 49. Skating Two skaters of unequal mass have the same speed and are moving in the same direction. If the ice exerts the same frictional force on each skater, how will the stopping distances of their bodies compare? The larger skater will have more kinetic energy. The kinetic energy of each skater will be dissipated by the negative work, W Fd, done by the friction of the ice. Since the frictional forces are equal, the larger skater will go farther before stopping. 50. You swing a 55-g mass on the end of a 0.75-m string around your head in a nearly horizontal circle at constant speed, as shown in Figure 11-16. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 0.75 m 55 g 51. Give specific examples that illustrate the following processes. a. Work is done on a system, thereby increasing kinetic energy with no change in potential energy. pushing a hockey puck horizontally across ice; system consists of hockey puck only b. Potential energy is changed to kinetic energy with no work done on the system. dropping a ball; system consists of ball and Earth c. Work is done on a system, increasing potential energy with no change in kinetic energy. compressing the spring in a toy pistol; system consists of spring only d. Kinetic energy is reduced, but potential energy is unchanged. Work is done by the system. A car, speeding on a level track, brakes and reduces its speed. 52. Roller Coaster You have been hired to make a roller coaster more exciting. The owners want the speed at the bottom of the first hill doubled. How much higher must the first hill be built? The hill must be made higher by a factor of 4. ■ Figure 11-16 a. How much work is done on the mass by the tension of the string in one revolution? No work is done by the tension force on the mass because the tension is pulling perpendicular to the motion of the mass. b. Is your answer to part a in agreement with the work-energy theorem? Explain. 53. Two identical balls are thrown from the top of a cliff, each with the same speed. One is thrown straight up, the other straight down. How do the kinetic energies and speeds of the balls compare as they strike the ground? Even though the balls are moving in opposite directions, they have the same kinetic energy and potential energy when they are thrown. Therefore, they will have the same mechanical energy and speed when they hit the ground. This does not violate the workenergy theorem because the kinetic energy of the mass is constant; it is moving at a constant speed. Physics: Principles and Problems Solutions Manual 257 Chapter 11 continued Mastering Problems Unless otherwise directed, assume that air resistance is negligible. 11.1 The Many Forms of Energy pages 307–308 Level 1 54. A 1600-kg car travels at a speed of 12.5 m/s. What is its kinetic energy? 1 1 KE mv 2 (1600 kg)(12.5 m/s)2 2 2 1.3105 J 55. A racing car has a mass of 1525 kg. What is its kinetic energy if it has a speed of 108 km/h? 1 KE mv 2 2 2 v1 (10.0)2 4 2 2 (5.0) v2 1 Twice the velocity gives four times the kinetic energy. The kinetic energy is proportional to the square of the velocity. 58. Katia and Angela each have a mass of 45 kg, and they are moving together with a speed of 10.0 m/s. a. What is their combined kinetic energy? 1 1 KEc mv 2 (mK mA)v 2 2 2 2 4.5103 J b. What is the ratio of their combined mass to Katia’s mass? 6.86105 J 56. Shawn and his bike have a combined mass of 45.0 kg. Shawn rides his bike 1.80 km in 10.0 min at a constant velocity. What is Shawn’s kinetic energy? 2 t 2 1 (1.80 km)(1000 mkm) (45 kg) (10.0 min)(60 smin) mK mA 45 kg 45 kg 45 kg mK 2 1 c. What is the ratio of their combined kinetic energy to Katia’s kinetic energy? Explain. 1 1 KEK mKv 2 (45 kg)(10.0 m/s)2 2 203 J 2 2.3103 J 57. Tony has a mass of 45 kg and is moving with a speed of 10.0 m/s. a. Find Tony’s kinetic energy. 1 1 KE mv 2 (45 kg)(10.0 m/s)2 2 2 2.3103 J b. Tony’s speed changes to 5.0 m/s. Now what is his kinetic energy? 1 1 KE mv 2 (45 kg)( 5.0 m/s)2 2 258 5.6102 Solutions Manual 2 1 (mK mA)v 2 mK mA KEC 2 1 mK KEK 2 m v 2 K 2 1 The ratio of their combined kinetic energy to Katia’s kinetic energy is the same as the ratio of their combined mass to Katia’s mass. Kinetic energy is proportional to mass. J Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. d 2 1 1 KE mv 2 m 2 1 (mv12) 2 1 (mv22) 2 1 (45 kg 45 kg)(10.0 m/s)2 1 (108 kmh)(1000 mkm) 2 (1525 kg) 2 3600 sh 2 c. What is the ratio of the kinetic energies in parts a and b? Explain. Chapter 11 continued 59. Train In the 1950s, an experimental train, which had a mass of 2.50104 kg, was powered across a level track by a jet engine that produced a thrust of 5.00105 N for a distance of 509 m. Fg Now m g So d a. Find the work done on the train. W Fd (5.00105 N)(509 m) 2.55108 J b. Find the change in kinetic energy. KE W 2.55108 J c. Find the final kinetic energy of the train if it started from rest. KE KEf KEi so KEf KE KEi 2.55108 J 0.00 J 2.55108 J d. Find the final speed of the train if there had been no friction. 1 KEf mvf2 2 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 60. Car Brakes A 14,700-N car is traveling at 25 m/s. The brakes are applied suddenly, and the car slides to a stop, as shown in Figure 11-17. The average braking force between the tires and the road is 7100 N. How far will the car slide once the brakes are applied? m 14,700 N Figure 11-17 1 W Fd mv 2 2 Physics: Principles and Problems 7100 N 66 m 61. A 15.0-kg cart is moving with a velocity of 7.50 m/s down a level hallway. A constant force of 10.0 N acts on the cart, and its velocity becomes 3.20 m/s. a. What is the change in kinetic energy of the cart? 1 KE KEf KEi m(vf2 vi2) 2 1 (15.0 kg)((3.20 m/s)2 W KE 345 J J 104 m2 /s2 143 m/s So vf 2.04 ■ 1 14,700 N (25.0 m/s)2 2 9.80 m/s2 b. How much work was done on the cart? (2.5010 kg) 2 v 25 m/s F 345 J 1 4 Before (initial) 1 Fg v 2 2 g (7.50 m/s)2) So vf2 1 m 2 F 2 KEf 2.55108 1 mv 2 2 After (final) v 0.0 m/s c. How far did the cart move while the force acted? W Fd W 345 J so d 34.5 m F 10.0 N 62. How much potential energy does DeAnna with a mass of 60.0 kg, gain when she climbs a gymnasium rope a distance of 3.5 m? PE mgh (60.0 kg)(9.80 m/s2)(3.5 m) 2.1103 J 63. Bowling A 6.4-kg bowling ball is lifted 2.1 m into a storage rack. Calculate the increase in the ball’s potential energy. PE mgh (6.4 kg)(9.80 m/s2)(2.1 m) 1.3102 J Solutions Manual 259 Chapter 11 continued 64. Mary weighs 505 N. She walks down a flight of stairs to a level 5.50 m below her starting point. What is the change in Mary’s potential energy? PE mgh Fgh (505 N)(5.50 m) 2.78103 J 65. Weightlifting A weightlifter raises a 180-kg barbell to a height of 1.95 m. What is the increase in the potential energy of the barbell? Level 2 69. Tennis It is not uncommon during the serve of a professional tennis player for the racket to exert an average force of 150.0 N on the ball. If the ball has a mass of 0.060 kg and is in contact with the strings of the racket, as shown in Figure 11-18, for 0.030 s, what is the kinetic energy of the ball as it leaves the racket? Assume that the ball starts from rest. PE mgh (180 kg)(9.80 m/s2)(1.95 m) 150.0 N 3.4103 J 66. A 10.0-kg test rocket is fired vertically from Cape Canaveral. Its fuel gives it a kinetic energy of 1960 J by the time the rocket engine burns all of the fuel. What additional height will the rocket rise? PE mgh KE KE 1960 h 2 mg (10.0 kg)(9.80 m/s ) 20.0 m PE mgh Fgh Fg(hf hi) (12.0 N)(2.15 m 0.75 m) 17 J 68. A hallway display of energy is constructed in which several people pull on a rope that lifts a block 1.00 m. The display indicates that 1.00 J of work is done. What is the mass of the block? W PE mgh 1.00 J W m (9.80 m/s2)(1.00 m) gh 0.102 kg Figure 11-18 Ft mv mvf mvi and vi 0 Ft (150.0 N)(3.0102 s) so vf 2 m 6.010 kg 75 m/s 1 KE mv 2 2 1 (6.0102 kg)(75 m/s)2 2 1.7102 J 70. Pam, wearing a rocket pack, stands on frictionless ice. She has a mass of 45 kg. The rocket supplies a constant force for 22.0 m, and Pam acquires a speed of 62.0 m/s. a. What is the magnitude of the force? 1 KEf mvf2 2 1 (45 kg)(62.0 ms)2 2 8.6104 J b. What is Pam’s final kinetic energy? Work done on Pam equals her change in kinetic energy. W Fd KE KEf KEi KEi 0 J KE 8.6104 J So, F f d 22.0 m 3.9103 N 260 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 67. Antwan raised a 12.0-N physics book from a table 75 cm above the floor to a shelf 2.15 m above the floor. What was the change in the potential energy of the system? ■ Chapter 11 continued 71. Collision A 2.00103-kg car has a speed of 12.0 m/s. The car then hits a tree. The tree doesn’t move, and the car comes to rest, as shown in Figure 11-19. Before (initial) vi 12.0 m/s After (final) vf 0.0 m/s W PE mgh Fgh (98.0 N)(50.0 m) b. What is the increase in potential energy of the sack of grain at this height? Figure 11-19 a. Find the change in kinetic energy of the car. 1 KE KEf KEi m(vf2 vi2) 2 1 (2.00103 kg)((0.0 m/s)2 2 a. How much work was done? 4.90103 J m 2.00103 kg ■ 11.2 Conservation of Energy pages 308–309 Level 1 73. A 98.0-N sack of grain is hoisted to a storage room 50.0 m above the ground floor of a grain elevator. (12.0 m/s)2) 1.44105 J b. Find the amount of work done as the front of the car crashes into the tree. W KE 1.44105 J PE W 4.90103 J c. The rope being used to lift the sack of grain breaks just as the sack reaches the storage room. What kinetic energy does the sack have just before it strikes the ground floor? KE PE 4.90103 J 74. A 20-kg rock is on the edge of a 100-m cliff, as shown in Figure 11-20. c. Find the size of the force that pushed in the front of the car by 50.0 cm. 20 kg Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. W Fd W so F 100 m d 1.4410 J 5 0.500 m 2.88105 N 72. A constant net force of 410 N is applied upward to a stone that weighs 32 N. The upward force is applied through a distance of 2.0 m, and the stone is then released. To what height, from the point of release, will the stone rise? W Fd (410 N)(2.0 m) 8.2102 J But W PE mgh, so W 8.2102 J h 26 m mg 32 N ■ Figure 11-20 a. What potential energy does the rock possess relative to the base of the cliff? PE mgh (20 kg)(9.80 m/s2)(100 m) 2104 J b. The rock falls from the cliff. What is its kinetic energy just before it strikes the ground? KE PE 2104 J c. What speed does the rock have as it strikes the ground? 1 KE mv 2 2 v (2)(210 J) m 20 kg 2KE 4 40 m/s Physics: Principles and Problems Solutions Manual 261 Chapter 11 continued 75. Archery An archer puts a 0.30-kg arrow to the bowstring. An average force of 201 N is exerted to draw the string back 1.3 m. a. Assuming that all the energy goes into the arrow, with what speed does the arrow leave the bow? Work done on the string increases the string’s elastic potential energy. W PE Fd All of the stored potential energy is transformed to the arrow’s kinetic energy. 1 KE mv 2 PE Fd 2 2Fd v2 m v m 0.30 kg 2Fd (2)(201 N)(1.3 m) 77. A physics book of unknown mass is dropped 4.50 m. What speed does the book have just before it hits the ground? KE PE 1 mv 2 mgh 2 The mass of the book divides out, so 1 v 2 gh 2 v 2gh m/s2)(4.50 2(9.80 m) 9.39 m/s 78. Railroad Car A railroad car with a mass of 5.0105 kg collides with a stationary railroad car of equal mass. After the collision, the two cars lock together and move off at 4.0 m/s, as shown in Figure 11-21. m 5.0 105 kg 42 m/s v 4.0 m/s b. If the arrow is shot straight up, how high does it rise? The change in the arrow’s potential energy equals the work done to pull the string. PE mgh Fd mg (0.30 kg)(9.80 ms ) 89 m a. Before the collision, the first railroad car was moving at 8.0 m/s. What was its momentum? mv (5.0105 kg)(8.0 m/s) 76. A 2.0-kg rock that is initially at rest loses 407 J of potential energy while falling to the ground. Calculate the kinetic energy that the rock gains while falling. What is the rock’s speed just before it strikes the ground? PEi KEi PEf KEf KEi 0 4.0106 kg m/s b. What was the total momentum of the two cars after the collision? Because momentum is conserved, it must be 4.0106 kg m/s c. What were the kinetic energies of the two cars before and after the collision? So, Before the collision: KEf PEi PEf 407 J 1 KEi mv 2 1 KEf mvf2 vf 2 1 (5.0105 kg)(8.0 m/s)2 2 2KE vf2 f m 2 1.6107 J 2KEf m Figure 11-21 (2)(407 J) (2.0 kg) After the collision: 1 KEf mv 2 2 2.0101 m/s 262 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Fd (201 N)(1.3 m) h 2 ■ Chapter 11 continued 1 (5.0105 kg 5.0105 kg) 2 (4.0 m/s)2 8.0106 J d. Account for the loss of kinetic energy. While momentum was conserved during the collision, kinetic energy was not. The amount not conserved was turned into thermal energy and sound energy. 79. From what height would a compact car have to be dropped to have the same kinetic energy that it has when being driven at 1.00102 km/h? km h 1000 m 1 km 1h 3600 s v 1.00102 27.8 m/s The work done by friction equals the change in mechanical energy. W PE KE 1 mg(hf hi) mvf2 2 (420 N)(0.40 m 1.00 m) 1 420 N (2.0 ms)2 2 9.80 ms2 1.7102 J 81. Hakeem throws a 10.0-g ball straight down from a height of 2.0 m. The ball strikes the floor at a speed of 7.5 m/s. What was the initial speed of the ball? KEf KEi PEi KE PE 1 1 mvf2 mvi2 mgh 2 2 1 mv 2 mgh 2 the mass of the ball divides out, so 1 v 2 gh 2 vi2 vf2 2gh, 2 2 v (27.8 m/s) h 2 2g 2(9.80 m/s ) 39.4 m Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. b. If Kelli moves through the lowest point at 2.0 m/s, how much work was done on the swing by friction? Level 2 80. Kelli weighs 420 N, and she is sitting on a playground swing that hangs 0.40 m above the ground. Her mom pulls the swing back and releases it when the seat is 1.00 m above the ground. a. How fast is Kelli moving when the swing passes through its lowest position? PE mgh mg(hf hi) vi vf2 2gh (7.5 m /s)2 (2)(9. 80 m/ s2)(2.0 m) 4.1 m/s 82. Slide Lorena’s mass is 28 kg. She climbs the 4.8-m ladder of a slide and reaches a velocity of 3.2 m/s at the bottom of the slide. How much work was done by friction on Lorena? The work done by friction on Lorena equals the change in her mechanical energy. W PE KE 1 1 KE m(vf2 vi2) mvf2 2 2 1 mg(hf hi) m(vf2 vi2) 2 By conservation of mechanical energy: (28 kg)(9.80 ms2)(0.0 m 4.8 m) PE KE 0 1 mg(hf hi) mvf2 0 2 vf 2g(hi hf) 0 ms2 )(1.00 m 0 .40 m) (2)(9.8 3.4 m/s Physics: Principles and Problems 1 (28 kg)((3.2 ms)2 (0.0 ms)2) 2 1.2103 J 83. A person weighing 635 N climbs up a ladder to a height of 5.0 m. Use the person and Earth as the system. Solutions Manual 263 Chapter 11 continued a. Draw energy bar graphs of the system before the person starts to climb the ladder and after the person stops at the top. Has the mechanical energy changed? If so, by how much? 3200 J 86. High Jump The world record for the men’s high jump is about 2.45 m. To reach that height, what is the minimum amount of work that a 73.0-kg jumper must exert in pushing off the ground? W E mgh (73.0 kg)(9.80 m/s2)(2.45 m) 1.75 kJ KEi PEi KEf PEf Yes. The mechanical energy has changed, increase in potential energy of (635 N)(5.0 m) 3200 J. b. Where did this energy come from? from the internal energy of the person Mixed Review pages 309–310 Level 1 84. Suppose a chimpanzee swings through the jungle on vines. If it swings from a tree on a 13-m-long vine that starts at an angle of 45°, what is the chimp’s velocity when it reaches the ground? The chimpanzee’s initial height is Conservation of mechanical energy: Level 2 88. Football A 110-kg football linebacker has a head-on collision with a 150-kg defensive end. After they collide, they come to a complete stop. Before the collision, which player had the greater momentum and which player had the greater kinetic energy? 1 mghi mvf2 0 2 The momentum after the collision is zero; therefore, the two players had equal and opposite momenta before the collision. That is, plinebacker mlinebackervlinebacker pend vf 2gh m/s2) (3.8 m) i 2(9.80 mendvend. After the collision, each had PE KE 0 1 mg(hf hi) m(vf2 vi2) 0 2 zero energy. The energy loss for each 8.6 m/s 85. An 0.80-kg cart rolls down a frictionless hill of height 0.32 m. At the bottom of the hill, the cart rolls on a flat surface, which exerts a frictional force of 2.0 N on the cart. How far does the cart roll on the flat surface before it comes to a stop? E mgh W Fd 2 (0.80 kg)(9.80 m/s )(0.32 m) mgh d F 1.3 m 264 Assume that the foam peanuts exert a constant force to slow him down, W Fd E mgh. If the height is increased five times, then the depth of the foam peanuts also should be increased five times to 5 m. Solutions Manual 2.0 N 1 1 m 2v 2 p2 player was mv 2 . 2 2 m 2m Because the momenta were equal but mlinebacker mend the linebacker lost more energy. 89. A 2.0-kg lab cart and a 1.0-kg lab cart are held together by a compressed spring. The lab carts move at 2.1 m/s in one direction. The spring suddenly becomes uncompressed and pushes the two lab carts apart. The 2-kg Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. h (13 m)(1 cos 45°) 3.8 m 87. A stuntwoman finds that she can safely break her fall from a one-story building by landing in a box filled to a 1-m depth with foam peanuts. In her next movie, the script calls for her to jump from a five-story building. How deep a box of foam peanuts should she prepare? Chapter 11 continued lab cart comes to a stop, and the 1.0-kg lab cart moves ahead. How much energy did the spring add to the lab carts? 1 1 Ei mv 2 (2.0 kg 1.0 kg)(2.1 m/s)2 2 2 6.6 J pi mv (2.0 kg 1.0 kg)(2.1 m/s) 91. An 0.80-kg cart rolls down a 30.0° hill from a vertical height of 0.50 m as shown in Figure 11-22. The distance that the cart must roll to the bottom of the hill is 0.50 m/sin 30.0° 1.0 m. The surface of the hill exerts a frictional force of 5.0 N on the cart. Does the cart roll to the bottom of the hill? m 0.80 kg F 5.0 N 6.3 kg m/s pf (1.0 kg)vf so, vf 6.3 m/s 1 1 Ef mvf2 (1.0 kg)(6.3 m/s)2 19.8 J 2 2 0.50 m E 19.8 J 6.6 J 13.2 J 30.0° 13.2 J was added by the spring. 90. A 55.0-kg scientist roping through the top of a tree in the jungle sees a lion about to attack a tiny antelope. She quickly swings down from her 12.0-m-high perch and grabs the antelope (21.0 kg) as she swings. They barely swing back up to a tree limb out of reach of the lion. How high is this tree limb? Ei mBgh Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. The velocity of the botanist when she reaches the ground is 1 Ei mBv 2 mBgh 2 v 2gh m m 2Ei 2mBgh B Momentum is conserved when the botanist grabs the antelope. Figure 11-22 Ei mgh (0.80 kg)(9.80 m/s2)(0.50 m) 3.9 J The work done by friction over 1.0 m would be W Fd (5.0 N)(1.0 m) 5.0 J. The work done by friction is greater than the energy of the cart. The cart would not reach the bottom of the hill. Level 3 92. Object A, sliding on a frictionless surface at 3.2 m/s, hits a 2.0-kg object, B, which is motionless. The collision of A and B is completely elastic. After the collision, A and B move away from each other at equal and opposite speeds. What is the mass of object A? pi mAv1 0 mBv (mB mA)vf m v (mB mA) ■ m mB mA B B so, vf 2gh The final energy of the two is 1 Ef (mB mA)vf2 2 mB 2 1 (mB mA) (2gh) 2 mB mA (mB mA)ghf mB 2 So, hf h mB mA 2 55.0 kg 55.0 kg 21.0 kg (12.0 m) pf mA(v2) mBv2 pi pf (conservation of momentum) therefore, mAv1 mA(v2) mBv 2 (mB mA)v2 mAv1 (m v ) (mB mA) A 1 v2 1 Ei mAv12 2 1 1 Ef mAv22 mBv22 2 2 6.28 m Physics: Principles and Problems Solutions Manual 265 Chapter 11 continued 1 Ef (mA mB)v22 2 (mAv1) 2 1 (mA mB) 2 (mB mA) From the conservation of momentum, mcvc1 mcvc2 mbvb2 m v b b2 Solve for vc2, vc2 vc1 m c Ei Ef (conservation of energy in elastic collision) therefore, From conservation of energy, 1 1 mAv12 (mA mB) 2 2 Multiply by two and substitute to get: (m v ) m v mc 2 b b2 mcvc12 mcvc1 mbvb22 2 A 1 (mB mA) After cancelling out common factors, (mA mB)mA (mB mA)2 mB2 2mAmB mA2 mB 2.00 kg mA 0.67 kg 3 3 1 1 1 mcvc12 mcvc22 mbvb22 2 2 2 93. Hockey A 90.0-kg hockey player moving at 5.0 m/s collides head-on with a 110-kg hockey player moving at 3.0 m/s in the opposite direction. After the collision, they move off together at 1.0 m/s. How much energy was lost in the collision? 1 1 Before: E mv12 mv22 2 2 or mcvc12 mcvc12 2mbvc2vc1 mb2vb22 mbvb22 mc Simplify and factor: mb2vb2 0 (mbvb2)2vc1 vb2 mc mbvb2 0 or m mc b 1vb2 0 2vc1 Ignoring the solution vb2 0, then vb2 2 1 (110 kg)(3.0 m/s)2 2 2(44 m/s) 0.0 46 kg 1 0.220 kg 73 m/s 1.6103 J 1 After: E (m m)vf2 2 1 (200.0 kg)(1.0 m/s)2 2 1.0102 J Energy loss 1.6103 J 1.0102 J 1.5103 J Thinking Critically page 310 94. Apply Concepts A golf ball with a mass of 0.046 kg rests on a tee. It is struck by a golf club with an effective mass of 0.220 kg and a speed of 44 m/s. Assuming that the collision is elastic, find the speed of the ball when it leaves the tee. 266 Solutions Manual 95. Apply Concepts A fly hitting the windshield of a moving pickup truck is an example of a collision in which the mass of one of the objects is many times larger than the other. On the other hand, the collision of two billiard balls is one in which the masses of both objects are the same. How is energy transferred in these collisions? Consider an elastic collision in which billiard ball m1 has velocity v1 and ball m2 is motionless. a. If m1 m2, what fraction of the initial energy is transferred to m2? If m1 m2, we know that m1 will be at rest after the collision and m2 will move with velocity v1. All of the energy will be transferred to m2. Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1 (90.0 kg)(5.0 m/s)2 2v c1 m b mc 1 Chapter 11 continued b. If m1 m2, what fraction of the initial energy is transferred to m2? If m1 m2, we know that the motion of m1 will be unaffected by the collision and that the energy transfer to m2 will be minimal. c. In a nuclear reactor, neutrons must be slowed down by causing them to collide with atoms. (A neutron is about as massive as a proton.) Would hydrogen, carbon, or iron atoms be more desirable to use for this purpose? The best way to stop a neutron is to have it hit a hydrogen atom, which has about the same mass as the neutron. 96. Analyze and Conclude In a perfectly elastic collision, both momentum and mechanical energy are conserved. Two balls, with masses mA and mB, are moving toward each other with speeds vA and vB, respectively. Solve the appropriate equations to find the speeds of the two balls after the collision. conservation of momentum (1) mAvA1 mBvB1 mAvA2 mBvB2 mAvA1 mAvA2 mBvB1 mBvB2 (2) mA(vA1 vA2) mB(vB1 vB2) conservation of energy 1 1 1 1 mAvA12 mBvB12 mAvA22 mBvB22 2 2 2 2 mAvA12 mAvA22 mBvB12 mBvB22 mA(vA12 vA22) mB(vB12 vB22) (3) mA(vA1 vA2)(vA1 vA2) mB(vB1 vB2)(vB1 vB2) Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Divide equation (3) by (2) to obtain (4) vA1 vA2 vB1 vB2 Solve equation (1) for vA2 and vB2 m B vA2 vA1 (v vB2) mA B1 mA vB2 vB1 (vA1 vA2) mB Substitute into (4) and solve for vB2 and vA2 m B vA1 vA1 (v vB2) vB1 vB2 mA B1 2mAvA1 mBvB1 mBvB2 mAvB1 mAvB2 2m m m B A A vB2 v vB1 mA mB mA mB A1 mA (vA1 vA2) vA1 vA2 vB1 vB1 m B mBvA1 mBvA2 2mBvB1 mAvA1 mAvA2 m m 2m A B B vA2 vA1 v mA mB mA mB B1 Physics: Principles and Problems Solutions Manual 267 Chapter 11 continued 97. Analyze and Conclude A 25-g ball is fired with an initial speed of v1 toward a 125-g ball that is hanging motionless from a 1.25-m string. The balls have a perfectly elastic collision. As a result, the 125-g ball swings out until the string makes an angle of 37.0° with the vertical. What is v1? Object 1 is the incoming ball. Object 2 is the one attached to the string. In the collision, momentum is conserved. p1i p1f p2f or m1v1i m1v1f m2v2f In the collision, kinetic energy is conserved. 1 1 1 m1v1i2 m1v1f2 m2v2f2 2 2 2 m1v1i2 m1v1f2 m2v2f2 m m m1 m m1 m2 1 1 2 (m1v1i2) (m1v1f2) (m2v2f2) m12v1i2 m12v1f2 m22v2f2 m1 m1 m2 p1f2 p2f2 p1i2 m1 m1 m2 m 1 p1i2 p1f2 p 2 m2 2f We don’t care about v1f, so get rid of p1f using p1f p1i p2f m 1 p1i2 (p1i p2f)2 p 2 m2 2f m m2 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1 p1i2 p1i2 2p1ip2f p2f2 p2f2 m m2 1 2p1ip2f 1 p2f2 m m2 1 1 p1i 1 p2f 2 1 m1v1i (m2 m1)v2f 2 m m1 1 2 v1i 1v2f 2 Now consider the pendulum. 1 m2v2f2 m2gh 2 or v2f 2gh where h L(1 cos ) co s ) Thus, v2f 2gL(1 v2f (2)(9.8 m/s2)(1.25 m)(1 cos 37.0°) 0 2.22 m/s 268 Solutions Manual Physics: Principles and Problems Chapter 11 continued 1 125 g v1i 1(2.22 m/s) 2 25 g 6.7 m/s Writing in Physics Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. page 310 98. All energy comes from the Sun. In what forms has this solar energy come to us to allow us to live and to operate our society? Research the ways that the Sun’s energy is turned into a form that we can use. After we use the Sun’s energy, where does it go? Explain. The Sun produces energy through nuclear fusion and releases that energy in the form of electromagnetic radiation, which is transferred through the vacuum of space to Earth. Earth absorbs that electromagnetic radiation in its atmosphere, land, and oceans in the form of thermal energy or heat. Part of the visible radiation also is converted by plants into chemical energy through photosynthesis. There are several other chemical reactions mediated by sunlight, such as ozone production. The energy then is transferred into various forms, some of which are the chemical processes that allow us to digest food and turn it into chemical energy to build tissues, to move, and to think. In the end, after we have used the energy, the remainder is dispersed as electromagnetic radiation back into the universe. make molecules and released when the molecules are broken up or rearranged. Separation of electric charges produces electric potential energy, as in a battery. Electric potential energy is converted to kinetic energy in the motion of electric charges in an electric current when a conductive path, or circuit, is provided. Biological processes are all chemical, and thus, biological energy is just a form of chemical energy. Solar energy is fusion energy converted to to electromagnetic radiation. (See the answer to the previous question.) Light is a wave form of electromagnetic energy whose frequency is in a range detectible by the human eye. Cumulative Review page 310 100. A satellite is placed in a circular orbit with a radius of 1.0107 m and a period of 9.9103 s. Calculate the mass of Earth. Hint: Gravity is the net force on such a satellite. Scientists have actually measured the mass of Earth this way. (Chapter 7) m v2 r Gm m r s s e Fnet 2 2r Since, v T m 2 2 Gm m 4 r s s e r T 2 r2 2 2 4 r me 2 GT 2 99. All forms of energy can be classified as either kinetic or potential energy. How would you describe nuclear, electric, chemical, biological, solar, and light energy, and why? For each of these types of energy, research what objects are moving and how energy is stored in those objects. Potential energy is stored in the binding of the protons and neutrons in the nucleus. The energy is released when a heavy nucleus is broken into smaller pieces (fission) or when very small nuclei are combined to make bigger nuclei (fusion). In the same way, chemical potential energy is stored when atoms are combined to Physics: Principles and Problems 7 3 4 (1.010 m) 11 2 2 2 (6.6710 N m /kg )(9.910 s) 6.01024 kg 101. A 5.00-g bullet is fired with a velocity of 100.0 m/s toward a 10.00-kg stationary solid block resting on a frictionless surface. (Chapter 9) a. What is the change in momentum of the bullet if it is embedded in the block? mbvb1 mbv2 mwv2 (mb mw)v2 Solutions Manual 269 Chapter 11 continued m v mb mw IMA de (2.0102)(33 cm) b b1 so v2 dr 66 m Then, pv mb(v2 vb1) m v b b1 vb1 mb m m b w m mb mw b mbvb1 1 m m mb mw b w vb1 (5.00103 kg)(10.00 kg) 5.0010 kg 10.00 kg Challenge Problem page 300 A bullet of mass m, moving at speed v1, goes through a motionless wooden block and exits with speed v2. After the collision, the block, which has mass mB, is moving. 3 (100.0 ms) Initial Final v1 v2 0.500 kgm/s vB b. What is the change in momentum of the bullet if it ricochets in the opposite direction with a speed of 99 m/s? pv mb(v2 vb1) (5.00103 kg) (99.0 m/s 100.0 m/s) 0.995 kg m/s c. In which case does the block end up with a greater speed? Conservation of momentum: mv1 mv2 mBvB mBvB m(v1 v2) m(v v ) 1 2 vB m B 2. How much energy was lost to the bullet? For the bullet alone: 1 KE1 mv12 2 102. An automobile jack must exert a lifting force of at least 15 kN. a. If you want to limit the effort force to 0.10 kN, what mechanical advantage is needed? 15 kN MA 150 0.10 kN b. If the jack is 75% efficient, over what distance must the effort force be exerted in order to raise the auto 33 cm? MA IMA 2.0102. e 1 KE2 mv22 2 1 KE m(v12 v22) 2 3. How much energy was lost to friction inside the block? Energy lost to friction KE1 KE2 KEblock 1 1 1 Elost mv12 mv22 mBvB2 2 2 2 d dr e Since IMA, 270 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. When the bullet ricochets, its change in momentum is larger in magnitude, and so is the block’s change in momentum, so the block ends up with a greater speed. 1. What is the final speed vB, of the block? CHAPTER 12 Thermal Energy Practice Problems 12.1 Temperature and Thermal Energy pages 313–322 page 317 1. Convert the following Kelvin temperatures to Celsius temperatures. a. 115 K TC TK 273 115 273 158°C b. 172 K TC TK 273 172 273 101°C c. 125 K TC TK 273 125 273 148°C d. 402 K TC TK 273 402 273 129°C e. 425 K TC TK 273 425 273 152°C Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. f. 212 K TC TK 273 212 273 61°C 2. Find the Celsius and Kelvin temperatures for the following. a. room temperature Room temperature is about 72°F, 22°C. TK TC 273 22 273 295 K b. a typical refrigerator A refrigerator is kept at about 4°C. TK TC 273 4 273 277 K c. a hot summer day in North Carolina A hot summer day is about 95°F, 35°C. TK TC 273 35 273 308 K d. a winter night in Minnesota A typical winter night in Minnesota is about 14°F, 10°C. TK TC 273 10 273 263 K Physics: Principles and Problems page 319 3. When you turn on the hot water to wash dishes, the water pipes have to heat up. How much heat is absorbed by a copper water pipe with a mass of 2.3 kg when its temperature is raised from 20.0°C to 80.0°C? Q mCT (2.3 kg)(385 J/kgK) (80.0°C 20.0°C) 5.3104 J 4. The cooling system of a car engine contains 20.0 L of water (1 L of water has a mass of 1 kg). a. What is the change in the temperature of the water if the engine operates until 836.0 kJ of heat is added? Q mCT (8.36105 J) (20.0 kg)(4180 J/kgK) Q T mC 10.0 K b. Suppose that it is winter, and the car’s cooling system is filled with methanol. The density of methanol is 0.80 g/cm3. What would be the increase in temperature of the methanol if it absorbed 836.0 kJ of heat? The mass of methanol would be 0.80 times the mass of 20.0 L of water, or 16 kg. Q mCT 5 8.3610 J Q T mC (16 kg)(2450 J/kgK) 21 K c. Which is the better coolant, water or methanol? Explain. For temperatures above 0°C, water is the better coolant because it can absorb heat without changing its temperature as much as methanol does. Solutions Manual 271 Chapter 12 continued 5. Electric power companies sell electricity by the kWh, where 1 kWh 3.6106 J. Suppose that it costs $0.08 per kWh to run an electric water heater in your neighborhood. How much does it cost to heat 75 kg of water from 15°C to 43°C to fill a bathtub? Q mCT (75 kg)(4180 J/kgK)(43°C 15°C) 8.8106 J 8.8106 J 2.4 kWh 3.6106 J/kWh (2.4 kWh)($0.15 per kWh) $0.36 page 321 6. A 2.00102-g sample of water at 80.0°C is mixed with 2.00102 g of water at 10.0°C. Assume that there is no heat loss to the surroundings. What is the final temperature of the mixture? mACA(Tf TAi) mBCB(Tf TBi) 0 Since mA mB and CA CB, there is cancellation in this particular case so that T T 2 80.0°C 10.0°C 2 Ai Bi 45.0°C Tf 7. A 4.00102-g sample of methanol at 16.0°C is mixed with 4.00102 g of water at 85.0°C. Assume that there is no heat loss to the surroundings. What is the final temperature of the mixture? Since in this particular case, mA mW, the masses cancel and CATAi CWTWi Tf CA CW (2450 J/kgK)(16.0°C) (4180 J/kgK)(85.0°C) 2450 J/kgK 4180 J/kgK 59.5°C 8. Three lead fishing weights, each with a mass of 1.00102 g and at a temperature of 100.0°C, are placed in 1.00102 g of water at 35.0°C. The final temperature of the mixture is 45.0°C. What is the specific heat of the lead in the weights? Heat gained by the water: Q mCT (0.100 kg)(4180 J/kg°C)(10.0°C) 4.18 kJ Thus, heat lost by the weights 4.18 kJ mweightsCweightsT (4.184 kJ)(1000 J/kJ) (0.100 kg)(55.0°C) hence, Cweights 2.53102 J/kg°C 272 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. mACA(Tf TAi) mWCW(Tf TWi) 0 Chapter 12 continued 9. A 1.00102-g aluminum block at 100.0°C is placed in 1.00102 g of water at 10.0°C. The final temperature of the mixture is 25.0°C. What is the specific heat of the aluminum? Heat gained by the water: Q mCT (0.100 kg)(4180 J/kg°C)(15.0°C) 6.27 kJ Thus, heat lost by the aluminum block 6.27 kJ mAluminumCAluminumT Q mAluminumT hence, CAluminum 6.27 kJ (0.100 kg)(75.0°C) 8.36102 J/kg°C Section Review Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 12.1 Temperature and Thermal Energy pages 313–322 page 322 10. Temperature Make the following conversions. a. 5°C to kelvins 278 K b. 34 K to degrees Celsius 239°C c. 212°C to kelvins 485 K d. 316 K to degrees Celsius 43°C 11. Conversions Convert the following Celsius temperatures to Kelvin temperatures. a. 28°C 301 K b. 154°C 427 K c. 568°C 841 K Physics: Principles and Problems d. 55°C 218 K e. 184°C 89 K 12. Thermal Energy Could the thermal energy of a bowl of hot water equal that of a bowl of cold water? Explain your answer. Thermal energy is the measure of the total energy of all the molecules in an object. The temperature (hot or cold) measures the amount of energy per molecule. If the bowls are identical and contain the same amount of water, they have the same number of molecules, but the bowl of hot water has more total thermal energy. However, if the cold water mass is slightly more than that of the hot water, the two energies could be equal. 13. Heat Flow On a dinner plate, a baked potato always stays hot longer than any other food. Why? A potato has a large specific heat and conducts heat poorly, so it loses its heat energy slowly. 14. Heat The hard tile floor of a bathroom always feels cold to bare feet even though the rest of the room is warm. Is the floor colder than the rest of the room? The floor is usually at the same temperature as the rest of the room, but the tile conducts heat more efficiently than most materials, so it conducts heat from a person’s feet, making them feel cold. 15. Specific Heat If you take a plastic spoon out of a cup of hot cocoa and put it in your mouth, you are not likely to burn your tongue. However, you could very easily burn your tongue if you put the hot cocoa in your mouth. Why? The plastic spoon has a lower specific heat than the cocoa, so it does not transmit much heat to your tongue as it cools. Solutions Manual 273 Chapter 12 continued 16. Heat Chefs often use cooking pans made of thick aluminum. Why is thick aluminum better than thin aluminum for cooking? Thick aluminum conducts heat better and does not have any “hot spots.” 17. Heat and Food It takes much longer to bake a whole potato than to cook french fries. Why? Potatoes do not conduct heat well. Increasing surface area by cutting a potato into small parts increases heat flow into the potato. Heat flow from hot oil to the potato is also more efficient than from hot air. 18. Critical Thinking As water heats in a pot on a stove, the water might produce some mist above its surface right before the water begins to roll. What is happening, and where is the coolest part of the water in the pot? The heat flows from the burner (the hottest part) to the top surface of the water (coldest). The water first transfers heat from bottom to top through conduction, and then convection begins to move hot water in currents to the top. Practice Problems 12.2 Changes of State and the Laws of Thermodynamics pages 323–331 page 325 19. How much heat is absorbed by 1.00102 g of ice at 20.0°C to become water at 0.0°C? (0.100 kg)(2060 J/kg°C)(20.0°C) (0.100 kg)(3.34105 J/kg) 3.75104 J 20. A 2.00102-g sample of water at 60.0°C is heated to steam at 140.0°C. How much heat is absorbed? Q mCwaterT mHv mCsteamT (0.200 kg)(4180 J/kg°C)(100.0°C 60.0°C) (0.200 kg)(2.26106 J/kg) (0.200 kg)(2020 J/kg°C)(140.0°C 100.0°C) 502 kJ 21. How much heat is needed to change 3.00102 g of ice at 30.0°C to steam at 130.0°C? Q mCiceT mHf mCwater T mHv mCsteamT (0.300 kg)(2060 J/kg°C)(0.0°C (30.0°C)) (0.300 kg) (3.34105 J/kg) (0.300 kg)(4180 J/kg°C)(100.0°C 0.0°C) (0.300 kg)(2.26106 J/kg) (0.300 kg)(2020 J/kg°C)(130.0°C 100.0°C) 9.40102 kJ 274 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Q mCT mHf Chapter 12 continued page 328 22. A gas balloon absorbs 75 J of heat. The balloon expands but stays at the same temperature. How much work did the balloon do in expanding? U Q W Since the balloon did not change temperature, U 0. Therefore, Q W. Thus, the balloon did 75 J of work in expanding. 23. A drill bores a small hole in a 0.40-kg block of aluminum and heats the aluminum by 5.0°C. How much work did the drill do in boring the hole? U Q Wblock; since Wdrill Wblock (0.15 kg)(4180 J/kg°C)(2.0°C) 1.3103 J. The number of stirs is 1.3103 J 2.6104 stirs 0.050 J 26. How can the first law of thermodynamics be used to explain how to reduce the temperature of an object? Since U Q W, it is possible to have a negative U and therefore, cool an object if Q 0 and the object does work, for instance, by expanding. Alternatively, have W 0 and Q negative by having it transfer heat to its surroundings. Any combination of these will work well. and assume no heat added to drill: 0 Wdrill mCT (0.40 kg)(897 J/kg°C)(5.0°C) 1.8103 J Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 24. How many times would you have to drop a 0.50-kg bag of lead shot from a height of 1.5 m to heat the shot by 1.0°C? U mCT (0.50 kg)(130 J/kg°C)(1.0°C) 65 J Each time the bag is raised its potential energy is PE mgh (0.50 kg)(9.80 m/s2)(1.5 m) 7.4 J When the bag hits the ground this energy is (mostly) transmitted as work on the lead shot. The number of drops 65 J is 9 drops 7.4 J 25. When you stir a cup of tea, you do about 0.050 J of work each time you circle the spoon in the cup. How many times would you have to stir the spoon to heat a 0.15-kg cup of tea by 2.0°C? Section Review 12.2 Changes of State and the Laws of Thermodynamics pages 323–331 page 331 27. Heat of Vaporization Old-fashioned heating systems sent steam into radiators in each room of a house. In the radiators, the steam condensed back to water. Analyze this process and explain how it heated a room. The condensing steam released its heat of vaporization into the room and was then circulated back to the boiler to receive the heat of vaporization again. 28. Heat of Vaporization How much heat is needed to change 50.0 g of water at 80.0°C to steam at 110.0°C? Q mCwaterT mHv mCsteamT (0.500 kg)(4180 J/kg°C)(100.0°C 80.0°C) (0.500 kg) (2.26106 J/kg) (0.500 kg) (2020 J/kg°C)(110.0°C 100.0°C) 1.18105 J U mCT Physics: Principles and Problems Solutions Manual 275 Chapter 12 continued 29. Heat of Vaporization The specific heat of mercury is 140 J/kg°C. Its heat of vaporization is 3.06105 J/kg. How much energy is needed to heat 1.0 kg of mercury metal from 10.0°C to its boiling point and vaporize it completely? The boiling point of mercury is 357°C. 32. Mechanical Energy and Thermal Energy Water flows over a fall that is 125.0 m high, as shown in Figure 12-17. If the potential energy of the water is all converted to thermal energy, calculate the temperature difference between the water at the top and the bottom of the fall. Q mCHgT mHv (1.0 kg)(140 J/kg°C) (357°C 10.0°C) 125.0 m (1.0 kg)(3.06105 J/kg) 3.5105 J 30. Mechanical Energy and Thermal Energy James Joule carefully measured the difference in temperature of water at the top and bottom of a waterfall. Why did he expect a difference? The water at the top has gravitational potential energy that is dissipated into thermal energy when the water splashes at the bottom. The water should be hotter at the bottom, but not by much. Part of the kinetic energy of the hammer is absorbed as thermal energy by the lead block. The hammer’s energy is 1 1 mv 2 (320 kg)(5.0 m/s)2 4.0 kJ. 2 Figure 12-17 PEgravity Qabsorbed by water mgh mCT gh C T (9.80 m/s2)(125.0 m) 4180 J/kg°C 0.293°C rise in temperature at the bottom 33. Entropy Evaluate why heating a home with natural gas results in an increased amount of disorder. The gas releases heat, Q, at its combustion temperature, T. The natural gas molecules break up and combust with oxygen. The heat is distributed in many new ways, and the natural gas molecules cannot readily be reassembled. 2 The change in thermal energy of the block is U mCT (3.0 kg)(130 J/kgK)(5.0°C) 2.0 kJ Hence, about half of the hammer’s energy went to the lead block. 34. Critical Thinking A new deck of cards has all the suits (clubs, diamonds, hearts, and spades) in order, and the cards are ordered by number within the suits. If you shuffle the cards many times, are you likely to return the cards to their original order? Explain. Of what physical law is this an example? The cards are very unlikely to return to their original order. This is an example of the second law of thermodynamics, in which disorder increases. 276 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 31. Mechanical Energy and Thermal Energy A man uses a 320-kg hammer moving at 5.0 m/s to smash a 3.0-kg block of lead against a 450-kg rock. When he measured the temperature he found that it had increased by 5.0°C. Explain how this happened. ■ Chapter 12 continued Chapter Assessment Concept Mapping page 336 35. Complete the following concept map using the following terms: heat, work, internal energy. First law of thermodynamics heat entropy internal energy temperature work external forces Mastering Concepts Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. page 336 36. Explain the differences among the mechanical energy of a ball, its thermal energy, and its temperature. (12.1) The mechanical energy is the sum of the potential and kinetic energies of the ball considered as one mass. The thermal energy is the sum of the potential and kinetic energies of the individual particles that make up the mass of the ball. The temperature is a measure of the internal energy of the ball. 37. Can temperature be assigned to a vacuum? Explain. (12.1) No, because there are no particles that have energy in a vacuum. 38. Do all of the molecules or atoms in a liquid have the same speed? (12.1) No. There is a distribution of velocities of the atoms or molecules. 39. Is your body a good judge of temperature? On a cold winter day, a metal doorknob feels much colder to your hand than a wooden door does. Explain why this is true. (12.1) Your skin measures heat flow to or from itself. The metal doorknob absorbs heat from your skin faster than the wooden door, so it feels colder. Physics: Principles and Problems 40. When heat flows from a warmer object in contact with a colder object, do the two have the same temperature changes? (12.1) The two objects will change temperatures depending on their masses and specific heats. The temperature changes are not necessarily the same for each. 41. Can you add thermal energy to an object without increasing its temperature? Explain. (12.2) When you melt a solid or boil a liquid, you add thermal energy without changing the temperature. 42. When wax freezes, does it absorb or release energy? (12.2) When wax freezes, it releases energy. 43. Explain why water in a canteen that is surrounded by dry air stays cooler if it has a canvas cover that is kept wet. (12.2) When the water in the cover evaporates into the dry air, it must absorb an amount of energy proportional to its heat of fusion. In doing so, it cools off the canteen. This works only if the air is dry; if the air is humid, then the water will not evaporate. 44. Which process occurs at the coils of a running air conditioner inside a house, vaporization or condensation? Explain. (12.2) Inside the house, the coolant is evaporating in the coils to absorb energy from the rooms. Applying Concepts page 336 45. Cooking Sally is cooking pasta in a pot of boiling water. Will the pasta cook faster if the water is boiling vigorously or if it is boiling gently? It should make no difference. Either way, the water is at the same temperature. Solutions Manual 277 Chapter 12 continued 46. Which liquid would an ice cube cool faster, water or methanol? Explain. Methanol, because it has a lower specific heat; for a given mass and heat transfer, it generates a bigger T since Q mCT. 47. Equal masses of aluminum and lead are heated to the same temperature. The pieces of metal are placed on a block of ice. Which metal melts more ice? Explain. The specific heat of aluminum is much greater than that of lead; therefore, it melts more ice. 48. Why do easily vaporized liquids, such as acetone and methanol, feel cool to the skin? As they evaporate, they absorb their heat of vaporization from the skin. 49. Explain why fruit growers spray their trees with water when frost is expected to protect the fruit from freezing. The water on the leaves will not freeze until it can release its heat of fusion. This process keeps the leaves warmer longer. The heat capacity of the ice slows down the cooling below 0°C. The cup with block A will be hotter because block A contains more thermal energy. 51. Windows Often, architects design most of the windows of a house on the north side. How does putting windows on the south side affect the heating and cooling of the house? In the northern hemisphere, the sunlight comes from the south. The Sun’s light would help heat the house in the winter but also would also heat the house in the summer. 278 Solutions Manual 12.1 Temperature and Thermal Energy pages 336–337 Level 1 52. How much heat is needed to raise the temperature of 50.0 g of water from 4.5°C to 83.0°C? Q mCT (0.0500 kg)(4180 J/kg°C) (83.0°C 4.5°C) 1.64104 J 53. A 5.00102-g block of metal absorbs 5016 J of heat when its temperature changes from 20.0°C to 30.0°C. Calculate the specific heat of the metal. Q mCT Q mT so C 5016 J 1 (5.0010 kg)(30.0°C 20.0°C) 1.00103 J/kg°C 1.00103 J/kgK 54. Coffee Cup A 4.00102-g glass coffee cup is 20.0°C at room temperature. It is then plunged into hot dishwater at a temperature of 80.0°C, as shown in Figure 12-18. If the temperature of the cup reaches that of the dishwater, how much heat does the cup absorb? Assume that the mass of the dishwater is large enough so that its temperature does not change appreciably. 20.0°C 80.0°C 4.00102 g ■ Figure 12-18 Q mCT (4.00101 kg)(840 J/kg°C) (80.0°C 20.0°C) 2.02104 J Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 50. Two blocks of lead have the same temperature. Block A has twice the mass of block B. They are dropped into identical cups of water of equal temperatures. Will the two cups of water have equal temperatures after equilibrium is achieved? Explain. Mastering Problems Chapter 12 continued 55. A 1.00102-g mass of tungsten at 100.0°C is placed in 2.00102 g of water at 20.0°C. The mixture reaches equilibrium at 21.6°C. Calculate the specific heat of tungsten. QT Q W 0 or m TC TTT m WC WTW mWCWTW (0.200 kg)(4180 J/kgK)(21.6°C 20.0°C) (0.100 kg)(21.6°C 100.0°C) C T mTTT 171 J/kgK 56. A 6.0102-g sample of water at 90.0°C is mixed with 4.00102 g of water at 22.0°C. Assume that there is no heat loss to the surroundings. What is the final temperature of the mixture? mACATAi mBCBTBi Tf mACA mBCB but CA CB because both liquids are water, and the C’s will divide out. m T m T mA mB (6.0102 g)(90.0°C) (4.00102 g)(22.0°C) 6.010 g 4.0010 g A Ai B Bi Tf 2 2 57. A 10.0-kg piece of zinc at 71.0°C is placed in a container of water, as shown in Figure 12-19. The water has a mass of 20.0 kg and a temperature of 10.0°C before the zinc is added. What is the final temperature of the water and the zinc? kg 20.0 kg 1 0 .0 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 63°C 10.0°C m AC ATAi m BC BTBi Tf m AC A m BC B ■ Figure 12-19 (10.0 kg)(388 J/kgK)(71.0°C) (20.0 kg)(4180 J/kgK)(10.0°C) (10.0 kg)(388 J/kgK) (20.0 kg)(4180 J/kgK) 12.7°C Level 2 58. The kinetic energy of a compact car moving at 100 km/h is 2.9105 J. To get a feeling for the amount of energy needed to heat water, what volume of water (in liters) would 2.9105 J of energy warm from room temperature (20.0°C) to boiling (100.0°C)? Q mCT VCT where is the density of the material 5 2.910 J Q so, V CT (1.00 kg/L)(4180 J/kg°C)(100.0°C 20.0°C) 0.87 L Physics: Principles and Problems Solutions Manual 279 Chapter 12 continued 59. Water Heater A 3.0102-W electric immersion heater is used to heat a cup of water, as shown in Figure 12-20. The cup is made of glass, and its mass is 3.00102 g. It contains 250 g of water at 15°C. How much time is needed to bring the water to the boiling point? Assume that the temperature of the cup is the same as the temperature of the water at all times and that no heat is lost to the air. 3.00102 W 15°C 250 g Q mGCGTG mWC WTW 3.00102 g but TG TW, so ■ Figure 12-20 Q (mGCG mWCW)T ((0.300 kg)(840 J/kg°C) (0.250 kg)(4180 J/kg°C))(100.0°C 15°C) 1.1105 J E t Q t Now P , so Q 1.110 J t 2 5 3.0010 J/s P 370 s 6.1 min 60. Car Engine A 2.50102-kg cast-iron car engine contains water as a coolant. Suppose that the engine’s temperature is 35.0°C when it is shut off, and the air temperature is 10.0°C. The heat given off by the engine and water in it as they cool to air temperature is 4.40106 J. What mass of water is used to cool the engine? Q miCiT (4.4106 J) ((2.50102 kg)(450 J/kg°C)(35.0°C 10.0°C)) (4180 J/kg°C)(35.0°C 10.0°C) mW CWT 15 kg 12.2 Changes of State and the Laws of Thermodynamics page 337 Level 1 61. Years ago, a block of ice with a mass of about 20.0 kg was used daily in a home icebox. The temperature of the ice was 0.0°C when it was delivered. As it melted, how much heat did the block of ice absorb? Q mHf (20.0 kg)(3.34105 J/kg) 6.68106 J 62. A 40.0-g sample of chloroform is condensed from a vapor at 61.6°C to a liquid at 61.6°C. It liberates 9870 J of heat. What is the heat of vaporization of chloroform? Q mHv Q 9870 J Hv 2.47105 J/kg m 280 Solutions Manual 0.0400 kg Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Q mWCWT miCiT Chapter 12 continued 63. A 750-kg car moving at 23 m/s brakes to a stop. The brakes contain about 15 kg of iron, which absorbs the energy. What is the increase in temperature of the brakes? During braking, the kinetic energy of the car is converted into heat energy. So, KEC QB 0.0, and KEC mBC BT 0.0 so, 1 mC(v f2 v i2) KEC 2 T mBCB mBCB 1 (750 kg)(0.02 (23 m/s)2) 2 (15 kg)(450 J/kg°C) 29°C Level 2 64. How much heat is added to 10.0 g of ice at 20.0°C to convert it to steam at 120.0°C? Amount of heat needed to heat ice to 0.0°C: Q mCT (0.0100 kg)(2060 J/kg°C) Q mCT (0.0100 kg)(2060 J/kg°C) (120.0°C 100.0°C) 404 J The total heat is 412 J 3.34103 J 4.18103 J 2.26104 J 404 J 3.09104 J 65. A 4.2-g lead bullet moving at 275 m/s strikes a steel plate and comes to a stop. If all its kinetic energy is converted to thermal energy and none leaves the bullet, what is its temperature change? Because the kinetic energy is converted to thermal energy, KE Q 0. So KE mBCBT and KE T m BC B T Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Amount of heat to melt ice: Q mHf (0.0100 kg)(3.34105 J/kg) 3.34103 J Amount of heat to heat water to 100.0°C: Q mCT (0.0100 kg)(4180 J/kg°C) (100.0°C 0.0°C) 4.18103 J Amount of heat to boil water: Q mHv (0.0100 kg)(2.26106 J/kg) 2.26104 J Amount of heat to heat steam to 120.0°C: Physics: Principles and Problems m C B B and the mass of the bullet divides out so (0.0°C (20.0°C)) 412 J 1 2 mB(vf2 vi2) 1 2 (vf2 vi2) C B 1 2 ((0.0 m/s)2 (275 m/s)2) 130 J/kg°C 290°C 66. Soft Drink A soft drink from Australia is labeled “Low-Joule Cola.” The label says “100 mL yields 1.7 kJ.” The can contains 375 mL of cola. Chandra drinks the cola and then wants to offset this input of food energy by climbing stairs. How high would Chandra have to climb if she has a mass of 65.0 kg? Chandra gained (3.75)(1.7 kJ) 6.4103 J of energy from the drink. To conserve energy, E PE 0 or 6.4103 J mgh so, 6.410 J 6.410 J h 2 3 mg 3 (65.0 kg)(9.80 m/s ) 1.0101 m, or about three flights of stairs Solutions Manual 281 Chapter 12 continued Mixed Review KE mC Therefore, T pages 337–338 Level 1 67. What is the efficiency of an engine that produces 2200 J/s while burning enough gasoline to produce 5300 J/s? How much waste heat does the engine produce per second? W 2200 J Efficiency 100 100 QH 5300 J 42% The heat loss is 5300 J 2200 J 2900 J 68. Stamping Press A metal stamping machine in a factory does 2100 J of work each time it stamps out a piece of metal. Each stamped piece is then dipped in a 32.0-kg vat of water for cooling. By how many degrees does the vat heat up each time a piece of stamped metal is dipped into it? U mCT, U mC therefore T 2100 J (32.0 kg)(4180 J/kg°C) ° 12°C 70. Iced Tea To make iced tea, you start by brewing the tea with hot water. Then you add ice. If you start with 1.0 L of 90°C tea, what is the minimum amount of ice needed to cool it to 0°C? Would it be better to let the tea cool to room temperature before adding the ice? Heat lost by the tea Q mCT (1.0 kg)(4180 J/kgK)(90°C) 376 kJ Amount of ice melted Q m Hf 376 kJ 1.1 kg 334 kJ Thus, you need slightly more ice than tea, but this ratio would make watery tea. Let the tea cool to room temperature before adding the ice. Level 2 71. A block of copper at 100.0°C comes in contact with a block of aluminum at 20.0°C, as shown in Figure 12-21. The final temperature of the blocks is 60.0°C. What are the relative masses of the blocks? 0.016°C. 69. A 1500-kg automobile comes to a stop from 25 m/s. All of the energy of the automobile is deposited in the brakes. Assuming that the brakes are about 45 kg of aluminum, what would be the change in temperature of the brakes? The energy change in the car is 1 2 KE (1500 kg)(25 m/s)2 4.7105 J. If all of this energy is transferred as work to the brakes, then U KE mCT. 282 Solutions Manual 100.0°C 20.0°C Copper Aluminum 60.0°C 60.0°C Copper Aluminum ■ Figure 12-21 Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. If we assume the 2100 J of work from the machine is absorbed as thermal energy in the stamped piece, then the vat must absorb 2100 J in the form of heat from each piece. No work is done on the water, only heat is transferred. The change in temperature of the water is given by 5 4.710 J (45 kg)(897 J/kg C) Chapter 12 continued The heat lost from the copper equals the heat gained by the aluminum. The T for the copper is 40.0°C and the aluminum heats by 40.0°C. therefore, mcopperCcopper maluminumCaluminum mcopper Caluminum maluminum Ccopper and 897 J/kgK 2.3 385 J/kgK The copper block has 2.3 times as much mass as the aluminum block. 72. A 0.35-kg block of copper sliding on the floor hits an identical block moving at the same speed from the opposite direction. The two blocks come to a stop together after the collision. Their temperatures increase by 0.20°C as a result of the collision. What was their velocity before the collision? The change in internal energy of the blocks is Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. U mCT KE m Hf 6.6 J 3.3410 J/kg 5 2.0105 kg Thinking Critically page 338 74. Analyze and Conclude A certain heat engine removes 50.0 J of thermal energy from a hot reservoir at temperature TH 545 K and expels 40.0 J of heat to a colder reservoir at temperature TL 325 K. In the process, it also transfers entropy from one reservoir to the other. a. How does the operation of the engine change the total entropy of the reservoirs? As the engine operates, it removes energy from the hot reservoir. QH Therefore, SH so that the TH entropy of the hot reservoir decreases. (0.70 kg)(385 J/kg°C)(0.20°C) The entropy of the cold reservoir 54 J SL increases. The net QL Therefore, 54 J equals the kinetic energy of the blocks before the collision. 54 J (2)mv 2 1 2 v Therefore, 6.6 J is added to the ice. The amount of ice melted is given by 54 J 0.35 kg 12 m/s Level 3 73. A 2.2-kg block of ice slides across a rough floor. Its initial velocity is 2.5 m/s and its final velocity is 0.50 m/s. How much of the ice block melted as a result of the work done by friction? The work done by friction equals the negative of the change in kinetic energy of the block, assuming not too much of the block melted. 1 2 KE (2.2 kg)(0.50 m/s)2 TL increase in entropy of the reservoirs together is S T SL SH QL QH TL TH 40.0 J 325 K 50.0 J 545 K S T 0.0313 J/K b. What would be the total entropy change in the reservoirs if TL 205 K? 40.0 J 205 K 50.0 J 545 K S T 0.103 J/K The total entropy change in the reservoirs, and in the universe, has increased approximately by a factor of three. 1 (2.2 kg)(2.5 m/s)2 6.6 J 2 Physics: Principles and Problems Solutions Manual 283 Chapter 12 continued 75. Analyze and Conclude During a game, the metabolism of basketball players often increases by as much as 30.0 W. How much perspiration must a player vaporize per hour to dissipate this extra thermal energy? The amount of thermal energy to be dissipated in 1.00 h is U (30.0 J/s)(3600 s/h) 1.08105 J. The amount of water this energy, transmitted as heat, would vaporize is Q m HV 5 1.0810 J 6 2.2610 J/kg 0.0478 kg 76. Analyze and Conclude Chemists use calorimeters to measure the heat produced by chemical reactions. For instance, a chemist dissolves 1.01022 molecules of a powdered substance into a calorimeter containing 0.50 kg of water. The molecules break up and release their binding energy to the water. The water temperature increases by 2.3°C. What is the binding energy per molecule for this substance? U mCT (0.50 kg)(4180 J/kg°C)(2.3°C) 4.8 kJ The energy per molecule is therefore, 4.8 kJ 4.81019 J/molecule 1022 molecules 77. Apply Concepts All of the energy on Earth comes from the Sun. The surface temperature of the Sun is approximately 104 K. What would be the effect on our world if the Sun’s surface temperature were 103 K? Student answers will vary. Answers should reflect changing average temperature on Earth, different weather patterns, plant and animal species dying out, etc. 284 Solutions Manual page 338 78. Our understanding of the relationship between heat and energy was influenced by a soldier named Benjamin Thompson, Count Rumford; and a brewer named James Prescott Joule. Both relied on experimental results to develop their ideas. Investigate what experiments they did and evaluate whether or not it is fair that the unit of energy is called the Joule and not the Thompson. In 1799, heat was thought to be a liquid that flowed from one object to another. However, Count Rumford thought that heat was caused by the motion of particles in the metal cannon. He did not do any quantitative measurements and his ideas were not widely accepted. In 1843, Joule, doing careful measurements, measured the change in temperature caused by adding heat or doing work on a quantity of water. He proved that heat is a flavor of energy and that energy is conserved. Joule deserves the credit and the eponymic unit. 79. Water has an unusually large specific heat and large heats of fusion and vaporization. Our weather and ecosystems depend upon water in all three states. How would our world be different if water’s thermodynamic properties were like other materials, such as methanol? The large specific heat and large heats of fusion and vaporization mean that water, ice, and water vapor can store a lot of thermal energy without changing their temperatures too much. The implications are many. The oceans and large lakes moderate the temperature changes in nearby regions on a daily and seasonal basis. The day-to-night temperature variation near a lake is much smaller than the day-to-night temperature variation in the desert. The large heat of fusion of water controls the change of seasons in the far north and south. The absorption of energy by freezing water in the fall and its release Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. The amount of energy added to the water is Writing in Physics Chapter 12 continued in the spring slows the temperature changes in the atmosphere. Water absorbs and stores a lot of energy as it vaporizes. This energy can be used to drive meteorological events, such as thunderstorms and hurricanes. Cumulative Review page 338 80. A rope is wound around a drum with a radius of 0.250 m and a moment of inertia of 2.25 kg m2. The rope is connected to a 4.00-kg block. (Chapter 8) a. Find the linear acceleration of the block. Solve Newton’s second law for the block: mg FT ma, where the positive direction is downward and where FT is the force of the rope on the drum. Newton’s second law for the drum is FT r I or FT r Ia/r. That is, FT Ia/r2. Therefore, mg (I/r 2)a ma. That is, a mg/(m I/r2) g/10.0 0.980 m/s2. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. b. Find the angular acceleration of the drum. 0.980 m/s2 0.250 m a r 3.92 rad/s2 c. Find the tension, FT, in the rope. I FT r (2.25 kg m2)(3.92 rad/s2) 0.250 m 81. A weight lifter raises a 180-kg barbell to a height of 1.95 m. How much work is done by the weight lifter in lifting the barbell? (Chapter 10) W mgh (180 kg)(9.80 m/s2)(1.95 m) 3.4103 J 82. In a Greek myth, the man Sisyphus is condemned by the gods to forever roll an enormous rock up a hill. Each time he reaches the top, the rock rolls back down to the bottom. If the rock has a mass of 215 kg, the hill is 33 m in height, and Sisyphus can produce an average power of 0.2 kW, how many times in 1 h can he roll the rock up the hill? (Chapter 11) The amount of work needed to roll the rock up once is W mgh (215 kg)(9.80 m/s2)(33 m) 70,000 J In one hour Sisyphus does an amount of work (0.2103 J)(3600 s) 720,000 J He pushes the rock up the hill (720,000)/(70,000) 10 times in one hour Challenge Problem page 329 Entropy has some interesting properties. Compare the following situations. Explain how and why these changes in entropy are different. 35.3 N d. Find the angular velocity of the drum after the block has fallen 5.00 m. 1 2 x at 2, so t Q a 3.19 s 1 kg 1 kg Ti Tf 2x Therefore, t (3.92 rad/s2)(3.19 s) 12.5 rad/s 1. Heating 1.0 kg of water from 273 K to 274 K. Q T mCT T S (1.0 kg)(4180 J/kgK)(274 K 273 K) 273 K 15 J/K Physics: Principles and Problems Solutions Manual 285 Chapter 12 continued 2. Heating 1.0 kg of water from 353 K to 354 K. Q T mCT T S (1.0 kg)(4180 J/kgK)(354 K 353 K) 353 K 12 J/K 3. Completely melting 1.0 kg of ice at 273 K. Q T mH T S f (1.0 kg)(3.34105 J/kg) 273 K 1.2103 J/K 4. Heating 1.0 kg of lead from 273 K to 274 K. Q T mCT T S (1.0 kg)(130 J/kgK)(274 K 273 K) 273 K 0.48 J/K Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 286 Solutions Manual Physics: Principles and Problems CHAPTER 13 States of Matter Practice Problems 13.1 Properties of Fluids pages 341–348 page 344 1. The atmospheric pressure at sea level is about 1.0105 Pa. What is the force at sea level that air exerts on the top of a desk that is 152 cm long and 76 cm wide? F PA Plw 1.2105 N (0.15)(1.0105 Pa) 1.5104 Pa 2. A car tire makes contact with the ground on a rectangular area of 12 cm by 18 cm. If the car’s mass is 925 kg, what pressure does the car exert on the ground as it rests on all four tires? Fg, car m g 4lw F car P A 2 (925 kg)(9.80 m/s ) Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. (4)(0.12 m)(0.18 m) 1.0102 kPa 3. A lead brick, 5.0 cm 10.0 cm 20.0 cm, rests on the ground on its smallest face. Lead has a density of 11.8 g/cm3. What pressure does the brick exert on the ground? mbrick V lwh (11.8 g/cm3)(5.0 cm) lwhg lw (11.8 g/cm3)(20.0 cm) 23 kPa Physics: Principles and Problems 5. In industrial buildings, large pieces of equipment must be placed on wide steel plates that spread the weight of the equipment over larger areas. If an engineer plans to install a 454-kg device on a floor that is rated to withstand additional pressure of 5.0104 Pa, how large should the steel support plate be? Fg mg A The maximum pressure P A mg P Therefore, A 6. A tank of helium gas used to inflate toy balloons is at a pressure of 15.5106 Pa and a temperature of 293 K. The tank’s volume is 0.020 m3. How large a balloon would it fill at 1.00 atmosphere and 323 K? hg 2.7104 N directed from the inside of the house outward 8.9102 m2 Fg, brick mbrickg P A lw (100 cm)2 (1 m)2 (1.5104 Pa)(1.95 m)(0.91 m) (454 kg)(9.80 m/s2) 5.010 Pa 1.18104 g 11.8 kg 1 kg (9.80 m/s2) 1000 g F PdiffA Pdifflw 4 (10.0 cm)(20.0 cm) Vg lw The pressure difference across the door is Pdiff (15%)(Patm) (1.0105 Pa)(1.52 m)(0.76 m) A 4. In a tornado, the pressure can be 15 percent below normal atmospheric pressure. Suppose that a tornado occurred outside a door that is 195 cm high and 91 cm wide. What net force would be exerted on the door by a sudden 15 percent drop in normal atmospheric pressure? In what direction would the force be exerted? P V1 P V2 T2P1V1 1 2 , so V2 T1 T2 P2T1 1.00 atm 1.013105 Pa Solutions Manual 287 Chapter 13 continued (323 K)(15.5106 Pa)(0.020 m3) (1.01310 Pa)(293 K) V2 5 13.1 Properties of Fluids pages 341–348 3.4 m3 7. What is the mass of the helium gas in the previous problem? The molar mass of helium gas is 4.00 g/mol. PV nRT (15.5106 Pa)(0.020 m3) (8.31 Pam /molK)(293 K) PV n 3 RT Section Review 127.3 mol m (127.3 mol)(4.00 g/mol) 5.1102 g 8. A tank containing 200.0 L of hydrogen gas at 0.0°C is kept at 156 kPa. The temperature is raised to 95°C, and the volume is decreased to 175 L. What is the new pressure of the gas? P1V1 P2V2 with T1 273 K and T1 T2 T2 95°C 273°C 368 K T P V V2T1 2 1 1 P2 (368 K)(156 kPa)(200.0 L) (175 L)(273 K) 9. The average molar mass of the components of air (mainly diatomic oxygen gas and diatomic nitrogen gas) is about 29 g/mol. What is the volume of 1.0 kg of air at atmospheric pressure and 20.0°C? PV nRT a. How does the pressure of the air on the outside of the two boxes compare? The pressure of the air is the same on the two boxes. b. How does the magnitude of the total force of the air on the two boxes compare? Because F PA the total force of the air is greater on the box with the greater area. The second box has twice the surface area, so it has twice the total force of the first box. 11. Meteorology A weather balloon used by meteorologists is made of a flexible bag that allows the gas inside to freely expand. If a weather balloon containing 25.0 m3 of helium gas is released from sea level, what is the volume of gas when the balloon reaches a height of 2100 m, where the pressure is 0.82105 Pa? Assume that the temperature is unchanged. P1V1 P2V2 P V P2 1 1 V2 (1.013105 Pa)(25.0 m3) 0.8210 Pa 5 nRT V P 3.1101 m3 1.0103 g 29 g/mol m where n M and T 20.0°C 273 293 K 1.0103 g V (8.31 Pam3/molK)(293 K) 29g/mol (1.013105 Pa) 0.83 m3 12. Gas Compression In a certain internalcombustion engine, 0.0021 m3 of air at atmospheric pressure and 303 K is rapidly compressed to a pressure of 20.1105 Pa and a volume of 0.0003 m3. What is the final temperature of the compressed gas? P1V1 P2V2 T1 T2 T P V P1V1 1 2 2 T2 288 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 2.4102 kPa page 348 10. Pressure and Force Suppose that you have two boxes. One is 20 cm 20 cm 20 cm. The other is 20 cm 20 cm 40 cm. Chapter 13 continued (303 K)(20.1105 Pa)(0.0003 m3) (1.01310 Pa)(0.0021 m ) 5 3 9102 K 13. Density and Temperature Starting at 0°C, how will the density of water change if it is heated to 4°C? To 8°C? As the water is heated from 0°C, the density will increase until it reaches a maximum at 4°C. On further heating to 8°C, the density of the water will decrease. 14. The Standard Molar Volume What is the volume of 1.00 mol of a gas at atmospheric pressure and a temperature of 273 K? nRT P V (1.00 mol)(8.31 Pam3/molK)(273 K) 1.01310 Pa 5 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 0.0224 m3 15. The Air in a Refrigerator How many moles of air are in a refrigerator with a volume of 0.635 m3 at a temperature of 2.00°C? If the average molar mass of air is 29 g/mol, what is the mass of the air in the refrigerator? PV RT n 5 3 (1.01310 Pa)(0.635 m ) 3 (8.31 Pam molK)(275 K) 28.1 mol m nM (28.1 mol)(29 g/mol) 0.81 kg 16. Critical Thinking Compared to the particles that make up carbon dioxide gas, the particles that make up helium gas are very small. What can you conclude about the number of particles in a 2.0-L sample of carbon dioxide gas compared to the number of particles in a 2.0-L sample of helium gas if both samples are at the same temperature and pressure? Physics: Principles and Problems There are an equal number of particles in the two samples. In an ideal gas, the size of the particles is not relevant to the volume of the gas or the pressure exerted by the gas. Section Review 13.2 Forces Within Liquids pages 349–351 page 351 17. Evaporation and Cooling In the past, when a baby had a high fever, the doctor might have suggested gently sponging off the baby with rubbing alcohol. Why would this help? Since alcohol evaporates easily, there is a very noticeable evaporative cooling effect. 18. Surface Tension A paper clip, which has a density greater than that of water, can be made to stay on the surface of water. What procedures must you follow for this to happen? Explain. The paper clip should be placed carefully and flatly onto the surface of the water. This will reduce the weight per unit area of water surface on which it will rest. The surface tension of the water then is sufficient to support the reduced weight per unit area of the paper clip. 19. Language and Physics The English language includes the terms adhesive tape and working as a cohesive group. In these terms, are adhesive and cohesive being used in the same context as their meanings in physics? Yes, adhesive tape is sticking to something different than tape. A cohesive group is a collection of people working together. 20. Adhesion and Cohesion In terms of adhesion and cohesion, explain why alcohol clings to the surface of a glass rod but mercury does not. Solutions Manual 289 Chapter 13 continued Adhesion is the force between unlike materials. Alcohol has a greater adhesive attraction to glass than mercury has. The cohesive forces of mercury are strong enough to overcome its adhesive force with glass. 21. Floating How can you tell that the paper clip in problem 18 was not floating? If the paper clip broke through the surface of the water, it sank. An object that floats would simply bob back to the surface. 22. Critical Thinking On a hot, humid day, Beth sat on the patio with a glass of cold water. The outside of the glass was coated with water. Her younger sister, Jo, suggested that the water had leaked through the glass from the inside to the outside. Suggest an experiment that Beth could do to show Jo where the water came from. Practice Problems 13.3 Fluids at Rest and in Motion pages 352–358 page 353 23. Dentists’ chairs are examples of hydrauliclift systems. If a chair weighs 1600 N and rests on a piston with a cross-sectional area of 1440 cm2, what force must be applied to the smaller piston, with a cross-sectional area of 72 cm2, to lift the chair? 290 Solutions Manual 1 8.0101 N 24. A mechanic exerts a force of 55 N on a 0.015 m2 hydraulic piston to lift a small automobile. The piston that the automobile sits on has an area of 2.4 m2. What is the weight of the automobile? (55 N)(2.4 m2) (0.015 m ) F A 1 2 8.8103 N F2 2 A 1 25. By multiplying a force, a hydraulic system serves the same purpose as a lever or seesaw. If a 400-N child standing on one piston is balanced by a 1100-N adult standing on another piston, what is the ratio of the areas of their pistons? F A 1 2 F2 A 1 A2 F2 400 N A1 F1 1100 N 0.4 The adult stands on the larger piston. 26. In a machine shop, a hydraulic lift is used to raise heavy equipment for repairs. The system has a small piston with a crosssectional area of 7.0102 m2 and a large piston with a cross-sectional area of 2.1101 m2. An engine weighing 2.7103 N rests on the large piston. a. What force must be applied to the small piston to lift the engine? F A 1 2 F2 A 1 (2.7103 N)(7.0102 m2) 2.110 m 1 2 9.0102 N b. If the engine rises 0.20 m, how far does the smaller piston move? V1 V2 and A1h1 A2h2 (2.1101 m2)(0.20 m) 7.010 m A h 1 1 h2 2 2 A 2 0.60 m Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Beth could weigh the glass before putting it in the refrigerator for a while to cool it down. Then she could remove it from the refrigerator and allow moisture to collect on the outside. Finally, she would weigh the glass a second time. If water simply leaks from the inside to the outside, the mass of the glass and water will be unchanged. However, if the moisture is condensation, there will be an increase in the mass at the second weighing. (1600 N)(72 cm2) 1440 cm F A 1 2 F2 2 A Chapter 13 continued page 356 27. Common brick is about 1.8 times denser than water. What is the apparent weight of a 0.20 m3 block of bricks under water? Fapparent Fg Fbuoyant brickVg waterVg (brick water)Vg (1.8103 kg/m3 1.00103 kg/m3) (0.20 m3)(9.80 1.6103 m/s2) N 28. A girl is floating in a freshwater lake with her head just above the water. If she weighs 610 N, what is the volume of the submerged part of her body? She is floating so she displaces a volume of water that weighs as much as she does. Fg Fbuoyant waterVg Fg V waterg 610 N 3 3 2 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. (1.0010 kg/m )(9.80 m/s ) 6.2102 m3 29. What is the tension in a wire supporting a 1250-N camera submerged in water? The volume of the camera is 16.5103 m3. To hold the camera in place the tension in the wire must equal the apparent weight of the camera. T Fapparent Fg Fbuoyant water and is barely submerged, leaving the bricks dry? The foam would displace V (1.0 m)(1.0 m)(0.10 m) 0.10 m3 of water. The weight of the foam is Fg, foam foamVg (1.0102 kg/m3)(0.10 m3) (9.80 m/s2) 98 N The buoyant force is Fbuoyant waterVg (1.00103 kg/m3) (0.10 m3)(9.80 m/s2) 980 N The weight of brick that you could stack is Fg, brick Fbuoyant Fg foam 980 N 98 N 8.8102 N 31. Canoes often have plastic foam blocks mounted under the seats for flotation in case the canoe fills with water. What is the approximate minimum volume of foam needed for flotation for a 480-N canoe? The buoyant force on the foam must equal 480 N. We are assuming the canoe is made of dense material. Fbuoyant waterVg Fbuoyant V waterg 480 N 3 3 2 (1.0010 kg/m )(9.80 m/s ) 4.9102 m3 Fg waterVg 1250 N (1.00103 kg/m3) (16.5103 m3)(9.80 m/s2) 1.09103 N 30. Plastic foam is about 0.10 times as dense as water. What weight of bricks could you stack on a 1.0 m 1.0 m 0.10 m slab of foam so that the slab of foam floats in Physics: Principles and Problems Section Review 13.3 Fluids at Rest and in Motion pages 352–358 page 358 32. Floating and Sinking Does a full soda pop can float or sink in water? Try it. Does it matter whether or not the drink is diet? All soda Solutions Manual 291 Chapter 13 continued pop cans contain the same volume of liquid, 354 mL, and displace the same volume of water. What is the difference between a can that sinks and one that floats? The difference is a lot of sugar. About one-fourth cup of sugar is dissolved in the regular drink, making it denser than water. The diet drink has a small amount of an artificial sweetener. The diet drink is less dense than the sugar-laden regular soft drink. 33. Floating and Density A fishing bobber made of cork floats with one-tenth of its volume below the water’s surface. What is the density of cork? The weight of the water displaced equals the weight of the bobber. Fg waterVwaterg corkVcork g water V Vcork cork water Therefore, 1 10 The cork is about one-tenth as dense as water. 34. Floating in Air A helium balloon rises because of the buoyant force of the air lifting it. The density of helium is 0.18 kg/m3, and the density of air is 1.3 kg/m3. How large a volume would a helium balloon need to lift a 10-N lead brick? Fapparent must equal 10 N to counteract the weight of the lead brick. Fapparent Fg Fbuoyant heliumVballoong airVballoong (helium air)Vballoong Fapparent Vballoon (helium air)g 10 N 3 3 2 (0.18 kg/m 1.3 kgm )(9.80 ms ) 0.9 m3 35. Transmission of Pressure A toy rocket launcher is designed so that a child stomps on a rubber cylinder, which increases the air pressure in a launching tube and pushes a foam rocket into the sky. If the child stomps with a force of 150 N on a 2.5103 m2 area piston, what is the additional force transmitted to the 4.0104 m2 launch tube? F1A2 F2 A 1 (150 N)(4.0104 m2) 3 2 2.510 m 24 N 292 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Thus, Chapter 13 continued 36. Pressure and Force An automobile weighing 2.3104 N is lifted by a hydraulic cylinder with an area of 0.15 m2. a. What is the pressure in the hydraulic cylinder? F A P 2.3104 N 2 0.15 m 1.5105 Pa b. The pressure in the lifting cylinder is produced by pushing on a 0.0082 m2 cylinder. What force must be exerted on this small cylinder to lift the automobile? F1A2 F2 A1 (2.3104 N)(0.0082 m2) 2 0.15 m 1.3103 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. a. A 1.0-kg block of aluminum or a 1.0-kg block of lead? Both aluminum and iron will sink to the bottom of the aquarium. Because aluminum is less dense than iron, 1 kg of aluminum has a greater volume than 1 kg of iron. Therefore, the block of aluminum will displace more water. 10-cm3 Both blocks will sink, and each will displace the same volume of water, 10 cm3. 38. Critical Thinking As you discovered in Practice Problem 4, a tornado passing over a house sometimes makes the house explode from the inside out. How might Bernoulli’s principle explain this phenomenon? What could be done to reduce the danger of a door or window exploding outward? Physics: Principles and Problems 13.4 Solids pages 359–363 page 362 39. A piece of aluminum house siding is 3.66 m long on a cold winter day of 28°C. How much longer is it on a very hot summer day at 39°C? L L1(T2 T1) 37. Displacement Which of the following displaces more water when it is placed in an aquarium? b. A block of aluminum or a block of lead? Practice Problems L2 L1 L1(T2 T1), so N 10-cm3 The fast-moving air of the tornado has a lower pressure than the still air inside the house. Therefore, the air inside the house is at a higher pressure and produces an enormous force on the windows, doors, and walls of the house. This pressure difference is reduced by opening doors and windows to let the air flow freely to the outside. (25106 °C1)(3.66 m) (39°C (28°C)) 6.1103 m 6.1 mm 40. A piece of steel is 11.5 cm long at 22°C. It is heated to 1221°C, close to its melting temperature. How long is it? L2 L1 L1(T2 T1) (0.115 m) (12106°C1) (0.115 m)(1221°C 22°C) 1.2101 m 12 cm 41. A 400-mL glass beaker at room temperature is filled to the brim with cold water at 4.4°C. When the water warms up to 30.0°C, how much water will spill from the beaker? At the beginning, 400 mL of 4.4°C water is in the beaker. Find the change in volume at 30.0°C. V V T (210106°C1)(400106 m3) (30.0°C 4.4°C) 2106 m3 2 mL Solutions Manual 293 Chapter 13 continued 42. A tank truck takes on a load of 45,725 L of gasoline in Houston, where the temperature is 28.0°C. The truck delivers its load in Minneapolis, where the temperature is 12.0°C. a. How many liters of gasoline does the truck deliver? By what percentage would the ruler be incorrect at 30.0°C? Because the steel shrinks, the distances between the millimeter marks decrease when cooled. % incorrect (100) L L V V V1 T (100)(Tf Ti) V 2 1 V1 T (100)(12106°C1) V2 V1 T V1 (950106 °C1)(45,725 (30.0°C 30.0°C) L) 0.072% (12.0°C 28.0°C) 45,725 L 4.4104 L b. What happened to the gasoline? The gasoline volume decreased because the temperature decreased. The mass of the gasoline remained the same. 43. A hole with a diameter of 0.85 cm is drilled into a steel plate. At 30.0°C, the hole exactly accommodates an aluminum rod of the same diameter. What is the spacing between the plate and the rod when they are cooled to 0.0°C? Laluminum L T (25106°C1) (0.85 cm)(0.0°C 30.0°C) 6.38104 cm For the steel, the diameter of the hole shrinks by Lsteel L T (12106°C1) (0.85 cm)(0.0°C 30.0°C) 3.06104 cm The spacing between the rod and the hole will be 12(6.4104 cm 3.1104 cm) 1.6104 cm 44. A steel ruler is marked in millimeters so that the ruler is absolutely correct at 30.0°C. 13.4 Solids pages 359–363 page 363 45. Relative Thermal Contraction On a hot day, you are installing an aluminum screen door in a concrete door frame. You want the door to fit well on a cold winter day. Should you make the door fit tightly in the frame or leave extra room? Fit the door tightly. Aluminum shrinks when cooled much more than concrete does. 46. States of Matter Why could candle wax be considered a solid? Why might it also be considered a viscous liquid? The wax could be considered a solid because it has a definite volume and shape. It could be considered a viscous liquid because the particles do not form a fixed crystalline pattern. 47. Thermal Expansion Can you heat a piece of copper enough to double its length? The thermal expansion coefficient for copper is 16106/°C. To double its length L L L T, which means that T1 1 T 1 1610 °C 6 1 63,000°C 294 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. The aluminum shrinks more than the steel. Let L be the diameter of the rod. Section Review Chapter 13 continued The copper would be vaporized at that temperature. States of Matter 48. States of Matter Does Table 13-2 provide a way to distinguish between solids and liquids? The coefficients of volume expansion are much greater for liquids than for solids. 49. Solids and Liquids A solid can be defined as a material that can be bent and will resist bending. Explain how these properties relate to the binding of atoms in a solid, but do not apply to a liquid. Particles in a solid are closer and, therefore, more tightly bound. They vibrate about a fixed position. This allows the solid to be bent, but it also resists bending. Particles in a liquid are farther apart and less tightly bound. Because the particles are free to flow past one another, a liquid cannot be bent. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 50. Critical Thinking The iron ring in Figure 13-23 was made by cutting a small piece from a solid ring. If the ring in the figure is heated, will the gap become wider or narrower? Explain your answer. fluids density viscosity solids density elasticity pressure Mastering Concepts page 368 52. How are force and pressure different? (13.1) Force depends only on the push or pull on an object. Pressure depends on the force as well as the area over which the force is applied. 53. A gas is placed in a sealed container, and some liquid is placed in a container of the same size. The gas and liquid both have definite volume. How do they differ? (13.1) The liquid’s volume will remain unchanged. The gas will expand to fill the volume of the container. 54. In what way are gases and plasmas similar? In what way are they different? (13.1) ■ Figure 13-23 The gap will become wider. All of the measurements of the ring increase when heated. Chapter Assessment Concept Mapping page 368 51. Complete the concept map below using the following terms: density, viscosity, elasticity, pressure. A term may be used more than once. Physics: Principles and Problems Both gases and plasmas have no definite volume and no definite shape. A gas is made of atoms. A plasma is made of positively charged ions and negatively charged electrons. The particles of a plasma are more energetic than the particles of a gas. Plasmas can conduct electricity. Gases cannot. 55. The Sun is made of plasma. How is this plasma different from the plasmas on Earth? (13.1) The Sun’s plasma is extremely hot, but more importantly, it is very dense— denser than most solids on Earth. 56. Lakes A frozen lake melts in the spring. What effect does this have on the temperature of the air above the lake? (13.2) Solutions Manual 295 Chapter 13 continued To melt, the ice must absorb energy in the amount of its heat of fusion from the air and water. It will cool the air above it. 57. Hiking Canteens used by hikers often are covered with canvas bags. If you wet the canvas bag covering a canteen, the water in the canteen will be cooled. Explain. (13.2) The water evaporates into the air, absorbing energy from the canteen and the water inside. 58. What do the equilibrium tubes in Figure 13-24 tell you about the pressure exerted by a liquid? (13.3) 61. Does Archimedes’ principle apply to an object inside a flask that is inside a spaceship in orbit? (13.3) No, it does not. The apparent weight of the displaced fluid is zero because the fluid is in free-fall. Thus, there is no buoyant force. 62. A stream of water goes through a garden hose into a nozzle. As the water speeds up, what happens to the water pressure? (13.3) The water pressure decreases because of Bernoulli’s principle. 63. How does the arrangement of atoms in a crystalline substance differ from that in an amorphous substance? (13.4) The atoms in a crystalline substance are arranged in an ordered pattern. In the amorphous substance, the atoms are randomly arranged. 64. Does the coefficient of linear expansion depend on the unit of length used? Explain. (13.4) Figure 13-24 The equilibrium tubes illustrate that pressure is independent of the shape of the container. 59. According to Pascal’s principle, what happens to the pressure at the top of a container if the pressure at the bottom is increased? (13.3) Changes in pressure are distributed equally to all parts of the container. The pressure at the top increases. 60. How does the water pressure 1 m below the surface of a small pond compare with the water pressure the same distance below the surface of a lake? (13.3) Size or shape of the body of water does not matter, only the depth. The pressure is the same in each case. 296 Solutions Manual Applying Concepts page 368–369 65. A rectangular box with its largest surface resting on a table is rotated so that its smallest surface is now on the table. Has the pressure on the table increased, decreased, or remained the same? The pressure increased. The weight stayed the same, but the weight per area increased. 66. Show that a pascal is equivalent to a kg/ms2. Pa N/m2 (kgm/s2)/m2 kg/ms2 Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. ■ No. The coefficient is a measure of the expansion of an object relative to its total length. Units and total length do not change the coefficient of linear expansion. Chapter 13 continued 67. Shipping Cargo Compared to an identical empty ship, would a ship filled with tabletennis balls sink deeper into the water or rise in the water? Explain. It would sink deeper into the water because it would have a greater weight. 68. Drops of mercury, water, ethanol, and acetone are placed on a smooth, flat surface, as shown in Figure 13-25. From this figure, what can you conclude about the cohesive forces in these liquids? the foil, will the size of the hole decrease or increase? Explain. As you heat the foil, the hole becomes larger. Heating transfers more energy to particles of the aluminum, causing the volume of the aluminum to increase. 72. Equal volumes of water are heated in two narrow tubes that are identical, except that tube A is made of soft glass and tube B is made of ovenproof glass. As the temperature increases, the water level rises higher in tube B than in tube A. Give a possible explanation. The ovenproof glass expands less than the soft glass when heated. The water does not rise as high in A because the soft glass tube is expanding in volume. 73. A platinum wire easily can be sealed in a glass tube, but a copper wire does not form a tight seal with the glass. Explain. ■ Figure 13-25 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. The cohesive forces are strongest in mercury and weakest in acetone. The stronger the cohesive force, the more spherical the drop will be. 69. How deep would a water container have to be to have the same pressure at the bottom as that found at the bottom of a 10.0-cm deep beaker of mercury, which is 13.55 times as dense as water? Pwater Pmercury waterhwaterg mercuryhmercuryg mercury hwater hmercury Platinum’s coefficient of thermal expansion is similar to that of glass, so it expands and contracts as the glass does. Copper has a much larger coefficient than glass. 74. Five objects with the following densities are put into a tank of water. a. 0.85 g/cm3 d. 1.15 g/cm3 3 b. 0.95 g/cm e. 1.25 g/cm3 c. 1.05 g/cm3 The density of water is 1.00 g/cm3. The diagram in Figure 13-26 shows six possible positions of these objects. Select a position, from 1 to 6, for each of the five objects. Not all positions need to be selected. water (13.55)(10.0 cm) 136 cm 1 2 3 4 5 70. Alcohol evaporates more quickly than water does at the same temperature. What does this observation allow you to conclude about the properties of the particles in the two liquids? The cohesive forces of water are greater than those of alcohol. 71. Suppose you use a hole punch to make a circular hole in aluminum foil. If you heat Physics: Principles and Problems 6 ■ Figure 13-26 The positions of the objects should be a–1, b–2, c–6, d–6, e–6 Solutions Manual 297 Chapter 13 continued Mastering Problems 13.1 Properties of Fluids pages 369–370 Level 1 75. Textbooks A 0.85-kg physics book with dimensions of 24.0 cm 20.0 cm is at rest on a table. a. What force does the book apply to the table? The force on the table is the weight of the book. W mg (0.85 kg)(9.80 m/s2) 8.3 N b. What pressure does the book apply? The pressure applied by the book is F P A mg lw dissolved in a 2-L bottle of soda. The molar mass of CO2 is 44 g/mol. a. How many moles of carbon dioxide are in the 2-L bottle? (1 L 0.001 m3) From the ideal gas law PV RT n (1.01105 Pa)(0.0080 m3) (8.31 Pam molK)(300.0 K) 3 0.32 moles b. What is the mass of the carbon dioxide in the 2-L bottle of soda? The molecular weight of carbon dioxide is M 12 2(16) 44 g/mol Therefore, the mass is m nM (0.32 mol)(44 g/mol) (0.85 kg)(9.80 ms2) 1 1 (2.4010 m)(2.0010 14 g m) 1.7102 Pa F A P mg r 2 (75 kg)(9.80 ms2) (0.070 m) 2 Cylinder 4.8104 Pa Piston 20 cm 77. What is the total downward force of the atmosphere on the top of your head right now? Assume that the top of your head has an area of about 0.025 m2. F PA (1.01105 Pa)(0.025 m2) 2.5103 N 78. Soft Drinks Sodas are made fizzy by the carbon dioxide (CO2) dissolved in the liquid. An amount of carbon dioxide equal to about 8.0 L of carbon dioxide gas at atmospheric pressure and 300.0 K can be 298 Solutions Manual Gas ■ Figure 13-27 Because the pressure is kept constant, V1/T 1 V2 /T 2. The height of the piston is directly proportional to the volume of the cylinder. Therefore, h1 h2 T1 T2 Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 76. A 75-kg solid cylinder that is 2.5 m long and has an end radius of 7.0 cm stands on one end. How much pressure does it exert? 79. As shown in Figure 13-27, a constantpressure thermometer is made with a cylinder containing a piston that can move freely inside the cylinder. The pressure and the amount of gas enclosed in the cylinder are kept constant. As the temperature increases or decreases, the piston moves up or down in the cylinder. At 0°C, the height of the piston is 20 cm. What is the height of the piston at 100°C? Chapter 13 continued h1T2 h2 T 1 (20 cm)(373 K) 273 K 3101 cm Level 2 80. A piston with an area of 0.015 m2 encloses a constant amount of gas in a cylinder with a volume of 0.23 m3. The initial pressure of the gas is 1.5105 Pa. A 150-kg mass is then placed on the piston, and the piston moves downward to a new position, as shown in Figure 13-28. If the temperature is constant, what is the new volume of the gas in the cylinder? Volume 0.23 m3 Piston area 0.015 m2 Volume ? (One pound per square inch equals 6.90103 Pa.) The term gauge pressure means the pressure above atmospheric pressure. Thus, the actual pressure in the tire is 1.01105 Pa (30.0 psi)(6.90103 Pa/psi) 3.08105 Pa. As the car is driven, the tire’s temperature increases, and the volume and pressure increase. Suppose you filled a car’s tire to a volume of 0.55 m3 at a temperature of 280 K. The initial pressure was 30.0 psi, but during the drive, the tire’s temperature increased to 310 K and the tire’s volume increased to 0.58 m3. a. What is the new pressure in the tire? P V1 P V2 1 2 T1 T2 P V T T1V2 1 1 2 P2 (3.08105 Pa)(0.55 m3)(310 K) 3 (280 K)(0.58 m ) 150 kg 3.2105 Pa b. What is the new gauge pressure? (30.0 psi)(0.55 m3)(310 K) Pgauge (280 K)(0.58 m3) 31 psi Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Gas Gas ■ 13.3 Fluids at Rest and in Motion page 370 Level 1 82. Reservoirs A reservoir behind a dam is 17-m deep. What is the pressure of the water at the following locations? Figure 13-28 P1V1 P2V2 P1V1 V2 a. the base of the dam P2 P hg P1V1 P1 mAg (1.00103 kg/m3)(17 m) (9.80 m/s2) (1.5105 Pa)(0.23 m3) (150 kg)(9.80 m/s2) 0.015 m 1.5105 Pa 2 0.14 m3 1.7105 Pa b. 4.0 m from the top of the dam P hg (1.00103 kg/m3)(4.0 m) (9.80 m/s2) Level 3 81. Automobiles A certain automobile tire is specified to be used at a gauge pressure of 30.0 psi, or 30.0 pounds per square inch. Physics: Principles and Problems 3.9104 Pa Solutions Manual 299 Chapter 13 continued 83. A test tube standing vertically in a test-tube rack contains 2.5 cm of oil ( 0.81 g/cm3) and 6.5 cm of water. What is the pressure exerted by the two liquids on the bottom of the test tube? P Poil Pwater oilhoilg waterhwaterg (810 kg/m3)(0.025 m)(9.80 m/s2) (1.00103 kg/m3)(0.065 m) (9.80 m/s2) 8.4102 Pa 84. Antiques An antique yellow metal statuette of a bird is suspended from a spring scale. The scale reads 11.81 N when the statuette is suspended in air, and it reads 11.19 N when the statuette is completely submerged in water. a. Find the volume of the statuette. Fbuoyant waterVg Fg Fapparent b. The rock is removed from the aquarium, and the amount of water is adjusted until the scale again reads 195 N. A fish weighing 2 N is added to the aquarium. What is the scale reading with the fish in the aquarium? Fg 195 N 2 N 197 N In each case the buoyant force is equal to the weight of the water displaced. 86. What is the size of the buoyant force on a 26.0-N ball that is floating in fresh water? If the ball is floating Fbuoyant Fg 26.0 N 87. What is the apparent weight of a rock submerged in water if the rock weighs 45 N in air and has a volume of 2.1103 m3? Fapparent Fg Fbuoyant Fg waterVg Thus, Fg Fapparent V g water 6.33105 m3 b. Is the bird made of gold ( 19.3103 kg/m3) or gold-plated aluminum ( 2.7103 kg/m3)? m V Fg Vg 11.81 N (6.33105 m3)(9.80 ms2) 19.0103 kg/m3 The statuette is made of gold. 85. During an ecology experiment, an aquarium half-filled with water is placed on a scale. The scale shows a weight of 195 N. a. A rock weighing 8 N is added to the aquarium. If the rock sinks to the bottom of the aquarium, what will the scale read? Solutions Manual 45 N (1.00103 kg/m3) (2.1103 m3)(9.80 m/s2) 24 N 88. What is the maximum weight that a balloon filled with 1.00 m3 of helium can lift in air? Assume that the density of air is 1.20 kg/m3 and that of helium is 0.177 kg/m3. Neglect the mass of the balloon. Fapparent Fg Fbuoyant heliumVg airVg (helium air)Vg (0.177 kg/m3 1.20 kg/m3) (1.00 m3)(9.80 m/s2) 10.0 N Level 2 89. If a rock weighs 54 N in air and has an apparent weight of 46 N when submerged in a liquid with a density twice that of water, what will be its apparent weight when it is submerged in water? Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 11.81 N 11.19 N (1.00103 kgm3)(9.80 ms2) 300 Fg 195 N 8 N 203 N Chapter 13 continued Fapparent, water Fg waterVg and Fapparent, liquid Fg 2waterVg Fg Fapparent, liquid or V 2 g water Substitute this into the first equation. Fg Fapparent, liquid Fapparent, water Fg waterg 2waterg 1 Fg (Fg Fapparent, liquid) 2 1 (Fg Fapparent, liquid) 2 1 (54 N 46 N) 2 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 5.0101 N Level 3 90. Oceanography As shown in Figure 13-29, a large buoy used to support an oceanographic research instrument is made of a cylindrical, hollow iron tank. The tank is 2.1 m in height and 0.33 m in diameter. The total mass of the buoy and the research instrument is about 120 kg. The buoy must float so that one end is above the water to support a radio transmitter. Assuming that the mass of the buoy is evenly distributed, how much of the buoy will be above the waterline when it is floating? ?m 2.1 m 0.33 m ■ Figure 13-29 The height of the buoy above water is water l total labove 1 V V buoy 1 m water r 2h l total m 1 l total r 2h water 1 120 kg 1 2 (0.33 m) (2.1 m)(1.00103 kgm3) 2 (2.1 m) 0.70 m Physics: Principles and Problems Solutions Manual 301 Chapter 13 continued 13.4 Solids pages 370–371 Level 1 91. A bar of an unknown metal has a length of 0.975 m at 45°C and a length of 0.972 m at 23°C. What is its coefficient of linear expansion? L2 L1 L1(T2 T1) 0.972 m 0.975 m (0.975 m)(23°C 45°C) 1.4104 °C1 92. An inventor constructs a thermometer from an aluminum bar that is 0.500 m in length at 273 K. He measures the temperature by measuring the length of the aluminum bar. If the inventor wants to measure a 1.0-K change in temperature, how precisely must he measure the length of the bar? T 1.0 K 1.0°C L L1 T (25106°C1)(0.500 m)(1.0°C) 1.3105 m L L1 T L1(T2 T1) (12106°C1)(300 m) (30°C (10°C)) 0.1 m 94. What is the change in length of a 2.00-m copper pipe if its temperature is raised from 23°C to 978°C? L L1 T L1(T2 T1) (16106°C1)(2.00 m) (978°C 23°C) 3.1102 m 95. What is the change in volume of a 1.0-m3 concrete block if its temperature is raised 45°C? 302 Solutions Manual (36106°C1)(1.0 m3)(45°C) 1.6103 m3 96. Bridges Bridge builders often use rivets that are larger than the rivet hole to make the joint tighter. The rivet is cooled before it is put into the hole. Suppose that a builder drills a hole 1.2230 cm in diameter for a steel rivet 1.2250 cm in diameter. To what temperature must the rivet be cooled if it is to fit into the rivet hole, which is at 20.0°C? L2 L1 L1(T2 T1) (L L ) 2 1 T2 T1 L 1 1.2230 cm 1.2250 cm 20.0°C (12106°C1)(1.2250 cm) 1.2102°C Level 2 97. A steel tank filled with methanol is 2.000 m in diameter and 5.000 m in height. It is completely filled at 10.0°C. If the temperature rises to 40.0°C, how much methanol (in liters) will flow out of the tank, given that both the tank and the methanol will expand? V V1 T (methanol steel)(r 2h)(T2 T1) (1200106°C1 35106°C1) ()(1.000 m)2(5.000 m) (40.0°C 10.0°C) 0.55 m3 98. An aluminum sphere is heated from 11°C to 580°C. If the volume of the sphere is 1.78 cm3 at 11°C, what is the increase in volume of the sphere at 580°C? V V1 T (75106°C1)(1.78 cm3) (580°C 11°C) 7.6102 cm3 Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 93. Bridges How much longer will a 300-m steel bridge be on a 30°C day in August than on a 10°C night in January? V V1 T Chapter 13 continued 99. The volume of a copper sphere is 2.56 cm3 after being heated from 12°C to 984°C. What was the volume of the copper sphere at 12°C? V2 V1 V1 T V1(1 T) V2 V1 1 T 3 2.56 cm 6 (1 (4810 °C1)(984°C 12°C)) 2.4 cm3 Level 3 100. A square of iron plate that is 0.3300 m on each side is heated from 0°C to 95°C. a. What is the change in the length of the sides of the square? L L1 T L1 (T2 T1) (12106°C1)(0.3300 m)(95°C 0°C) 3.8104 m b. What is the relative change in area of the square? A relative change A1 A A A1 2 1 L L L1 2 2 2 1 2 (L L)2 L L1 2 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1 1 2 (0.3300 m 3.8104 m)2 (0.3300 m)2 (0.3300 m) 2 2.3103 101. An aluminum cube with a volume of 0.350 m3 at 350.0 K is cooled to 270.0 K. a. What is its volume at 270.0 K? V2 V1 V1 T V1(1 T ) V1(1 (T2 T1)) (0.350 m3)(1 (75106°C1)(270.0 K 350.0 K)) 0.348 m3 b. What is the length of a side of the cube at 270.0 K? 1 L (V 2) 3 1 (0.348 m3) 3 0.703 m Physics: Principles and Problems Solutions Manual 303 Chapter 13 continued 102. Industry A machinist builds a rectangular mechanical part for a special refrigerator system from two rectangular pieces of steel and two rectangular pieces of aluminum. At 293 K, the part is a perfect square, but at 170 K, the part becomes warped, as shown in Figure 13-30. Which parts were made of steel and which were made of aluminum? 2 2 T 293 K 1 3 1 T 170 K 4 3 4 ■ Figure 13-30 Parts 1 and 2 experienced a greater reduction in length than parts 3 and 4; therefore, parts 1 and 2 must have been made of aluminum, which has a larger coefficient of expansion than steel. Mixed Review page 371 Level 1 103. What is the pressure on the hull of a submarine at a depth of 65 m? P Patmosphere watergh (1.01105 Pa) (1.00103 kg/m3)(9.80 m/s2)(65 m) 7.4105 Pa P1V1 P2V2 P1V1 V2 P 2 (Patmosphere watergh)V1 Patmosphere (1.01105 Pa (1.00103 kgm3)(9.80 ms2)(5.0 m))(4.2106 m3) 5 1.0110 Pa 6.2106 m3 105. An 18-N bowling ball floats with about half of the ball submerged. a. What is the diameter of the bowling ball? Vball Fg Vwaterg g, 2 d 3 d 3 where Vball r 3 3 3 2 6 4 4 1 d Then Fg g, 3 2 304 Solutions Manual 6 Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 104. Scuba Diving A scuba diver swimming at a depth of 5.0 m under water exhales a 4.2106 m3 bubble of air. What is the volume of that bubble just before it reaches the surface of the water? Chapter 13 continued so that d g 3 12Fg (12)(18 N) (1.0010 kgm )(9.80 ms ) 3 3 3 2 0.19 m b. What would be the approximate apparent weight of a 36-N bowling ball? Half the ball sank when the weight was 18 N. The apparent weight of the 36-N ball should be near zero. Level 2 106. An aluminum bar is floating in a bowl of mercury. When the temperature is increased, does the aluminum float higher or sink deeper into the mercury? The volume coefficient of expansion of mercury is greater than the volume coefficient of expansion of aluminum. Therefore, as they are heated, the aluminum becomes denser relative to the mercury and would sink deeper into the mercury. 107. There is 100.0 mL of water in an 800.0-mL soft-glass beaker at 15.0°C. How much will the water level have dropped or risen when the bottle and water are heated to 50.0°C? The water expands: V V T (210106°C1)(100.0 mL)(35.0°C) 0.735 mL Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. The bottle expands: V V T (27106°C1)(800.0 mL)(35.0°C) 0.756 mL The water level will go down slightly, but not enough to notice. 108. Auto Maintenance A hydraulic jack used to lift cars for repairs is called a three-ton jack. The large piston is 22 mm in diameter, and the small one is 6.3 mm in diameter. Assume that a force of 3 tons is 3.0104 N. a. What force must be exerted on the small piston to lift a 3-ton weight? F A 1 2 F2 A1 F r r 1 2 1 2 2 d 2 d1 2 F1 2 6.3 mm 2 22 mm (3.0104 N) 2.5103 N Physics: Principles and Problems Solutions Manual 305 Chapter 13 continued b. Most jacks use a lever to reduce the force needed on the small piston. If the resistance arm is 3.0 cm, how long must the effort arm of an ideal lever be to reduce the force to 100.0 N? FrLr FeLe The pressure in the water is (8.7107 Pa)/(1.01105 Pa) 860 times greater than standard air pressure. Therefore, the density of air would be 860 times greater than the density of the air on the surface of the ocean. FrLr Le Fe Thinking Critically (2.5103 N)(3.0 cm) 100.0 N 75 cm 109. Ballooning A hot-air balloon contains a fixed volume of gas. When the gas is heated, it expands and pushes some gas out at the lower, open end. As a result, the mass of the gas in the balloon is reduced. Why would the air in a balloon have to be hotter to lift the same number of people above Vail, Colorado, which has an altitude of 2400 m, than above the tidewater flats of Virginia, which have an altitude of 6 m? 110. The Living World Some plants and animals are able to live in conditions of extreme pressure. a. What is the pressure exerted by the water on the skin of a fish or worm that lives near the bottom of the Puerto Rico Trench, 8600 m below the surface of the Atlantic Ocean? Use 1030 kg/m3 for the density of seawater. The pressure is P gh (1030 kg/m3)(9.80 m/s2)(8600 m) 8.7107 Pa b. What would be the density of air at that pressure, relative to its density above the surface of the ocean? 306 Solutions Manual When it was floating, the bowl displaced a volume of water that weighed as much as it did. When it sank, it displaced a volume of water that weighed less than the bowl, because the buoyancy force is the weight of the water displaced. In the second case, the bowl displaces less water and the level in the sink goes down. 112. Apply Concepts Persons confined to bed are less likely to develop bedsores if they use a waterbed rather than an ordinary mattress. Explain. The surface of the waterbed conforms more than the surface of a mattress to the contours of your body. One “sinks” more easily into a waterbed. Because H O mattress, the buoyant force from 2 a waterbed is less. 113. Analyze and Conclude One method of measuring the percentage of body fat is based on the fact that fatty tissue is less dense than muscle tissue. How can a person’s average density be assessed with a scale and a swimming pool? What measurements does a physician need to record to find a person’s average percentage of body fat? The physician weighs the person normally and then weighs the person totally submerged. Weight has to be Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Atmospheric pressure is lower at higher altitudes. Therefore, the mass of the volume of fluid displaced by a balloon of the same volume is less at higher altitudes. To obtain the same buoyant force at higher altitudes, a balloon must expel more gas, requiring higher temperatures. page 372 111. Apply Concepts You are washing dishes in the sink. A serving bowl has been floating in the sink. You fill the bowl with water from the sink, and it sinks to the bottom. Did the water level in the sink go up or down when the bowl was submerged? Chapter 13 continued added to the weighing device because the density of a human is normally less than the density of water. The volume of water displaced by the person also should be measured. The average density of the person can be calculated from the balance of forces that hold the person in equilibrium underwater. 114. Analyze and Conclude A downward force of 700 N is required to fully submerge a plastic foam sphere, as shown in Figure 13-31. The density of the foam is 95 kg/m3. a. What percentage of the sphere would be submerged if the sphere were released to float freely? The density of the foam relative to the 3 95 kgm water is 0.095, so 3 3 1.0010 kgm 700 N 9.5 percent of the floating sphere would be submerged. b. What is the weight of the sphere in air? ■ Figure 13-31 The weight of water displaced by 9.5 percent of the sphere’s volume balances the entire weight of the sphere, Fg. An additional 700 N is needed to submerge the remaining 90.5 percent of the sphere’s volume. Thus, Fg 700 N 0.095 0.905 Fg 7101 N Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. c. What is the volume of the sphere? Fbuoyant Fg Fdown waterVg foamVg Fdown Fdown V (water down)g 700 N 3 3 3 2 (1.0010 kgm 95 kgm )(9.80 ms ) 8102 m3 115. Apply Concepts Tropical fish for aquariums are often transported home from pet shops in transparent plastic bags filled mostly with water. If you placed a fish in its unopened transport bag in a home aquarium, which of the cases in Figure 13-32 best represents what would happen? Explain your reasoning. Bag water level Aquarium water level Bag water level Bag water level Aquarium water level ■ Physics: Principles and Problems Aquarium water level Figure 13-32 Solutions Manual 307 Chapter 13 continued The density of the water in the bag, the fish, and the plastic are all near the density of the water in the aquarium. Therefore, the bag should float with the water level in the bag at the same height as the water level in the aquarium. Writing in Physics page 372 116. Some solid materials expand when they are cooled. Water between 4° and 0°C is the most common example, but rubber bands also expand in length when cooled. Research what causes this expansion. 117. Research Joseph Louis Gay-Lussac and his contributions to the gas laws. How did Gay-Lussac’s work contribute to the discovery of the formula for water? Gay-Lussac was a French scientist who also was interested in highaltitude balloon ascents. He discovered that when gases are at the same temperature and pressure, their volumes react in ratios of small, whole numbers. Gay-Lussac’s work contributed to the discovery of water’s formula by showing that two volumes of hydrogen gas react with one volume 308 Solutions Manual Cumulative Review page 372 118. Two blocks are connected by a string over a frictionless, massless pulley such that one is resting on an inclined plane and the other is hanging over the top edge of the plane, as shown in Figure 13-33. The hanging block has a mass of 3.0 kg and the block on the plane has a mass of 2.0 kg. The coefficient of kinetic friction between the block and the inclined plane is 0.19. Answer the following questions assuming the blocks are released from rest. (Chapter 5) a. What is the acceleration of the blocks? 2.0 kg 3.0 kg 45° ■ Figure 13-33 For the 2-kg block, T kFN mg sin ma where FN mg cos is the normal force. Thus, T kmg cos mg sin ma T (mg)(k cos sin ) ma For the 3-kg block, Mg T Ma or, T Mg Ma Substitute this for T in the 2-kg block equation. Mg Ma (mg)(k cos sin ) ma Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Rubber bands are made of long rubber molecules called polymers that act like chains with many long links. The properties of rubber come from the ability of the chain links to twist and turn. When the rubber is colder, the polymer links are stretched out in a straight line, like the links in a steel chain that you hold at one end and let hang freely. Because the links are ordered that way, the polymers have relatively little disorder, or entropy. Adding heat to the polymers increases their thermal motion. The links begin to shake about and their disorder increases. If you shake a chain like this, you will see that its average length becomes less than if the chain were hanging motionless. of oxygen gas. Avogadro built on Gay-Lussac’s work to develop the relationship between moles of a gas and volume. Chapter 13 continued Solve for acceleration. Mg (mg)( cos sin ) mM k a (3.0 kg)(9.80 ms2) (2.0 kg)(9.80 ms2)((0.19)(cos 45°) sin 45°) 2.0 kg 3.0 kg 2.6 m/s2 b. What is the tension in the string connecting the blocks? To find T, use the equation for the 3-kg block. T Mg Ma M(g a) (3.0 kg)(9.80 m/s2 2.6 m/s2) 22 N 119. A compact car with a mass of 875 kg, moving south at 15 m/s, is struck by a full-sized car with a mass of 1584 kg, moving east at 12 m/s. The two cars stick together, and momentum is conserved. (Chapter 9) a. Sketch the situation, assigning coordinate axes and identifying “before” and “after.” y-axis Before A (State 1) B Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. x-axis mB 1584 kg vB 12 m/s mA 875 kg vA 15 m/s After (State 2) v2 ? p2 ? m2 2459 kg b. Find the direction and speed of the wreck immediately after the collision, remembering that momentum is a vector quantity. pA1 mAvA (875 kg)(15 m/s) 1.31104 kgm/s south pB1 mBvB (1584 kg)(12 m/s) 1.90104 kgm/s east p2 pA12 pB12 (1.31 104 kg m/s )2 (1.901 04 kg m/s)2 2.3104 kgm/s p pA1 B1 tan Physics: Principles and Problems Solutions Manual 309 Chapter 13 continued p pA1 Challenge Problem B1 tan1 1.90104 kgm/s 1.3110 kgm/s tan1 4 55° east of south p2 2.3104 kgm/s v2 m2 2459 kg 9.4 m/s c. The wreck skids along the ground and comes to a stop. The coefficient of kinetic friction while the wreck is skidding is 0.55. Assume that the acceleration is constant. How far does the wreck skid after impact? To find the distance, use the equation of motion: v 2 vi2 2a(d di) where the final velocity is zero and di 0. Solve for d: v i2 page 363 You need to make a 1.00-m-long bar that expands with temperature in the same way as a 1.00-m-long bar of copper would. As shown in the figure at the right, your bar must be made from a bar of iron and a bar of aluminum attached end to end. How long should each of them be? Lcopper Laluminum Liron and copperLcopper T (aluminumLaluminum ironLiron) T Substituting Laluminum Lcopper Liron, this gives ( iron aluminum (16106°C1 25106°C1)(1.00 m) °C 2510 °C 6 1 6 1 1210 d 2a )L copper aluminum copper Liron 0.69 m To find acceleration, notice that the force that slows the cars equals the frictional force. 1.00 m 0.69 m 0.31 m a kg The distance is then v 2 2kg 0 d (9.4 ms)2 (2)(0.55)(9.80 ms ) 2 8.2 m 120. A 188-W motor will lift a load at the rate (speed) of 6.50 cm/s. How great a load can the motor lift at this rate? (Chapter 10) v 6.50 cm/s 0.0650 m/s P F Fv W t Fd t d t P Fgv P v 188 W 0.0650 m/s Fg 2.89103 N 310 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. (ma mb)a k(ma mb)g Laluminum Lcopper Liron CHAPTER 14 Vibrations and Waves PEsp kx 2 14.1 Periodic Motion pages 375–380 x page 378 1. How much force is necessary to stretch a spring 0.25 m when the spring constant is 95 N/m? F kx (95 N/m)(0.25 m) 24 N 2. A spring has a spring constant of 56 N/m. How far will it stretch when a block weighing 18 N is hung from its end? F kx 18 N F x 0.32 m 56 N/m k 3. What is the spring constant of a spring that stretches 12 cm when an object weighing 24 N is hung from it? Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1 2 Practice Problems F kx F k 0.61 m 256 N/m k 2PEsp (2)(48 J) page 379 6. What is the period on Earth of a pendulum with a length of 1.0 m? g /s 9.80 m l 1.0 m T 2 2 2 2.0 s 7. How long must a pendulum be on the Moon, where g 1.6 m/s2, to have a period of 2.0 s? g l T 2 T 2 2 2.0 s 2 2 l g (1.6 m/s2) 0.16 m 8. On a planet with an unknown value of g, the period of a 0.75-m-long pendulum is 1.8 s. What is g for this planet? g l T 2 2 2 2 2 g l (0.75 m) 9.1 m/s2 x T 24 N 0.12 m 2.0102 N/m 4. A spring with a spring constant of 144 N/m is compressed by a distance of 16.5 cm. How much elastic potential energy is stored in the spring? 1 2 PEsp kx 2 1 2 (144 N/m)(0.165 m)2 1.96 J 5. A spring has a spring constant of 256 N/m. How far must it be stretched to give it an elastic potential energy of 48 J? Physics: Principles and Problems 1.8 s Section Review 14.1 Periodic Motion pages 375–380 page 380 9. Hooke’s Law Two springs look alike but have different spring constants. How could you determine which one has the greater spring constant? Hang the same object from both springs. The one that stretches less has the greater spring constant. 10. Hooke’s Law Objects of various weights are hung from a rubber band that is suspended from a hook. The weights of the objects are plotted on a graph against the Solutions Manual 311 Chapter 14 continued stretch of the rubber band. How can you tell from the graph whether or not the rubber band obeys Hooke’s law? If the graph is a straight line, the rubber band obeys Hooke’s law. If the graph is curved, it does not. 11. Pendulum How must the length of a pendulum be changed to double its period? How must the length be changed to halve the period? 1 PEsp kx 2, so 2 4.0 The energy of the first spring is 4.0 times greater than the energy of the second spring. 12. Energy of a Spring What is the difference between the energy stored in a spring that is stretched 0.40 m and the energy stored in the same spring when it is stretched 0.20 m? l l2 1 To double the period: l l 2 2, so 2 4 l1 l1 The length must be quadrupled. To halve the period: T2 T1 1 1 , so 2 4 l l l2 l2 1 1 The length is reduced to one-fourth its original length. 13. Resonance If a car’s wheel is out of balance, the car will shake strongly at a specific speed, but not when it is moving faster or slower than that speed. Explain. At that speed, the tire’s rotation frequency matches the resonant frequency of the car. 312 Solutions Manual a. What is the speed of sound of the clock’s chime in air? v d t 515 m 1.50 s 343 m/s b. The sound wave has a frequency of 436 Hz. What is the period of the wave? 1 f T 1 436 Hz 2.29103 s c. What is the wave’s wavelength? v f 343 m/s 436 Hz 0.787 m 16. A hiker shouts toward a vertical cliff 465 m away. The echo is heard 2.75 s later. a. What is the speed of sound of the hiker’s voice in air? (2)(465 m) 2.75 s d v 338 m/s t Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. l 2 T 2 , so T2 T1 Practice Problems page 386 15. A sound wave produced by a clock chime is heard 515 m away 1.50 s later. (0.40 m)2 (0.20 m)2 T T1 Both are periodic motions. In uniform circular motion, the accelerating force is not proportional to the displacement. Also, simple harmonic motion is onedimensional and uniform circular motion is two-dimensional. 14.2 Wave Properties pages 381–386 PE1 x 2 1 PE2 x22 g 14. Critical Thinking How is uniform circular motion similar to simple harmonic motion? How are they different? Chapter 14 continued b. The wavelength of the sound is 0.750 m. What is its frequency? v 338 m/s 0.750 m v f, so f 451 Hz c. What is the period of the wave? 1 1 T 2.22103 s 451 Hz f 17. If you want to increase the wavelength of waves in a rope, should you shake it at a higher or lower frequency? at a lower frequency, because wavelength varies inversely with frequency 18. What is the speed of a periodic wave disturbance that has a frequency of 3.50 Hz and a wavelength of 0.700 m? v f (0.700 m)(3.50 Hz) 2.45 m/s 19. The speed of a transverse wave in a string is 15.0 m/s. If a source produces a disturbance that has a frequency of 6.00 Hz, what is its wavelength? v f 15.0 m/s 6.00 Hz Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. v f, so 2.50 m 20. Five pulses are generated every 0.100 s in a tank of water. What is the speed of propagation of the wave if the wavelength of the surface wave is 1.20 cm? 0.10 0 s 0.0200 s/pulse, so 5 pulses T 0.0200 s vT, so v T 1.20 cm 0.0200 s 60.0 cm/s 0.600 m/s 21. A periodic longitudinal wave that has a frequency of 20.0 Hz travels along a coil spring. If the distance between successive compressions is 0.600 m, what is the speed of the wave? v f (0.600 m)(20.0 Hz) 12.0 m/s Section Review 14.2 Wave Properties pages 381–386 page 386 22. Speed in Different Media If you pull on one end of a coiled-spring toy, does the pulse reach the other end instantaneously? What happens if you pull on a rope? What happens if you hit the end of a metal rod? Compare and contrast the pulses traveling through these three materials. It takes time for the pulse to reach the other end in each case. It travels faster on the rope than on the spring, and fastest in the metal rod. 23. Wave Characteristics You are creating transverse waves in a rope by shaking your hand from side to side. Without changing the distance that your hand moves, you begin to shake it faster and faster. What happens to the amplitude, wavelength, frequency, period, and velocity of the wave? The amplitude and velocity remain unchanged, but the frequency increases while the period and the wavelength decrease. 24. Waves Moving Energy Suppose that you and your lab partner are asked to demonstrate that a transverse wave transports energy without transferring matter. How could you do it? Tie a piece of yarn somewhere near the middle of a rope. With your partner holding one end of the rope, shake the other end up and down to create a transverse wave. Note that while the wave moves down the rope, the yarn moves up and down but stays in the same place on the rope. 25. Longitudinal Waves Describe longitudinal waves. What types of media transmit longitudinal waves? In longitudinal waves, the particles of the medium vibrate in a direction parallel to the motion of the wave. Physics: Principles and Problems Solutions Manual 313 Chapter 14 continued Nearly all media—solids, liquids, and gases—transmit longitudinal waves. 26. Critical Thinking If a raindrop falls into a pool, it creates waves with small amplitudes. If a swimmer jumps into a pool, waves with large amplitudes are produced. Why doesn’t the heavy rain in a thunderstorm produce large waves? The energy of the swimmer is transferred to the wave in a small space over a short time, whereas the energy of the raindrops is spread out in area and time. Section Review 14.3 Wave Behavior pages 387–391 page 391 27. Waves at Boundaries Which of the following wave characteristics remain unchanged when a wave crosses a boundary into a different medium: frequency, amplitude, wavelength, velocity, and/or direction? 28. Refraction of Waves Notice in Figure 14-17a how the wave changes direction as it passes from one medium to another. Can two-dimensional waves cross a boundary between two media without changing direction? Explain. Yes, if they strike the boundary while traveling normal to its surface, or if they have the same speed in both media. 29. Standing Waves In a standing wave on a string fixed at both ends, how is the number of nodes related to the number of antinodes? The number of nodes is always one greater than the number of antinodes. 314 Solutions Manual Figure 14-14a behaves like a rigid wall because the reflected wave is inverted; 14-14b behaves like an open end because the boundary is an antinode and the reflected wave is not inverted. Chapter Assessment Concept Mapping page 396 31. Complete the concept map using the following terms and symbols: amplitude, frequency, v, , T. Waves speed amplitude period frequency wavelength v A T f Mastering Concepts page 396 32. What is periodic motion? Give three examples of periodic motion. (14.1) Periodic motion is motion that repeats in a regular cycle. Examples include oscillation of a spring, swing of a simple pendulum, and uniform circular motion. 33. What is the difference between frequency and period? How are they related? (14.1) Frequency is the number of cycles or repetitions per second, and period is the time required for one cycle. Frequency is the inverse of the period. Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Frequency remains unchanged. In general, amplitude, wavelength, and velocity will change when a wave enters a new medium. Direction may or may not change, depending on the original direction of the wave. 30. Critical Thinking As another way to understand wave reflection, cover the righthand side of each drawing in Figure 14-13a with a piece of paper. The edge of the paper should be at point N, the node. Now, concentrate on the resultant wave, shown in darker blue. Note that it acts like a wave reflected from a boundary. Is the boundary a rigid wall, or is it open-ended? Repeat this exercise for Figure 14-13b. Chapter 14 continued 34. What is simple harmonic motion? Give an example of simple harmonic motion. (14.1) Simple harmonic motion is periodic motion that results when the restoring force on an object is directly proportional to its displacement. A block bouncing on the end of a spring is one example. 35. If a spring obeys Hooke’s law, how does it behave? (14.1) The spring stretches a distance that is directly proportional to the force applied to it. 36. How can the spring constant of a spring be determined from a graph of force versus displacement? (14.1) The spring constant is the slope of the graph of F versus x. 37. How can the potential energy in a spring be determined from the graph of force versus displacement? (14.1) Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. The potential energy is the area under the curve of the graph of F versus x. 38. Does the period of a pendulum depend on the mass of the bob? The length of the string? Upon what else does the period depend? (14.1) no; yes; the acceleration of gravity, g 39. What conditions are necessary for resonance to occur? (14.1) Resonance will occur when a force is applied to an oscillating system at the same frequency as the natural frequency of the system. 40. How many general methods of energy transfer are there? Give two examples of each. (14.2) Two. Energy is transferred by particle transfer and by waves. There are many examples that can be given of each: a baseball and a bullet for particle transfer; sound waves and light waves. Physics: Principles and Problems 41. What is the primary difference between a mechanical wave and an electromagnetic wave? (14.2) The primary difference is that mechanical waves require a medium to travel through and electromagnetic waves do not need a medium. 42. What are the differences among transverse, longitudinal, and surface waves? (14.2) A transverse wave causes the particles of the medium to vibrate in a direction that is perpendicular to the direction in which the wave is moving. A longitudinal wave causes the particles of the medium to vibrate in a direction parallel with the direction of the wave. Surface waves have characteristics of both. 43. Waves are sent along a spring of fixed length. (14.2) a. Can the speed of the waves in the spring be changed? Explain. Speed of the waves depends only on the medium and cannot be changed. b. Can the frequency of a wave in the spring be changed? Explain. Frequency can be changed by changing the frequency at which the waves are generated. 44. What is the wavelength of a wave? (14.2) Wavelength is the distance between two adjacent points on a wave that are in phase. 45. Suppose you send a pulse along a rope. How does the position of a point on the rope before the pulse arrives compare to the point’s position after the pulse has passed? (14.2) Once the pulse has passed, the point is exactly as it was prior to the advent of the pulse. Solutions Manual 315 Chapter 14 continued 46. What is the difference between a wave pulse and a periodic wave? (14.2) A pulse is a single disturbance in a medium, whereas a periodic wave consists of several adjacent disturbances. 47. Describe the difference between wave frequency and wave velocity. (14.2) Frequency is the number of vibrations per second of a part of the medium. Velocity describes the motion of the wave through the medium. 48. Suppose you produce a transverse wave by shaking one end of a spring from side to side. How does the frequency of your hand compare with the frequency of the wave? (14.2) They are the same. 49. When are points on a wave in phase with each other? When are they out of phase? Give an example of each. (14.2) 50. What is the amplitude of a wave and what does it represent? (14.2) Amplitude is the maximum displacement of a wave from the rest or equilibrium position. The amplitude of the wave represents the amount of energy transferred. 51. Describe the relationship between the amplitude of a wave and the energy it carries. (14.2) The energy carried by a wave is proportional to the square of its amplitude. 52. When a wave reaches the boundary of a new medium, what happens to it? (14.3) ■ Figure 14-18 The frequency depends only on the rate at which the thin rope is shaken and the thin rope causes the vibrations in the thick rope. 54. How does a spring pulse reflected from a rigid wall differ from the incident pulse? (14.3) The reflected pulse will be inverted. 55. Describe interference. Is interference a property of only some types of waves or all types of waves? (14.3) The superposition of two or more waves is interference. The superposition of two waves with equal but opposite amplitudes results in destructive interference. The superposition of two waves with amplitudes in the same direction results in constructive interference; all waves; it is a prime test for wave nature. 56. What happens to a spring at the nodes of a standing wave? (14.3) Nothing, the spring does not move. 57. Violins A metal plate is held fixed in the center and sprinkled with sugar. With a violin bow, the plate is stroked along one edge and made to vibrate. The sugar begins to collect in certain areas and move away from others. Describe these regions in terms of standing waves. (14.3) Bare areas are antinodal regions where there is maximum vibration. Sugarcovered areas are nodal regions where there is no vibration. Part of the wave can be reflected and part of the wave can be transmitted into the new medium. 316 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Points are in phase when they have the same displacement and the same velocity. Otherwise, the points are out of phase. Two crests are in phase with each other. A crest and a trough are out of phase with each other. 53. When a wave crosses a boundary between a thin and a thick rope, as shown in Figure 14-18, its wavelength and speed change, but its frequency does not. Explain why the frequency is constant. (14.3) Chapter 14 continued 58. If a string is vibrating in four parts, there are points where it can be touched without disturbing its motion. Explain. How many of these points exist? (14.3) A standing wave exists and the string can be touched at any of its five nodal points. In the first case, longitudinal waves; in the second case, transverse waves. 59. Wave fronts pass at an angle from one medium into a second medium, where they travel with a different speed. Describe two changes in the wave fronts. What does not change? (14.3) 63. Suppose you repeatedly dip your finger into a sink full of water to make circular waves. What happens to the wavelength as you move your finger faster? The wavelength and direction of the wave fronts change. The frequency does not change. The frequency of the waves will increase; the speed will remain the same; the wavelength will decrease. Applying Concepts page 397 60. A ball bounces up and down on the end of a spring. Describe the energy changes that take place during one complete cycle. Does the total mechanical energy change? Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 62. Suppose you hold a 1-m metal bar in your hand and hit its end with a hammer, first, in a direction parallel to its length, and second, in a direction at right angles to its length. Describe the waves produced in the two cases. At the bottom of the motion, the elastic potential energy is at a maximum, while gravitational potential energy is at a minimum and the kinetic energy is zero. At the equilibrium position, the KE is at a maximum and the elastic potential energy is zero. At the top of the bounce, the KE is zero, the gravitational potential energy is at a maximum, and the elastic potential energy is at a maximum. The total mechanical energy is conserved. 61. Can a pendulum clock be used in the orbiting International Space Station? Explain. No, the space station is in free-fall, and therefore, the apparent value of g is zero. The pendulum will not swing. Physics: Principles and Problems 64. What happens to the period of a wave as the frequency increases? As the frequency increases, the period decreases. 65. What happens to the wavelength of a wave as the frequency increases? As the frequency increases, the wavelength decreases. 66. Suppose you make a single pulse on a stretched spring. How much energy is required to make a pulse with twice the amplitude? approximately two squared, or four times the energy 67. You can make water slosh back and forth in a shallow pan only if you shake the pan with the correct frequency. Explain. The period of the vibration must equal the time for the wave to go back and forth across the pan to create constructive interference. Solutions Manual 317 Chapter 14 continued 68. In each of the four waves in Figure 14-19, the pulse on the left is the original pulse moving toward the right. The center pulse is a reflected pulse; the pulse on the right is a transmitted pulse. Describe the rigidity of the boundaries at A, B, C, and D. 70. Car Shocks Each of the coil springs of a car has a spring constant of 25,000 N/m. How much is each spring compressed if it supports one-fourth of the car’s 12,000-N weight? F kx, F so x A k 14(12,000 N) 25,000 N/m B 0.12 m 71. How much potential energy is stored in a spring with a spring constant of 27 N/m if it is stretched by 16 cm? C 1 PEsp kx 2 2 D (27 N/m)(0.16 m)2 0.35 J 1 2 ■ Figure 14-19 Boundary A is more rigid; boundary B is less rigid; boundary C is less rigid; boundary D is more rigid. Mastering Problems 1 PEsp kx 2, 2 so x k 35 N/m 2PEsp (2)(1.5 J) 0.29 m Level 3 73. Force-versus-length data for a spring are plotted on the graph in Figure 14-21. Force (N) 12.0 3.2 N 8.0 4.0 0.0 ■ Figure 14-20 F kx, 0.20 0.40 0.60 Length (m) ■ Figure 14-21 F 3.2 N 27 N/m so k x 318 0.12 m Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 14.1 Periodic Motion pages 397–398 Level 1 69. A spring stretches by 0.12 m when some apples weighing 3.2 N are suspended from it, as shown in Figure 14-20. What is the spring constant of the spring? Level 2 72. Rocket Launcher A toy rocket-launcher contains a spring with a spring constant of 35 N/m. How far must the spring be compressed to store 1.5 J of energy? Chapter 14 continued a. What is the spring constant of the spring? 1 T f 1 1 T 8.3 s k slope F 12.0 N 4.0 N x 0.6 m 0.2 m 20 N/m b. What is the energy stored in the spring when it is stretched to a length of 50.0 cm? 1 PEsp area bh 0.12 Hz f 76. Ocean Waves An ocean wave has a length of 12.0 m. A wave passes a fixed location every 3.0 s. What is the speed of the wave? v f (12.0 m) 1 T 1 3.0 s 4.0 m/s 2 (0.500 m)(10.0 N) 2.50 J 1 2 74. How long must a pendulum be to have a period of 2.3 s on the Moon, where g 1.6 m/s2? g T 2g 4 l T 2 , so l 2 (2.3 s)2(1.6 m/s2) 4 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 0.21 m 2 14.2 Wave Properties page 398 Level 1 75. Building Motion The Sears Tower in Chicago, shown in Figure 14-22, sways back and forth in the wind with a frequency of about 0.12 Hz. What is its period of vibration? 77. Water waves in a shallow dish are 6.0-cm long. At one point, the water moves up and down at a rate of 4.8 oscillations/s. a. What is the speed of the water waves? v f (0.060 m)(4.8 Hz) 0.29 m/s b. What is the period of the water waves? 1 1 T 0.21 s f 4.8 Hz 78. Water waves in a lake travel 3.4 m in 1.8 s. The period of oscillation is 1.1 s. a. What is the speed of the water waves? d 3.4 m v 1.9 m/s t 1.8 s b. What is their wavelength? v f vT (1.9 m/s)(1.1 s) 2.1 m Level 2 79. Sonar A sonar signal of frequency 1.00106 Hz has a wavelength of 1.50 mm in water. a. What is the speed of the signal in water? v f (1.50103 m)(1.00106 Hz) 1.50103 m/s b. What is its period in water? 1 f 1 1.0010 Hz T 6 1.00106 s ■ Figure 14-22 Physics: Principles and Problems Solutions Manual 319 Chapter 14 continued c. What is its period in air? 1.00106 (6.0 m)(0.60 Hz) s The period and frequency remain unchanged. 80. A sound wave of wavelength 0.60 m and a velocity of 330 m/s is produced for 0.50 s. a. What is the frequency of the wave? v f v v f 330 m/s 0.60 m f 550 Hz b. How many complete waves are emitted in this time interval? ft (550 Hz)(0.50 s) 280 complete waves c. After 0.50 s, how far is the front of the wave from the source of the sound? 3.6 m/s 83. Earthquakes The velocity of the transverse waves produced by an earthquake is 8.9 km/s, and that of the longitudinal waves is 5.1 km/s. A seismograph records the arrival of the transverse waves 68 s before the arrival of the longitudinal waves. How far away is the earthquake? d vt. We do not know t, only the difference in time, t. The transverse distance, dT vTt, is the same as the longitudinal distance, dL vL(t t). Use vTt vL(t t), and solve for t: v t vT vL L t (5.1 km/s)(68 s) 8.9 km/s 5.1 km/s t 91 s d vt (330 m/s)(0.50 s) Then putting t back into d T vTt (8.9 km/s)(91 s) 1.6102 m The time for the wave to travel down and back up is 1.80 s. The time one way is half 1.80 s or 0.900 s. d vt (1498 m/s)(0.900 s) 8.1102 km 14.3 Wave Behavior pages 398–399 Level 1 84. Sketch the result for each of the three cases shown in Figure 14-23, when the centers of the two approaching wave pulses lie on the dashed line so that the pulses exactly overlap. 1350 m Level 3 82. Pepe and Alfredo are resting on an offshore raft after a swim. They estimate that 3.0 m separates a trough and an adjacent crest of each surface wave on the lake. They count 12 crests that pass by the raft in 20.0 s. Calculate how fast the waves are moving. 1 2 3 (2)(3.0 m) 6.0 m 12 waves 20.0 s f 0.60 Hz 320 Solutions Manual ■ Figure 14-23 Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 81. The speed of sound in water is 1498 m/s. A sonar signal is sent straight down from a ship at a point just below the water surface, and 1.80 s later, the reflected signal is detected. How deep is the water? Chapter 14 continued 15 cm from the other end, where the distances traveled are the same. 1. The amplitude is doubled. 2. The amplitudes cancel each other. 87. Sketch the result for each of the four cases shown in Figure 14-24, when the centers of each of the two wave pulses lie on the dashed line so that the pulses exactly overlap. 3. If the amplitude of the first pulse is one-half of the second, the resultant pulse is one-half the amplitude of the second. 1 2 3 85. If you slosh the water in a bathtub at the correct frequency, the water rises first at one end and then at the other. Suppose you can make a standing wave in a 150-cm-long tub with a frequency of 0.30 Hz. What is the velocity of the water wave? 4 ■ Figure 14-24 2(1.5 m) 3.0 m v f (3.0 m)(0.30 Hz) 1 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 0.90 m/s Level 2 86. Guitars The wave speed in a guitar string is 265 m/s. The length of the string is 63 cm. You pluck the center of the string by pulling it up and letting go. Pulses move in both directions and are reflected off the ends of the string. 2 3 4 a. How long does it take for the pulse to move to the string end and return to the center? (2)(63 cm) 2 d 63 cm 0.63 m d so t 2.4103 s 265 m/s v b. When the pulses return, is the string above or below its resting location? Pulses are inverted when reflected from a more dense medium, so returning pulse is down (below). c. If you plucked the string 15 cm from one end of the string, where would the two pulses meet? Physics: Principles and Problems Mixed Review page 399–400 Level 1 88. What is the period of a pendulum with a length of 1.4 m? g l T 2 9.80 m/s 1.4 m 2 2 2.4 s 89. The frequency of yellow light is 5.11014 Hz. Find the wavelength of yellow light. The speed of light is 3.00108 m/s. Solutions Manual 321 Chapter 14 continued c f c. In the course of 15 s you count ten waves that pass you. What is the period of the waves? 3.00108 m/s 5.110 Hz 14 15 s T 1.5 s 10 waves 5.9107 m d. What is the frequency of the waves? 90. Radio Wave AM-radio signals are broadcast at frequencies between 550 kHz (kilohertz) and 1600 kHz and travel 3.0108 m/s. a. What is the range of wavelengths for these signals? v f 8 3.010 m/s v 1 5 5.510 Hz f1 1 T 1 1.5 s f 0.67 Hz e. You estimate that the wave crests are 3 m apart. What is the velocity of the waves? v f (3 m)(0.67 Hz) 2 m/s f. After returning to the beach, you learn that the waves are moving at 1.8 m/s. What is the actual wavelength of the waves? v f 1.8 m/s 0.67 Hz 2.7 m 550 m 8 3.010 m/s v 2 6 1.610 Hz f2 190 m Range is 190 m to 550 m. b. FM frequencies range between 88 MHz (megahertz) and 108 MHz and travel at the same speed. What is the range of FM wavelengths? 8 3.010 m/s v 7 3.4 m (68 kg)(9.80 m/s2) 1710 m 540 m mg x F k x 0.57 N/m 8 3.010 m/s v 8 f 1.0810 Hz 2.8 m Range is 2.8 m to 3.4 m. 91. You are floating just offshore at the beach. Even though the waves are steadily moving in toward the beach, you don’t move any closer to the beach. a. What type of wave are you experiencing as you float in the water? transverse waves b. Explain why the energy in the wave does not move you closer to shore. The displacement is perpendicular to the direction of the wave—in this case, up and down. 322 Solutions Manual 93. The time needed for a water wave to change from the equilibrium level to the crest is 0.18 s. a. What fraction of a wavelength is this? 1 wavelength 4 b. What is the period of the wave? T (4)(0.18 s) 0.72 s c. What is the frequency of the wave? 1 1 f 1.4 Hz T 0.72 s 94. When a 225-g mass is hung from a spring, the spring stretches 9.4 cm. The spring and mass then are pulled 8.0 cm from this new equilibrium position and released. Find the spring constant of the spring and the maximum speed of the mass. Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 8.810 Hz f Level 2 92. Bungee Jumper A high-altitude bungee jumper jumps from a hot-air balloon using a 540-m-bungee cord. When the jump is complete and the jumper is just suspended from the cord, it is stretched 1710 m. What is the spring constant of the bungee cord if the jumper has a mass of 68 kg? Chapter 14 continued New: mg x F k x 2 (0.225 kg)(9.80 0.094 m m/s2) 23 N/m Maximum velocity occurs when the mass passes through the equilibrium point, where all the energy is kinetic energy. Using the conservation of energy: PEsp KEmass 2 kx v2 m kx 2 m (23 N/m)(0.080 m)2 0.225 kg Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 0.81 m/s 95. Amusement Ride You notice that your favorite amusement-park ride seems bigger. The ride consists of a carriage that is attached to a structure so it swings like a pendulum. You remember that the carriage used to swing from one position to another and back again eight times in exactly 1 min. Now it only swings six times in 1 min. Give your answers to the following questions to two significant digits. a. What was the original period of the ride? 1 T 7.5 s 8 swings f 1 60.0 s b. What is the new period of the ride? 1 1.0101 s T 6 swings f 1 4 (1.0101 s)2 4 (9.80 m/s2) 2 25 m The arm on the new structure is 11 m longer. e. If the park owners wanted to double the period of the ride, what percentage increase would need to be made to the length of the pendulum? 1 1 kx 2 mv 2 2 2 v T l g 2 60.0 s c. What is the new frequency? 1 1 0.10 Hz f T 1.0101 s d. How much longer is the arm supporting the carriage on the larger ride? Original: 2 T l g 2 4 s)2 (7.5 (9.80 m/s2) 4 2 Because of the square relationship, there would need to be a 4 times increase in the length of the pendulum, or a 300% increase. 96. Clocks The speed at which a grandfather clock runs is controlled by a swinging pendulum. a. If you find that the clock loses time each day, what adjustment would you need to make to the pendulum so it will keep better time? The clock must be made to run faster. The period of the pendulum can be shortened, thus increasing the speed of the clock, by shortening the length of the pendulum. b. If the pendulum currently is 15.0 cm, by how much would you need to change the length to make the period lessen by 0.0400 s? g g l l T 2 2 2 1 T 2 g g T 2 gl2 gl1 l2 1 l1 1 T 1 1 l2 l1 2 g g T g 2 l2 l1 14 m Physics: Principles and Problems Solutions Manual 323 Chapter 14 continued T g l2 l1 2 T g 2 2 l2 l1 9.80 m /s2 (0.0400 s) 2 2 0.150 m 0.135 m The length would need to shorten by l1 l2 0.150 m 0.135 m 0.015 m 97. Bridge Swinging In the summer over the New River in West Virginia, several teens swing from bridges with ropes, then drop into the river after a few swings back and forth. a. If Pam is using a 10.0-m length of rope, how long will it take her to reach the peak of her swing at the other end of the bridge? 1 swing to peak T 2 g 9.80 m/s 10.0 m l 2 3.17 s b. If Mike has a mass that is 20 kg more than Pam, how would you expect the period of his swing to differ from Pam’s? There should be no difference. T is not affected by mass. c. At what point in the swing is KE at a maximum? At the bottom of the swing, KE is at a maximum. d. At what point in the swing is PE at a maximum? e. At what point in the swing is KE at a minimum? At the top of the swing, KE is at a minimum. f. At what point in the swing is PE at a minimum? At the bottom of the swing, PE is at a minimum. 98. You have a mechanical fish scale that is made with a spring that compresses when weight is added to a hook attached below the scale. Unfortunately, the calibrations have completely worn off of the scale. However, you have one known mass of 500.0 g that displaces the spring 2.0 cm. a. What is the spring constant for the spring? F mg kx mg x k (0.5000 kg)(9.80 m/s2) 0.020 m 2.4102 N/m b. If a fish displaces the spring 4.5 cm, what is the mass of the fish? F mg kx 324 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. At the top of the swing, PE is at a maximum. Chapter 14 continued kx g Now v (2.4102 N/m)(4.5102 m) 9.80 m/s FT v 2 m 2 1.1 kg 99. a. What is the spring constant for each of the springs? F mg (45 kg)(9.80 m/s2) 440 N force per spring 220 N F x F kx, so k 220 N 0.010 m k 22,000 N/m b. How much additional potential energy is stored in each of the car springs after loading the trunk? 1 PE kx 2 2 (22,000 N/m)(0.010 m)2 1 2 1.1 J Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. FT (1.50102 m/s)2(2.83103 kg/m) 63.7 N Car Springs When you add a 45-kg load to the trunk of a new small car, the two rear springs compress an additional 1.0 cm. Level 3 100. The velocity of a wave on a string depends on how tightly the string is stretched, and on the mass per unit length of the string. If FT is the tension in the string, and is the mass/unit length, then the velocity, v, can be determined by the following equation. v , so Thinking Critically page 400 101. Analyze and Conclude A 20-N force is required to stretch a spring by 0.5 m. a. What is the spring constant? F x b. How much energy is stored in the spring? 1 PEsp kx 2 2 (40 N/m)(0.5 m)2 5 J 1 2 c. Why isn’t the work done to stretch the spring equal to the force times the distance, or 10 J? The force is not constant as the spring is stretched. The average force, 10 N, times the distance does give the correct work. 102. Make and Use Graphs Several weights were suspended from a spring, and the resulting extensions of the spring were measured. Table 14-1 shows the collected data. FT A piece of string 5.30-m long has a mass of 15.0 g. What must the tension in the string be to make the wavelength of a 125-Hz wave 120.0 cm? v f (1.200 m)(125 Hz) 1.50102 m/s m and L 20 N 0.5 m F kx, so k 40 N/m Table 14-1 Weights on a Spring Force, F (N) Extension, x (m) 2.5 0.12 5.0 0.26 7.5 0.35 10.0 0.50 12.5 0.60 15.0 0.71 1.50102 kg 5.30 m 2.83103 kg/m Physics: Principles and Problems Solutions Manual 325 Chapter 14 continued a. Make a graph of the force applied to the spring versus the spring length. Plot the force on the y-axis. Force (N) 20.0 15.0 10.0 5.0 0.0 0.20 0.60 0.40 0.80 Stretch (m) b. Determine the spring constant from the graph. The spring constant is the slope. 15.0 N 2.5 N 0.71 m 0.12 m F x k slope 21 N/m c. Using the graph, find the elastic potential energy stored in the spring when it is stretched to 0.50 m. The potential energy is the area under the graph. 1 PEsp area bh 2 (0.50 m)(10.0 N) 1 2 2.5 J ■ Figure 14-25 v f v 5 m/s 1.5 m f 3 Hz 326 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 103. Apply Concepts Gravel roads often develop regularly spaced ridges that are perpendicular to the road, as shown in Figure 14-25. This effect, called washboarding, occurs because most cars travel at about the same speed and the springs that connect the wheels to the cars oscillate at about the same frequency. If the ridges on a road are 1.5 m apart and cars travel on it at about 5 m/s, what is the frequency of the springs’ oscillation? Chapter 14 continued Writing in Physics page 400 104. Research Christiaan Huygens’ work on waves and the controversy between him and Newton over the nature of light. Compare and contrast their explanations of such phenomena as reflection and refraction. Whose model would you choose as the best explanation? Explain why. Huygens proposed the wave theory of light and Newton proposed the particle theory of light. The law of reflection can be explained using both theories. Huygen’s principle and Newton’s particle theory are opposed, however, in their explanation of the law of refraction. Cumulative Review page 400 105. A 1400-kg drag racer automobile can complete a one-quarter mile (402 m) course in 9.8 s. The final speed of the automobile is 250 mi/h (112 m/s). (Chapter 11) Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. a. What is the kinetic energy of the automobile? 1 KE mv 2 2 (1400 kg)(112 m/s)2 1 2 8.8106 J b. What is the minimum amount of work that was done by its engine? Why can’t you calculate the total amount of work done? The minimum amount of work must equal KE, or 8.8106 J. The engine had to do more work than was dissipated in work done against friction. c. What was the average acceleration of the automobile? v a t 112 m/s 9.8 s 11 m/s2 106. How much water would a steam engine have to evaporate in 1 s to produce 1 kW of power? Assume that the engine is 20 percent efficient. (Chapter 12) W 1000 J/s t If the engine is only 20 percent efficient it must use five times more heat to produce the 1000 J/s. mHv Q 5000 J/s t t m t 5000 J/s Hv Therefore, 5000 J/s 2.2610 J/kg 6 2103 kg/s Challenge Problem page 380 A car of mass m rests at the top of a hill of height h before rolling without friction into a crash barrier located at the bottom of the hill. The crash barrier contains a spring with a spring constant, k, which is designed to bring the car to rest with minimum damage. 1. Determine, in terms of m, h, k, and g, the maximum distance, x, that the spring will be compressed when the car hits it. Conservation of energy implies that the gravitational potential energy of the car at the top of the hill will be equal to the elastic potential energy in the spring when it has brought the car to rest. The equations for these energies can be set equal and solved for x. 1 2 PEg PEsp, so mgh kx 2 x Physics: Principles and Problems k 2mgh Solutions Manual 327 Chapter 14 continued 2. If the car rolls down a hill that is twice as high, how much farther will the spring be compressed? The height is doubled and x is proportional to the square root of the height, so x will increase by 2 . 3. What will happen after the car has been brought to rest? In the case of an ideal spring, the spring will propel the car back to the top of the hill. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 328 Solutions Manual Physics: Principles and Problems CHAPTER 15 Sound Practice Problems vs 24.6 m/s 15.1 Properties and Detection of Sound pages 403–410 fd 524 Hz page 405 1. Find the wavelength in air at 20°C of an 18-Hz sound wave, which is one of the lowest frequencies that is detectable by the human ear. v f 343 m/s 18 Hz 19 m 2. What is the wavelength of an 18-Hz sound wave in seawater at 25°C? v 1533 m/s 85 m f 18 Hz 3. Find the frequency of a sound wave moving through iron at 25°C with a wavelength of 1.25 m. v 5130 m/s 1.25 m Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. f 4.10103 Hz 4. If you shout across a canyon and hear the echo 0.80 s later, how wide is the canyon? d t v so d vt (343 m/s)(0.40 s) 140 m 5. A 2280-Hz sound wave has a wavelength of 0.655 m in an unknown medium. Identify the medium. v f so v f (0.655 m)(2280 Hz) 1490 m/s This speed corresponds to water at 25°C. page 409 6. Repeat Example Problem 1, but with the car moving away from you. What frequency would you hear? Physics: Principles and Problems 1 (24.6 m/s) 1 343 m/s 489 Hz 7. You are in an auto traveling at 25.0 m/s toward a pole-mounted warning siren. If the siren’s frequency is 365 Hz, what frequency do you hear? Use 343 m/s as the speed of sound. v 343 m/s, fs 365 Hz, vs 0, vd 25.0 m/s vv v vs d fd fs 343 m/s 25.0 m/s (365 Hz) 343 m/s 392 Hz 8. You are in an auto traveling at 55 mph (24.6 m/s). A second auto is moving toward you at the same speed. Its horn is sounding at 475 Hz. What frequency do you hear? Use 343 m/s as the speed of sound. v 343 m/s, fs 475 Hz, vs 24.6 m/s, vd 24.6 m/s vv v vs d fd fs 343 m/s 24.6 m/s 343 m/s 24.6 m/s (475 Hz) 548 Hz 9. A submarine is moving toward another submarine at 9.20 m/s. It emits a 3.50-MHz ultrasound. What frequency would the second sub, at rest, detect? The speed of sound in water is 1482 m/s. v 1482 m/s, fs 3.50 MHz, vs 9.20 m/s, vd 0 m/s vv v vs d fd fs Solutions Manual 329 Chapter 15 continued (3.50 MHz) 1482 m/s 1482 m/s 9.20 m/s 3.52 MHz 10. A sound source plays middle C (262 Hz). How fast would the source have to go to raise the pitch to C sharp (271 Hz)? Use 343 m/s as the speed of sound. v 343 m/s, fs 262 Hz, fd 271 Hz, vd 0 m/s, vs is unknown Solve this equation for vs. f fd vs v s(v vd) (343 m/s 0 m/s) 343 m/s 271 Hz 262 Hz 11.4 m/s Section Review 15.1 Properties and Detection of Sound pages 403–410 Displacement Time (ms) 1.0 1.5 2.0 1.0 kHz 2.0 kHz 12. Effect of Medium List two sound characteristics that are affected by the medium through which the sound passes and two characteristics that are not affected. affected: speed and wavelength; unaffected: period and frequency 330 Solutions Manual 14. Decibel Scale How much greater is the sound pressure level of a typical rock band’s music (110 dB) than a normal conversation (50 dB)? 15. Early Detection In the nineteenth century, people put their ears to a railroad track to get an early warning of an approaching train. Why did this work? The velocity of sound is greater in solids than in gases. Therefore, sound travels faster in steel rails than in air, and the rails help focus the sound so it does not die out as quickly as in air. 16. Bats A bat emits short pulses of highfrequency sound and detects the echoes. a. In what way would the echoes from large and small insects compare if they were the same distance from the bat? They would differ in intensity. Larger insects would reflect more of the sound energy back to the bat. b. In what way would the echo from an insect flying toward the bat differ from that of an insect flying away from the bat? An insect flying toward the bat would return an echo of higher frequency (Doppler shift). An insect flying away from the bat would return an echo of lower frequency. 17. Critical Thinking Can a trooper using a radar detector at the side of the road determine the speed of a car at the instant the car passes the trooper? Explain. No. The car must be approaching or Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. page 410 11. Graph The eardrum moves back and forth in response to the pressure variations of a sound wave. Sketch a graph of the displacement of the eardrum versus time for two cycles of a 1.0-kHz tone and for two cycles of a 2.0-kHz tone. 0.5 frequency; amplitude The sound pressure level increases by a factor of 10 for every 20-dB increase in sound level. Therefore, 60 dB corresponds to a 1000-fold increase in SPL. v vd fd fs v vs 13. Sound Properties What physical characteristic of a sound wave should be changed to change the pitch of the sound? To change the loudness? Chapter 15 continued v 343 m/s f1 64.7 Hz receding from the detector for the Doppler effect to be observed. Transverse motion produces no Doppler effect. 1 b. Find the next two resonant frequencies for the bugle. v v 343 m/s f2 129 Hz 2 Practice Problems 3 page 416 18. A 440-Hz tuning fork is held above a closed pipe. Find the spacing between the resonances when the air temperature is 20°C. v Resonance spacing so using f the resonance spacing is v 343 m/s 0.39 m 2f (2)(440 Hz) 2 19. A 440-Hz tuning fork is used with a resonating column to determine the velocity of sound in helium gas. If the spacings between resonances are 110 cm, what is the velocity of sound in helium gas? Resonance spacing 1.1 m so 2.2 m 2 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. v f (2.2 m)(440 Hz) 970 m/s 20. The frequency of a tuning fork is unknown. A student uses an air column at 27°C and finds resonances spaced by 20.2 cm. What is the frequency of the tuning fork? Use the speed calculated in Example Problem 2 for the speed of sound in air at 27°C. v 347 m/s at 27°C Resonance spacing gives 0.202 m, 2 or 0.404 m v 347 m/s f 859 Hz 0.404 m 21. A bugle can be thought of as an open pipe. If a bugle were straightened out, it would be 2.65-m long. a. If the speed of sound is 343 m/s, find the lowest frequency that is resonant for a bugle (ignoring end corrections). 1 2L (2)(2.65 m) 5.30 m The lowest frequency is Physics: Principles and Problems L 2.65 m (3)(343 m/s) (2)(2.65 m) v 3v f3 194 Hz 15.2 The Physics of Music pages 411–419 2 5.30 m 2L Section Review 15.2 The Physics of Music pages 411–419 page 419 22. Origins of Sound What is the vibrating object that produces sounds in each of the following? a. a human voice vocal cords b. a clarinet a reed c. a tuba the player’s lips d. a violin a string 23. Resonance in Air Columns Why is the tube from which a tuba is made much longer than that of a cornet? The longer the tube, the lower the resonant frequency it will produce. 24. Resonance in Open Tubes How must the length of an open tube compare to the wavelength of the sound to produce the strongest resonance? The length of the tube should be one-half the wavelength. 25. Resonance on Strings A violin sounds a note of F sharp, with a pitch of 370 Hz. What are the frequencies of the next three harmonics produced with this note? A string’s harmonics are whole number multiples of the fundamental, so the frequencies are: Solutions Manual 331 Chapter 15 continued f2 2f1 (2)(370 Hz) 740 Hz f3 3f1 (3)(370 Hz) 1110 Hz 1100 Hz f4 4f1 (4)(370 Hz) 1480 Hz 1500 Hz 26. Resonance in Closed Pipes One closed organ pipe has a length of 2.40 m. a. What is the frequency of the note played by this pipe? 4L (4)(2.40 m) 9.60 m v f v 343 m/s f 35.7 Hz greatly when it is pressed against other objects because they resonate like a sounding board. They sound different because they resonate with different harmonics; therefore, they have different timbres. Chapter Assessment Concept Mapping page 424 30. Complete the concept map below using the following terms: amplitude, perception, pitch, speed. 9.60 m Sound b. When a second pipe is played at the same time, a 1.40-Hz beat note is heard. By how much is the second pipe too long? f 35.7 Hz 1.40 Hz 34.3 Hz frequency v 343 m/s 10.0 m f perception properties pitch loudness 34.3 Hz 4L 10.0 m L 2.50 m 4 4 The difference in lengths is 27. Timbre Why do various instruments sound different even when they play the same note? Each instrument produces its own set of fundamental and harmonic frequencies, so they have different timbres. 28. Beats A tuning fork produces three beats per second with a second, 392-Hz tuning fork. What is the frequency of the first tuning fork? It is either 389 Hz or 395 Hz. You can’t tell which without more information. 29. Critical Thinking Strike a tuning fork with a rubber hammer and hold it at arm’s length. Then press its handle against a desk, a door, a filing cabinet, and other objects. What do you hear? Why? amplitude Mastering Concepts page 424 31. What are the physical characteristics of sound waves? (15.1) Sound waves can be described by frequency, wavelength, amplitude, and speed. 32. When timing the 100-m run, officials at the finish line are instructed to start their stopwatches at the sight of smoke from the starter’s pistol and not at the sound of its firing. Explain. What would happen to the times for the runners if the timing started when sound was heard? (15.1) Light travels at 3.00108 m/s, while sound travels at 343 m/s. Officials would see the smoke before they would hear the pistol fire. The times would be less than actual if sound were used. The tuning fork’s sound is amplified 332 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 2.50 m 2.40 m 0.10 m speed Chapter 15 continued 33. Name two types of perception of sound and the physical characteristics of sound waves that correspond to them. (15.1) pitch—frequency, loudness—amplitude 34. Does the Doppler shift occur for only some types of waves or for all types of waves? (15.1) 39. Musical Instruments Why don’t most musical instruments sound like tuning forks? (15.2) Tuning forks produce simple, singlefrequency waves. Musical instruments produce complex waves containing many different frequencies. This gives them their timbres. all types of waves 35. Sound waves with frequencies higher than can be heard by humans, called ultrasound, can be transmitted through the human body. How could ultrasound be used to measure the speed of blood flowing in veins or arteries? Explain how the waves change to make this measurement possible. (15.1) Doctors can measure the Doppler shift from sound reflected by the moving blood cells. Because the blood is moving, sound gets Doppler shifted, the compressions either get piled up or spaced apart. This alters the frequency of the wave. 40. Musical Instruments What property distinguishes notes played on both a trumpet and a clarinet if they have the same pitch and loudness? (15.2) the sound quality or timbre 41. Trombones Explain how the slide of a trombone, shown in Figure 15-21, changes the pitch of the sound in terms of a trombone being a resonance tube. (15.2) 36. What is necessary for the production and transmission of sound? (15.2) ■ Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. a vibrating object and a material medium 37. Singing How can a certain note sung by an opera singer cause a crystal glass to shatter? (15.2) The frequency of the note is the same as the natural resonance of the crystal, causing its molecules to increase their amplitude of vibration as energy from the sound is accepted. 38. Marching In the military, as marching soldiers approach a bridge, the command “route step” is given. The soldiers then walk out-of-step with each other as they cross the bridge. Explain. (15.2) While marching in step, a certain frequency is established that could resonate the bridge into destructive oscillation. No single frequency is maintained under “route step.” Physics: Principles and Problems Figure 15-21 The slide of a trombone varies pitch by changing the length of the resonating column of vibrating air. Applying Concepts pages 424–425 42. Estimation To estimate the distance in kilometers between you and a lightning flash, count the seconds between the flash and the thunder and divide by 3. Explain how this rule works. Devise a similar rule for miles. The speed of sound 343 m/s 0.343 km/s (1/2.92) km/s; or, sound travels approximately 1 km in 3 s. Therefore, divide the number of seconds by three. For miles, sound travels approximately 1 mile in 5 s. Therefore, divide the number of seconds by five. Solutions Manual 333 Chapter 15 continued 43. The speed of sound increases by about 0.6 m/s for each degree Celsius when the air temperature rises. For a given sound, as the temperature increases, what happens to the following? a. the frequency There is no change in frequency. b. the wavelength First, if you had heard a sound, you would have heard it after you saw the explosion. Sound waves travel much more slowly than electromagnetic waves. Second, in space the density of matter is so small that the sound waves do not propagate. Consequently, no sound should have been heard. ■ b. the wavelength Wavelength will decrease. c. the wave velocity 7107 m Figure 15-23 Red light has a longer wavelength and therefore, a lower frequency than other colors. The Doppler shift of their light to lower frequencies indicates that distant galaxies are moving away from us. Amplitude will remain the same. 48. The speed of sound increases with temperature. Would the pitch of a closed pipe increase or decrease when the temperature of the air rises? Assume that the length of the pipe does not change. 4l and v f so v 4fl. If v increases and l remains unchanged, f increases and pitch increases. 49. Marching Bands Two flutists are tuning up. If the conductor hears the beat frequency increasing, are the two flute frequencies getting closer together or farther apart? The frequencies are getting farther apart. 50. Musical Instruments A covered organ pipe plays a certain note. If the cover is removed to make it an open pipe, is the pitch increased or decreased? The pitch is increased; the frequency is twice as high for an open pipe as for a closed pipe. 51. Stringed Instruments On a harp, Figure 15-23a, long strings produce low notes and short strings produce high notes. On a guitar, Figure 15-23b, the strings are all the same length. How can they produce notes of different pitches? 46. Does a sound of 40 dB have a factor of 100 (102) times greater pressure variation than the threshold of hearing, or a factor of 40 times greater? A 40-dB sound has sound pressures 100 times greater. 334 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 45. The Redshift Astronomers have observed that the light coming from distant galaxies appears redder than light coming from nearer galaxies. With the help of Figure 15-22, which shows the visible spectrum, explain why astronomers conclude that distant galaxies are moving away from Earth. 6107 m Frequency will increase. d. the amplitude of the wave 44. Movies In a science-fiction movie, a satellite blows up. The crew of a nearby ship immediately hears and sees the explosion. If you had been hired as an advisor, what two physics errors would you have noticed and corrected? 5107 m a. the frequency Wave velocity will remain the same. The wavelength increases. 4107 m 47. If the pitch of sound is increased, what are the changes in the following? Chapter 15 continued ■ Figure 15-23 The strings have different tensions and masses per unit length. Thinner, tighter strings produce higher notes than do thicker, looser strings. Mastering Problems 15.1 Properties and Detection of Sound pages 425–426 Level 1 52. You hear the sound of the firing of a distant cannon 5.0 s after seeing the flash. How far are you from the cannon? d vt (343 m/s)(5.0 s) 1.7 km 53. If you shout across a canyon and hear an echo 3.0 s later, how wide is the canyon? d vt (343 m/s)(3.0 s) is the total distance traveled. The distance to the wall 1 is (343 m/s)(3.0 s) 5.1102 m ■ Figure 15-24 The total distance the sound must travel is 6.00 m. d v t 6.00 m d so t 0.0175 s v 343 m/s 57. Sound with a frequency of 261.6 Hz travels through water at 25°C. Find the sound’s wavelength in water. Do not confuse sound waves moving through water with surface waves moving through water. v 1493 m/s 5.707 m f 261.6 Hz 58. If the wavelength of a 4.40102-Hz sound in freshwater is 3.30 m, what is the speed of sound in freshwater? v f (3.30 m)(4.40102 Hz) 1.45103 m/s 2 54. A sound wave has a frequency of 4700 Hz and travels along a steel rod. If the distance between compressions, or regions of high pressure, is 1.1 m, what is the speed of the wave? Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. v f (1.1 m)(4700 Hz) 5200 m/s 55. Bats The sound emitted by bats has a wavelength of 3.5 mm. What is the sound’s frequency in air? v 343 m/s f 9.8104 Hz 0.0035 m 56. Photography As shown in Figure 15-24, some cameras determine the distance to the subject by sending out a sound wave and measuring the time needed for the echo to return to the camera. How long would it take the sound wave to return to such a camera if the subject were 3.00 m away? 59. Sound with a frequency of 442 Hz travels through an iron beam. Find the wavelength of the sound in iron. v 5130 m/s 11.6 m f 442 Hz 60. Aircraft Adam, an airport employee, is working near a jet plane taking off. He experiences a sound level of 150 dB. a. If Adam wears ear protectors that reduce the sound level to that of a typical rock concert, what decrease in dB is provided? A typical rock concert is 110 dB, so 40 dB reduction is needed. b. If Adam then hears something that sounds like a barely audible whisper, what will a person not wearing the ear protectors hear? A barely audible whisper is 10 dB, so the actual level would be 50 dB, or that of an average classroom. 61. Rock Music A rock band plays at an 80-dB sound level. How many times greater is the sound pressure from another rock band playing at each of the following sound levels? 3.00 m Physics: Principles and Problems Solutions Manual 335 Chapter 15 continued a. 100 dB Each 20 dB increases pressure by a factor of 10, so 10 times greater pressure. b. 120 dB (10)(10) 100 times greater pressure 62. A coiled-spring toy is shaken at a frequency of 4.0 Hz such that standing waves are observed with a wavelength of 0.50 m. What is the speed of propagation of the wave? v f (0.50 m)(4.0 s1) 2.0 m/s 63. A baseball fan on a warm summer day (30°C) sits in the bleachers 152 m away from home plate. a. What is the speed of sound in air at 30°C? The speed increases 0.6 m/s per °C, so the increase from 20°C to 30°C is 6 m/s. Thus, the speed is 343 6 349 m/s. b. How long after seeing the ball hit the bat does the fan hear the crack of the bat? d 152 m t 0.436 s v Level 2 65. Medical Imaging Ultrasound with a frequency of 4.25 MHz can be used to produce images of the human body. If the speed of sound in the body is the same as in salt water, 1.50 km/s, what is the length of a 4.25-MHz pressure wave in the body? v 1.50103 m/s 3.53104 m 6 4.2510 Hz f 0.353 mm 66. Sonar A ship surveying the ocean bottom sends sonar waves straight down into the seawater from the surface. As illustrated in Figure 15-26, the first reflection, off of the mud at the sea floor, is received 1.74 s after it was sent. The second reflection, from the bedrock beneath the mud, returns after 2.36 s. The seawater is at a temperature of 25°C, and the speed of sound in mud is 1875 m/s. t 1.74 s t 2.36 s 349 m Seawater Mud Bedrock ■ Figure 15-26 a. How deep is the water? The speed of sound in the seawater is 1533 m/s and the time for a oneway trip is 0.87 s, so dw vtw (1533 m/s)(0.87 s) 1300 m d ■ Figure 15-25 At 15°C, the speed of sound is 3 m/s slower than at 20°C. Thus, the speed of sound is 340 m/s. v 340 m/s and 2t 2.0 s b. How thick is the mud? The round-trip time in the mud is 2.36 s 1.74 s 0.62 s The one-way time in the mud is 0.31 s, so dm vtm (1875 m/s)(0.31 s) 580 m d vt (340 m/s)(1.0 s) 3.4102 m 336 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 64. On a day when the temperature is 15°C, a person stands some distance, d, as shown in Figure 15-25, from a cliff and claps his hands. The echo returns in 2.0 s. How far away is the cliff? Chapter 15 continued (305 Hz)(343 m/s 0) 67. Determine the variation in sound pressure of a conversation being held at a sound level of 60 dB. The pressure variation at 0 dB is 2105 Pa. For every 20-dB increase, the pressure variation increases by a factor of 10. Therefore, 60 dB has a pressure variation amplitude of 2102 Pa. 68. A fire truck is moving at 35 m/s, and a car in front of the truck is moving in the same direction at 15 m/s. If a 327-Hz siren blares from the truck, what frequency is heard by the driver of the car? vs 35 m/s, v 343 m/s, vd 15 m/s, fs 327 Hz vv v vs d fd fs 343 15 343 35 (327 Hz) 350 Hz Level 3 69. A train moving toward a sound detector at 31.0 m/s blows a 305-Hz whistle. What frequency is detected on each of the following? Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. a. a stationary train vv v vs d fd fs (305 Hz)(343 m/s 0) 343 m/s 31.0 m/s 335 Hz b. a train moving toward the first train at 21.0 m/s vv v vs d fd fs (305 Hz)(343 m/s (21.0 m/s)) 343 m/s 31.0 m/s 356 Hz 70. The train in the previous problem is moving away from the detector. What frequency is now detected on each of the following? a. a stationary train vv v vs d fd fs Physics: Principles and Problems 343 m/s (31.0 m/s) 2.80102 Hz b. a train moving away from the first train at a speed of 21.0 m/s vv v vs d fd fs (305 Hz)(343 m/s 21.0 m/s) 343 m/s (31.0 m/s) 2.63102 Hz 15.2 The Physics of Music pages 426–427 Level 1 71. A vertical tube with a tap at the base is filled with water, and a tuning fork vibrates over its mouth. As the water level is lowered in the tube, resonance is heard when the water level has dropped 17 cm, and again after 49 cm of distance exists from the water to the top of the tube. What is the frequency of the tuning fork? 49 cm 17 cm 32 cm or 0.32 m 1 exists between points of resonance 2 1 0.32 m 2 0.64 m v 343 m/s f 540 Hz 0.64 m 72. Human Hearing The auditory canal leading to the eardrum is a closed pipe that is 3.0 cm long. Find the approximate value (ignoring end correction) of the lowest resonance frequency. 4 L v f v 343 m/s f 2.9 kHz 4L (4)(0.030 m) 73. If you hold a 1.2-m aluminum rod in the center and hit one end with a hammer, it will oscillate like an open pipe. Antinodes of pressure correspond to nodes of molecular motion, so there is a pressure antinode in the center of the bar. The speed of sound Solutions Manual 337 Chapter 15 continued in aluminum is 5150 m/s. What would be the bar’s lowest frequency of oscillation? 1 The rod length is , so 2.4 m 2 v 5150 m/s f 2.1 kHz 2.4 m 74. One tuning fork has a 445-Hz pitch. When a second fork is struck, beat notes occur with a frequency of 3 Hz. What are the two possible frequencies of the second fork? 445 Hz 3 Hz 442 Hz and 445 Hz 3 Hz 448 Hz 75. Flutes A flute acts as an open pipe. If a flute sounds a note with a 370-Hz pitch, what are the frequencies of the second, third, and fourth harmonics of this pitch? f2 2f1 (2)(370 Hz) 740 Hz f3 3f1 (3)(370 Hz) 1110 Hz 1100 Hz f4 4f1 (4)(370 Hz) 1480 Hz 1500 Hz 3f1 (3)(370 Hz) 1110 Hz 1100 Hz a. What is the shortest open organ pipe that will resonate at this frequency? v 343 m/s 20.9 m and f 16.4 Hz L , so 2 v 343 m/s L 10.5 m 2f (2)(16.4Hz) b. What is the pitch if the same organ pipe is closed? Since a closed pipe produces a fundamental with a wavelength twice as long as that of an open pipe of the same length, the frequency would be 1 (16.4 Hz) 8.20 Hz. 2 79. Musical Instruments Two instruments are playing musical A (440.0 Hz). A beat note with a frequency of 2.5 Hz is heard. Assuming that one instrument is playing the correct pitch, what is the frequency of the pitch played by the second instrument? It could be either 440.0 2.5 442.5 Hz or 440.0 2.5 437.5 Hz. 80. A flexible, corrugated, plastic tube, shown in Figure 15-27, is 0.85 m long. When it is swung around, it creates a tone that is the lowest pitch for an open pipe of this length. What is the frequency? 0.85 m 5f1 (5)(370 Hz) 1850 Hz 1800 Hz 7f1 (7)(370 Hz) 2590 Hz 2600 Hz 77. String Instruments A guitar string is 65.0 cm long and is tuned to produce a lowest frequency of 196 Hz. a. What is the speed of the wave on the string? 1 2L (2)(0.650 m) 1.30 m v f (1.30 m)(196 Hz) 255 m/s b. What are the next two higher resonant frequencies for this string? f2 2f1 (2)(196 Hz) 392 Hz f3 3f1 (3)(196 Hz) 588 Hz 338 Solutions Manual ■ Figure 15-27 L 0.85 m , so 1.7 m 2 v 343 m/s f 2.0102 Hz 1.7 m 81. The tube from the previous problem is swung faster, producing a higher pitch. What is the new frequency? f2 2f1 (2)(2.0102 Hz) 4.0102 Hz Level 2 82. During normal conversation, the amplitude of a pressure wave is 0.020 Pa. Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 76. Clarinets A clarinet sounds the same note, with a pitch of 370 Hz, as in the previous problem. The clarinet, however, acts as a closed pipe. What are the frequencies of the lowest three harmonics produced by this instrument? 78. Musical Instruments The lowest note on an organ is 16.4 Hz. Chapter 15 continued a. If the area of an eardrum is 0.52 cm2, what is the force on the eardrum? F PA (0.020 N/m2)(0.52104 m2) 1.0106 N 1 other is wavelength. Therefore, 4 4L 4(0.024 m) 0.096 m, and v f (1760 Hz)(0.096 m) b. The mechanical advantage of the three bones in the middle ear is 1.5. If the force in part a is transmitted undiminished to the bones, what force do the bones exert on the oval window, the membrane to which the third bone is attached? F MA r so Fr (MA)(Fe) F e Fr (1.5)(1.0106 N) 1.5106 N c. The area of the oval window is 0.026 cm2. What is the pressure increase transmitted to the liquid in the cochlea? 6 F 1.510 N P 4 2 0.58 Pa A The length of the steel finger clamped at one end and free to vibrate at the 0.02610 1.7102 m/s Mixed Review pages 427–428 Level 1 85. An open organ pipe is 1.65 m long. What fundamental frequency note will it produce if it is played in helium at 0°C? An open pipe has a length equal to one-half its fundamental wavelength. Therefore, 3.30 m. The speed of sound in helium is 972 m/s. Therefore, v 972 m/s f 295 Hz m 83. Musical Instruments One open organ pipe has a length of 836 mm. A second open pipe should have a pitch that is one major third higher. How long should the second pipe be? 3.30 m 86. If you drop a stone into a well that is 122.5 m deep, as illustrated in Figure 15-29, how soon after you drop the stone will you hear it hit the bottom of the well? v L , so 2L and f Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 2 v 343 m/s f 205 Hz 2L (2)(0.836 m) The ratio of a frequency one major third 122.5 m 5 higher is 5:4, so (205 Hz) 256 Hz. 4 The length of the second pipe is v 343 m/s L 6.70102 mm 2f (2)(256 Hz) 84. As shown in Figure 15-28, a music box contains a set of steel fingers clamped at one end and plucked on the other end by pins on a rotating drum. What is the speed of a wave on a finger that is 2.4 cm long and plays a note of 1760 Hz? ■ First find the time it takes the stone to 1 fall down the shaft by d gt 2, so 2 t d 1 g 2 Steel fingers Figure 15-29 122.5 m 5.00 s 12(9.80 m/s2) The time it takes the sound to comes back up is found with d vst, so d 122.5 m t 0.357 s vs 343 m/s The total time is 5.00 s 0.357 s 5.36 s. ■ Figure 15-28 Physics: Principles and Problems Solutions Manual 339 Chapter 15 continued 87. A bird on a newly discovered planet flies toward a surprised astronaut at a speed of 19.5 m/s while singing at a pitch of 945 Hz. The astronaut hears a tone of 985 Hz. What is the speed of sound in the atmosphere of this planet? fd 985 Hz, fs 945 Hz, vs 19.5 m/s, v? f 1 v d vs v vs fs 1 v f fd v v 90. When a wet finger is rubbed around the rim of a glass, a loud tone of frequency 2100 Hz is produced. If the glass has a diameter of 6.2 cm and the vibration contains one wavelength around its rim, what is the speed of the wave in the glass? v f df 985 Hz (0.062 m)(2100 Hz) 4.1102 m/s 88. In North America, one of the hottest outdoor temperatures ever recorded is 57°C and one of the coldest is 62°C. What are the speeds of sound at those two temperatures? v(T) v(0°C) (0.6 m/s)T, where v(0°C) 331 m/s. So, v(57°C) (331 m/s) (57°C) 0.6 m/s °C v(62°C) (331 m/s) (62°C) 0.6 m/s °C fd 440 Hz 3.0 Hz 443 Hz vv d fd fs v vs so (v vs)fd (v vd)fs and Level 2 89. A ship’s sonar uses a frequency of 22.5 kHz. The speed of sound in seawater is 1533 m/s. What is the frequency received on the ship that was reflected from a whale traveling at 4.15 m/s away from the ship? Assume that the ship is at rest. Part 1. From ship to whale: vd 4.15 m/s, v 1533 m/s, fs 22.5 kHz, vs 0 vv v vs 91. History of Science In 1845, Dutch scientist Christoph Buys-Ballot developed a test of the Doppler effect. He had a trumpet player sound an A note at 440 Hz while riding on a flatcar pulled by a locomotive. At the same time, a stationary trumpeter played the same note. Buys-Ballot heard 3.0 beats per second. How fast was the train moving toward him? 1533 4.15 1533 d fd fs (22.5 kHz) 22.4 kHz Part 2. From whale to ship: vs 4.15 m/s, v 1533 m/s, (v v )f fd d s vs v (343 m/s 0)(440 Hz) 443 Hz 343 m/s 2.3 m/s 92. You try to repeat Buys-Ballot’s experiment from the previous problem. You plan to have a trumpet played in a rapidly moving car. Rather than listening for beat notes, however, you want to have the car move fast enough so that the moving trumpet sounds one major third above a stationary trumpet. a. How fast would the car have to move? 5 4 major third ratio Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 365 m/s Solutions Manual 22.3 kHz Therefore, the speed is 4.80102 m/s 340 1533 1533 4.15 circumference of the glass rim, d vs 19.5 m/s or v fs 945 Hz 1 1 294 m/s vv v vs d fd fs (22.4 kHz) The wavelength is equal to the s So 1 s , fd fs 22.4 kHz, vd 0 Chapter 15 continued vv 367 Hz d fd fs v vs Part 2. vd 37.5 m/s, v 343 m/s, fs 367 Hz so (v vs)fd (v vd)fs (v v )f fd vv v vs 343 (37.5) 343 d fd fs (367 Hz) d s and vs v 4 5 343 m/s (343 m/s 0) 68.6 m/s b. Should you try the experiment? Explain. 407 Hz Thinking Critically page 428 95. Make and Use Graphs The wavelengths of the sound waves produced by a set of tuning forks with given frequencies are shown in Table 15-2 below. v (68.6 m/s) 1 mi 3600 s 1609 m 1h 153 mph, so the car would be moving dangerously fast. Do not try the experiment. Table 15-2 Tuning Forks Level 3 93. Guitar Strings The equation for the speed Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 0.65 m 2L 2(0.65 m) 1.30 m v f (147 Hz)(1.30 m) 191 m/s (191 m/s)2(4.9103 kg/m) 180 N 94. A train speeding toward a tunnel at 37.5 m/s sounds its horn at 327 Hz. The sound bounces off the tunnel mouth. What is the frequency of the reflected sound heard on the train? Hint: Solve the problem in two parts. First, assume that the tunnel is a stationary observer and find the frequency. Then, assume that the tunnel is a stationary source and find the frequency measured on the train. Part 1. vs 37.5 m/s, v 343 m/s, fs 327 Hz v vd 3 43 fd fs (327 Hz) v vs 343 37.5 Physics: Principles and Problems 2.62 147 2.33 165 2.08 196 1.75 220 1.56 247 1.39 where FT is 0.0032 kg 4.9103 kg/m FT 131 FT , the tension in the string and is the mass per unit length of the string. A guitar string has a mass of 3.2 g and is 65 cm long. What must be the tension in the string to produce a note whose fundamental frequency is 147 Hz? v 2 Wavelength (m) a. Plot a graph of the wavelength versus the frequency (controlled variable). What type of relationship does the graph show? The graph shows an inverse relationship between frequency and wavelength. 2.60 2.20 (m) of a wave on a string is v Frequency (Hz) 1.80 1.40 1.00 100 140 180 240 260 f (Hz) b. Plot a graph of the wavelength versus the inverse of the frequency (1/f). What kind of graph is this? Determine the Solutions Manual 341 Chapter 15 continued speed of sound from this graph. The graph shows a direct relationship between period (1/f) and wavelength. The speed of sound is represented by the slope, ~343 m/s. 2.60 (m) 2.20 1.40 6.0 8.0 1 – (ms) f 96. Make Graphs Suppose that the frequency of a car horn is 300 Hz when it is stationary. What would the graph of the frequency versus time look like as the car approached and then moved past you? Complete a rough sketch. 97. Analyze and Conclude Describe how you could use a stopwatch to estimate the speed of sound if you were near the green on a 200m golf hole as another group of golfers hit their tee shots. Would your estimate of the speed of sound be too large or too small? You could start the watch when you saw the hit and stop the watch when the sound reached you. The speed would be calculated by dividing the distance, 200 m, by the measured time. The measured time would be too large because you could anticipate the impact by sight, but you could not anticipate the sound. The calculated speed would be too small. 98. Apply Concepts A light wave coming from a point on the left edge of the Sun is found by astronomers to have a slightly higher Solutions Manual Measure the mass and length of the string to determine . Then clamp the string to a table, hang one end over the table edge, and stretch the string by hanging weights on its end to obtain FT. Calculate the speed of the wave using the formula. Next, pluck the string in its middle and determine the frequency by matching it to a frequency generator, using beats to tune the generator. Multiply the frequency by twice the string length, which is equal to the wavelength, to obtain the speed from the wave equation. Compare the results. Repeat for different tensions and other strings with different masses per unit length. Consider possible causes of error. Writing in Physics page 428 100. Research the construction of a musical instrument, such as a violin or French horn. What factors must be considered besides the length of the strings or tube? What is the difference between a quality instrument and a cheaper one? How are they tested for tone quality? Answers will vary. A report on violin construction might include information about the bridge as a link between the strings and body and information about the role of the body in causing air molecules around the violin to vibrate. Students also might discuss Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. The graph should show a fairly steady frequency above 300 Hz as it approaches and a fairly steady frequency below 300 Hz as it moves away. 342 The Sun must be rotating on its axis in the same manner as Earth. The Doppler shift indicates that the left side of the Sun is coming toward us, while the right side is moving away. 99. Design an Experiment Design an experiment that could test the formula for the speed of a wave on a string. Explain what measurements you would make, how you would make them, and how you would use them to test the formula. 1.80 1.00 4.0 frequency than light from the right side. What do these measurements tell you about the Sun’s motion? Chapter 15 continued the ways in which the woods and finishes used in making violins affect the quality of the sound produced by the instruments. 101. Research the use of the Doppler effect in the study of astronomy. What is its role in the big bang theory? How is it used to detect planets around other stars? To study the motions of galaxies? Students should discuss the work of Edwin Hubble, the redshift and the expanding universe, spectroscopy, and the detection of wobbles in the motion of planet-star systems. Horizontal: mAvA1 mBvB2 So mBvB2 (1.0 kg)(3.0 m/s) 3.0 kgm/s Vertical: 0 mAvA2 mBvB2 So mBvB2 (1.0 kg)(2.0 m/s) 2.0 kgm/s The vector sum is mv (3.0 kgm/s )2 (2.0 s)2 kgm/ 2.0 kgm/s 3.6 kgm/s and tan 3.0 kgm/s Cumulative Review so 34° page 428 102. Ball A, rolling west at 3.0 m/s, has a mass of 1.0 kg. Ball B has a mass of 2.0 kg and is stationary. After colliding with ball B, ball A moves south at 2.0 m/s. (Chapter 9) Therefore, mBvB2 3.6 kgm/s at a. Sketch the system, showing the velocities and momenta before and after the collision. Westward and southward are positive. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. b. Calculate the momentum and velocity of ball B after the collision. Before B A pB1 = 0 vA1 = 3.0 m/s west pA1 = 3.0 kgm/s west After A A 34° north of west 3.6 kgm/s vB2 2.0 kg 1.8 m/s at 34° north of west 103. Chris carries a 10-N carton of milk along a level hall to the kitchen, a distance of 3.5 m. How much work does Chris do? (Chapter 10) No work, because the force and the displacement are perpendicular. 104. A movie stunt person jumps from a fivestory building (22 m high) onto a large pillow at ground level. The pillow cushions her fall so that she feels a deceleration of no more than 3.0 m/s2. If she weighs 480 N, how much energy does the pillow have to absorb? How much force does the pillow exert on her? (Chapter 11) The energy to be absorbed equals the mechanical energy that she had, which equals her initial potential energy. U mgh (480 N)(22 m) 11 kJ. The force on her is vB2 = ? pB2 = ? vA2 = 2.0 m/s south pA2 = 2.0 kgm/s south Physics: Principles and Problems Fg 480 N 2 F ma (a) 2 (3.0 m/s ) g 9.80 m/s 150 N Solutions Manual 343 Chapter 15 continued Challenge Problem page 417 1. Determine the tension, FT, in a violin string of mass m and length L that will play the fundamental note at the same frequency as a closed pipe also of length L. Express your answer in terms of m, L, and the speed of sound in air, v. The equation for the speed of a wave on a string is u . where FT is FT the tension string and is the mass per unit length of the string. The wavelength of the fundamental in a closed pipe is equal to 4L, so v the frequency is f . The wavelength of the fundamental on a string is 4L u equal to 2L, so the frequency of the string is f , where u is the speed of the wave on the string, u 2L FT . The mass per unit length of the string m/L. Squaring the frequencies and setting them equal gives FT F L FT v2 u2 T2 2 16L 4L2 4L2 4L m 4Lm mv 2 Finally, rearranging for the tension gives FT 4L 2. What is the tension in a string of mass 1.0 g and 40.0 cm long that plays the same note as a closed pipe of the same length? For a string of mass 1.0 g and length 0.400 m, the tension is (0.0010 kg)(343 m/s)2 mv 2 FT 74 N 4L 4(0.400 m) Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 344 Solutions Manual Physics: Principles and Problems CHAPTER 16 Fundamentals of Light 4. A 64-cd point source of light is 3.0 m above the surface of a desk. What is the illumination on the desk’s surface in lux? Practice Problems 16.1 Illumination pages 431–438 P 4(64 cd) 256 lm page 436 1. A lamp is moved from 30 cm to 90 cm above the pages of a book. Compare the illumination on the book before and after the lamp is moved. P Eafter Ebefore 4dafter2 P 4dbefore2 d dafter 2 before 2 (30 cm)2 1 2 ; therefore, after 9 (90 cm) the lamp is moved the illumination is one-ninth of the original illumination. 2. What is the illumination on a surface that is 3.0 m below a 150-W incandescent lamp that emits a luminous flux of 2275 lm? P 2275 lm E 2 2 2.0101 lx 4(3.0 m) 3. Draw a graph of the illuminance produced by a 150-W incandescent lamp between 0.50 m and 5.0 m. 4(3.0 m) 4d 5. A public school law requires a minimum illuminance of 160 lx at the surface of each student’s desk. An architect’s specifications call for classroom lights to be located 2.0 m above the desks. What is the minimum luminous flux that the lights must produce? P E 2 4d P 4Ed 2 4(160 lm/m2)(2.0 m)2 8.0103 lm 6. A screen is placed between two lamps so that they illuminate the screen equally, as shown in Figure 16-7. The first lamp emits a luminous flux of 1445 lm and is 2.5 m from the screen. What is the distance of the second lamp from the screen if the luminous flux is 2375 lm? Illuminance of a 150-W bulb Screen P 2275, d 0.50, 0.75, …, 5.0 P E(d) 2 4d P 2375 lm 720 2.5 m P 1445 lm ■ E (lx) Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 4d 256 lm P so E 2 2 7.1 lx Figure 16-7 (Not to scale) E1 E2 P 2275 lm P d1 P d2 1 2 So 2 2 7.2 0.50 5.0 r (m) P 2 or d2 d1 P 1 1445 2375 (2.5 m) 3.2 m Physics: Principles and Problems Solutions Manual 345 Chapter 16 continued Section Review 16.1 Illumination pages 431–438 page 438 7. Use of Material Light Properties Why might you choose a window shade that is translucent? Opaque? You would use a translucent window shade to keep people from looking in or out, while still allowing daylight in. You would use an opaque window shade to keep the daylight out. 8. Illuminance Does one lightbulb provide more illuminance than two identical lightbulbs at twice the distance? Explain. One lightbulb provides an illuminance that is four times larger than two of the same lightbulb at twice the distance, P because E 2. d d Since the illumination is equal, E1 E2 I d2 1 2 So 2 2 2 1 2 or I2 2 2 (5.0 m) 10. Distance of a Light Source Suppose that a lightbulb illuminating your desk provides only half the illuminance that it should. If it is currently 1.0 m away, how far should it be to provide the correct illuminance? Illumination depends on 1/d2, E df2 1 so i 2 di 1 0.71 m 11. Distance of Light Travel How far does light travel in the time it takes sound to travel 1 cm in air at 20°C? Sound velocity is 343 m/s, so sound takes 3105 s to travel 1 cm. In that time, light travels 9 km. vsound 343 m/s d tsound v sound 2 110 m 343 m/s 3105 s vlight 3.00108 m/s (3.00108 m/s)(3105 s) 9103 m 9 km 12. Distance of Light Travel The distance to the Moon can be found with the help of mirrors left on the Moon by astronauts. A pulse of light is sent to the Moon and returns to Earth in 2.562 s. Using the measured value of the speed of light to the same precision, calculate the distance from Earth to the Moon. d ct (75 cd)(3.0 m) 27 cd 2 Ef 2 m 2 (299,800,000 m/s) (2.562 s) 1 2 3.840108 m 13. Critical Thinking Use the correct time taken for light to cross Earth’s orbit, 16.5 min, and the diameter of Earth’s orbit, 2.981011 m, to calculate the speed of light using Roemer’s method. Does this method appear to be accurate? Why or why not? 11 d 3.010 v t (16 min)(60 s/min) 3.1108 m/s 346 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. I E 2 I d d1 df dlight vlighttsound 9. Luminous Intensity Two lamps illuminate a screen equally—lamp A at 5.0 m, lamp B at 3.0 m. If lamp A is rated 75 cd, what is lamp B rated? I d1 df2 1 2 (1.0 m)2 Chapter 16 continued Practice Problems v c (obs ) 16.2 The Wave Nature of Light pages 439–447 v obs 1 c page 447 14. What is the frequency of oxygen’s spectral line if its wavelength is 513 nm? c Use and solve for f. f c f 8 3.0010 m/s 7 5.1310 m 5.851014 Hz Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 15. A hydrogen atom in a galaxy moving with a speed of 6.55106 m/s away from Earth emits light with a frequency of 6.161014 Hz. What frequency of light from that hydrogen atom would be observed by an astronomer on Earth? The relative speed along the axis is much less than the speed of light. Thus, you can use the observed light frequency equation. Because the astronomer and the galaxy are moving away from each other, use the negative form of the observed light frequency equation. v fobs f 1 c 6.5510 m/s (6.161014 Hz)1 8 6 3.0010 m/s 6.031014 Hz 16. A hydrogen atom in a galaxy moving with a speed of 6.551016 m/s away from Earth emits light with a wavelength of 4.86107 m. What wavelength would be observed on Earth from that hydrogen atom? 6.55106 3.00108 (4.86107 m)1 m/s m/s 4.97107 m 17. An astronomer is looking at the spectrum of a galaxy and finds that it has an oxygen spectral line of 525 nm, while the laboratory value is measured at 513 nm. Calculate how fast the galaxy would be moving relative to Earth. Explain whether the galaxy is moving toward or away from Earth and how you know. Assume that the relative speed along the axis is much less than the speed of light. Thus, you can use the Doppler shift equation. v c (obs ) The observed (apparent) wavelength appears to be longer than the known (actual) wavelength of the oxygen spectral line. This means that the astronomer and the galaxy are moving away from each other. So use the positive form of the Doppler shift equation. v c (obs ) Solve for the unknown variable. ( ) obs v c (525 nm 513 nm) 513 nm (3.00108 m/s) 7.02106 m/s The relative speed along the axis is much less than the speed of light. Thus, you can use the observed Doppler shift equation. Because the astronomer and the galaxy are moving away from each other, use the positive form of the Doppler shift equation. Physics: Principles and Problems Solutions Manual 347 Chapter 16 continued Section Review 16.2 The Wave Nature of Light pages 439–447 page 447 18. Addition of Light Colors What color of light must be combined with blue light to obtain white light? yellow (a mixture of the other two primaries, red and green) 19. Combination of Pigments What primary pigment colors must be mixed to produce red? Explain how red results using color subtraction for pigment colors. 22. Polarization Describe a simple experiment that you could do to determine whether sunglasses in a store are polarizing. See if the glasses reduce glare from the reflective surfaces, such as windows or roadways. Polarization of light allows photographers to photograph objects while eliminating glare. 23. Critical Thinking Astronomers have determined that Andromeda, a neighboring galaxy to our own galaxy, the Milky Way, is moving toward the Milky Way. Explain how they determined this. Can you think of a possible reason why Andromeda is moving toward our galaxy? The spectral lines of the emissions of known atoms are blue-shifted in the light we see coming from Andromeda. Andromeda would be moving toward us due to gravitational attraction. This gravitational attraction could be due to the mass of the Milky Way or other objects located near the Milky Way. Yellow and magenta pigments are used to produce red. Yellow pigment subtracts blue and magenta pigment subtracts green, neither subtracts red so the mixture would reflect red. 20. Light and Pigment Interaction What color will a yellow banana appear to be when illuminated by each of the following? a. white light yellow yellow c. blue light black Concept Mapping page 452 24. Complete the following concept map using the following terms: wave, c, Doppler effect, polarization. 21. Wave Properties of Light The speed of red light is slower in air and water than in a vacuum. The frequency, however, does not change when red light enters water. Does the wavelength change? If so, how? Yes, because v f and v/f, when v decreases, so does . Light Models wave ray straight path c diffraction polarization 348 Solutions Manual Doppler effect 0 c/f Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. b. green and red light Chapter Assessment Chapter 16 continued Mastering Concepts page 452 25. Sound does not travel through a vacuum. How do we know that light does? (16.1) Light comes through the vacuum of space from the Sun. 26. Distinguish between a luminous source and an illuminated source. (16.1) A luminous body emits light. An illuminated body is one on which light falls and is reflected. 27. Look carefully at an ordinary, frosted, incandescent bulb. Is it a luminous or an illuminated source? (16.1) It is mainly illuminated. The filament is luminous; the frosted glass is illuminated. You barely can see the hot filament through the frosted glass. 28. Explain how you can see ordinary, nonluminous classroom objects. (16.1) Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Ordinary nonluminous objects are illuminated by reflected light, allowing them to be seen. 29. Distinguish among transparent, translucent, and opaque objects. (16.1) A transparent object is a material through which light can pass without distortion. A translucent object allows light to pass but distorts the light to the point where images are not discernable. An opaque object does not allow light to pass through. 30. To what is the illumination of a surface by a light source directly proportional? To what is it inversely proportional? (16.1) The illumination on a surface is directly proportional to the intensity of the source and inversely proportional to the square of the distance of the surface from the source. Physics: Principles and Problems 31. What did Galileo assume about the speed of light? (16.1) The speed of light is very fast, but finite. 32. Why is the diffraction of sound waves more familiar in everyday experience than is the diffraction of light waves? (16.2) Diffraction is most pronounced around obstacles approximately the same size as the wavelength of the wave. We are more accustomed to obstacles of the size that diffract the much larger wavelengths of sound. 33. What color of light has the shortest wavelength? (16.2) Violet light has the shortest wavelength. 34. What is the range of the wavelengths of light, from shortest to longest? (16.2) 400 nm to 700 nm 35. Of what colors does white light consist? (16.2) White light is a combination of all the colors, or at least the primary colors. 36. Why does an object appear to be black? (16.2) An object appears to be black because little, if any, light is being reflected from it. 37. Can longitudinal waves be polarized? Explain. (16.2) No. They have no vertical or horizontal components. 38. If a distant galaxy were to emit a spectral line in the green region of the light spectrum, would the observed wavelength on Earth shift toward red light or toward blue light? Explain. (16.2) Because the galaxy is distant, it is most likely moving away from Earth. The wavelength actually would shift away from the wavelength of green light toward a longer red wavelength. If it Solutions Manual 349 Chapter 16 continued shifted toward the blue wavelength, the wavelength would be shorter, not longer. This would indicate the galaxy is getting closer to Earth, and no galaxy outside the Local Group has been discovered moving toward us. 39. What happens to the wavelength of light as the frequency increases? (16.2) As the frequency increases, the wavelength decreases. Applying Concepts page 452 40. A point source of light is 2.0 m from screen A and 4.0 m from screen B, as shown in Figure 16-21. How does the illuminance at screen B compare with the illuminance at screen A? Screen A Screen B Source 2m 43. Eye Sensitivity The eye is most sensitive to yellow-green light. Its sensitivity to red and blue light is less than 10 percent as great. Based on this knowledge, what color would you recommend that fire trucks and ambulances be painted? Why? Fire trucks should be painted yellowgreen, 550 nm, because less light has to be reflected to the eye for the fire truck to be seen. 44. Streetlight Color Some very efficient streetlights contain sodium vapor under high pressure. They produce light that is mainly yellow with some red. Should a community that has these lights buy dark blue police cars? Why or why not? Blue pigment of a police car will absorb the red and yellow light. Dark blue police cars would not be very visible. If a community wants its police cars to be visible, they should buy yellow cars. Refer to Figure 16-22 for problems 45 and 46. Figure 16-21 Illumination E l/r 2; therefore, the illumination at screen B will be one-fourth of that at screen A because it is twice as far from the source. 41. Reading Lamp You have a small reading lamp that is 35 cm from the pages of a book. You decide to double the distance. a. Is the illuminance at the book the same? No. b. If not, how much more or less is it? Distance is doubled, so the illumination of the page is one-fourth as great. 42. Why are the insides of binoculars and cameras painted black? ■ Figure 16-22 45. What happens to the illuminance at a book as the lamp is moved farther away from the book? Illuminance decreases, as described by the inverse-square law. 46. What happens to the luminous intensity of the lamp as it is moved farther away from the book? No change; distance does not affect the luminous intensity of the lamp. The insides are painted black because black does not reflect any light, and thus there is no interference while 350 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. ■ 4m observing or photographing objects. Chapter 16 continued 47. Polarized Pictures Photographers often put polarizing filters over camera lenses to make clouds in the sky more visible. The clouds remain white, while the sky looks darker. Explain this based on your knowledge of polarized light. Light scattered from the atmosphere is polarized, but light scattered from the clouds is not. By rotating the filter, the photographer can reduce the amount of polarized light reaching the film. 48. An apple is red because it reflects red light and absorbs blue and green light. 52. Traffic Violation Suppose that you are a traffic officer and you stop a driver for going through a red light. Further suppose that the driver draws a picture for you (Figure 16-23) and explains that the light looked green because of the Doppler effect when he went through it. Explain to him using the Doppler shift equation just how fast he would have had to be going for the red light ( 645 nm), to appear green ( 545 nm). Hint: For the purpose of this problem, assume that the Doppler shift equation is valid at this speed. a. Why does red cellophane look red in reflected light? Cellophane reflects red light and absorbs blue and green light. b. Why does red cellophane make a white lightbulb look red when you hold the cellophane between your eye and the lightbulb? Cellophane transmits red light. c. What happens to the blue and green light? Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Blue and green light are absorbed. 49. You put a piece of red cellophane over one flashlight and a piece of green cellophane over another. You shine the light beams on a white wall. What color will you see where the two flashlight beams overlap? yellow 50. You now put both the red and green cellophane pieces over one of the flashlights in Problem 49. If you shine the flashlight beam on a white wall, what color will you see? Explain. Black; almost no light would get through because the light transmitted through the first filter would be absorbed by the second. 51. If you have yellow, cyan, and magenta pigments, how can you make a blue pigment? Explain. Mix cyan and magenta. Physics: Principles and Problems ■ Figure 16-23 645 nm 545 nm (3.00108 m/s) 645 nm 4.65107 m/s That is over 100 million mph. If he does not get a ticket for running a red light, he will get a ticket for speeding. Mastering Problems 16.1 Illumination page 453 Level 1 53. Find the illumination 4.0 m below a 405-lm lamp. P 405 lm E 2 2 2.0 lx 4d 4(4.0 m) 54. Light takes 1.28 s to travel from the Moon to Earth. What is the distance between them? d vt (3.00108 m/s)(1.28 s) 3.84108 m Level 2 55. A three-way bulb uses 50, 100, or 150 W of electric power to deliver 665, 1620, or 2285 lm in its three settings. The bulb is placed 80 cm above a sheet of paper. If an Solutions Manual 351 Chapter 16 continued illumination of at least 175 lx is needed on the paper, what is the minimum setting that should be used? P d1 P d2 1 2 So 2 2 P d d1 2 1 2 or P2 2 P E 4d 2 P 4Ed 2 4(175 lx)(0.80 m)2 1.4103 lm Thus, the 100-W (1620-lm) setting is needed. 56. Earth’s Speed Ole Roemer found that the average increased delay in the disappearance of Io from one orbit around Jupiter to the next is 13 s. a. How far does light travel in 13 s? 3.9109 m b. Each orbit of Io takes 42.5 h. Earth travels the distance calculated in part a in 42.5 h. Find the speed of Earth in km/s. d v 2 (1750 lm)(1.08 m) 2 (1.25 m) 1.31103 lm 58. Suppose that you wanted to measure the speed of light by putting a mirror on a distant mountain, setting off a camera flash, and measuring the time it takes the flash to reflect off the mirror and return to you, as shown in Figure 16-24. Without instruments, a person can detect a time interval of about 0.10 s. How many kilometers away would the mirror have to be? Compare this distance with that of some known distances. Mirror You t d 9 1 km 3.910 m 5 1.5310 s 1000 m 25 km/s ■ 8 d 2(1.510 km) 1d v t 365 d 24 h 1h 3600 s 3.0101 km/s, so fairly accurate Level 3 57. A student wants to compare the luminous flux of a lightbulb with that of a 1750-lm lamp. The lightbulb and the lamp illuminate a sheet of paper equally. The 1750-lm lamp is 1.25 m away from the sheet of paper; the lightbulb is 1.08 m away. What is the lightbulb’s luminous flux? P E 2 4d Since the illumination is equal, E1 E2 352 Solutions Manual Figure 16-24 d vt (3.00108 m/s)(0.1 s) 1 km 1000 m 3104 km The mirror would be half this distance, or 15,000 km away. Earth is 40,000 km in circumference, so this is threeeighths of the way around Earth. 16.2 The Wave Nature of Light pages 453–454 Level 1 59. Convert 700 nm, the wavelength of red light, to meters. 1109 m 1 nm (700 nm) 7107 m Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. c. Check to make sure that your answer for part b is reasonable. Calculate Earth’s speed in orbit using the orbital radius, 1.5108 km, and the period, 1.0 yr. Chapter 16 continued 60. Galactic Motion How fast is a galaxy moving relative to Earth if a hydrogen spectral line of 486 nm is red-shifted to 491 nm? Assume that the relative speed along the axis is much less than the speed of light. Thus, you can use the Doppler shift equation. v (app ) c The light is red-shifted, so the astronomer and the galaxy are moving away from each other. So use the positive form of the apparent light wavelength equation. v c (app ) Solve for the unknown variable. (app ) v c 491 nm 486 nm 486 nm (3.00108 m/s) 3.09106 m/s Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. The original assumption was valid. 61. Suppose that you are facing due east at sunrise. Sunlight is reflected off the surface of a lake, as shown in Figure 16-25. Is the reflected light polarized? If so, in what direction? Level 2 62. Polarizing Sunglasses In which direction should the transmission axis of polarizing sunglasses be oriented to cut the glare from the surface of a road: vertically or horizontally? Explain. The transmission axis should be oriented vertically, since the light reflecting off the road will be partially polarized in the horizontal direction. A vertical transmission axis will filter horizontal waves. 63. Galactic Motion A hydrogen spectral line that is known to be 434 nm is red-shifted by 6.50 percent in light coming from a distant galaxy. How fast is the galaxy moving away from Earth? Assume that the relative speed along the axis is much less than the speed of light. Thus, you can use the Doppler shift equation. v c (app ) The light is red-shifted, so the astronomer and the galaxy are moving away from each other. So use the positive form of the apparent light wavelength equation. v c (app ) Solve for the unknown variable. (app ) v c (3.00108 m/s) (1.065)(434 nm) 434 nm 434 nm 1.95107 m/s ■ Figure 16-25 The reflected light is partially polarized in the direction parallel to the surface of the lake and perpendicular to the path of travel of the light from the lake to your eyes. Physics: Principles and Problems The original assumption was valid. Level 3 64. For any spectral line, what would be an unrealistic value of the apparent wavelength for a galaxy moving away from Earth? Why? An unrealistic value would make the galaxy seem to be moving away from us at a speed close to or greater than the speed of light, or v c. If this were the Solutions Manual 353 Chapter 16 continued case, use of the low-speed Doppler shift equation would give a wavelength c difference of (app ) . When c solved, this would give an apparent wavelength of app 2. It would be twice as large as the actual wavelength. So any apparent wavelength close to or greater than twice the actual wavelength would be unrealistic. Mixed Review page 454 Level 1 65. Streetlight Illumination A streetlight contains two identical bulbs that are 3.3 m above the ground. If the community wants to save electrical energy by removing one bulb, how far from the ground should the streetlight be positioned to have the same illumination on the ground under the lamp? P E 2 4d If P is reduced by a factor of 2, so must d 2. Thus, d is reduced by a factor of 2 , becoming 66. An octave in music is a doubling of frequency. Compare the number of octaves that correspond to the human hearing range to the number of octaves in the human vision range. Humans hear over a range of about nine or ten octaves (20 Hz to 10,240 or 20,480 Hz); however, human vision is less than one “octave.” Level 2 67. A 10.0-cd point-source lamp and a 60.0-cd point-source lamp cast equal intensities on a wall. If the 10.0-cd lamp is 6.0 m from the wall, how far from the wall is the 60.0-cd lamp? I E 2 and since the intensities on the d E1 E2 I d2 I d1 1 2 So 2 2 I I 60.0 cd or d 2 d1 2 (6.0 m) 10.0 c d 1 15 m 68. Thunder and Lightning Explain why it takes 5 s to hear thunder when lightning is 1.6 km away. The time for light to travel 1.6 km is a small fraction of a second (5.3 s). The sound travels about 340 m/s, which is about one-fifth of the 1.6 km every second, and takes about 4.7 s to travel 1.6 km. Level 3 69. Solar Rotation Because the Sun rotates on its axis, one edge of the Sun moves toward Earth and the other moves away. The Sun rotates approximately once every 25 days, and the diameter of the Sun is 1.4109 m. Hydrogen on the Sun emits light of frequency 6.161014 Hz from the two sides of the Sun. What changes in wavelength are observed? Speed of rotation is equal to circumference times period of rotation. 9 (1.410 m) vrot (25 days)(24 h/day)(3600 s/h) 2.04103 m/s c f 3.0010 m/s 14 8 6.1610 Hz 4.87107 m v c v c rot (2.04103 m/s) (3.0010 m/s) (4.87107 m) 8 3.31012 m wall are equal, the wall is equally 354 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. (3.3 m) 2.3 m 2 illuminated and Chapter 16 continued Thinking Critically page 454 70. Research Why did Galileo’s method for measuring the speed of light not work? It was not precise enough. He was not able to measure the small time intervals involved in a terrestrial measurement. E (lm) 71. Make and Use Graphs A 110-cd light source is 1.0 m from a screen. Determine the illumination on the screen originally and for every meter of increasing distance up to 7.0 m. Graph the data. 110 100 90 80 70 60 50 40 30 20 10 0.0 2.0 3.0 4.0 page 454 75. A 2.0-kg object is attached to a 1.5-m long string and swung in a vertical circle at a constant speed of 12 m/s. (Chapter 7) 5.0 6.0 7.0 r (m) a. What is the shape of the graph? hyperbola Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Answers will vary. Begin with the element thorium. Heat it to the melting point of platinum. At this temperature, the thorium will glow. Surround the thorium with an opaque material that can take the high temperature. Leave an opening that is one-sixtieth of a square centimeter in size. The candela is defined as the amount of steady flow of light energy that is emitted by the thorium through the opening under these conditions. Cumulative Review P 1.4103 lm 1.0 74. Look up information on the SI unit candela, cd, and explain in your own words the standard that is used to set the value of 1 cd. b. What is the relationship between illuminance and distance shown by the graph? inverse square 72. Analyze and Conclude If you were to drive at sunset in a city filled with buildings that have glass-covered walls, the setting Sun reflected off the building’s walls might temporarily blind you. Would polarizing glasses solve this problem? Yes. Light reflected off glass is partially polarized, so polarizing sunglasses will reduce much of the glare, if the sunglasses are aligned correctly. Writing in Physics page 454 73. Write an essay describing the history of human understanding of the speed of light. Include significant individuals and the contribution that each individual made. a. What is the tension in the string when the object is at the bottom of its path? mv 2 (2.0 kg)(12 m/s)2 Fnet r 1.9101 1.5 m N Fg mg (2.0 kg)(9.80 m/s2) 2.0102 N Fnet T Fg T Fnet Fg 1.9102 N 0.20102 N 2.1102 N b. What is the tension in the string when the object is at the top of its path? Fnet T Fg T Fnet Fg 1.9102 N 0.20102 N 1.7102 N Answers will vary. Physics: Principles and Problems Solutions Manual 355 Chapter 16 continued 76. A space probe with a mass of 7.600103 kg is traveling through space at 125 m/s. Mission control decides that a course correction of 30.0° is needed and instructs the probe to fire rockets perpendicular to its present direction of motion. If the gas expelled by the rockets has a speed of 3.200 km/s, what mass of gas should be released? (Chapter 9) mgvg tan 30.0° mpvp1 mpvp1(tan 30.0°) mg vg (7.600103 kg)(125 m/s)(tan 30.0°) 3.200103 m/s 171 kg 77. When a 60.0-cm-long guitar string is plucked in the middle, it plays a note of frequency 440 Hz. What is the speed of the waves on the string? (Chapter 14) 2L 2(0.600 m) 1.20 m v f (1.20 m)(440 Hz) 530 m/s v 1493 m/s 0.0878 m 8.8 cm f The analyzer filter allows some light to pass through, since its polarizing axis is not perpendicular to the polarizing axis of the first filter. The last polarizing filter now can pass light from the analyzer filter, since the polarizing axis of the analyzer filter is not perpendicular to the polarizing axis of the last polarizing filter. 2. The analyzer filter is placed at an angle of relative to the polarizing axes of filter 1. Derive an equation for the intensity of light coming out of filter 2 compared to the intensity of light coming out of filter 1. I1 is the light intensity out of the first filter, Ianalyzer is the light intensity out of the analyzer filter, and I2 is the light intensity out of the last filter. Ianalyzer I1 cos2 I2 Ianalyzer cos2(90° ) I2 I1cos2() cos2(90° ) Filter 1 17,000 Hz Analyzer Challenge Problem page 444 You place an analyzer filter between the two cross-polarized filters, such that its polarizing axis is not parallel to either of the two filters, as shown in the figure to the right. 356 Solutions Manual Polarized light Filter 2 90° Polarizing axes Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 78. What is the wavelength of a sound wave with a frequency of 17,000 Hz in water at 25°C? (Chapter 15) 1. You observe that some light passes through filter 2, though no light passed through filter 2 previous to inserting the analyzer filter. Why does this happen? CHAPTER 17 Reflection and Mirrors Practice Problems 17.1 Reflection from Plane Mirrors pages 457– 463 page 460 1. Explain why the reflection of light off ground glass changes from diffuse to specular if you spill water on it. Water fills in the rough areas and makes the surface smoother. 2. If the angle of incidence of a ray of light is 42°, what is each of the following? a. the angle of reflection r i 42° b. the angle the incident ray makes with the mirror r1 i1 30° i2 90° r1 90° 30° 60° Section Review 17.1 Reflection from Plane Mirrors pages 457–463 page 463 6. Reflection A light ray strikes a flat, smooth, reflecting surface at an angle of 80° to the normal. What is the angle that the reflected ray makes with the surface of the mirror? r i 80° r, mirror 90° r 90° 80° i, mirror 90° i 90° 42° 48° Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. c. the angle between the incident ray and the reflected ray i r 2i 84° 3. If a light ray reflects off a plane mirror at an angle of 35° to the normal, what was the angle of incidence of the ray? i r 35° 4. Light from a laser strikes a plane mirror at an angle of 38° to the normal. If the laser is moved so that the angle of incidence increases by 13°, what is the new angle of reflection? i i, initial 13° 38° 13° 51° r i 51° 5. Two plane mirrors are positioned at right angles to one another. A ray of light strikes one mirror at an angle of 30° to the normal. It then reflects toward the second mirror. What is the angle of reflection of the light ray off the second mirror? Physics: Principles and Problems 10° 7. Law of Reflection Explain how the law of reflection applies to diffuse reflection. The law of reflection applies to individual rays of light. Rough surfaces make the light rays reflect in many different directions. 8. Reflecting Surfaces Categorize each of the following as a specular or a diffuse reflecting surface: paper, polished metal, window glass, rough metal, plastic milk jug, smooth water surface, and ground glass. Specular: window glass, smooth water, polished metal. Diffuse: paper, rough metal, ground glass, plastic milk jug. 9. Image Properties A 50-cm-tall dog stands 3 m from a plane mirror and looks at its image. What is the image position, height, and type? di do 3m Solutions Manual 357 Chapter 17 continued hi ho 13. An object is 36.0 cm in front of a concave mirror with a 16.0-cm focal length. Determine the image position. 50 cm The image is virtual. 10. Image Diagram A car is following another car down a straight road. The first car has a rear window tilted 45°. Draw a ray diagram showing the position of the Sun that would cause sunlight to reflect into the eyes of the driver of the second car. The Sun Rear window of first car 45° 45° Driver of second car 1 1 1 do di f d f do f di o (36.0 cm)(16.0 cm) 36.0 cm 16.0 cm 28.8 cm 14. A 3.0-cm-tall object is 20.0 cm from a 16.0-cm-radius concave mirror. Determine the image position and image height. 1 1 1 do di f 45° Road grade The Sun’s position directly overhead would likely reflect light into the driver’s eyes, according to the law of reflection. 11. Critical Thinking Explain how diffuse reflection of light off an object enables you to see an object from any angle. Practice Problems (20.0 cm) 16.0 cm 2 16.0 cm 20.0 cm 2 hi di ho do m diho (13.3 cm)(3.0 cm) 20.0 cm hi do 2.0 cm 15. A concave mirror has a 7.0-cm focal length. A 2.4-cm-tall object is 16.0 cm from the mirror. Determine the image height. 17.2 Curved Mirrors pages 464–473 1 1 1 do di f page 469 12. Use a ray diagram, drawn to scale, to solve Example Problem 2. di o O1 Ray 1 F I1 d f do f (16.0 cm)(7.0 cm) 16.0 cm 7.0 cm 12.4 cm Ray 2 Horizontal scale: C 1 block 1.0 cm Vertical scale: 3 blocks 1.0 cm 13.3 cm hi di ho do m diho hi do (12.4 cm)(2.4 cm) 16.0 cm 1.9 cm 358 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. The incoming light reflects off the surface of the object in all directions. This enables you to view the object from any location. d f do f di o Chapter 17 continued 16. An object is near a concave mirror of 10.0-cm focal length. The image is 3.0 cm tall, inverted, and 16.0 cm from the mirror. What are the object position and object height? 1 1 1 do di f df (16.0 cm)(10.0 cm) 16.0 cm 10.0 cm 1.1 cm 19. A convex mirror is needed to produce an image that is three-fourths the size of an object and located 24 cm behind the mirror. What focal length should be specified? dohi ho di (26.7 cm)(3.0 cm) 16.0 cm 1 1 1 f di do 5.0 cm Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. page 472 17. An object is located 20.0 cm in front of a convex mirror with a 15.0-cm focal length. Find the image position using both a scale diagram and the mirror equation. dodi di do di do f and m di so do m di 24 cm and m 0.75, so (24 cm) 0.75 Ray 1 do I1 Ray 2 Horizontal scale: 1 block 1.0 cm do hi mho (0.178)(6.0 cm) m O1 di 0.178 26.7 cm do hi ho m di 10.7 cm (10.7 cm) 60.0 cm i hi (60.0 cm)(13.0 cm) 60.0 cm (13.0 cm) m i do d f ho d f do f di o F di 8.6 cm 1 1 1 do di f d f so di o do f (20.0 cm)(15.0 cm) 20.0 cm (15.0 cm) 8.57 cm 18. A convex mirror has a focal length of 13.0 cm. A lightbulb with a diameter of 6.0 cm is placed 60.0 cm from the mirror. What is the lightbulb’s image position and diameter? 32 cm (32 cm)(24 cm) 32 cm (24 cm) f 96 cm 20. A 7.6-cm-diameter ball is located 22.0 cm from a convex mirror with a radius of curvature of 60.0 cm. What are the ball’s image position and diameter? 1 1 1 do di f d f do f di o (22.0 cm)(30.0 cm) 22.0 cm (30.0 cm) 12.7 cm 1 1 1 do di f Physics: Principles and Problems Solutions Manual 359 Chapter 17 continued hi di ho do m diho hi do (12.7 cm)(7.6 cm) 22.0 cm 1 1 1 do di f d f do f di o (20.0 cm)(9.0 cm) 20.0 cm 9.0 cm 4.4 cm 16.4 cm 21. A 1.8-m-tall girl stands 2.4 m from a store’s security mirror. Her image appears to be 0.36 m tall. What is the focal length of the mirror? hi di m ho do dohi di ho (2.4 m)(0.36 m) 1.8 m 0.48 m 1 1 1 do di f dd di do o f i (0.48 m)(2.4 m) 0.48 m 2.4 m Section Review 17.2 Curved Mirrors pages 464–473 page 473 22. Image Properties If you know the focal length of a concave mirror, where should you place an object so that its image is upright and larger compared to the object? Will this produce a real or virtual image? do 16.4 cm 20.0 cm 0.82 24. Object Position The placement of an object in front of a concave mirror with a focal length of 12.0 cm forms a real image that is 22.3 cm from the mirror. What is the object position? 1 1 1 do di f df i do d f i (22.3 cm)(12.0 cm) 22.3 cm 12.0 cm 26.0 cm 25. Image Position and Height A 3.0-cm-tall object is placed 22.0 cm in front of a concave mirror having a focal length of 12.0 cm. Find the image position and height by drawing a ray diagram to scale. Verify your answer using the mirror and magnification equations. 360 Solutions Manual Ray 1 F C I1 You should place the object between the mirror and the focal point. The image will be virtual. 23. Magnification An object is placed 20.0 cm in front of a concave mirror with a focal length of 9.0 cm. What is the magnification of the image? O1 hi 3.6 cm di 26.4 cm Ray 2 Horizontal scale: 1 block 1.0 cm Vertical scale: 1 block 1.0 cm 1 1 1 do di f d f do f di o Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 0.60 m di m Chapter 17 continued (22.0 cm)(12.0 cm) 22.0 cm 12.0 cm 26.4 cm hi di ho do hi di ho do m dohi di ho m (16.4 cm)(2.8 cm) 6.0 cm diho hi 7.7 cm do 1 1 1 do di f 3.6 cm f (26.4 cm)(3.0 cm) 22.0 cm dodi do di 26. Ray Diagram A 4.0-cm-tall object is located 14.0 cm from a convex mirror with a focal length of 12.0 cm. Draw a scale ray diagram showing the image position and height. Verify your answer using the mirror and magnification equations. (7.7 cm)(16.4 cm) 7.7 cm 16.4 cm 14.5 cm r 2f (2)(14.5 cm) 29 cm O1 Ray 1 Ray 2 I1 Horizontal scale: 1 block 1.0 cm Vertical scale: 3 blocks 2.0 cm F hi 1.8 cm di 6.5 cm 28. Focal Length A convex mirror is used to produce an image that is two-thirds the size of an object and located 12 cm behind the mirror. What is the focal length of the mirror? hi di ho do Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. m 1 1 1 di do f di do m d f do f di o (14.0 cm)(12.0 cm) 14.0 cm (12.0 cm) 6.46 cm hi di ho do m diho hi do (6.46 cm)(4.0 cm) 14.0 cm 1.8 cm 27. Radius of Curvature A 6.0-cm-tall object is placed 16.4 cm from a convex mirror. If the image of the object is 2.8 cm tall, what is the radius of curvature of the mirror? Physics: Principles and Problems (12 cm) 23 18 cm 1 1 1 do di f dodi f do di (12 cm)(18 cm) 12 cm 18 cm 36 cm 29. Critical Thinking Would spherical aberration be less for a mirror whose height, compared to its radius of curvature, is small or large? Explain. It would be less for a mirror whose height is relatively small compared to Solutions Manual 361 Chapter 17 continued its radius of curvature; diverging light rays from an object that strike the mirror are more paraxial so they converge more closely to create an image that is not blurred. Chapter Assessment Concept Mapping page 478 30. Complete the following concept map using the following terms: convex, upright, inverted, real, virtual. concave A plane mirror is a flat, smooth surface from which light is reflected by specular reflection. The images created by plane mirrors are virtual, upright, and as far behind the mirror as the object is in front of it. 35. A student believes that very sensitive photographic film can detect a virtual image. The student puts photographic film at the location of a virtual image. Does this attempt succeed? Explain. (17.1) No, the rays do not converge at a virtual image. No image forms and the student would not get a picture. Some virtual images are behind the mirror. Mirrors plane 34. Describe the properties of a plane mirror. (17.1) convex 36. How can you prove to someone that an image is a real image? (17.1) virtual real virtual virtual upright inverted upright upright Mastering Concepts When parallel light is reflected from a smooth surface, the rays are reflected parallel to each other. The result is an image of the origin of the rays. When light is reflected from a rough surface, it is reflected in many different directions. The rays are diffused or scattered. No image of the source results. 32. What is meant by the phrase “normal to the surface”? (17.1) any line that is perpendicular to the surface at any point 33. Where is the image produced by a plane mirror located? (17.1) The image is on a line that is perpendicular to the mirror and the same distance behind the mirror as the object is in front of the mirror. 362 Solutions Manual 37. An object produces a virtual image in a concave mirror. Where is the object located? (17.2) Object must be located between F and the mirror. 38. What is the defect that all concave spherical mirrors have and what causes it? (17.2) Rays parallel to the axis that strike the edges of a concave spherical mirror are not reflected through the focal point. This effect is called spherical aberration. 39. What is the equation relating the focal point, object position, and image position? (17.2) 1 1 1 di do f 40. What is the relationship between the center of curvature and the focal length of a concave mirror? (17.2) C 2f Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. page 478 31. How does specular reflection differ from diffuse reflection? (17.1) Place a sheet of plain paper or photographic film at the image location and you should be able to find the image. Chapter 17 continued 41. If you know the image position and object position relative to a curved mirror, how can you determine the mirror’s magnification? (17.2) The magnification is equal to the negative of the image distance divided by the object distance. 42. Why are convex mirrors used as rearview mirrors? (17.2) Convex mirrors are used as rearview mirrors because they allow for a wide range of view, allowing the driver to see a much larger area than is afforded by ordinary mirrors. 43. Why is it impossible for a convex mirror to form a real image? (17.2) The light rays always diverge. Applying Concepts Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. pages 478–479 44. Wet Road A dry road is more of a diffuse reflector than a wet road. Based on Figure 17-16, explain why a wet road appears blacker to a driver than a dry road does. 46. Locate and describe the physical properties of the image produced by a concave mirror when the object is located at the center of curvature. The image will be at C, the center of curvature, inverted, real, and the same size as the object. 47. An object is located beyond the center of curvature of a spherical concave mirror. Locate and describe the physical properties of the image. The image will be between C and F, and will be inverted, real, and smaller than the object. 48. Telescope You have to order a large concave mirror for a telescope that produces high-quality images. Should you order a spherical mirror or a parabolic mirror? Explain. You should order a parabolic mirror to eliminate spherical aberrations. 49. Describe the properties of the image seen in the single convex mirror in Figure 17-17. Wet asphalt Dry asphalt ■ Figure 17-16 Less light is reflected back to the car from a wet road. 45. Book Pages Why is it desirable that the pages of a book be rough rather than smooth and glossy? The smoother and glossier the pages are, the lesser the diffuse reflection of light and the greater the glare from the pages. Physics: Principles and Problems ■ Figure 17-17 The image in a single convex mirror is always virtual, erect, smaller than the object, and located closer to the mirror than the object. Solutions Manual 363 Chapter 17 continued 50. List all the possible arrangements in which you could use a spherical mirror, either concave or convex, to form a real image. You can use only a concave mirror with the object beyond the focal point. A convex mirror will not form a real image. 51. List all possible arrangements in which you could use a spherical mirror, either concave or convex, to form an image that is smaller compared to the object. You may use a concave mirror with the object beyond the center of curvature or a convex mirror with the object anywhere. 52. Rearview Mirrors The outside rearview mirrors of cars often carry the warning “Objects in the mirror are closer than they appear.” What kind of mirrors are these and what advantage do they have? 55. A ray of light incident upon a mirror makes an angle of 36° with the mirror. What is the angle between the incident ray and the reflected ray? i 90° 36° 54° r i 54° i r 54° 54° 108° Level 2 56. Picture in a Mirror Penny wishes to take a picture of her image in a plane mirror, as shown in Figure 17-18. If the camera is 1.2 m in front of the mirror, at what distance should the camera lens be focused? Convex mirror; it provides a wider field of view. Mastering Problems r i 38° 54. A ray of light strikes a mirror at an angle of 53° to the normal. a. What is the angle of reflection? r i 53° b. What is the angle between the incident ray and the reflected ray? i r ■ Figure 17-18 The image is 1.2 m behind the mirror, so the camera lens should be set to 2.4 m. 57. Two adjacent plane mirrors form a right angle, as shown in Figure 17-19. A light ray is incident upon one of the mirrors at an angle of 30° to the normal. 53° 53° 106° 364 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 17.1 Reflection from Plane Mirrors page 479 Level 1 53. A ray of light strikes a mirror at an angle of 38° to the normal. What is the angle that the reflected angle makes with the normal? Chapter 17 continued the feet hits the mirror halfway between the eyes and the feet. The distance between the point the two rays hit the mirror is half the total height. 30° ■ Figure 17-19 a. What is the angle at which the light ray is reflected from the other mirror? Reflection from the first mirror: r1 i1 30° Reflection from the second mirror: i2 90° r1 90° 30° 60° r2 i2 60° Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. b. A retroreflector is a device that reflects incoming light rays back in a direction opposite to that of the incident rays. Draw a diagram showing the angle of incidence on the first mirror for which the mirror system acts as a retroreflector. Level 3 59. Two plane mirrors are connected at their sides so that they form a 45° angle between them. A light ray strikes one mirror at an angle of 30° to the normal and then reflects off the second mirror. Calculate the angle of reflection of the light ray off the second mirror. Reflection from the first mirror is r, 1 i, 1 30°. The angle the ray forms with the mirror is thus 90° 30° 60°. Because the two mirrors form a 45° angle, the angle the ray reflecting off the first mirror forms with the second mirror is 180° 60° 45° 75°. The angle the ray forms with the second mirror is thus i, 2 90° 75° 15°. The angle of reflection from the second mirror is r, 2 i, 2 15°. 60. A ray of light strikes a mirror at an angle of 60° to the normal. The mirror is then rotated 18° clockwise, as shown in Figure 17-20. What is the angle that the reflected ray makes with the mirror? Incident light 45° 18° 58. Draw a ray diagram of a plane mirror to show that if you want to see yourself from your feet to the top of your head, the mirror must be at least half your height. Eye level Mirror 60° Normal Image Mirror Feet The ray from the top of the head hits the mirror halfway between the eyes and the top of the head. The ray from Physics: Principles and Problems ■ Figure 17-20 i i, old 18° 60° 18° 42° Solutions Manual 365 Chapter 17 continued r i 42° r, mirror 90° r 90° 42° 48° 17.2 Curved Mirrors page 480 Level 1 61. A concave mirror has a focal length of 10.0 cm. What is its radius of curvature? r 2f 2(10.0 cm) 20.0 cm 62. An object located 18 cm from a convex mirror produces a virtual image 9 cm from the mirror. What is the magnification of the image? di C F ■ Figure 17-21 real; inverted; larger Level 2 65. Star Image Light from a star is collected by a concave mirror. How far from the mirror is the image of the star if the radius of curvature is 150 cm? Stars are far enough away that the light coming into the mirror can be considered to be parallel and parallel light will converge at the focal point. 0.5 Since r 2f, m do (9 cm) 18 cm r 2 150 cm 2 f 75 cm 66. Find the image position and height for the object shown in Figure 17-22. 1 3 mirror is , what is the boy’s height? hi m ho hi ho 3.8 cm m F 16 cm 0.60 m 13 31 cm 1.8 m 64. Describe the image produced by the object in Figure 17-21 as real or virtual, inverted or upright, and smaller or larger than the object. ■ Figure 17-22 1 1 1 f do di d f do f di o (31 cm)(16 cm) 31 cm 16 cm 366 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 63. Fun House A boy is standing near a convex mirror in a fun house at a fair. He notices that his image appears to be 0.60 m tall. If the magnification of the Chapter 17 continued 33 cm what is the magnification of the image? hi di ho do m diho hi do (33 cm)(3.8 cm) 31 cm 4.1 cm 67. Rearview Mirror How far does the image of a car appear behind a convex mirror, with a focal length of 6.0 m, when the car is 10.0 m from the mirror? 1 1 1 do di f d f do f di o (10.0 m)(6.0 m) 10.0 m (6.0 m) Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 3.8 m 68. An object is 30.0 cm from a concave mirror of 15.0 cm focal length. The object is 1.8 cm tall. Use the mirror equation to find the image position. What is the image height? 1 1 1 do di f d f do f di o (30.0 cm)(15.0 cm) 30.0 cm 15.0 cm 30.0 cm hi di m ho do diho hi do (30.0 cm)(1.8 cm) (30.0 cm) 1.8 cm 69. Dental Mirror A dentist uses a small mirror with a radius of 40 mm to locate a cavity in a patient’s tooth. If the mirror is concave and is held 16 mm from the tooth, Physics: Principles and Problems (40 mm) 2 r 2 f 20 mm 1 1 1 do di f d f do f (16 mm)(20 mm) 16 mm 20 mm di o 80 mm di (80 mm) 16 mm m 5 do 70. A 3.0-cm-tall object is 22.4 cm from a concave mirror. If the mirror has a radius of curvature of 34.0 cm, what are the image position and height? r 2 f 34.0 cm 2 17.0 cm 1 1 1 do di f d f do f di o (22.4 cm)(17.0 cm) 22.4 cm 17.0 cm 70.5 cm hi di ho do m diho hi do (70.5 cm)(3.0 cm) 22.4 cm 9.4 cm Level 3 71. Jeweler’s Mirror A jeweler inspects a watch with a diameter of 3.0 cm by placing it 8.0 cm in front of a concave mirror of 12.0-cm focal length. a. Where will the image of the watch appear? 1 1 1 do di f Solutions Manual 367 Chapter 17 continued d f do f (8.0 cm)(12.0 cm) 8.0 cm 12.0 cm di o 24 cm 1 1 1 do di f fdo b. What will be the diameter of the image? hi di ho do di do f (10.0 cm)(150 cm) 150 cm (10.0 cm) 9.4 cm diho (24 cm)(3.0 cm) 8.0 cm hi do di (9.4 cm) 150 cm m 0.063 do hi mho (0.063)(12 cm) 0.75 cm 9.0 cm 72. Sunlight falls on a concave mirror and forms an image that is 3.0 cm from the mirror. An object that is 24 mm tall is placed 12.0 cm from the mirror. a. Sketch the ray diagram to show the location of the image. O1 Ray 1 Ray 2 C F Horizontal scale: 1 block 1.0 cm Vertical scale: 1 block 4 mm I1 b. Use the mirror equation to calculate the image position. fd do f pages 480–481 Level 1 74. A light ray strikes a plane mirror at an angle of 28° to the normal. If the light source is moved so that the angle of incidence increases by 34°, what is the new angle of reflection? i i, initial 34° 28° 34° 62° r i 62° 75. Copy Figure 17-23 on a sheet of paper. Draw rays on the diagram to determine the height and location of the image. (3.0 cm)(12.0 cm) 12.0 cm 3.0 cm o di 4.0 cm c. How tall is the image? di 4.0 cm m 0.33 do 12.0 cm 3.0 cm 8.0 cm F 4.0 cm hi mho (0.33)(24 mm) 8.0 mm 73. Shiny spheres that are placed on pedestals on a lawn are convex mirrors. One such sphere has a diameter of 40.0 cm. A 12-cm-tall robin sits in a tree that is 1.5 m from the sphere. Where is the image of the robin and how tall is the image? ■ Figure 17-23 r 20.0 cm, f 10.0 cm 368 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1 1 1 do di f Mixed Review Chapter 17 continued b. What is the image height? O1 I1 Horizontal scale: 1 block 1.0 cm Vertical scale: 2 blocks 1.0 cm hi 1.0 cm di 2.7 cm hi m ho F diho hi do (22.9 cm)(2.4 cm) 30.0 cm The image height is 1.0 cm, and its location is 2.7 cm from the mirror. Level 2 76. An object is located 4.4 cm in front of a concave mirror with a 24.0-cm radius. Locate the image using the mirror equation. 78. What is the radius of curvature of a concave mirror that magnifies an object by a factor of 3.2 when the object is placed 20.0 cm from the mirror? hi m r f 2 ho 24.0 cm 2 12.0 cm 1 1 1 do di f d f do f di o (4.4 cm)(12.0 cm) 4.4 cm 12.0 cm Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1.8 cm 6.9 cm 77. A concave mirror has a radius of curvature of 26.0 cm. An object that is 2.4 cm tall is placed 30.0 cm from the mirror. a. Where is the image position? r 2 f 26.0 cm 2 13.0 cm 1 1 1 do di f d f di o do f (30.0 cm)(13.0 cm) 30.0 cm 13.0 cm 22.9 cm di mdo (3.2)(20.0 cm) 64 cm 1 1 1 do di f d d do di i f o (20.0 cm)(64 cm) 20.0 cm (64 cm) 29 cm r 2f (2)(29 cm) 58 cm 79. A convex mirror is needed to produce an image one-half the size of an object and located 36 cm behind the mirror. What focal length should the mirror have? hi di ho do m diho do hi (36 cm)h o ho 2 72 cm 1 1 1 do di f Physics: Principles and Problems Solutions Manual 369 Chapter 17 continued dodi a. What kind of mirror would do this job? f do di An enlarged, upright image results only from a concave mirror, with the object inside the focal length. (72 cm)(36 cm) 72 cm (36 cm) b. What is its radius of curvature? 72 cm di m 80. Surveillance Mirror A convenience store uses a surveillance mirror to monitor the store’s aisles. Each mirror has a radius of curvature of 3.8 m. do di mdo (7.5)(14.0 mm) 105 mm a. What is the image position of a customer who stands 6.5 m in front of the mirror? 1 1 1 do di f A mirror that is used for surveillance is a convex mirror. So the focal length is the negative of half the radius of curvature. dodi di do 16 mm r 2f (2)(16mm) 32 mm r 2 f 3.8 m 2 1.9 m 1 1 1 do di f (14.0 mm)(105 mm) 14.0 mm (105 mm) f 82. The object in Figure 17-24 moves from position 1 to position 2. Copy the diagram onto a sheet of paper. Draw rays showing how the image changes. d f do f di o 1 C 2 F 1.5 m 1.0 m b. What is the image height of a customer who is 1.7 m tall? hi di ho do 1.5 m 2.0 m 2.5 m m ■ diho hi do O1 (1.5 m)(1.7 m) 6.5 m Level 3 81. Inspection Mirror A production-line inspector wants a mirror that produces an image that is upright with a magnification of 7.5 when it is located 14.0 mm from a machine part. 370 Solutions Manual O2 1 Ray 1 Ray 2 Ray 2 0.38 m Ray 1 Figure 17-24 2 C F I1 I2 Horizontal scale: 1 block 10 cm 83. A ball is positioned 22 cm in front of a spherical mirror and forms a virtual image. If the spherical mirror is replaced with a plane mirror, the image appears 12 cm closer to the mirror. What kind of spherical mirror was used? Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. (6.5 m)(1.9 m) 6.5 m (1.9 m) Chapter 17 continued The object position for both mirrors is 22 cm. So, the image position for the plane mirror is 22 cm. Because the spherical mirror forms a virtual image, the image is located behind the mirror. Thus, the image position for the spherical mirror is negative. 34 cm 1 1 1 do di f d d do di i f o (22 cm)(34 cm) 22 cm (34 cm) hi 2.4 m di 14 m C F I1 The image is 2.4 m tall, and it is 14 m from the mirror. 86. A 4.0-cm-tall object is placed 12.0 cm from a convex mirror. If the image of the object is 2.0 cm tall, and the image is located at 6.0 cm, what is the focal length of the mirror? Draw a ray diagram to answer the question. Use the mirror equation and the magnification equation to verify your answer. O1 Ray 1 62 cm I1 Ray 2 The focal length is positive, so the spherical mirror is a concave mirror. hi di ho do d d do di i f o hido di ho (12.0 cm)(6.0 cm) 12.0 cm (6.0 cm) (0.28 m)(3.2 m) 1.6 m 1 1 1 do di f d d do di i f o (3.2 m)(0.56 m) 3.2 m (0.56 m) f 12 cm 1 1 1 do di f m 0.56 m F Horizontal scale: 1 block 1.0 cm Vertical scale: 3 blocks 2.0 cm 84. A 1.6-m-tall girl stands 3.2 m from a convex mirror. What is the focal length of the mirror if her image appears to be 0.28 m tall? Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Horizontal scale: 1 block 1.0 m Vertical scale: 2 blocks 1.0 m Ray 2 di di, plane 12 cm 22 cm 12 cm Ray 1 O1 12 cm Thinking Critically pages 481–482 87. Apply Concepts The ball in Figure 17-25 slowly rolls toward the concave mirror on the right. Describe how the size of the ball’s image changes as it rolls along. 0.68 m 85. Magic Trick A magician uses a concave mirror with a focal length of 8.0 m to make a 3.0-m-tall hidden object, located 18.0 m from the mirror, appear as a real image that is seen by his audience. Draw a scale ray diagram to find the height and location of the image. Physics: Principles and Problems C ■ F Figure 17-25 Solutions Manual 371 Chapter 17 continued Beyond C, the image is smaller than the ball. As the ball rolls toward the mirror, the image size increases. The image is the same size as the ball when the ball is at C. The image size continues to increase until there is no image when the ball is at F. Past F, the size of the image decreases until it equals the ball’s size when the ball touches the mirror. mirror, what is the focal length of the concave mirror? di, initial do, initial 6.0 cm di di, initial (8.0 cm) 6.0 cm (8.0 cm) 14.0 cm 1 1 1 do di f 88. Analyze and Conclude The object in Figure 17-26 is located 22 cm from a concave mirror. What is the focal length of the mirror? d d do di i f o (6.0 cm)(14.0 cm) 6.0 cm (14.0 cm) f f 1.0101 cm 22 cm ■ 91. Analyze and Conclude The layout of the two-mirror system shown in Figure 17-11 is that of a Gregorian telescope. For this question, the larger concave mirror has a radius of curvature of 1.0 m, and the smaller mirror is located 0.75 m away. Why is the secondary mirror concave? r 2 f do 2 22 cm 2 11 cm 89. Use Equations Show that as the radius of curvature of a concave mirror increases to infinity, the mirror equation reduces to the relationship between the object position and the image position for a plane mirror. 92. Analyze and Conclude An optical arrangement used in some telescopes is the Cassegrain focus, shown in Figure 17-27. This telescope uses a convex secondary mirror that is positioned between the primary mirror and the focal point of the primary mirror. Convex secondary mirror As f → , 1/f → 0. The mirror equation then becomes 1/do 1/di, or do di. 90. Analyze and Conclude An object is located 6.0 cm from a plane mirror. If the plane mirror is replaced with a concave mirror, the resulting image is 8.0 cm farther behind the mirror. Assuming that the object is located between the focal point and the concave 372 Solutions Manual Concave primary mirror F Telescope tube Eyepiece ■ Figure 17-27 a. A single convex mirror produces only virtual images. Explain how the convex Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. The smaller mirror is concave to produce a real image at the eyepiece that is upright. The light rays are inverted by the first concave mirror and then inverted again by the secondary concave mirror. Figure 17-26 Chapter 17 continued mirror in this telescope functions within the system of mirrors to produce real images. The convex mirror is placed to intercept the rays from a concave mirror before they converge. The convex mirror places the point of convergence in the opposite direction back toward the concave mirror, and lengthens the total distance the light travels before converging. This effectively increases the focal length compared to using the concave mirror by itself, thus increasing the total magnification. b. Are the images produced by the Cassegrain focus upright or inverted? How does this relate to the number of times that the light crosses? Inverted; each time the light rays cross the image inverts. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Writing in Physics page 482 93. Research a method used for grinding, polishing, and testing mirrors used in reflecting telescopes. You may report either on methods used by amateur astronomers who make their own telescope optics, or on a method used by a project at a national laboratory. Prepare a one-page report describing the method, and present it to the class. Answers will vary depending on the mirrors and methods chosen by the students. Amateur methods usually involve rubbing two “blanks” against each other with varying grits between them. Methods used at national labs vary. 94. Mirrors reflect light because of their metallic coating. Research and write a summary of one of the following: a. the different types of coatings used and the advantages and disadvantages of each Answers will vary. Student answers should include information about shininess as well as tarnish resistance. Physics: Principles and Problems b. the precision optical polishing of aluminum to such a degree of smoothness that no glass is needed in the process of making a mirror Answers will vary. Student answers might include information about deformation of a mirror from its own weight as size increases and how a mirror made of aluminum could impact this problem. Cumulative Review page 482 95. A child runs down the school hallway and then slides on the newly waxed floor. He was running at 4.7 m/s before he started sliding and he slid 6.2 m before stopping. What was the coefficient of friction of the waxed floor? (Chapter 11) The work done by the waxed floor equals the child’s initial kinetic energy. 1 KE mv 2 W Fd kmgd 2 The mass of the child cancels out, giving v2 2gd k (4.7 m/s)2 (2)(9.80 m/s )(6.2 m) 2 0.18 96. A 1.0 g piece of copper falls from a height of 1.0104 m from an airplane to the ground. Because of air resistance it reaches the ground moving at a velocity of 70.0 m/s. Assuming that half of the energy lost by the piece was distributed as thermal energy to the copper, how much did it heat during the fall? (Chapter 12) Potential energy of the piece E mgh (0.0010 kg)(9.80 m/s2)(1.0104 m) 9.8 J Final energy 1 2 Ef mv 2 1 2 (0.0010 kg)(70.0 m/s)2 2.4 J Solutions Manual 373 Chapter 17 continued 1 2 99. Organ pipes An organ builder must design a pipe organ that will fit into a small space. (Chapter 15) 1 2 a. Should he design the instrument to have open pipes or closed pipes? Explain. Heat added to the piece Q (E Ef) (9.8 J 2.4 J) 3.7 J Q T mc 3.7 J (0.0010 kg)(385 J/kg°C) 9.5°C 97. It is possible to lift a person who is sitting on a pillow made from a large sealed plastic garbage bag by blowing air into the bag through a soda straw. Suppose that the cross-sectional area of the person sitting on the bag is 0.25 m2 and the person’s weight is 600 N. The soda straw has a cross-sectional area of 2105 m2. With what pressure must you blow into the straw to lift the person that is sitting on the sealed garbage bag? (Chapter 13) Apply Pascal’s principle. A1 F2 F1 A2 2105 m2 0.25 m (600 N) 2 0.048 N A2 210 m or 2 kPa to one significant digit not a very large pressure at all 98. What would be the period of a 2.0-m-long pendulum on the Moon’s surface? The Moon’s mass is 7.341022 kg, and its radius is 1.74106 m. What is the period of this pendulum on Earth? (Chapter 14) Gmm gm d 2m (6.671011 Nm2/kg2)(7.341022 kg) 1.62 m/s2 g 1.62 m /s g 9.80 m /s l 2.0 m TMoon 2 2 2 7.0 s l 2.0 m TEarth 2 2 2 2.8 s 374 Solutions Manual b. Will an organ constructed with open pipes sound the same as one constructed with closed pipes? Explain. No. While the two organs will have the same fundamental tones, closed pipes produce only the odd harmonics, so they will have different timbres than open pipes. 100. Filters are added to flashlights so that one shines red light and the other shines green light. The beams are crossed. Explain in terms of waves why the light from both flashlights is yellow where the beams cross, but revert back to their original colors beyond the intersection point. (Chapter 16) Waves can interfere, add, and then pass through unaffected. Chapter 14 showed the amplitude of waves adding. In this case, the waves retain their color information as they cross through each other. Challenge Problem page 470 An object of height ho is located at do relative to a concave mirror with focal length f. 1. Draw and label a ray diagram showing the focal length and location of the object if the image is located twice as far from the mirror as the object. Prove your answer mathematically. Calculate the focal length as a function of object position for this placement. Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. F2 0.048 N P 5 2 2.4 kPa The resonant frequency of an open pipe is twice that of a closed pipe of the same length. Therefore, the pipes of a closed-pipe organ need be only half as long as open pipes to produce the same range of fundamental frequencies. Chapter 17 continued Ray 1 O1 di 2dO Ray 2 fd di f i do F C f 1 1 1 do di f 2dO 3 f(2f) 2f f 2f 1 1 1 f do di do d h do (2f)ho 2do 3 2f ho 2. Draw and label a ray diagram showing the location of the object if the image is located twice as far from the mirror as the focal point. Prove your answer mathematically. Calculate the image height as a function of the object height for this placement. O1 Ra Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. di i o hi do(2do) do 2do di 2f hi hO hi ho m d di f o do di 3. Where should the object be located so that no image is formed? The object should be placed at the focal point. Ray 1 y2 C F I1 Physics: Principles and Problems Solutions Manual 375 CHAPTER 18 Refraction and Lenses Practice Problems 18.1 Refraction of Light pages 485–492 page 487 1. A laser beam in air is incident upon ethanol at an angle of incidence of 37.0°. What is the angle of refraction? n1 sin 1 n2 sin 2 n sin n2 1 1 2 sin1 sin1 (1.00)(sin 37.0°) 1.36 26.3° n1 sin 1 n2 sin 2 n sin n2 1 1 2 sin1 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. n1 sin 1 n2 sin 2 n sin sin 2 1 1 n2 (1.33)(sin 31°) sin 27° 1.5 Section Review 2. Light in air is incident upon a piece of crown glass at an angle of incidence of 45.0°. What is the angle of refraction? sin1 (1.00)(sin 45.0°) 1.52 27.7° 3. Light passes from air into water at 30.0° to the normal. Find the angle of refraction. n1 sin 1 n2 sin 2 2 5. A block of unknown material is submerged in water. Light in the water is incident on the block at an angle of incidence of 31°. The angle of refraction of the light in the block is 27°. What is the index of refraction of the material of the block? n1 sin 1 sin1 n2 sin1 (1.00)(sin 30.0°) 1.33 22.1° 18.1 Refraction of Light pages 485–492 page 492 6. Index of Refraction You notice that when a light ray enters a certain liquid from water, it is bent toward the normal, but when it enters the same liquid from crown glass, it is bent away from the normal. What can you conclude about the liquid’s index of refraction? nwater nliquid ncrown glass, therefore, nliquid must be between 1.33 and 1.52. 7. Index of Refraction A ray of light has an angle of incidence of 30.0° on a block of unknown material and an angle of refraction of 20.0°. What is the index of refraction of the material? n1 sin 1 n2 sin 2 4. Light is incident upon a diamond facet at 45.0°. What is the angle of refraction? n sin sin 2 1 1 n2 n1 sin 1 n2 sin 2 (1.00)(sin 30.0°) n1 sin 1 2 sin1 n2 1.46 sin 20.0° sin1 17.0° (1.00)(sin 45.0°) 2.42 Physics: Principles and Problems Solutions Manual 377 Chapter 18 continued 8. Speed of Light Could an index of refraction ever be less than 1? What would this imply about the speed of light in that medium? No, it would mean the speed of light in the medium is faster than it is in a vacuum. 9. Speed of Light What is the speed of light in chloroform (n 1.51)? c n v c vchloroform nchloroform 8 3.0010 m/s 1.51 1.99108 m/s 10. Total Internal Reflection If you were to use quartz and crown glass to make an optical fiber, which would you use for the cladding layer? Why? crown glass because it has a lower index of refraction and would produce total internal reflection n1 sin 1 n2 sin 2 n sin n2 1 1 2 sin1 sin1 (1.33)(sin 57.5°) 1.50 48.4° 12. Critical Angle Is there a critical angle for light traveling from glass to water? From water to glass? Yes, because nglass nwater. No. 13. Dispersion Why can you see the image of the Sun just above the horizon when the Sun itself has already set? because of bending of light rays in the atmosphere; refraction 378 Solutions Manual In the east, because the Sun sets in the west and sunlight must shine from behind you in order for you to see a rainbow. Practice Problems 18.2 Convex and Concave Lenses pages 493–499 page 496 15. A 2.25-cm-tall object is 8.5 cm to the left of a convex lens of 5.5-cm focal length. Find the image position and height. 1 1 1 do di f d f do f o di (8.5 cm)(5.5 cm) 8.5 cm 5.5 cm 15.6 cm, or 16 cm h ho d do m i i d h do i o hi (15.6 cm)(2.25 cm) 8.5 cm 4.1 cm 16. An object near a convex lens produces a 1.8-cm-tall real image that is 10.4 cm from the lens and inverted. If the focal length of the lens is 6.8 cm, what are the object position and height? 1 1 1 do di f df di f i do (10.4 cm)(6.8 cm) 10.4 cm 6.8 cm 2.0101 cm Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 11. Angle of Refraction A beam of light passes from water into polyethylene with n 1.50. If i 57.5°, what is the angle of refraction in the polyethylene? 14. Critical Thinking In what direction can you see a rainbow on a rainy late afternoon? Explain. Chapter 18 continued d do h ho d f do f m i i o di d h di (25 cm)(5.0 cm) o i ho 25 cm 5.0 cm 6.2 cm (19.6 cm)(1.8 cm) 10.4 cm 3.4 cm 17. An object is placed to the left of a convex lens with a 25-mm focal length so that its image is the same size as the object. What are the image and object positions? 1 1 1 f di do with do di because d do Therefore, (6.2 cm)(2.0 cm) 25 cm 0.50 cm (inverted image) 1 1 1 f di do 1 2 f di d f do f o So di di 2f 2(25 mm) (6.0 cm)(20.0 cm) 6.0 cm 20.0 cm 5.0101 mm 8.6 cm do di 5.0101 mm Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. d h do i o hi page 497 20. A newspaper is held 6.0 cm from a convex lens of 20.0-cm focal length. Find the image position of the newsprint image. m i and m 1 18. Use a scale ray diagram to find the image position of an object that is 30 cm to the left of a convex lens with a 10-cm focal length. di 15 cm O1 d do h ho m i i 1 1 1 f di do Ray 1 di 15 cm Ray 2 Horizontal scale: 1 block 1 cm 21. A magnifying glass has a focal length of 12.0 cm. A coin, 2.0 cm in diameter, is placed 3.4 cm from the lens. Locate the image of the coin. What is the diameter of the image? F F I1 19. Calculate the image position and height of a 2.0-cm-tall object located 25 cm from a convex lens with a focal length of 5.0 cm. What is the orientation of the image? d f do f o di (3.4 cm)(12.0 cm) 3.4 cm 12.0 cm 4.7 cm hodi (2.0 cm)(4.7 cm) hi do 3.4 cm 2.8 cm 1 1 1 f do di Physics: Principles and Problems Solutions Manual 379 Chapter 18 continued 22. A convex lens with a focal length of 22.0 cm is used to view a 15.0-cm-long pencil located 10.0 cm away. Find the height and orientation of the image. 1 1 1 f do di Section Review d f do f o di 18.2 Convex and Concave Lenses pages 493–499 (10.0 cm)(22.0 cm) 10.0 cm 22.0 cm 18.3 cm page 499 25. Magnification Magnifying glasses normally are used to produce images that are larger than the related objects, but they also can produce images that are smaller than the related objects. Explain. d do h ho The location should be about 15 cm on the same side of the lens (15 cm) and the image should be upright and about 1.5 cm tall. m i i d h do i o hi (18.3 cm)(15.0 cm) 10.0 cm 27.5 cm (upright image) 23. A stamp collector wants to magnify a stamp by 4.0 when the stamp is 3.5 cm from the lens. What focal length is needed for the lens? d do m i 14 cm 26. Image Position and Height A 3.0-cm-tall object is located 2.0 cm from a convex lens having a focal length of 6.0 cm. Draw a ray diagram to determine the location and size of the image. Use the thin lens equation and the magnification equation to verify your answer. 1 1 1 f di do I1 O1 dodi (3.5 cm)(14 cm) f do di F 3.5 cm (14 cm) 4.7 cm 24. A magnifier with a focal length of 30 cm is used to view a 1-cm-tall object. Use ray tracing to determine the location and size of the image when the magnifier is positioned 10 cm from the object. d f do f o di 2.0 cm 6.0 cm O1 hi 1.5 cm di 15 cm 380 1 1 1 f do di (2.0 cm)(6.0 cm) I1 F hi 4.5 cm di 3.0 cm F Horizontal scale: 2 blocks 1.0 cm Vertical scale: 1 block 1.0 cm Solutions Manual F Horizontal scale: 1 block 2 cm Vertical scale: 3 blocks 1 cm 3.0 cm h ho d do m i i Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. di mdo (4.0)(3.5 cm) If the object is located farther than twice the focal length from the lens, the size of the image is smaller than the size of the object. Chapter 18 continued d h do i o hi Water (3.0 cm)(3.0 cm) 2.0 cm 4.5 cm Light rays Air 27. Types of Lenses The cross sections of four different thin lenses are shown in Figure 18-16. Water ■ Figure 18-17 The light rays will diverge. Water ■ Figure 18-16 a. Which of these lenses, if any, are convex, or converging, lenses? Lenses a and c are converging. Light rays n 1.3 n 1.0 n 1.3 b. Which of these lenses, if any, are concave, or diverging, lenses? Lenses b and d are diverging. Water Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 28. Chromatic Aberration All simple lenses have chromatic aberration. Explain, then, why you do not see this effect when you look through a microscope. All precision optical instruments use a combination of lenses, called an achromatic lens, to minimize chromatic aberration. 29. Chromatic Aberration You shine white light through a convex lens onto a screen and adjust the distance of the screen from the lens to focus the red light. Which direction should you move the screen to focus the blue light? closer to the lens 30. Critical Thinking An air lens constructed of two watch glasses is placed in a tank of water. Copy Figure 18-17 and draw the effect of this lens on parallel light rays incident on the lens. Physics: Principles and Problems Section Review 18.3 Applications of Lenses pages 500–503 page 503 31. Refraction Explain why the cornea is the primary focusing element in the eye. The difference in index of refraction between the air and the cornea is greater than any other difference that light rays encounter when traveling toward the retina. 32. Lens Types Which type of lens, convex or concave, should a nearsighted person use? Which type should a farsighted person use? A nearsighted person should use a concave lens. A farsighted person should use a convex lens. Solutions Manual 381 Chapter 18 continued 33. Focal Length Suppose your camera is focused on a person who is 2 m away. You now want to focus it on a tree that is farther away. Should you move the lens closer to the film or farther away? Closer; real images are always farther from the lens than the focal point. The farther away the object is, the closer the image is to the focal point. Chapter Assessment Concept Mapping page 508 37. Complete the following concept map using the following terms: inverted, larger, smaller, virtual. Lenses 34. Image Why is the image that you observe in a refracting telescope inverted? convex After the light rays pass through the objective lens, they cross, forming an image that is inverted. The eyepiece maintains this orientation when it uses this image as its object. 35. Prisms What are three benefits of having prisms in binoculars? The prisms extend the light’s path length to make the binoculars more compact, invert light rays so that the viewer sees an upright image, and increase separation between objective lenses to improve the three-dimensional view. You are using the light that strikes only a small area of the object. A brighter lamp could be used. real virtual virtual inverted upright upright larger smaller unchanged size larger Mastering Concepts page 508 38. How does the angle of incidence compare with the angle of refraction when a light ray passes from air into glass at a nonzero angle? (18.1) The angle of incidence is larger than the angle of refraction, because air has a smaller index of refraction. 39. How does the angle of incidence compare with the angle of refraction when a light ray leaves glass and enters air at a nonzero angle? (18.1) The angle of incidence is smaller than the angle of refraction, because glass has a larger index of refraction. 40. Regarding refraction, what is the critical angle? (18.1) The term critical angle refers to the incident angle that causes the refracted ray to lie right along the boundary of the substance when a ray is passing from a region of higher index of refraction to a region of lower index of refraction. If the incident angle exceeds the critical angle, total internal reflection will occur. 382 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 36. Critical Thinking When you use the highest magnification on a microscope, the image is much darker than it is at lower magnifications. What are some possible reasons for the darker image? What could you do to obtain a brighter image? smaller concave Chapter 18 continued 41. Although the light coming from the Sun is refracted while passing through Earth’s atmosphere, the light is not separated into its spectrum. What does this indicate about the speeds of different colors of light traveling through air? (18.1) The speeds of the different colors of light traveling through air are the same. 42. Explain why the Moon looks red during a lunar eclipse. (18.1) During a lunar eclipse, Earth blocks the Sun’s rays from the Moon. However, sunlight refracting off Earth’s atmosphere is directed inward toward the Moon. Because blue wavelengths of light are dispersed more, red wavelengths of light reflect off the Moon toward Earth. 43. How do the shapes of convex and concave lenses differ? (18.2) Convex lenses are thicker at the center than at the edges. Concave lenses are thinner in the middle than at the edges. Another lens is included in the optics system of the projector to invert the image again. As a result, the image is upright compared to the original object. 47. Describe why precision optical instruments use achromatic lenses. (18.2) All lenses have chromatic aberration, which means different wavelengths of light are bent at slightly different angles near their edges. An achromatic lens is a combination of two or more lenses with different indices of refraction that reduce this effect. 48. Describe how the eye focuses light. (18.3) Light entering the eye is primarily focused by the cornea. Fine focusing occurs when muscles change the shape of the lens, allowing the eye to focus on either near or far objects. 49. What is the condition in which the focal length of the eye is too short to focus light on the retina? (18.3) Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. nearsightedness 44. Locate and describe the physical properties of the image produced by a convex lens when an object is placed some distance beyond 2F. (18.2) It is a real image that is located between F and 2F, and that is inverted and smaller compared to the object. 45. What factor, other than the curvature of the surfaces of a lens, determines the location of the focal point of the lens? (18.2) The index of refraction of the material from which the lens is made also determines the focus. 46. To project an image from a movie projector onto a screen, the film is placed between F and 2F of a converging lens. This arrangement produces an image that is inverted. Why does the filmed scene appear to be upright when the film is viewed? (18.2) Physics: Principles and Problems 50. What type of image is produced by the objective lens in a refracting telescope? (18.3) real image, inverted 51. The prisms in binoculars increase the distance between the objective lenses. Why is this useful? (18.3) It improves the three-dimensional view. 52. What is the purpose of a camera’s reflex mirror? (18.3) The reflex mirror diverts the image onto a prism so that it can be viewed before taking a photograph. When the shutter release button is pressed, the reflex mirror moves out of the way so that the lens focuses the image onto the film or other photodetector. Solutions Manual 383 Chapter 18 continued Applying Concepts pages 508—509 53. Which substance, A or B, in Figure 18-24 has a larger index of refraction? Explain. A ■ B c, glass/air sin1 1.00 1.52 41.1° Air and glass have the smaller critical angle of 41.1°. The critical angle for air and water is 48.8°. 58. Cracked Windshield If you crack the windshield of your car, you will see a silvery line along the crack. The glass has separated at the crack, and there is air in the crack. The silvery line indicates that light is reflecting off the crack. Draw a ray diagram to explain why this occurs. What phenomenon does this illustrate? From outside Figure 18-24 The angle in substance A is smaller, so it has the larger index of refraction. 54. A light ray strikes the boundary between two transparent media. What is the angle of incidence for which there is no refraction? 55. How does the speed of light change as the index of refraction increases? As the index of refraction of a material increases, the speed of light in that material decreases. 56. How does the size of the critical angle change as the index of refraction increases? The critical angle decreases as the index of refraction increases. 57. Which pair of media, air and water or air and glass, has the smaller critical angle? n 2 c sin1 n 1 c nglass 1 c nglass To the driver’s eyes This illustrates light reflected at angles larger than the critical angle, or total internal reflection. 59. Legendary Mirage According to legend, Eric the Red sailed from Iceland and discovered Greenland after he had seen the island in a mirage. Describe how the mirage might have occurred. Even though Greenland is below the horizon, it is visible as a mirage due to the refraction of light. 1 c, water/air sin1 1.00 1.33 48.8° 384 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. An angle of incidence of 0° allows the light to go through unchanged. Or if the angle of incidence is greater than the critical angle there is total internal reflection. Crack n 1.0 Chapter 18 continued The magnification is much less in water than in air. The difference in the indices of refraction for water and glass is much less than the difference for air and glass. Mirage Cold, de ns e a ir m ar W ir a 64. Why is there chromatic aberration for light that goes through a lens but not for light that reflects from a mirror? Iceland Greenland 60. A prism bends violet light more than it bends red light. Explain. Violet light travels slower in a prism than red light does. 61. Rainbows Why would you never see a rainbow in the southern sky if you were in the northern hemisphere? In which direction should you look to see rainbows if you are in the southern hemisphere? Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. You can see a rainbow only when the Sun’s rays come from behind you at an angle not greater than 42° with the horizon. When you are facing south in the northern hemisphere, the Sun is never behind you at an angle of 42° or less. 62. Suppose that Figure 18-14 is redrawn with a lens of the same focal length but a larger diameter. Explain why the location of the image does not change. Would the image be affected in any way? The location of the image depends on the focal length of the lens and the distance of the object from the lens. Therefore, the location of the image doesn’t change. 63. A swimmer uses a magnifying glass to observe a small object on the bottom of a swimming pool. She discovers that the magnifying glass does not magnify the object very well. Explain why the magnifying glass is not functioning as it would in air. Physics: Principles and Problems Chromatic aberration for lenses is due to the dispersion of light (different wavelengths of light have different speeds in the lens and refract with slightly different angles). Mirrors reflect, and reflection is independent of wavelength. 65. When subjected to bright sunlight, the pupils of your eyes are smaller than when they are subjected to dimmer light. Explain why your eyes can focus better in bright light. Eyes can focus better in bright light because rays that are refracted into larger angles are cut off by the iris. Therefore, all rays converge at a narrow angle, so there is less spherical aberration. 66. Binoculars The objective lenses in binoculars form real images that are upright compared to their objects. Where are the images located relative to the eyepiece lenses? Each side of the binoculars is like a refracting telescope. Therefore, the objective lens image must be between the eyepiece lens and its focal point to magnify the image. Mastering Problems 18.1 Refraction of Light pages 509—510 Level 1 67. A ray of light travels from air into a liquid, as shown in Figure 18-25. The ray is incident upon the liquid at an angle of 30.0°. The angle of refraction is 22.0°. Solutions Manual 385 Chapter 18 continued b. At what angle does the beam enter the water? 30° ng sin g nw sin w ng sin g Air w sin1 nw Liquid sin1 (1.50)(sin 25.4°) 1.33 22° ■ Figure 18-25 a. Using Snell’s law, calculate the index of refraction of the liquid. n1 sin 1 n2 sin 2 n1 sin 1 n2 sin 2 (1.00)(sin 30.0°) sin 22.0° 28.9° 70. Refer to Table 18-1. Use the index of refraction of diamond to calculate the speed of light in diamond. c n v c vdiamond ndiamond 8 3.0010 m/s 2.42 1.33 b. Compare the calculated index of refraction to those in Table 18-1. What might the liquid be? water 71. Refer to Table 18-1. Find the critical angle for a diamond in air. n n1 2 c, diamond/air sin1 sin1 1.00 2.42 24.4° n1 sin 1 n2 sin 2 n sin 2 2 1 sin1 n 1 sin1 (1.36)(sin 25.0°) 1.62 20.8° 69. A beam of light strikes the flat, glass side of a water-filled aquarium at an angle of 40.0° to the normal. For glass, n 1.50. a. At what angle does the beam enter the glass? nA sin A ng sin g n sin ng A A g sin1 sin1 (1.00)(sin 40.0°) 1.50 Level 2 72. Aquarium Tank A thick sheet of plastic, n 1.500, is used as the side of an aquarium tank. Light reflected from a fish in the water has an angle of incidence of 35.0°. At what angle does the light enter the air? n1 sin 1 n2 sin 2 nwater sin water nplastic sin plastic n sin nplastic water water plastic sin1 sin1 (1.33)(sin 35.0°) 1.500 30.57° nplastic sin plastic nair sin air 25.4° 386 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 68. Light travels from flint glass into ethanol. The angle of refraction in the ethanol is 25.0°. What is the angle of incidence in the glass? 1.24108 m/s Chapter 18 continued nplastic sin plastic air sin1 nair sin1 (1.500)(sin 30.57°) 1.00 49.7° 73. Swimming-Pool Lights A light source is located 2.0 m below the surface of a swimming pool and 1.5 m from one edge of the pool, as shown in Figure 18-26. The pool is filled to the top with water. 74. A diamond’s index of refraction for red light, 656 nm, is 2.410, while that for blue light, 434 nm, is 2.450. Suppose that white light is incident on the diamond at 30.0°. Find the angles of refraction for red and blue light. nA sin A nd sin d n sin nd A A d sin1 For red light d sin1 (1.00)(sin 30.0°) 2.410 12.0° For blue light 2m d sin1 (1.00)(sin 30.0°) 2.450 1.5 m ■ 11.8° 75. The index of refraction of crown glass is 1.53 for violet light, and it is 1.51 for red light. Figure 18-26 a. At what angle does the light reaching the edge of the pool leave the water? i tan1 3.00108 m/s c v 1.5 m 2.0 m n Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 37° nA sin A nw sin w 1.53 1.96108 m/s Then find the angle in air. A a. What is the speed of violet light in crown glass? nw sin w sin1 nA c 3.00108 m/s v (1.33)(sin 37°) sin1 1.00 b. What is the speed of red light in crown glass? n 53° b. Does this cause the light viewed from this angle to appear deeper or shallower than it actually is? side opposite tan 53° side adjacent side opposite side adjacent tan 53° 1.5 m tan 53° 1.1 m, shallower 1.51 1.99108 m/s 76. The critical angle for a special glass in air is 41.0°. What is the critical angle if the glass is immersed in water? n A sin c, air n g n sin c, air 1.00 A 1.524 ng sin 41.0° n w sin c, water n g n ng w c, water sin1 sin1 1.33 1.524 60.8° Physics: Principles and Problems Solutions Manual 387 Chapter 18 continued Level 3 77. A ray of light in a tank of water has an angle of incidence of 55.0°. What is the angle of refraction in air? n1 sin 1 n2 sin 2 nair sin 2 nglass sin 1 nglass sin 1 2 sin1 nair n sin n2 sin1 (1.5)(sin 32°) 1.00 1 1 2 sin1 53° sin1 (1.33)(sin 55.0°) 1.00 sin1(1.09) The value sin 2 1.09 is not defined. Therefore, total internal reflection occurs. 78. The ray of light shown in Figure 18-27 is incident upon a 60°-60°-60° glass prism, n 1.5. Air 79. The speed of light in a clear plastic is 1.90108 m/s. A ray of light strikes the plastic at an angle of 22.0°. At what angle is the ray refracted? c nair sin air np sin p and np , so vp c sin p nair sin air vp vpnair sin air sin p c Air vpnair sin air p sin1 c 60° 1 45° P (1.90108 m/s)(1.00)(sin 22.0°) 3.0010 m/s sin1 8 Q 2 1 R nglass 1.5 2 Glass 60° Figure 18-27 13.7° 80. A light ray enters a block of crown glass, as illustrated in Figure 18-28. Use a ray diagram to trace the path of the ray until it leaves the glass. a. Using Snell’s law of refraction, determine the angle, 2, to the nearest degree. 45° nair sin 1 nglass sin 2 n sin nglass air 1 2 sin1 1.00 sin 45° sin1 1.5 28° b. Using elementary geometry, determine the value of 1. ■ nA sin A ng sin g n sin ng A A g sin1 P 90° 28° 62° sin1 Q 180° 62° 60° 58° 28° 1 90° 58° 32° Figure 18-28 (1.00)(sin 45°) 1.52 Find the critical angle for crown glass. n ng A sin c 388 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 60° ■ c. Determine 2. Chapter 18 continued 83. If an object is 10.0 cm from a converging lens that has a focal length of 5.00 cm, how far from the lens will the image be? n ng A c sin1 sin1 1.00 1.52 1 1 1 f do di 41.1° When the light ray in the glass strikes the surface at a 62° angle, total internal reflection occurs. d f do f o di (10.0 cm)(5.00 cm) 10.0 cm 5.00 cm 10.0 cm 62° 28° 62° 28° 45° 45° 18.2 Convex and Concave Lenses page 510 Level 1 81. The focal length of a convex lens is 17 cm. A candle is placed 34 cm in front of the lens. Make a ray diagram to locate the image. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. di 34 cm O1 F Horizontal scale: 1 block 2 cm di 34 cm F I1 82. A converging lens has a focal length of 25.5 cm. If it is placed 72.5 cm from an object, at what distance from the lens will the image be? 1 1 1 f do di d f do f o di (72.5 cm)(25.5 cm) 72.5 cm 25.5 cm 39.3 cm The image is 39.3 cm from the lens. Physics: Principles and Problems d do h ho m i i d m do i (24 cm) 0.75 32 cm 1 1 1 f do di Ray 1 Ray 2 Level 2 84. A convex lens is needed to produce an image that is 0.75 times the size of the object and located 24 cm from the lens on the other side. What focal length should be specified? d d do di o i f (32 cm)(24 cm) 32 cm 24 cm 14 cm 85. An object is located 14.0 cm from a convex lens that has a focal length of 6.0 cm. The object is 2.4 cm high. a. Draw a ray diagram to determine the location, size, and orientation of the image. O1 Ray 1 hi 1.8 cm di 10.5 cm Ray 2 F Horizontal scale: 1 block 1.0 cm Vertical scale: 1 block 0.4 cm F I1 Solutions Manual 389 Chapter 18 continued b. Solve the problem mathematically. 1 1 1 f do di fnew 2f d f do f o di 2(6.00 cm) (14.0 cm)(6.0 cm) 14.0 cm 6.0 cm 10.5 cm d do h ho b. If the original lens is replaced with a lens having twice the focal length, what are the image position, size, and orientation? m i i d h do i o hi (10.5 cm)(2.4 cm) 14.0 cm 1.8 cm, so the image is inverted 12.0 cm 1 1 1 f do di d f do fnew o new di, new (15.0 cm)(12.0 cm) 15.0 cm 12.0 cm 60.0 cm d do h ho m i i d 86. A 3.0-cm-tall object is placed 22 cm in front of a converging lens. A real image is formed 11 cm from the lens. What is the size of the image? h d m i i ho do d h do h i, new o hi, new do (60.0 cm)(3.0 cm) 15 cm 12 cm The image is inverted compared to the object. i o hi 22 cm 1.5 cm The image is 1.5 cm tall. Level 3 87. A 3.0-cm-tall object is placed 15.0 cm in front of a converging lens. A real image is formed 10.0 cm from the lens. a. What is the focal length of the lens? a. What are the object position and object height? 1 1 1 f do di df i do d f i (5.0 cm)(15.0 cm) 5.0 cm (15.0 cm) 1 1 1 f do di 7.5 cm d d do di m i i o i f (15.0 cm)(10.0 cm) 15.0 cm 10.0 cm 6.00 cm d do h ho d h di o i ho (7.5 cm)(2.0 cm) 5.0 cm 3.0 cm 390 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. (11 cm)(3.0 cm) 88. A diverging lens has a focal length of 15.0 cm. An object placed near it forms a 2.0-cm-high image at a distance of 5.0 cm from the lens. Chapter 18 continued b. The diverging lens is now replaced by a converging lens with the same focal length. What are the image position, height, and orientation? Is it a virtual image or a real image? fnew f (15.0 cm) 1 1 1 f di do d f do f o So di (125 m)(1.0000 m) 15.0 cm 125 m 1.00 m 1 1 1 fnew do di, new d f do fnew o new di, new (7.5 cm)(15 cm) 7.5 cm 15 cm 15 cm h ho b. A 1000.0-mm lens is focused on an object 125 m away. What is the image position? d do m i i 1.01 m 1.01103 mm 90. Eyeglasses To clearly read a book 25 cm away, a farsighted girl needs the image to be 45 cm from her eyes. What focal length is needed for the lenses in her eyeglasses? 1 1 1 f di do d d do di o i So f d h i, new o hi, new do (15 cm)(3.0 cm) (25 cm)(45 cm) 25 cm (45 cm) 56 cm 7.5 cm 6.0 cm Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. This is a virtual image that is upright compared to the object. 18.3 Applications of Lenses pages 510–511 Level 1 89. Camera Lenses Camera lenses are described in terms of their focal length. A 50.0-mm lens has a focal length of 50.0 mm. a. A camera with a 50.0-mm lens is focused on an object 3.0 m away. What is the image position? 1 1 1 f di do do f So di do f (3.0103 mm)(50.0 mm) 3 3.010 mm 50.0 mm 51 mm Physics: Principles and Problems Level 2 91. Copy Machine The convex lens of a copy machine has a focal length of 25.0 cm. A letter to be copied is placed 40.0 cm from the lens. a. How far from the lens is the copy paper? 1 1 1 f di do d f do f o di (40.0 cm)(25.0 cm) 40.0 cm 25.0 cm 66.7 cm b. How much larger will the copy be? h d i i ho do d h do (66.7 cm)(h ) 40.0 cm i o o hi 1.67ho The copy is enlarged and inverted. Solutions Manual 391 Chapter 18 continued 92. Camera A camera lens with a focal length of 35 mm is used to photograph a distant object. How far from the lens is the real image of the object? Explain. 35 mm; for a distant object, do can be considered at , thus 1/do is zero. According to the thin lens equation, di f. Level 3 93. Microscope A slide of an onion cell is placed 12 mm from the objective lens of a microscope. The focal length of the objective lens is 10.0 mm. a. How far from the lens is the image formed? 1 1 1 f di do do f So di do f (12 mm)(10.0 mm) 12 mm 10.0 mm 6.0101 mm b. What is the magnification of this image? di 6.0101 mm 5.0 mo do 12 mm 1 1 1 f di do d f do f o di mm)(20.0 mm) (10.0 10.0 mm 20.0 mm 20.0 mm, or 20.0 mm beneath the eyepiece d. What is the final magnification of this compound system? d (20.0 mm) i me 2.00 do 10.0 mm mtotal mome (5.0)(2.00) 1.0101 392 Solutions Manual a. What are the image position, height, and orientation as formed by the objective lens? Is this a real or virtual image? 1 1 1 f do di d f do f o di (425 cm)(20.0 cm) 425 cm 20.0 cm 21.0 cm d do h ho m i i d h do i o hi (21.0 cm)(10.0 cm) 425 cm 0.494 cm This is a real image that is inverted compared to the object. b. The objective lens image becomes the object for the eyepiece lens. What are the image position, height, and orientation that a person sees when looking into the telescope? Is this a real or virtual image? do, new 25.0 cm di 25.0 cm 21.0 cm 4.0 cm 1 1 1 fnew do, new di, new d f do, new fnew o, new new di, new (4.0 cm)(4.05 cm) 4.0 cm 4.05 cm 3.2102 cm Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. c. The real image formed is located 10.0 mm beneath the eyepiece lens. If the focal length of the eyepiece is 20.0 mm, where does the final image appear? 94. Telescope The optical system of a toy refracting telescope consists of a converging objective lens with a focal length of 20.0 cm, located 25.0 cm from a converging eyepiece lens with a focal length of 4.05 cm. The telescope is used to view a 10.0-cm-high object, located 425 cm from the objective lens. Chapter 18 continued ho, new hi 0.494 cm d do h ho m i i d h do, new i, new o, new hi, new (3.210 cm)(0.494 cm) 2 4.0 cm 4.0101 cm This is a virtual image that is inverted compared to the object. c. What is the magnification of the telescope? h i, new m ho 4.010 cm 1 10.0 cm 4.0 Mixed Review Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. pages 511—512 Level 1 95. A block of glass has a critical angle of 45.0°. What is its index of refraction? n n1 2 sin c n 2 , n2 1.00 for air n1 sin c 1.00 n1 sin 45.0° 1.41 96. Find the speed of light in antimony trioxide if it has an index of refraction of 2.35. c n v c v n 3.00108 m/s 2.35 1.28108 m/s 97. A 3.0-cm-tall object is placed 20 cm in front of a converging lens. A real image is formed 10 cm from the lens. What is the focal length of the lens? 1 1 1 do f di d d do di o i f (20 cm)(10 cm) 20 cm 10 cm 7 cm Level 2 98. Derive n sin 1/sin 2 from the general form of Snell’s law of refraction, n1 sin 1 n2 sin 2. State any assumptions and restrictions. The angle of incidence must be in air. If we let substance 1 be air, then n1 1.000. Let n2 n. Therefore, n1 sin 1 n2 sin 2 sin 1 n sin 2 sin 1 n sin 2 99. Astronomy How many more minutes would it take light from the Sun to reach Earth if the space between them were filled with water rather than a vacuum? The Sun is 1.5108 km from Earth. Time through vacuum 8 (1.510 km)(1000 m/1 km) d t 8 c 3.0010 m/s 5.0102 s Speed through water 3.00108 m/s c v n 1.33 2.26108 m/s Time through water 8 (1.510 km)(1000 m/1 km) d t 8 v 2.2610 m/s 660 s t 660 s 500 s 160 s (160 s)(1 min/60 s) 2.7 min Physics: Principles and Problems Solutions Manual 393 Chapter 18 continued 100. What is the focal length of the lenses in your eyes when you read a book that is 35.0 cm from them? The distance from each lens to the retina is 0.19 mm. 1 1 1 do f di b. At what depth does the bottom of the tank appear to be if you look into the water? Divide this apparent depth into the true depth and compare it to the index of refraction. Using right triangle geometry, (actual depth)(tan 1) (apparent depth)(tan 2) d d do di o i f apparent depth (12 cm) tan 5.0° tan 6.7° (350 mm)(0.19 mm) 350 mm 0.19 mm 8.9 cm 0.19 mm Level 3 101. Apparent Depth Sunlight reflects diffusively off the bottom of an aquarium. Figure 18-29 shows two of the many light rays that would reflect diffusively from a point off the bottom of the tank and travel to the surface. The light rays refract into the air as shown. The red dashed line extending back from the refracted light ray is a sight line that intersects with the vertical ray at the location where an observer would see the image of the bottom of the tank. 2 The refracted rays appear to intersect 8.9 cm below the surface; this is the apparent depth. apparent depth 8.9 cm 0.74 true depth 12 cm nair 1.0 Also, 0.75 nwater 1.33 Therefore, nair apparent depth true depth nwater 12 cm nglass 1.5 5.0° ■ 1 90° Figure 18-29 a. Compute the direction that the refracted ray will travel above the surface of the water. Air 2 1 n1 sin 1 n2 sin 2 n sin n2 1 1 2 sin1 (1.33)(sin 5.0°) sin1 1.0 6.7° 394 Solutions Manual Air 2 90° ■ Figure 18-30 The light ray enters the glass at an angle 1 and is refracted to an angle 2. Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 102. It is impossible to see through adjacent sides of a square block of glass with an index of refraction of 1.5. The side adjacent to the side that an observer is looking through acts as a mirror. Figure 18-30 shows the limiting case for the adjacent side to not act like a mirror. Use your knowledge of geometry and critical angles to show that this ray configuration is not achievable when nglass 1.5. Chapter 18 continued n sin ng Thinking Critically A A 2 sin1 (1.00)(sin 90°) sin1 1.5 42° Therefore, 1 48°. But the critical angle for glass is n ng A c sin1 Find the angles of refraction for red and blue light, and find the difference in those angles in degrees. sin1 1.00 1.5 42° Because 1 c, the light reflects back into the glass and one cannot see out of an adjacent side. 103. Bank Teller Window A 25-mm-thick sheet of plastic, n 1.5, is used in a bank teller’s window. A ray of light strikes the sheet at an angle of 45°. The ray leaves the sheet at 45°, but at a different location. Use a ray diagram to find the distance between the ray that leaves and the one that would have left if the plastic were not there. 8 mm Use Snell’s law, n1 sin 1 n2 sin 2. n sin n2 1 1 Thus, 2 sin1 For red light 2 sin1 (1.0003)(sin 45.000°) 1.7273 24.173° For blue light 2 sin1 (1.0003)(sin 45.000°) 1.7708 23.543° Difference 24.173° 23.543° 0.630° m m 25 mm 8 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. page 512 104. Recognize Spatial Relationships White light traveling through air (n 1.0003) enters a slab of glass, incident at exactly 45°. For dense flint glass, n 1.7708 for blue light ( 435.8 nm) and n 1.7273 for red light ( 643.8 nm). What is the angular dispersion of the red and blue light? 105. Compare and Contrast Find the critical angle for ice (n 1.31). In a very cold world, would fiber-optic cables made of ice or those made of glass do a better job of keeping light inside the cable? Explain. n nice air sin c 45° 28° 45° Physics: Principles and Problems n nice air c sin1 sin1 49.8° 1.00 1.31 In comparison, the critical angle for glass, n 1.54, is 40.5°. The larger critical angle means that fewer rays would be totally internally reflected in an ice core than in a glass core. Thus, they would not be able to transmit as much light. Fiber optic cables made of glass would work better. Solutions Manual 395 Chapter 18 continued 106. Recognize Cause and Effect Your lab partner used a convex lens to produce an image with di 25 cm and hi 4.0 cm. You are examining a concave lens with a focal length of 15 cm. You place the concave lens between the convex lens and the original image, 10 cm from the image. To your surprise, you see a real image on the wall that is larger than the object. You are told that the image from the convex lens is now the object for the concave lens, and because it is on the opposite side of the concave lens, it is a virtual object. Use these hints to find the new image position and image height and to predict whether the concave lens changed the orientation of the original image. The new do 10 cm. Thus, fdo (15 cm)(10 cm) di do f 10 cm (15 cm) 30 cm d 30 cm m i 3 do 10 cm Writing in Physics page 512 109. The process of accommodation, whereby muscles surrounding the lens in the eye contract or relax to enable the eye to focus on close or distant objects, varies for different species. Investigate this effect for different animals. Prepare a report for the class showing how this fine focusing is accomplished for different eye mechanisms. Answers will vary depending on the animals selected by the students. 110. Investigate the lens system used in an optical instrument such as an overhead projector or a particular camera or telescope. Prepare a graphics display for the class explaining how the instrument forms images. Answers will vary. Students may find that it is necessary to simplify their chosen system for explanation purposes. Cumulative Review The image orientation is not changed. page 512 111. If you drop a 2.0 kg bag of lead shot from a height of 1.5 m, you could assume that half of the potential energy will be converted into thermal energy in the lead. The other half would go to thermal energy in the floor. How many times would you have to drop the bag to heat it by 10°C? (Chapter 12) 107. Define Operationally Name and describe the effect that causes the rainbow-colored fringe commonly seen at the edges of a spot of white light from a slide or overhead projector. The light that passes through a lens near the edges of the lens is slightly dispersed, since the edges of a lens resemble a prism and refract different wavelengths of light at slightly different angles. The result is that white light is dispersed into its spectrum. The effect is called chromatic aberration. PE mgh (2.0 kg)(9.80 m/s2)(1.5 m) 29.4 J To heat the bag Q mCT (2.0 kg)(130 J/kg °C)(10°C) 108. Think Critically A lens is used to project the image of an object onto a screen. Suppose that you cover the right half of the lens. What will happen to the image? It will get dimmer, because fewer light rays will converge, but you will see a complete image. 396 Solutions Manual 2600 J N Q 1 PE 2 2600 J 1 (29.4 J) 2 180 times Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. hi mho (3)(4.0 cm) 10 cm Chapter 18 continued 112. A blacksmith puts an iron hoop or tire on the outer rim of a wooden carriage wheel by heating the hoop so that it expands to a diameter greater than the wooden wheel. When the hoop cools, it contracts to hold the rim in place. If a blacksmith has a wooden wheel with a 1.0000-m diameter and wants to put a rim with a 0.9950-m diameter on the wheel, what is the minimum temperature change the iron must experience? (iron 12106/°C) (Chapter 13) L iron LT where L is the diameter of the iron hoop. We want L greater than 0.0050 m. Therefore, L T ironL 0.0050 6 (1210 /°C)(0.9950 m) 420°C Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Actually he would heat it much hotter to give room to fit over the wheel easily. 113. A car sounds its horn as it approaches a pedestrian in a crosswalk. What does the pedestrian hear as the car brakes to allow him to cross the street? (Chapter 15) The pitch of the horn heard by the pedestrian will decrease as the car slows down. 114. Suppose that you could stand on the surface of the Sun and weigh yourself. Also suppose that you could measure the illuminance on your hand from the Sun’s visible spectrum produced at that position. Next, imagine yourself traveling to a position 1000 times farther away from the center of the Sun as you were when standing on its surface. (Chapter 16) a. How would the force of gravity on you from the Sun at the new position compare to what it was at the surface? b. How would the illuminance on your hand from the Sun at the new position compare to what it was when you were standing on its surface? (For simplicity, assume that the Sun is a point source at both positions.) 1 1 6 It is 2 110 1,000,000 1000 the value it was originally. c. Compare the effect of distance upon the gravitational force and illuminance. They both follow the inverse square law of distance. 115. Beautician’s Mirror The nose of a customer who is trying some face powder is 3.00-cm high and is located 6.00 cm in front of a concave mirror having a 14.0-cm focal length. Find the image position and height of the customer’s nose by means of the following. (Chapter 17) a. a ray diagram drawn to scale I1 O1 hi 5.25 cm di 10.5 cm Horizontal scale: 1 block 1.0 cm Vertical scale: 1 block 1.0 cm b. the mirror and magnification equations 1 1 1 di f do d f do f o di (6.00 cm)(14.0 cm) 6.00 cm 14.0 cm 10.5 cm h ho d do m i i d h do i o hi 1 1 6 It is 2 110 (10.5 cm)(3.00 cm) the value it was originally. 5.25 cm (1000) 1,000,000 Physics: Principles and Problems 6.00 cm Solutions Manual 397 Chapter 18 continued Challenge Problem page 501 As light enters the eye, it first encounters the air/cornea interface. Consider a ray of light that strikes the interface between the air and a person’s cornea at an angle of 30.0° to the normal. The index of refraction of the cornea is approximately 1.4. 1. Use Snell’s law to calculate the angle of refraction. n1 sin 1 n2 sin 2 n sin n2 1 1 2 sin1 (1.0)(sin 30.0°) 1.4 sin1 21° 4. If you want the angle of refraction for the light ray in water to be the same as it is for air, what should the new angle of incidence be? n1 sin 1 n2 sin 2 n sin n1 2 2 1 sin1 (1.4)(sin 21°) 1.33 sin1 22° Cornea Air 30.0° 2 2. What would the angle of refraction be if the person was swimming underwater? n1 sin 1 n2 sin 2 n sin n2 1 1 2 sin1 (1.33)(sin 30.0°) 1.4 sin1 28° Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 3. Is the refraction greater in air or in water? Does this mean that objects under water seem closer or more distant than they would in air? Refraction is greater in air because the angle to the normal is smaller. Objects seem closer in water. 398 Solutions Manual Physics: Principles and Problems Interference and Diffraction CHAPTER 19 Practice Problems 19.1 Interference pages 515–523 page 519 1. Violet light falls on two slits separated by 1.90105 m. A first-order bright band appears 13.2 mm from the central bright band on a screen 0.600 m from the slits. What is ? xd L 3 5 (13.210 m)(1.9010 m) 0.600 m 4. Yellow-orange light with a wavelength of 596 nm passes through two slits that are separated by 2.25105 m and makes an interference pattern on a screen. If the distance from the central line to the first-order yellow band is 2.00102 m, how far is the screen from the slits? xd L xd L (2.00102 m)(2.25105 m) 59610 m 9 0.755 m 418 nm Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 2. Yellow-orange light from a sodium lamp of wavelength 596 nm is aimed at two slits that are separated by 1.90105 m. What is the distance from the central band to the first-order yellow band if the screen is 0.600 m from the slits? page 522 5. In the situation in Example Problem 2, what would be the thinnest film that would create a reflected red ( 635 nm) band? 1 2 noil 2t m For the thinnest film, m 0. L x d (13.2103 m)(1.90105 m) 0.600 m 1.88102 m 18.8 mm 3. In a double-slit experiment, physics students use a laser with 632.8 nm. A student places the screen 1.000 m from the slits and finds the first-order bright band 65.5 mm from the central line. What is the slit separation? 1 t 4 noil 635 nm (4)(1.45) 109 nm 6. A glass lens has a nonreflective coating placed on it. If a film of magnesium fluoride, n 1.38, is placed on the glass, n 1.52, how thick should the layer be to keep yellow-green light from being reflected? xd Because nfilm nair, there is a phase L d inversion on the first reflection. Because nglass nfilm, there is a phase L x 9 (632.810 m)(1.000 m) 3 65.510 m 9.66106 m 9.66 m Physics: Principles and Problems inversion on the second reflection. For destructive interference to keep yellow-green from being reflected: 1 2 nfilm 2t m Solutions Manual 399 Chapter 19 continued For the thinnest film, m 0. For the thinnest film, m 1. 1 t t 4 nfilm 2nfilm 555 nm 521 nm 101 nm 196 nm (4)(1.38) 7. A silicon solar cell has a nonreflective coating placed on it. If a film of sodium monoxide, n 1.45, is placed on the silicon, n 3.5, how thick should the layer be to keep yellow-green light ( 555 nm) from being reflected? Because nfilm nair, there is a phase inversion on the first reflection. Because nsilicon nfilm, there is a phase inversion on the second reflection. For destructive interference to keep yellow-green from being reflected: 1 2 nfilm 2t m For the thinnest film, m 0. 1 t 4 nfilm 555 nm (4)(1.45) 8. You can observe thin-film interference by dipping a bubble wand into some bubble solution and holding the wand in the air. What is the thickness of the thinnest soap film at which you would see a black stripe if the light illuminating the film has a wavelength of 521 nm? Use n 1.33. Because nfilm nair, there is a phase change on the first reflection. Because nair nfilm, there is no phase change on the second reflection. For destructive interference to get a black stripe m n film 2t 9. What is the thinnest soap film (n 1.33) for which light of wavelength 521 nm will constructively interfere with itself? For constructive interference 1 2 nfilm 2t m For the thinnest film, m 0. 1 t 4 nfilm 521 nm (4)(1.33) 97.9 nm Section Review 19.1 Interference pages 515–523 page 523 10. Film Thickness Lucien is blowing bubbles and holds the bubble wand up so that a soap film is suspended vertically in the air. What is the second thinnest width of the soap film at which he could expect to see a bright stripe if the light illuminating the film has a wavelength of 575 nm? Assume the soap solution has an index of refraction of 1.33. There is one phase inversion, so constructive interference will be when 1 2 nfilm 2t m For the second thinnest thickness, m 1. 3 4 nfilm t (3)(575 nm) (4)(1.33) 324 nm 400 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 95.7 nm (2)(1.33) Chapter 19 continued 11. Bright and Dark Patterns Two very narrow slits are cut close to each other in a large piece of cardboard. They are illuminated by monochromatic red light. A sheet of white paper is placed far from the slits, and a pattern of bright and dark bands is seen on the paper. Describe how a wave behaves when it encounters a slit, and explain why some regions are bright while others are dark. When a wave encounters a slit, the wave bends. Light is diffracted by the slits. Light from one slit interferes with light from the other. If interference is constructive, there is a bright band; if destructive, the region is dark. 12. Interference Patterns Sketch the pattern described in problem 11. Repeats and fades Repeats and fades reflection. For constructive interference to reflect yellow-green light: 1 2 nfilm 2t m For the thinnest film, m 0. 1 4 nfilm color t 555 nm (4)(1.83) 75.8 nm b. Unfortunately, a film this thin cannot be manufactured. What is the nextthinnest film that will produce the same effect? For the next thinnest film, m 1. 3 color 4 n film t (3)(555 nm) Black Red Black Red Black Red Black (4)(1.83) 227 nm Black Blue Black Red Black Blue Black Red Blue Black Blue Black Red Black Repeats and fades Black Repeats and fades Blue Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 13. Interference Patterns Sketch what happens to the pattern in problem 11 when the red light is replaced by blue light. The light bands become more closely spaced. 14. Film Thickness A plastic reflecting film (n 1.83) is placed on an auto glass window (n 1.52). a. What is the thinnest film that will reflect yellow-green light? Because nfilm nair, there is a phase change on the first reflection. Because nglass nfilm, there is not 15. Critical Thinking The equation for wavelength from a double-slit experiment uses the simplification that is small so that sin tan . Up to what angle is this a good approximation when your data has two significant figures? Would the maximum angle for a valid approximation increase or decrease as you increase the precision of your angle measurement? sin tan to two significant digits up to 9.9°. An increase in the precision of the measurement reduces this angle to 2.99°. Practice Problems 19.2 Diffraction pages 524–531 page 526 16. Monochromatic green light of wavelength 546 nm falls on a single slit with a width of 0.095 mm. The slit is located 75 cm from a screen. How wide will the central bright band be? a phase change on the second Physics: Principles and Problems Solutions Manual 401 Chapter 19 continued xminw L L xmin w (5.46107 m)(0.75 m) 9.5105 m 20. White light falls on a single slit that is 0.050 mm wide. A screen is placed 1.00 m away. A student first puts a blue-violet filter ( 441 nm) over the slit, then a red filter ( 622 nm). The student measures the width of the central bright band. a. Which filter produced the wider band? 4.3 mm 17. Yellow light with a wavelength of 589 nm passes through a slit of width 0.110 mm and makes a pattern on a screen. If the width of the central bright band is 2.60102 m, how far is it from the slits to the screen? 2L w 2x1 Red, because central peak width is proportional to wavelength. b. Calculate the width of the central bright band for each of the two filters. 2 L 2x1 w For blue, 7 2(4.4110 m)(1.00 m) 2x1 5 (2x1)w L 2 5.010 (2.60102 m)(0.110103 m) (2)(589109 m) 18 mm For red, 7 2(6.2210 m)(1.00 m) 2x1 5 2.43 m 5.010 18. Light from a He-Ne laser ( 632.8 nm) falls on a slit of unknown width. A pattern is formed on a screen 1.15 m away, on which the central bright band is 15 mm wide. How wide is the slit? 2L 2x1 w (2)(632.8109 m)(1.15 m) 1510 m 3 9.7105 m 97 m 19. Yellow light falls on a single slit 0.0295 mm wide. On a screen that is 60.0 cm away, the central bright band is 24.0 mm wide. What is the wavelength of the light? 2L w 2x1 25 mm page 529 21. White light shines through a grating onto a screen. Describe the pattern that is produced. A full spectrum of color is seen. Because of the variety of wavelengths, dark fringes of one wavelength would be filled by bright fringes of another color. 22. If blue light of wavelength 434 nm shines on a diffraction grating and the spacing of the resulting lines on a screen that is 1.05 m away is 0.55 m, what is the spacing between the slits in the grating? d sin sin d where tan1 3 (2)(60.010 3 5.90102 nm Solutions Manual m) x sin tan1 L 0.55 m sin tan1 1.05 m (2x1)w (24.010 m)(0.029510 m) 2 x L 2L m 434109 9.410–7 m Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 2L w 2x1 402 m Chapter 19 continued 23. A diffraction grating with slits separated by 8.60107 m is illuminated by violet light with a wavelength of 421 nm. If the screen is 80.0 cm from the grating, what is the separation of the lines in the diffraction pattern? d sin sin d x tan L x L tan L tan sin1 d 421109 m 8.6010 m (0.800 m)tan sin1 7 Section Review 19.2 Diffraction pages 524–531 page 531 26. Distance Between First-Order Dark Bands Monochromatic green light of wavelength 546 nm falls on a single slit of width 0.080 mm. The slit is located 68.0 cm from a screen. What is the separation of the first dark bands on each side of the central bright band? 2L 2xmin w (2)(546109 m)(68.0102 m) 0.08010 m 3 0.449 m 9.3 mm 24. Blue light shines on the DVD in Example Problem 3. If the dots produced on a wall that is 0.65 m away are separated by 58.0 cm, what is the wavelength of the light? d sin d sin tan1 x L (7.41107 m)sin tan1 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 0.58 m 0.65 m 490 nm 27. Diffraction Patterns Many narrow slits are close to each other and equally spaced in a large piece of cardboard. They are illuminated by monochromatic red light. A sheet of white paper is placed far from the slits, and a pattern of bright and dark bands is visible on the paper. Sketch the pattern that would be seen on the screen. Pattern repeats Pattern repeats 25. Light of wavelength 632 nm passes through a diffraction grating and creates a pattern on a screen that is 0.55 m away. If the first bright band is 5.6 cm from the central bright band, how many slits per centimeter does the grating have? Bright red lines d sin 1 There is one slit per distance d, so d gives slits per centimeter. sin d sin tan1 x L 632109 m 0.05 6m sin tan1 0.55 m 6.210–6 m 6.210–4 cm 1 slit 1.6103 slits/cm 6.210–4 cm Physics: Principles and Problems Band spacing is exactly the same as in the pattern produced by the two slits, but now light bands are much thinner and separated by wider dark bands. 28. Line Spacing You shine a red laser light through one diffraction grating and form a pattern of red dots on a screen. Then you substitute a second diffraction grating for the first one, forming a different pattern. The dots produced by the first grating are spread out more than those produced by the second. Which grating has more lines Solutions Manual 403 Chapter 19 continued Complete the following concept map using , L, and d to indicate how you could vary them to produce the indicated change in the spacing between adjacent bright bands, x. per millimeter? xd L , so the greater the dot spacings, x, the narrower the slit spacing, d, and thus more lines per millimeter. 29. Rayleigh Criterion The brightest star in the winter sky in the northern hemisphere is Sirius. In reality, Sirius is a system of two stars that orbit each other. If the Hubble Space Telescope (diameter 2.4 m) is pointed at the Sirius system, which is 8.44 light-years from Earth, what is the minimum separation there would need to be between the stars in order for the telescope to be able to resolve them? Assume that the average light coming from the stars has a wavelength of 550 nm. 1.22Lobj xobj D 9 16 1.22(33010 m)(7.9910 m) 2.4 m 2.21010 m 30. Critical Thinking You are shown a spectrometer, but do not know whether it produces its spectrum with a prism or a grating. By looking at a white-light spectrum, how could you tell? Chapter Assessment Concept Mapping page 536 31. Monochromatic light of wavelength illuminates two slits in a Young’s double-slit experiment setup that are separated by a distance, d. A pattern is projected onto a screen a distance, L, away from the slits. 404 Solutions Manual to increase x, increase L to decrease x, decrease increase d d decrease L Mastering Concepts page 536 32. Why is it important that monochromatic light was used to make the interference pattern in Young’s interference experiment? (19.1) When monochromatic light is used, you get a sharp interference pattern; if you use white light, you get sets of colored bands. 33. Explain why the position of the central bright band of a double-slit interference pattern cannot be used to determine the wavelength of the light waves. (19.1) All wavelengths produce the line in the same place. 34. Describe how you could use light of a known wave-length to find the distance between two slits. (19.1) Let the light fall on the double slit, and let the interference pattern fall on a sheet of paper. Measure the spacings between the bright bands, x, and use L x the equation d . 35. Describe in your own words what happens in thin-film interference when a dark band is produced by light shining on a soap film suspended in air. Make sure you include in your explanation how the wavelength of the light and thickness of the film are related. (19.1) Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Determine if the violet or the red end of the spectrum makes the largest angle with the direction of the beam of incident white light. A prism bends the violet end of the spectrum the most, whereas a grating diffracts red wavelengths the most. xd L Chapter 19 continued When the light strikes the front of the film, some reflects off this surface and some passes through the film and reflects off the back surface of the film. When light reflects off a medium with a higher index of refraction, it undergoes a phase shift of one-half wavelength; this happens to the light that initially reflects. In order for a dark band to be produced, the two light rays must be one-half wavelength out of phase. If the thickness of the film is such that the ray reflecting off the back surface goes through a whole number of cycles while passing through the film, the light rays arriving at your eye will be out of phase and destructively interfere. Remember that the wavelength is altered by the index of refraction of the film, so that the thickness of the film must equal a multiple of half a wavelength of the light, divided by the film’s index of refraction. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 36. White light shines through a diffraction grating. Are the resulting red lines spaced more closely or farther apart than the resulting violet lines? Why? (19.2) The spacing is directly proportional to the wavelength, and because red light has a longer wavelength than violet, the red lines will be spaced farther apart than the violet lines. 37. Why do diffraction gratings have large numbers of slits? Why are these slits so close together? (19.2) The large number of grooves in diffraction gratings increases the intensity of the diffraction patterns. The grooves are close together, producing sharper images of light. 38. Why would a telescope with a small diameter not be able to resolve the images of two closely spaced stars? (19.2) Small apertures have large diffraction patterns that limit resolution. 39. For a given diffraction grating, which color of visible light produces a bright line closest to the central bright band? (19.2) violet light, the color with the smallest wavelength Applying Concepts page 536 40. For each of the following examples, indicate whether the color is produced by thin-film interference, refraction, or the presence of pigments. a. soap bubbles interference b. rose petals pigments c. oil films interference d. a rainbow refraction 41. How can you tell whether a pattern is produced by a single slit or a double slit? A double-slit interference pattern consists of equally spaced lines of almost equal brightness. A single-slit diffraction pattern has a bright, broad central band and dimmer side bands. 42. Describe the changes in a single-slit diffraction pattern as the width of the slit is decreased. The bands get wider and dimmer. 43. Science Fair At a science fair, one exhibition is a very large soap film that has a fairly consistent width. It is illuminated by a light with a wavelength of 432 nm, and nearly the entire surface appears to be a lovely shade of purple. What would you see in the following situations? a. the film thickness was doubled complete destructive interference b. the film thickness was increased by half a wavelength of the illuminating light complete constructive interference Physics: Principles and Problems Solutions Manual 405 Chapter 19 continued c. the film thickness was decreased by one quarter of a wavelength of the illuminating light complete destructive interference 44. What are the differences in the characteristics of the diffraction patterns formed by diffraction gratings containing 104 lines/cm and 105 lines/cm? The lines in the diffraction pattern are narrower for the 105 lines/cm grating. 45. Laser Pointer Challenge You have two laser pointers, a red one and a green one. Your friends Mark and Carlos disagree about which has the longer wavelength. Mark insists that red light has a longer wavelength, while Carlos is sure that green has the longer wavelength. You have a diffraction grating handy. Describe what demonstration you would do with this equipment and how you would explain the results to Carlos and Mark to settle their disagreement. 46. Optical Microscope Why is blue light used for illumination in an optical microscope? Less diffraction results from the short wavelength of blue light. Mastering Problems 19.1 Interference pages 536—537 Level 1 47. Light falls on a pair of slits 19.0 m apart and 80.0 cm from a screen, as shown in Figure 19-17. The first-order bright band is 1.90 cm from the central bright band. What is the wavelength of the light? 80.0 cm Double slit ■ Screen Figure 19-17 (Not to scale) xd L (19.0106 m)(1.90102 m) 80.010 m 2 451 nm 48. Oil Slick After a short spring shower, Tom and Ann take their dog for a walk and notice a thin film of oil (n 1.45) on a puddle of water, producing different colors. What is the minimum thickness of a place where the oil creates constructive interference for light with a wavelength of 545 nm? There is one phase inversion, so constructive interference will be when 1 2 nfilm 2t m For the minimum thickness, m 0. 1 4 nfilm t 545 nm (4)(1.45) 94.0 nm Level 2 49. Light of wavelength 542 nm falls on a double slit. First-order bright bands appear 4.00 cm from the central bright band. The screen is 1.20 m from the slits. How far apart are the slits? xd L L x d (5.42107 m)(1.20 m) 4.0010 m 2 16.3 m 50. Insulation Film Winter is approaching and Alejandro is helping to cover the windows 406 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Shine each laser pointer through the grating onto a nearby wall. The color with the longer wavelength will produce dots with a greater spacing on the wall because the spacing is directly proportional to the wavelength. (Mark is correct; red light has a longer wavelength than green light.) 19.0 m Chapter 19 continued in his home with thin sheets of clear plastic (n 1.81) to keep the drafts out. After the plastic is taped up around the windows such that there is air between the plastic and the glass panes, the plastic is heated with a hair dryer to shrink-wrap the window. The thickness of the plastic is altered during this process. Alejandro notices a place on the plastic where there is a blue stripe of color. He realizes that this is created by thin-film interference. What are three possible thicknesses of the portion of the plastic where the blue stripe is produced if the wavelength of the light is 4.40102 nm? There is one phase inversion, so constructive interference will be when 1 2 nfilm 2t m Three possible thicknesses occur at m 0, 1, and 2. 1 4 nfilm t for m 0 2 4.4010 nm (4)(1.81) xd L L d x Because is the same for each setup, x calculate to compare the setups. x L d Setup A: 0.60 m 1.5010 m 4 4.0103 Setup B: 0.80 m 1.7510 m 4 4.6103 Setup C: 0.80 m 1.5010 m 4 5.3103 xC xB xA 60.8 nm 3 4 nfilm t for m 1 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. first-order bright band, from least to most separation. Specifically indicate any ties. 2 (3)(4.4010 nm) (4)(1.81) 182 nm 5 4 nfilm t for m 2 2 (5)(4.4010 nm) (4)(1.81) 19.2 Diffraction page 537 Level 1 52. Monochromatic light passes through a single slit with a width of 0.010 cm and falls on a screen 100 cm away, as shown in Figure 19-18. If the width of the central band is 1.20 cm, what is the wavelength of the light? 304 nm Level 3 51. Samir shines a red laser pointer through three different double-slit setups. In setup A, the slits are separated by 0.150 mm and the screen is 0.60 m away from the slits. In setup B, the slits are separated by 0.175 mm and the screen is 0.80 m away. Setup C has the slits separated by 0.150 mm and the screen a distance of 0.80 m away. Rank the three setups according to the separation between the central bright band and the Physics: Principles and Problems 0.010 cm 100 cm Single slit Screen ■ Figure 19-18 (Not to scale) 2L w 2x1 xw L Solutions Manual 407 Chapter 19 continued (0.60 cm)(0.010 cm) 100 cm 600 nm xobj 1.22 Lobj D 1.22Lobj 53. A good diffraction grating has 2.5103 lines per cm. What is the distance between two lines in the grating? 1 d 3 2.510 lines/cm 4.0104 cm Level 2 54. Light with a wavelength of 4.5105 cm passes through a single slit and falls on a screen 100 cm away. If the slit is 0.015 cm wide, what is the distance from the center of the pattern to the first dark band? 2L 2x1 w 2x1 is the width of the bright band, so to get the distance from the center to the first dark band, divide by 2. L w xobj D (1.22)(5.1107 m)(1.0105 m) 2.4 m 2.6102 m 2.6 cm Level 3 56. Monochromatic light with a wavelength of 425 nm passes through a single slit and falls on a screen 75 cm away. If the central bright band is 0.60 cm wide, what is the width of the slit? 2L w 2x1 L 2L w x1 2x1 x1 (2x1) 0.30 cm 1 2 5 (4.2510 cm)(75 cm) x1 0.30 cm 5 (4.510 cm)(100 m) 0.015 cm 0.3 cm 57. Kaleidoscope Jennifer is playing with a kaleidoscope from which the mirrors have been removed. The eyehole at the end is 7.0 mm in diameter. If she can just distinguish two bluish-purple specks on the other end of the kaleidoscope separated by 40 m, what is the length of the kaleidoscope? Use 650 nm and assume that the resolution is diffraction limited through the eyehole. 1.22Lobj xobj D xobjD Lobj 1.22 6 3 (4010 m)(7.010 m) 9 (1.22)(65010 m) 0.4 m ■ 408 Figure 19-19 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 55. Hubble Space Telescope Suppose the Hubble Space Telescope, 2.4 m in diameter, is in orbit 1.0105 m above Earth and is turned to view Earth, as shown in Figure 19-19. If you ignore the effect of the atmosphere, how large an object can the telescope resolve? Use 5.1107 m. 1.1102 cm Chapter 19 continued 58. Spectroscope A spectroscope uses a grating with 12,000 lines/cm. Find the angles at which red light, 632 nm, and blue light, 421 nm, have first-order bright lines. 1.2104 m 1.2102 cm 1 d 8.33105 cm 1 1 83 ridges/cm d 1.2102 cm 12,000 lines/cm d sin b. Marie checks her results by noting that the ridges represent a song that lasts 4.01 minutes and takes up 16 mm on the record. How many ridges should there be in a centimeter? sin d For red light, 6.32105 cm 8.3310 cm sin1 5 Number of ridges is 49.3° (4.01 min)(33.3 rev/min) 134 ridges For blue light, 134 ridges 84 ridges/cm 1.6 cm 4.21105 cm 8.3310 cm sin1 5 30.3° Mixed Review page 538 Level 1 1 59. Record Marie uses an old 33 rpm record 3 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 632.8109 m 21103 m sin tan1 4.0 m as a diffraction grating. She shines a laser, 632.8 nm, on the record, as shown in Figure 19-20. On a screen 4.0 m from the record, a series of red dots 21 mm apart are visible. Level 2 60. An anti-reflective coating, n 1.2, is applied to a lens. If the thickness of the coating is 125 nm, what is (are) the color(s) of light for which complete destructive interference will occur? Hint: Assume the lens is made out of glass. Because nfilm nair, there is a phase inversion on the first reflection. Because nlens 1.52 nfilm, there is a phase inversion on the second reflection. For destructive interference: Laser 21 mm 21 mm Screen ■ Record a. How many ridges are there in a centimeter along the radius of the record? d sin sin 2dn film m 12 (2)(125 nm)(1.2) m 12 1 4.0 m Figure 19-20 (Not to scale) d 1 2 nfilm 2d m 1 m 2 (3.0102 nm) For m 0 1 1 (3.0102 nm) 2 6.0102 nm sin tan1 x L Physics: Principles and Problems The light is reddish-orange. For other values of m, the wavelength is shorter than that of light. Solutions Manual 409 Chapter 19 continued Level 3 61. Camera When a camera with a 50-mm lens is set at f/8, its aperture has an opening 6.25 mm in diameter. a. For light with 550 nm, what is the resolution of the lens? The film is 50.0 mm from the lens. 1.22Lobj xobj D 4 (1.22)(5.510 mm)(50.0 mm) 63. Apply Concepts Blue light of wavelength passes through a single slit of width w. A diffraction pattern appears on a screen. If you now replace the blue light with a green light of wavelength 1.5, to what width should you change the slit to get the original pattern back? The angle of diffraction depends on the ratio of slit width to wavelength. Thus, you would increase the width to 1.5w. 6.25 mm 5.4103 mm b. The owner of a camera needs to decide which film to buy for it. The expensive one, called fine-grained film, has 200 grains/mm. The less costly, coarsegrained film has only 50 grains/mm. If the owner wants a grain to be no smaller than the width of the central bright spot calculated in part a, which film should he purchase? Central bright band width 2x 10.7103 mm 1.22Lobj xobj D xobjD 1.22 between grains 5103 mm, so this film will work. 50 mm between grains 20103 mm, so this film won’t work. Thinking Critically page 538 62. Apply Concepts Yellow light falls on a diffraction grating. On a screen behind the grating, you see three spots: one at zero degrees, where there is no diffraction, and one each at 30° and 30°. You now add a blue light of equal intensity that is in the same direction as the yellow light. What pattern of spots will you now see on the screen? A green spot at 0°, yellow spots at 30° and 30°, and two blue spots slightly closer in. (8.0103 m)(1.80 m) (1.22)(5.2510 m) 7 2.2104 m 22 km b. Can you actually see a car’s headlights at the distance calculated in part a? Does diffraction limit your eyes’ sensing ability? Hypothesize as to what might be the limiting factors. No; a few hundred meters, not several kilometers, is the limit. Diffraction doesn’t limit the sensing ability of your eyes. More probable factors are the refractive effects of the atmosphere, like those that cause stars to twinkle, or the limitations of the retina and the optic area of the brain to separate two dim sources. Writing in Physics page 538 65. Research and describe Thomas Young’s contributions to physics. Evaluate the impact of his research on the scientific thought about the nature of light. Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1 The 50 grains/mm has Solutions Manual a. Based upon Rayleigh’s criterion, how far away can the human eye distinguish the two headlights at night? Hint: Assume a wavelength of 525 nm. Lobj 1 The 200 grains/mm film has 200 mm 410 64. Analyze and Conclude At night, the pupil of a human eye has an aperture diameter of 8.0 mm. The diameter is smaller in daylight. An automobile’s headlights are separated by 1.8 m. Chapter 19 continued Student answers will vary. Answers should include Young’s two-slit experiment that allowed him to precisely measure the wavelength of light. 66. Research and interpret the role of diffraction in medicine and astronomy. Describe at least two applications in each field. Student answers will vary. Answers could include diffraction in telescopes and microscopes, as well as spectroscopy. Cumulative Review page 538 67. How much work must be done to push a 0.5-m3 block of wood to the bottom of a 4-m-deep swimming pool? The density of wood is 500 kg/m3. (Chapter 13) The block would float, but to submerge it would require an extra force downward. 1.50 m b. If the speed of sound is 330 m/s, what is the frequency of the source? v f v 330 m/s 1.50 m f 220 Hz c. What is the speed of the airplane? The plane moves forward 0.50 m for every 1.50 m that the sound wave travels, so the plane’s speed is onethird the speed of sound, or 110 m/s. 70. A concave mirror has a 48.0-cm radius. A 2.0-cm-tall object is placed 12.0 cm from the mirror. Calculate the image position and image height. (Chapter 17) r 2 f 48.0 cm 2 W Fd F Fbuoyancy Fg 24.0 cm Fg Vg (500 kg/m3)(0.5 m3)(9.80 m/s2) 2450 N Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. a. What is the wavelength of the sound in still air? Fbuoyancy Vg (1000 kg/m3)(0.5 m3) (9.80 m/s2) 4900 N Work (4900 N 2450 N)(4 m) 10 kJ 68. What are the wavelengths of microwaves in an oven if their frequency is 2.4 GHz? (Chapter 14) c f 1 1 1 do di f d f do f o di (12.0 cm)(24.0 cm) 12.0 cm 24.0 cm 24.0 cm h ho d do m i i d h do i o hi (24.0 cm)(2.0 cm) 12.0 cm 4.0 cm 3.00108 m/s 2.410 Hz c 9 f 0.12 m 69. Sound wave crests that are emitted by an airplane are 1.00 m apart in front of the plane, and 2.00 m apart behind the plane. (Chapter 15) Physics: Principles and Problems 71. The focal length of a convex lens is 21.0 cm. A 2.00-cm-tall candle is located 7.50 cm from the lens. Use the thin-lens equation to calculate the image position and image height. (Chapter 18) 1 1 1 do di f Solutions Manual 411 Chapter 19 continued x d f do f min Use (1) , (2) vsubstance f, (7.50 cm)(21.0 cm) 7.50 cm 21.0 cm and (3) nsubstance . v Combine (2) and (3). o di w L c 11.7 cm f substance substance vacuum vacuum nsubstance f (4), h d m i i ho do because the frequency remains constant as the light crosses a boundary. diho hi do Rewrite (1) in terms of a substance in the space between the slits and the screen. (11.7 cm)(2.00 cm) 7.50 cm 3.11 cm x w min substance (5) L Combine (4) and (5) and solve for x. Challenge Problem page 526 You have several unknown substances and wish to use a single-slit diffraction apparatus to determine what each one is. You decide to place a sample of an unknown substance in the region between the slit and the screen and use the data that you obtain to determine the identity of each substance by calculating its index of refraction. vacuum xminw L L nsubstancew vacuum xmin 2. If the source you used had a wavelength of 634 nm, the slit width was 0.10 mm, the distance from the slit to the screen was 1.15 m, and you immersed the apparatus in water (nsubstance 1.33), then what would you expect the width of the center band to be? L nsubstanced vacuum x x1 9 (63410 m)(1.15 m) 3 (1.33)(0.1010 m) 5.510–3 m L Incoming light 1. Come up with a general formula for the index of refraction of an unknown substance in terms of the wavelength of the light, vacuum, the width of the slit, w, the distance from the slit to the screen, L, and the distance between the central bright band and the first dark band, x1. 412 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Unknown substance nsubstance CHAPTER 20 Static Electricity Section Review 20.1 Electric Charge pages 541–545 page 545 1. Charged Objects After a comb is rubbed on a wool sweater, it is able to pick up small pieces of paper. Why does the comb lose that ability after a few minutes? The comb loses its charge to its surroundings and becomes neutral once again. 2. Types of Charge In the experiments described earlier in this section, how could you find out which strip of tape, B or T, is positively charged? Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Bring a positively charged glass rod near the two strips of tape. The one that is repelled by the rod is positive. 3. Types of Charge A pith ball is a small sphere made of a light material, such as plastic foam, often coated with a layer of graphite or aluminum paint. How could you determine whether a pith ball that is suspended from an insulating thread is neutral, is charged positively, or is charged negatively? Bring an object of known charge, such as a negatively charged hard rubber rod, near the pith ball. If the pith ball is repelled, it has the same charge as the rod. If it is attracted, it may have the opposite charge or be neutral. To find out which, bring a positively charged glass rod near the pith ball. If they repel, the pith ball is positive; if they attract, the pith ball must be neutral. 4. Charge Separation A rubber rod can be charged negatively when it is rubbed with wool. What happens to the charge of the wool? Why? Physics: Principles and Problems The wool becomes positively charged because it gives up electrons to the rubber rod. 5. Conservation of Charge An apple contains trillions of charged particles. Why don’t two apples repel each other when they are brought together? Each apple contains equal numbers of positive and negative charges, so they appear neutral to each other. 6. Charging a Conductor Suppose you hang a long metal rod from silk threads so that the rod is isolated. You then touch a charged glass rod to one end of the metal rod. Describe the charges on the metal rod. The glass rod attracts electrons off the metal rod, so the metal becomes positively charged. The charge is distributed uniformly along the rod. 7. Charging by Friction You can charge a rubber rod negatively by rubbing it with wool. What happens when you rub a copper rod with wool? Because the copper is a conductor, it remains neutral as long as it is in contact with your hand. 8. Critical Thinking It once was proposed that electric charge is a type of fluid that flows from objects with an excess of the fluid to objects with a deficit. Why is the current two-charge model better than the single-fluid model? The two-charge model can better explain the phenomena of attraction and repulsion. It also explains how objects can become charged when they are rubbed together. The single-fluid model indicated that the charge should be equalized on objects that are in contact with each other. Solutions Manual 413 Chapter 20 continued Practice Problems 20.2 Electric Force pages 546–553 page 552 9. A negative charge of 2.0104 C and a positive charge of 8.0104 C are separated by 0.30 m. What is the force between the two charges? KqAqB (9.0109 Nm2/C2)(2.0104 C)(8.0104 C) F 2 2 (0.30 m) dAB 1.6104 N 10. A negative charge of 6.0106 C exerts an attractive force of 65 N on a second charge that is 0.050 m away. What is the magnitude of the second charge? Kq q A B F 2 dAB Fd 2 KqA 2 (65 N)(0.050 m) AB qB 9 2 2 6 (9.010 Nm /C )(6.010 C) 3.0106 C 11. The charge on B in Example Problem 1 is replaced by a charge of 3.00 C. Diagram the new situation and find the net force on A. Magnitudes of all forces remain the same. The direction changes to 42° above the x axis, or 138°. (2.0106 C)(3.6106 C) (0.60 m) q q A B 9 2 2 FB on A K 0.18 N 2 2 (9.010 Nm /C ) dAB direction: toward the right (2.0106 C)(4.0106 C) (0.80 m) q q A C 9 2 2 FC on A K 0.1125 N 2 2 (9.010 Nm /C ) dAC direction: toward the left Fnet FB on A FC on A (0.18 N) (0.1125 N) 0.068 N toward the right 13. Determine the net force on sphere B in the previous problem. q q A B FA on B K 2 dAB q q A B FC on B K 2 dAB Fnet FC on B FA on B q q q q dBC dAB B C A B K 2 K 2 414 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 12. Sphere A is located at the origin and has a charge of 2.0106 C. Sphere B is located at 0.60 m on the x-axis and has a charge of 3.6106 C. Sphere C is located at 0.80 m on the x-axis and has a charge of 4.0106 C. Determine the net force on sphere A. Chapter 20 continued (9.0109 Nm2/C2) (3.6106 C)(4.0106 C) (0.20 m)2 (9.0109 Nm2/C2) (2.0106 C)(3.6106 C) (0.60 m)2 3.1 N toward the right Section Review 20.2 Electric Force pages 546–553 page 553 14. Force and Charge How are electric force and charge related? Describe the force when the charges are like charges and the force when the charges are opposite charges. Electric force is directly related to each charge. It is repulsive between like charges and attractive between opposite charges. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 15. Force and Distance How are electric force and distance related? How would the force change if the distance between two charges were tripled? Electric force is inversely related to the square of the distance between charges. If the distance is tripled, the force will be one-ninth as great. 16. Electroscopes When an electroscope is charged, the leaves rise to a certain angle and remain at that angle. Why do they not rise farther? As the leaves move farther apart, the electric force between them decreases until it is balanced by the gravitational force pulling down on the leaves. 17. Charging an Electroscope Explain how to charge an electroscope positively using a. a positive rod. Touch the positive rod to the electroscope. Negative charges will move to the rod, leaving the electroscope positively charged. Physics: Principles and Problems b. a negative rod. Bring the negative rod near, but not touching the electroscope. Touch (ground) the electroscope with your finger, allowing electrons to be repelled off of the electroscope into your finger. Remove your finger and then remove the rod. 18. Attraction of Neutral Objects What two properties explain why a neutral object is attracted to both positively and negatively charged objects? Charge separation, caused by the attraction of opposite charges and the repulsion of like charges, moves the opposite charges in the neutral body closer to the charged object and the like charges farther away. The inverse relation between force and distance means that the nearer, opposite charges will attract more than the more distant, like charges will repel. The overall effect is attraction. 19. Charging by Induction In an electroscope being charged by induction, what happens when the charging rod is moved away before the ground is removed from the knob? Charge that had been pushed into the ground by the rod would return to the electroscope from the ground, leaving the electroscope neutral. 20. Electric Forces Two charged spheres are held a distance, r, apart. One sphere has a charge of 3C, and the other sphere has a charge of 9C. Compare the force of the 3C sphere on the 9C sphere with the force of the 9C sphere on the 3C sphere. The forces are equal in magnitude and opposite in direction. 21. Critical Thinking Suppose that you are testing Coulomb’s law using a small, positively charged plastic sphere and a large, positively charged metal sphere. According to Coulomb’s law, the force depends on 1/r 2, where r is the distance between the centers of the spheres. As the spheres get Solutions Manual 415 Chapter 20 continued close together, the force is smaller than expected from Coulomb’s law. Explain. Some charges on the metal sphere will be repelled to the opposite side from the plastic sphere, making the effective distance between the charges greater than the distance between the spheres’ centers. Chapter Assessment Concept Mapping page 558 22. Complete the concept map below using the following terms: conduction, distance, elementary charge. Electric Force Coulomb’s law charge elementary charge charging distance coulomb conduction induction page 558 23. If you comb your hair on a dry day, the comb can become positively charged. Can your hair remain neutral? Explain. (20.1) No. By conservation of charge, your hair must become negatively charged. 24. List some insulators and conductors. (20.1) Student answers will vary but may include dry air, wood, plastic, glass, cloth, and deionized water as insulators; and metals, tap water, and your body as conductors. 25. What property makes metal a good conductor and rubber a good insulator? (20.1) Metals contain free electrons; rubber has bound electrons. 416 Solutions Manual They have been charged by contact as they rub against other clothes, and thus, are attracted to clothing that is neutral or has an opposite charge. 27. Compact Discs If you wipe a compact disc with a clean cloth, why does the CD then attract dust? (20.2) Rubbing the CD charges it. Neutral particles, such as dust, are attracted to a charged object. 28. Coins The combined charge of all electrons in a nickel is hundreds of thousands of coulombs. Does this imply anything about the net charge on the coin? Explain. (20.2) No. Net charge is the difference between positive and negative charges. The coin still can have a net charge of zero. 29. How does the distance between two charges impact the force between them? If the distance is decreased while the charges remain the same, what happens to the force? (20.2) Electric force is inversely proportional to the distance squared. As distance decreases and charges remain the same, the force increases as the square of the distance. 30. Explain how to charge a conductor negatively if you have only a positively charged rod. (20.2) Bring the conductor close to, but not touching, the rod. Ground the conductor in the presence of the charged rod; then, remove the ground before removing the charged rod. The conductor will have a net negative charge. Applying Concepts page 558 31. How does the charge of an electron differ from the charge of a proton? How are they similar? Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Mastering Concepts 26. Laundry Why do socks taken from a clothes dryer sometimes cling to other clothes? (20.2) Chapter 20 continued The charge of the proton is exactly the same size as the electron, but has the opposite sign. 32. Using a charged rod and an electroscope, how can you find whether or not an object is a conductor? Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Use a known insulator to hold one end of the object against the electroscope. Touch the other end with the charged rod. If the electroscope indicates a charge, the object is a conductor. Law of Universal Gravitation Coulomb’s Law m m r q q r B F GA 2 mA B F KA 2 mB r qA qB r ■ Figure 20-13 (Not to scale) 33. A charged rod is brought near a pile of tiny plastic spheres. Some of the spheres are attracted to the rod, but as soon as they touch the rod, they are flung off in different directions. Explain. Similar: inverse-square dependence on distance, force proportional to product of two masses or two charges; different: only one sign of mass, so gravitational force is always attractive; two signs of charge, so electric force can be either attractive or repulsive. The natural spheres are initially attracted to the charged rod, but they acquire the same charge as the rod when they touch it. As a result, they are repelled from the rod. 37. The constant, K, in Coulomb’s equation is much larger than the constant, G, in the universal gravitation equation. Of what significance is this? 34. Lightning Lightning usually occurs when a negative charge in a cloud is transported to Earth. If Earth is neutral, what provides the attractive force that pulls the electrons toward Earth? The charge in the cloud repels electrons on Earth, causing a charge separation by induction. The side of Earth closest to the cloud is positive, resulting in an attractive force. 35. Explain what happens to the leaves of a positively charged electroscope when rods with the following charges are brought close to, but not touching, the electroscope. a. positive The leaves will move farther apart. b. negative The leaves will drop slightly. 36. As shown in Figure 20-13, Coulomb’s law and Newton’s law of universal gravitation appear to be similar. In what ways are the electric and gravitational forces similar? How are they different? Physics: Principles and Problems The electric force is much larger than the gravitational force. 38. The text describes Coulomb’s method for charging two spheres, A and B, so that the charge on B was exactly half the charge on A. Suggest a way that Coulomb could have placed a charge on sphere B that was exactly one-third the charge on sphere A. After changing spheres A and B equally, sphere B is touched to two other equally sized balls that are touching each other. The charge on B will be divided equally among all three balls, leaving one-third the total charge on it. 39. Coulomb measured the deflection of sphere A when spheres A and B had equal charges and were a distance, r, apart. He then made the charge on B one-third the charge on A. How far apart would the two spheres then have had to be for A to have had the same deflection that it had before? To have the same force with one-third the charge, the distance would have to be decreased such that d 2 1/3, or 0.58 times as far apart. Solutions Manual 417 Chapter 20 continued 40. Two charged bodies exert a force of 0.145 N on each other. If they are moved so that they are one-fourth as far apart, what force is exerted? 1 d F 2 and F 1 , so F 16 times the original force. 2 14 41. Electric forces between charges are enormous in comparison to gravitational forces. Yet, we normally do not sense electric forces between us and our surroundings, while we do sense gravitational interactions with Earth. Explain. Gravitational forces only can be attractive. Electric forces can be either attractive or repulsive, and we can sense only their vector sums, which are generally small. The gravitational attraction to Earth is larger and more noticeable because of Earth’s large mass. Mastering Problems 20.2 Electric Force page 559 Level 1 42. Two charges, qA and qB, are separated by a distance, r, and exert a force, F, on each other. Analyze Coulomb’s law and identify what new force would exist under the following conditions. a. qA is doubled 2qA, then new force 2F b. qA and qB are cut in half 1 1 1 1 1 qA and qB, then new force F F 2 2 2 2 4 1 F 3d, then new force 2 F (3) 9 d. r is cut in half 1 d, then new force 2 F 2 12 3 F 4F 4 e. qA is tripled and r is doubled (3)F 3 3qA and 2d, then new force 2 F (2) 4 43. Lightning A strong lightning bolt transfers about 25 C to Earth. How many electrons are transferred? (25 C) 1.61020 electrons 19 1 electron 1.6010 C 44. Atoms Two electrons in an atom are separated by 1.51010 m, the typical size of an atom. What is the electric force between them? Kq qB (9.0109 Nm2/C2)(1.601019 C)(1.601019 C) F A 10 2 2 d (1.510 m) 1.0108 N, away from each other 418 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. c. r is tripled Chapter 20 continued 45. A positive and a negative charge, each of magnitude 2.5105 C, are separated by a distance of 15 cm. Find the force on each of the particles. Kq q (9.0109 Nm2/C2)(2.5105 C)(2.5105 C) (1.510 m) A B F 1 2 2 d 2.5102 N, toward the other charge 46. A force of 2.4102 N exists between a positive charge of 8.0105 C and a positive charge of 3.0105 C. What distance separates the charges? Kq q A B F 2 d d KqAqB F (9.0109 Nm2/C2)(8.0105 C)(3.0105 C) 2.4102 N 0.30 m 47. Two identical positive charges exert a repulsive force of 6.4109 N when separated by a distance of 3.81010 m. Calculate the charge of each. Kq q Kq 2 d B F A 2 2 d Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. q 3.210 (6.4109 N)(3.81010 m)2 9.0109 Nm2/C2 Fd 2 K 19 Level 2 48. A positive charge of 3.0 C is pulled on by two negative charges. As shown in Figure 20-14, one 2.0 C negative charge, 2.0 C, is 0.050 m to the west, and the other, 4.0 C, is 0.030 m to the east. What total force is exerted on the positive charge? C 4.0 C 3.0 C 0.050 m (9.0109 Nm2/C2)(3.0106 C)(2.0106 C) F1 (0.050 m)2 0.030 m ■ Figure 20-14 22 N west (9.0109 Nm2/C2)(3.0106 C)(4.0106 C) (0.030 m) F2 2 120 N east Fnet F2 F1 (1.2102 N) (2.2101 N) 98 N, east 49. Figure 20-15 shows two positively charged spheres, one with three times the charge of the other. The spheres are 16 cm apart, and the force between them is 0.28 N. What are the charges on the two spheres? q q q 3q d d q 3q 16 cm ■ Figure 20-15 A A B F K A 2 K 2 Physics: Principles and Problems Solutions Manual 419 Chapter 20 continued qA Fd 2 3K (0.28 N)(0.16 m)2 5.2107 C 3(9.0109 Nm2/C2) qB 3qA 1.5106 C 50. Charge in a Coin How many coulombs of charge are on the electrons in a nickel? Use the following method to find the answer. a. Find the number of atoms in a nickel. A nickel has a mass of about 5 g. A nickel is 75 percent Cu and 25 percent Ni, so each mole of the coin’s atoms will have a mass of about 62 g. 5g A coin is 0.08 mole. 62 g Thus, it has (0.08)(6.021023) 51022 atoms b. Find the number of electrons in the coin. On average, each atom has 28.75 electrons. (51022 atoms)(28.75 electrons/atom) 11024 electrons c. Find the coulombs on the electrons. (1.61019 coulombs/electron)(11024 electrons) 2105 coulombs 51. Three particles are placed in a line. The left particle has a charge of 55 C, the middle one has a charge of 45 C, and the right one has a charge of 78 C. The middle particle is 72 cm from each of the others, as shown in Figure 20-16. 55 C 45 C 78 C 72 cm 72 cm ■ Figure 20-16 Let left be the negative direction Kqmql Kq q m r Fnet Fl (Fr) 2 2 d (9.0109 d Nm2/C2)(45106 C)(55106 C) 2 (0.72 m) (9.0109 Nm2/C2)(45106 C)(78106 C) (0.72 m)2 18 N, right b. Find the net force on the right particle. Kqlqr Kqmqr (2d) d Fnet Fl (Fm) 2 2 9 2 2 6 6 (9.010 Nm /C )(5510 C)(7810 C) 2 (2(0.72 m)) (9.0109 Nm2/C2)(45106 C)(78106 C) (0.72 m)2 42 N, left 420 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. a. Find the net force on the middle particle. Chapter 20 continued Mixed Review page 559 Level 1 52. A small metal sphere with charge 1.2105 C is touched to an identical neutral sphere and then placed 0.15 m from the second sphere. What is the electric force between the two spheres? The two spheres share the charge equally, so (6.0106 C)(6.0106 C) (0.15 m) q q A B 9 2 2 F K 14 N 2 2 (9.010 Nm C ) d 53. Atoms What is the electric force between an electron and a proton placed 5.31011 m apart, the approximate radius of a hydrogen atom? (1.601019 C)(1.601019 C) (5.310 m) q q A B 9 2 2 F K 8.2108 N 11 2 2 (9.010 Nm C ) d 54. A small sphere of charge 2.4 C experiences a force of 0.36 N when a second sphere of unknown charge is placed 5.5 cm from it. What is the charge of the second sphere? q q A B F K 2 d 2 (0.36 N)(5.5102 m)2 (9.010 Nm C )(2.410 Fd 5.0108 C qB 6 9 2 2 KqA C) 55. Two identically charged spheres placed 12 cm apart have an electric force of 0.28 N between them. What is the charge of each sphere? q q A B F K , where qA qB d2 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. q (0.28 N)(1.2101 m)2 (9.0109 Nm2C2) Fd 2 K 6.7107 C 56. In an experiment using Coulomb’s apparatus, a sphere with a charge of 3.6108 C is 1.4 cm from a second sphere of unknown charge. The force between the spheres is 2.7102 N. What is the charge of the second sphere? q q A B F K 2 d 2 Fd qB KqA (2.7102 N)(1.4102 m)2 (9.0109 Nm2C2)(3.6108 C) 1.6108 C 57. The force between a proton and an electron is 3.51010 N. What is the distance between these two particles? q q A B F K 2 d qAqB K F2 (1.6010 C)(1.610 C) (9.0 10 N m /C ) 8.110 3.510 N d 9 2 Physics: Principles and Problems 2 19 19 10 10 m Solutions Manual 421 Chapter 20 continued Thinking Critically page 560 58. Apply Concepts Calculate the ratio of the electric force to the gravitational force between the electron and the proton in a hydrogen atom. qeqp K d2 Fe Fg m m p G e 2 d Kq q p e Gmemp 9 2 19 2 2 (9.010 Nm /C )(1.6010 C) 2.31039 11 31 27 2 2 (6.6710 Nm /kg )(9.1110 kg)(1.6710 kg) 59. Analyze and Conclude Sphere A, with a charge of 64 C, is positioned at the origin. A second sphere, B, with a charge of 16 C, is placed at 11.00 m on the x-axis. a. Where must a third sphere, C, of charge 12 C be placed so there is no net force on it? The attractive and repulsive forces must cancel, so q q q q AC BC A C B C FAC K K FBC, so d 2 d 2 qA qB , and 16dAC2 64dBC2, or dAC2 dBC2 dAC2 4dBC2, so dAC 2dBC The third sphere must be placed at 2.00 m on the x-axis so it is twice as far from the first sphere as from the second sphere. b. If the third sphere had a charge of 6 C, where should it be placed? c. If the third sphere had a charge of 12 C, where should it be placed? As in part b, the magnitude and sign of the third charge, qc, do not matter. 60. Three charged spheres are located at the positions shown in Figure 20-17. Find the total force on sphere B. y 4.5 C 8.2 C A 4.0 cm x B 3.0 cm 6.0 C C ■ Figure 20-17 F1 FA on B Kq q 9 2 2 6 6 (9.010 Nm /C )(4.510 C)(8.210 C) A B 2 2 d 422 Solutions Manual (0.040 m) Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. The third charge, qc, cancels from the equation, so it doesn’t matter what its magnitude or sign is. Chapter 20 continued 208 N 208 N, to left The distance between the other two charges is m)2 (0.03 0 m)2 0.050 m (0.040 1 tan1 0.030 m 0.040 m 37° below the negative x-axis, or 217° from the positive x-axis. F2 FC on B Kq q (9.0109 Nm2/C)(8.2106 C)(6.0106 C) (0.050 m) C B 2 2 d 177 N 177 N at 217° from the positive x-axis (37° 180°) The components of F2 are: F2x F2 cos (177 N)(cos 217°) 142 N 142 N to the left F2y F2 sin (177 N)(sin 217°) 106 N 106 N down The components of the net (resultant) force are: Fnet, x 208 N 142 N 350 N 350 N, to left Fnet, y 106 N, down (350 N )2 (1 06 N)2 366 N 3.7102 N Fnet 2 tan1 106 N 350 N 17° below the negative x-axis Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Fnet 3.7102 N at 197° from the positive x-axis 61. The two pith balls in Figure 20-18 each have a mass of 1.0 g and an equal charge. One pith ball is suspended by an insulating thread. The other is brought to 3.0 cm from the suspended ball. The suspended ball is now hanging with the thread forming an angle of 30.0° with the vertical. The ball is in equilibrium with FE, Fg, and FT. Calculate each of the following. 30.0° FE 3.0 cm ■ Figure 20-18 a. Fg on the suspended ball Fg mg (1.0103 kg)(9.80 m/s2) 9.8103 N b. FE F E tan 30.0° Fg Physics: Principles and Problems Solutions Manual 423 Chapter 20 continued FE mg tan 30.0° (1.0103 kg)(9.80 m/s2)(tan 30.0°) 5.7103 N c. the charge on the balls KqAqB F 2 d Kq 2 d F 2 q 2.410 (5.7103 N)(3.0102 m2) (9.0109 Nm2/C2) Fd 2 K 8 C 62. Two charges, qA and qB, are at rest near a positive test charge, qT, of 7.2 C. The first charge, qA, is a positive charge of 3.6 C located 2.5 cm away from qT at 35°; qB is a negative charge of 6.6 C located 6.8 cm away at 125°. a. Determine the magnitude of each of the forces acting on qT. KqTqA 9 2 6 2 6 (9.010 Nm /C )(7.210 C)(3.610 C) FA 2 2 (0.025 m) d 3.7102 N, away (toward qT) KqTqA 9 2 6 2 6 (9.010 Nm /C )(7.210 C)(6.610 C) FB 2 2 (0.068 m) d 92 N, toward (away from qT) b. Sketch a force diagram. qA qB FA 125° FB 92 N 35° qT c. Graphically determine the resultant force acting on qT. 21° 35° F 3.8102 N FA 3.7102 N FB 92 N 424 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 3.7102 N Chapter 20 continued Writing in Physics page 560 63. History of Science Research several devices that were used in the seventeenth and eighteenth centuries to study static electricity. Examples that you might consider include the Leyden jar and the Wimshurst machine. Discuss how they were constructed and how they worked. Student answers will vary, but should include information such as the following. The Leyden jar, invented in the mid-1740s, was the earliest capacitor. It was used throughout the eighteenth and nineteenth centuries to store charges for electricity-related experiments and demonstrations. The Wimshurst machine was a device used in the nineteenth and early twentieth centuries to produce and discharge static charges. Wimshurst machines, which were replaced by the Van de Graaff generator in the twentieth century, used Leyden jars to store the charges prior to discharge. Measure the length and period of the pendulum, and use the equation for the period of a pendulum to solve for g. 66. A submarine that is moving 12.0 m/s sends a sonar ping of frequency 1.50103 Hz toward a seamount that is directly in front of the submarine. It receives the echo 1.800 s later. (Chapter 15) a. How far is the submarine from the seamount? d vt (1533 m/s)(0.900 s) 1380 m b. What is the frequency of the sonar wave that strikes the seamount? v vd 3 fd fs v v (1.5010 Hz) s 1533 m/s 0.0 m/s 1533 m/s 12.0 m/s 1510 Hz c. What is the frequency of the echo received by the submarine? v vd fd fs (1510 Hz) vv Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. s 64. In Chapter 13, you learned that forces exist between water molecules that cause water to be denser as a liquid between 0°C and 4°C than as a solid at 0°C. These forces are electrostatic in nature. Research electrostatic intermolecular forces, such as van der Waals forces and dipole-dipole forces, and describe their effects on matter. Answers will vary, but students should describe the interactions between positive and negative charges at the molecular level. Students should note that the strength of these forces accounts for differences in melting and boiling points and for the unusual behavior of water between 0°C and 4°C. Cumulative Review page 560 65. Explain how a pendulum can be used to determine the acceleration of gravity. (Chapter 14) Physics: Principles and Problems 1533 m/s (12.0 m/s) 1533 m/s 0.0 m/s 1520 Hz 67. Security Mirror A security mirror is used to produce an image that is three-fourths the size of an object and is located 12.0 cm behind the mirror. What is the focal length of the mirror? (Chapter 17) d m i do d do i m (12.0 cm) 3 4 16.0 cm 1 1 1 do di f d d do di i f o Solutions Manual 425 Chapter 20 continued (16.0 cm)(12.0 cm) Challenge Problem 48.0 cm page 552 As shown in the figure below, two spheres of equal mass, m, and equal positive charge, q, are a distance, r, apart. 16.0 cm (12.0 cm) 68. A 2.00-cm-tall object is located 20.0 cm away from a diverging lens with a focal length of 24.0 cm. What are the image position, height, and orientation? Is this a real or a virtual image? (Chapter 18) 1 1 1 do di f dof di do f (20.0 cm)(24.0 cm) 20.0 cm (24.0 cm) 10.9 cm d do h ho m i i d h do i o h i r mass m charge q mass m charge q ■ Figure 20-11 1. Derive an expression for the charge, q, that must be on each sphere so that the spheres are in equilibrium; that is, so that the attractive and repulsive forces between them are balanced. The attractive force is gravitation, and the repulsive force is electrostatic, so their expressions may be set equal. mAmB qAqB d d Fg G K 2 2 Fe (10.9 cm)(2.00 cm) 20.0 cm 1.09 cm This is a virtual image that is upright in orientation, relative to the object. The number of centimeters per slit is the slit separation distance, d. 1 slit 11,500 slits/cm d d 8.70105 cm d sin d sin1 527109 m 8.7010 m sin1 3 0.00347° Gm2 Kq 2, and K G q m m (6.671011 Nm2/kg2) (9.0109 Nm2/C2) (8.611011 C/kg)m 2. If the distance between the spheres is doubled, how will that affect the expression for the value of q that you determined in the previous problem? Explain. The distance does not affect the value of q because both forces are inversely related to the square of the distance, and the distance cancels out of the expression. 3. If the mass of each sphere is 1.50 kg, determine the charge on each sphere needed to maintain the equilibrium. q (8.611011 C/kg)(1.50 kg) 1.291010 C 426 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 69. Spectrometer A spectrometer contains a grating of 11,500 slits/cm. Find the angle at which light of wavelength 527 nm has a first-order bright band. (Chapter 19) The masses and charges are equal, and the distance cancels, so CHAPTER 21 Electric Fields Practice Problems 21.1 Creating and Measuring Electric Fields pages 563–568 page 565 1. A positive test charge of 5.0106 C is in an electric field that exerts a force of 2.0104 N on it. What is the magnitude of the electric field at the location of the test charge? F 2.0104 N E 4.0101 N/C 6 q 5.010 C 2. A negative charge of 2.0108 C experiences a force of 0.060 N to the right in an electric field. What are the field’s magnitude and direction at that location? F 0.060 N E 3.0106 N/C 8 q 2.010 C Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. directed to the left 3. A positive charge of 3.0107 C is located in a field of 27 N/C directed toward the south. What is the force acting on the charge? F E q F Eq (27 N/C)(3.0107 C) 8.1106 N 2.1103 N is placed 4. A pith ball weighing in a downward electric field of 6.5104 N/C. What charge (magnitude and sign) must be placed on the pith ball so that the electric force acting on it will suspend it against the force of gravity? The electric force and the gravitational force algebraically sum to zero because the ball is suspended, i.e. not in motion: Fg Fe 0, so Fe Fg F q e E F E F E 3 2.110 N g e q 4 6.510 N/C Physics: Principles and Problems 3.2108 C The electric force is upward (opposite the field), so the charge is negative. 5. You are probing the electric field of a charge of unknown magnitude and sign. You first map the field with a 1.0106-C test charge, then repeat your work with a 2.0106-C test charge. a. Would you measure the same forces at the same place with the two test charges? Explain. No. The force on the 2.0-C charge would be twice that on the 1.0-C charge. b. Would you find the same field strengths? Explain. Yes. You would divide the force by the strength of the test charge, so the results would be the same. page 566 6. What is the magnitude of the electric field strength at a position that is 1.2 m from a point charge of 4.2106 C? q d F E K 2 q (4.2106 C) (1.2 m) (9109 Nm2/C2) 2 2.6104 N/C 7. What is the magnitude of the electric field strength at a distance twice as far from the point charge in problem 6? Because the field strength varies as the square of the distance from the point charge, the new field strength will be one-fourth of the old field strength, or 6.5103 N/C. 8. What is the electric field at a position that is 1.6 m east of a point charge of 7.2106 C? Solutions Manual 427 Chapter 21 continued q d F q E K 2 (7.2106 C) (1.6 m) (9.0109 Nm2/C2) 2 2.5104 N/C The direction of the field is east (away from the positive point charge). 9. The electric field that is 0.25 m from a small sphere is 450 N/C toward the sphere. What is the charge on the sphere? q d F q E K 2 2 Ed q K 2 (450 N/C)(0.25 m) 9 C 9 2 2 3.110 (9.010 Nm /C ) The charge is negative, because the field is directed toward it. 10. How far from a point charge of 2.4106 C must a test charge be placed to measure a field of 360 N/C? q d F E K 2 q d (9.0109 Nm2/C2)(2.4106 C) 360 N/C 7.7 m Section Review 21.1 Creating and Measuring Electric Fields pages 563–568 page 568 11. Measuring Electric Fields Suppose you are asked to measure the electric field in space. How do you detect the field at a point? How do you determine the magnitude of the field? How do you choose the magnitude of the test charge? What do you do next? 428 Solutions Manual To determine the magnitude of the field, divide the magnitude of the force on the test charge by the magnitude of the test charge. The magnitude of the test charge must be chosen so that it is very small compared to the magnitudes of the charges producing the field. The next thing you should do is measure the direction of the force on the test charge. The direction of the field is the same as the direction of the force if the test charge is positive; otherwise, it is in the opposite direction. 12. Field Strength and Direction A positive test charge of magnitude 2.40108 C experiences a force of 1.50103 N toward the east. What is the electric field at the position of the test charge? 3 F 1.5010 N east E 8 q 2.4010 C 6.25104 N/C east 13. Field Lines In Figure 21-4, can you tell which charges are positive and which are negative? What would you add to complete the field lines? No. The field lines must have arrowheads indicating their directions, from positive to negative charges. 14. Field Versus Force How does the electric field, E, at the test charge differ from the force, F, on it? The field is a property of that region of space, and does not depend on the test charge used to measure it. The force depends on the magnitude and sign of the test charge. 15. Critical Thinking Suppose the top charge in Figure 21-2c is a test charge measuring the field resulting from the two negative charges. Is it small enough to produce an accurate measure? Explain. Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Kq E To detect a field at a point, place a test charge at that point and determine if there is a force on it. Chapter 21 continued No. This charge is large enough to distort the field produced by the other charges with its own field. Compare with Figure 21-4b. page 572 21. What work is done when 3.0 C is moved through an electric potential difference of 1.5 V? W qV (3.0 C)(1.5 V) 4.5 J Practice Problems 21.2 Applications of Electric Fields pages 569–579 page 571 16. The electric field intensity between two large, charged, parallel metal plates is 6000 N/C. The plates are 0.05 m apart. What is the electric potential difference between them? V Ed (6000 N/C)(0.05 m) 300 J/C 3102 V 17. A voltmeter reads 400 V across two charged, parallel plates that are 0.020 m apart. What is the electric field between them? V Ed 400 V V E 2104 N/C Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. d 0.020 m V Ed (2.50103 N/C)(0.200 m) 19. When a potential difference of 125 V is applied to two parallel plates, the field between them is 4.25103 N/C. How far apart are the plates? V Ed 125 V V d 2.94102 m 3 4.2510 N/C 20. A potential difference of 275 V is applied to two parallel plates that are 0.35 cm apart. What is the electric field between the plates? 275 V V E 7.9104 N/C 3 3.510 1.7107 J 23. An electron in a television picture tube passes through a potential difference of 18,000 V. How much work is done on the electron as it passes through that potential difference? W qV (1.601019 C)(1.8104 V) 2.91015 J 24. If the potential difference in problem 18 is between two parallel plates that are 2.4 cm apart, what is the magnitude of the electric field between them? d 0.024 m 25. The electric field in a particle-accelerator machine is 4.5105 N/C. How much work is done to move a proton 25 cm through that field? W qV qEd 5.00102 V d W qV (1.44106 C)(12 V) V 5.00102 V E 2.1104 N/C 18. What electric potential difference is applied to two metal plates that are 0.200 m apart if the electric field between them is 2.50103 N/C? E 22. A 12-V car battery can store 1.44106 C when it is fully charged. How much work can be done by this battery before it needs recharging? m Physics: Principles and Problems (1.601019 C)(4.5105 N/C)(0.25 m) 1.81014 J page 574 26. A drop is falling in a Millikan oil-drop apparatus with no electric field. What forces are acting on the oil drop, regardless of its acceleration? If the drop is falling at a constant velocity, describe the forces acting on it. Gravitational force (weight) downward, friction force of air upward. The two forces are equal in magnitude if the drop falls at constant velocity. Solutions Manual 429 Chapter 21 continued 27. An oil drop weighs 1.91015 N. It is suspended in an electric field of 6.0103 N/C. What is the charge on the drop? How many excess electrons does it carry? q C Fg Eq F E V , so the smaller capacitor has the larger potential difference. 15 N 1.910 g q 3 4 6.010 N/C 3.510 C V 1.1102 V 6 3.510 3.21019 C 19 q 3.210 C 2 # electrons 19 qe 1.6010 C 28. An oil drop carries one excess electron and weighs 6.41015 N. What electric field strength is required to suspend the drop so it is motionless? 15 F 6.410 N 4.0104 N/C E 19 q 1.6010 C 29. A positively charged oil drop weighing 1.21014 N is suspended between parallel plates separated by 0.64 cm. The potential difference between the plates is 240 V. What is the charge on the drop? How many electrons is the drop missing? 240 V V E 3.8104 N/C d 6.4103 m 14 F 1.210 N q 3.21019 C 4 3.810 N/C 19 q C 3.410 # electrons 2 19 qe 1.6010 C page 578 30. A 27-F capacitor has an electric potential difference of 45 V across it. What is the charge on the capacitor? q CV (27106 F)(45 V) 1.2103 C 31. Both a 3.3-F and a 6.8-F capacitor are connected across a 24-V electric potential difference. Which capacitor has a greater charge? What is it? q CV, so the larger capacitor has a greater charge. F 33. A 2.2-F capacitor first is charged so that the electric potential difference is 6.0 V. How much additional charge is needed to increase the electric potential difference to 15.0 V? q CV q C(V2 V1) (2.2106 F)(15.0 V 6.0 V) 2.0105 C 34. When a charge of 2.5105 C is added to a capacitor, the potential difference increases from 12.0 V to 14.5 V. What is the capacitance of the capacitor? 5 q 2.510 C C V2 V1 14.5 V 12.0 V 1.0105 F Section Review 21.2 Applications of Electric Fields pages 569–579 page 579 35. Potential Difference What is the difference between electric potential energy and electric potential difference? Electric potential energy changes when work is done to move a charge in an electric field. It depends on the amount of charge involved. Electric potential difference is the work done per unit charge to move a charge in an electric field. It is independent of the amount of charge that is moved. q (6.8106 F)(24 V) 1.6104 C 430 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. F E q E 32. The same two capacitors as in problem 31 are each charged to 3.5104 C. Which has the larger electric potential difference across it? What is it? Chapter 21 continued 36. Electric Field and Potential Difference Show that a volt per meter is the same as a newton per coulomb. V/m J/Cm Nm/Cm N/C 37. Millikan Experiment When the charge on an oil drop suspended in a Millikan apparatus is changed, the drop begins to fall. How should the potential difference on the plates be changed to bring the drop back into balance? The potential difference should be increased. 38. Charge and Potential Difference In problem 37, if changing the potential difference has no effect on the falling drop, what does this tell you about the new charge on the drop? The charges on the metal dome produce no field inside the dome. The charges from the belt are transferred immediately to the outside of the dome, where they have no effect on new charges arriving at point B. Chapter Assessment Concept Mapping page 584 42. Complete the concept map below using the following terms: capacitance, field strength, J/C, work. Electric Field 39. Capacitance How much charge is stored on a 0.47-F capacitor when a potential difference of 12 V is applied to it? test charge field strength q CV (4.7107 F)(12 V) Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. a. the potentials of the two spheres The spheres will have equal potentials. b. the charges on the two spheres The large sphere will have more charge than the small sphere, but they will be the same sign. The sign of the charge will depend on which sphere began with more charge. 41. Critical Thinking Referring back to Figure 21-3a, explain how charge continues to build up on the metal dome of a Van de Graaff generator. In particular, why isn’t charge repelled back onto the belt at point B? Physics: Principles and Problems potential difference capacitance 5.6106 C 40. Charge Sharing If a large, positively charged, conducting sphere is touched by a small, negatively charged, conducting sphere, what can be said about the following? work force The drop is electrically neutral (no electron excess or deficiency). J/C N/C C/V Mastering Concepts page 584 43. What are the two properties that a test charge must have? (21.1) The test charge must be small in magnitude relative to the magnitudes of the charges producing the field and be positive. 44. How is the direction of an electric field defined? (21.1) The direction of an electric field is the direction of the force on a positive charge placed in the field. This would be away from a positive object and toward a negative object. Solutions Manual 431 Chapter 21 continued 45. What are electric field lines? (21.1) lines of force 48. In Figure 21-15, where do the electric field lines leave the positive charge end? (21.1) 46. How is the strength of an electric field indicated with electric field lines? (21.1) The closer together the electric field lines are, the stronger the electric field. 47. Draw some of the electric field lines between each of the following. (21.1) a. two like charges of equal magnitude ■ Figure 21-15 They end on distant negative charges somewhere beyond the edges of the diagram. b. two unlike charges of equal magnitude 49. What SI unit is used to measure electric potential energy? What SI unit is used to measure electric potential difference? (21.2) electric potential energy: joule; electric potential: volt c. a positive charge and a negative charge having twice the magnitude of the positive charge 50. Define volt in terms of the change in potential energy of a charge moving in an electric field. (21.2) 51. Why does a charged object lose its charge when it is touched to the ground? (21.2) The charge is shared with the surface of Earth, which is an extremely large object. d. two oppositely charged parallel plates 52. A charged rubber rod that is placed on a table maintains its charge for some time. Why is the charged rod not discharged immediately? (21.2) The table is an insulator, or at least a very poor conductor. 53. A metal box is charged. Compare the concentration of charge at the corners of the box to the charge concentration on the sides of the box. (21.2) The concentration of charge is greater at the corners. 432 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. A volt is the change in electric potential energy, PE, resulting from moving a unit test charge, q, a distance, d, of 1 m in an electric field, E, of 1 N/C. V PE/q Ed Chapter 21 continued 54. Computers Delicate parts in electronic equipment, such as those pictured in Figure 21-16, are contained within a metal box inside a plastic case. Why? (21.2) E x A ■ Applying Concepts pages 584–585 55. What happens to the strength of an electric field when the charge on the test charge is halved? Nothing. Because the force on the test charge also would be halved, the ratio F/q and the electric field would remain the same. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. z ■ Figure 21-16 56. Does it require more energy or less energy to move a constant positive charge through an increasing electric field? Energy is proportional to the force, and the force is proportional to the electric field. Therefore, it requires more energy. 57. What will happen to the electric potential energy of a charged particle in an electric field when the particle is released and free to move? The electric potential energy of the particle will be converted into kinetic energy of the particle. 58. Figure 21-17 shows three spheres with charges of equal magnitude, with their signs as shown. Spheres y and z are held in place, but sphere x is free to move. Initially, sphere x is equidistant from spheres y and z. Choose the path that sphere x will begin to follow. Assume that no other forces are acting on the spheres. C B y The metal box shields the parts from external electric fields, which do not exist inside a hollow conductor. Physics: Principles and Problems D Figure 21-17 Sphere x will follow path C. It will experience forces shown by D and B. The vector sum is C. 59. What is the unit of electric potential difference in terms of m, kg, s, and C? V J/C Nm/C (kgm/s2)(m/C) kgm2/s2C 60. What do the electric field lines look like when the electric field has the same strength at all points in a region? They are parallel, equally spaced lines. 61. Millikan Oil-Drop Experiment When doing a Millikan oil-drop experiment, it is best to work with drops that have small charges. Therefore, when the electric field is turned on, should you try to find drops that are moving rapidly or slowly? Explain. Slowly. The larger the charge, the stronger the force, and thus, the larger the (terminal) velocity. 62. Two oil drops are held motionless in a Millikan oil-drop experiment. a. Can you be sure that the charges are the same? No. Their masses could be different. b. The ratios of which two properties of the oil drops have to be equal? charge to mass ratio, q/m (or m/q) 63. José and Sue are standing on an insulating platform and holding hands when they are given a charge, as in Figure 21-18. José is larger than Sue. Who has the larger amount of charge, or do they both have the same amount? Solutions Manual 433 Chapter 21 continued 0.30 N F E 3.0104 N/C 5 q 1.010 C in the same direction as the force 68. A test charge experiences a force of 0.30 N on it when it is placed in an electric field intensity of 4.5105 N/C. What is the magnitude of the charge? F q E ■ Figure 21-18 José has a larger surface area, so he will have a larger amount of charge. 64. Which has a larger capacitance, an aluminum sphere with a 1-cm diameter or one with a 10-cm diameter? The 10-cm diameter sphere has a larger capacitance because the charges can be farther apart, reducing potential rise as it is charged. 65. How can you store different amounts of charge in a capacitor? Change the voltage across the capacitor. Mastering Problems E 4.510 N/C 69. The electric field in the atmosphere is about 150 N/C downward. a. What is the direction of the force on a negatively charged particle? upward b. Find the electric force on an electron with charge 1.61019 C. F E q F qE (1.61019 C)(150 N/C) 2.41017 N F 2.41017 N directed upward c. Compare the force in part b with the force of gravity on the same electron (mass 9.11031 kg). F mg (9.11031 kg)(9.80 m/s2) 8.91030 N F 8.91030 N (downward), more than one trillion times smaller 70. Carefully sketch each of the following. a. the electric field produced by a 1.0-C charge F E q 8 F 1.410 N q 2.8105 C 4 E 5.010 N/C 67. A positive charge of 1.0105 C, shown in Figure 21-19, experiences a force of 0.30 N when it is located at a certain point. What is the electric field intensity at that point? 434 Solutions Manual 0.30 N 1.0 C 1.0105 C ■ Figure 21-19 Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 21.1 Creating and Measuring Electric Fields pages 585–586 The charge of an electron is 1.601019 C. Level 1 66. What charge exists on a test charge that experiences a force of 1.4108 N at a point where the electric field intensity is 5.0104 N/C? 0.30 N F q 6.7107 C 5 Chapter 21 continued b. the electric field resulting from a 2.0-C charge (Make the number of field lines proportional to the change in charge.) b. Charge Z now has a small positive charge on it. Draw an arrow representing the force on it. F 2.0 C Fy Fx Z 71. A positive test charge of 6.0106 C is placed in an electric field of 50.0-N/C intensity, as in Figure 21-20. What is the strength of the force exerted on the test charge? 73. In a television picture tube, electrons are accelerated by an electric field having a value of 1.00105 NC. a. Find the force on an electron. F E q q 6.0106 C F Eq (1.601019 C)(1.00105 N/C) 1.601014 N b. If the field is constant, find the acceleration of the electron (mass 9.111031 kg). E 50.0 N/C ■ Figure 21-20 F ma Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. F E q F qE (6.0106 C)(50.0 N/C) 3.0104 F 1.601014 N a 31 m N Level 2 72. Charges X, Y, and Z all are equidistant from each other. X has a 1.0-C charge, Y has a 2.0-C charge, and Z has a small negative charge. a. Draw an arrow representing the force on charge Z. 9.1110 1.761016 kg m/s2 74. What is the electric field strength 20.0 cm from a point charge of 8.0107 C? Kqq d F E , and F 2 q Kq d so E 2 9 2 2 7 (9.010 Nm /C )(8.010 C) 2 (0.200 m) Z 1.8105 N/C Fx Fy F X Physics: Principles and Problems Y Solutions Manual 435 Chapter 21 continued 75. The nucleus of a lead atom has a charge of 82 protons. a. What are the direction and magnitude of the electric field at 1.01010 m from the nucleus? 7.21017 J KQ 1 KqQ F E 2 2 d d (9.0109 Nm2/C2)(1.311017 C) (1.010 m) 10 2 1.21013 N/C, outward F Eq (1.21013 N/C)(1.601019 C) 1.9106 N, toward the nucleus ■ Figure 21-21 W 120 J V 5.0101 V q 2.4 C 77. How much work is done to transfer 0.15 C of charge through an electric potential difference of 9.0 V? W V q W qV (0.15 C)(9.0 V) 1.4 J W V W 1200 J q 1.0102 C V 12 V 80. The electric field intensity between two charged plates is 1.5103 N/C. The plates are 0.060 m apart. What is the electric potential difference, in volts, between the plates? V Ed (1.5103 N/C)(0.060 m) 9.0101 V 81. A voltmeter indicates that the electric potential difference between two plates is 70.0 V. The plates are 0.020 m apart. What electric field intensity exists between them? V Ed 70.0 V V E 3500 V/m d 0.020 m 3500 N/C 82. A capacitor that is connected to a 45.0-V source contains 90.0 C of charge. What is the capacitor’s capacitance? q 90.0106 C C 2.00 µF V 45.0 V 83. What electric potential difference exists across a 5.4-F capacitor that has a charge of 8.1104 C? q C V q 8.1104 C V 6 C 5.410 F 1.5102 V 436 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 21.2 Applications of Electric Fields pages 586–587 Level 1 76. If 120 J of work is performed to move 2.4 C of charge from the positive plate to the negative plate shown in Figure 21-21, what potential difference exists between the plates? 2.4 C 79. A 12-V battery does 1200 J of work transferring charge. How much charge is transferred? q b. What are the direction and magnitude of the force exerted on an electron located at this distance? q (1.601019 C)(450 V) 1.311017 C q W V W qV Q (82 protons) (1.601019 C/proton) q 78. An electron is moved through an electric potential difference of 450 V. How much work is done on the electron? Chapter 21 continued 84. The oil drop shown in Figure 21-22 is negatively charged and weighs 4.51015 N. The drop is suspended in an electric field intensity of 5.6103 N/C. 87. Photoflash The energy stored in a capacitor with capacitance C, and an electric potential difference, V, is represented by 1 W CV 2. One application of this is in 2 the electronic photoflash of a strobe light, like the one in Figure 21-24. In such a unit, a capacitor of 10.0 F is charged to 3.0102 V. Find the energy stored. 8.51015 N ■ Figure 21-22 a. What is the charge on the drop? F E q F 4.51015 N q 3 E 5.610 N/C ■ 8.01019 C 1 W CV 2 b. How many excess electrons does it carry? (8.01019 C) 19 1 electron 1.6010 C 5 electrons 85. What is the charge on a 15.0-pF capacitor when it is connected across a 45.0-V source? V Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 2 1 (10.0106 F)(3.0102 V)2 2 0.45 J 88. Suppose it took 25 s to charge the capacitor in problem 87. a. Find the average power required to charge the capacitor in this time. q C q CV (15.01012 F)(45.0 V) 6.751010 C Level 2 86. A force of 0.065 N is required to move a charge of 37 C a distance of 25 cm in a uniform electric field, as in Figure 21-23. What is the size of the electric potential difference between the two points? 37 C 0.053 N 25 cm ■ Figure 21-24 Figure 21-23 W 0.45 J P 1.8102 W t 25 s b. When this capacitor is discharged through the strobe lamp, it transfers all its energy in 1.0104 s. Find the power delivered to the lamp. 0.45 J W P 4.5103 W 4 t 1.010 s c. How is such a large amount of power possible? Power is inversely proportional to the time. The shorter the time for a given amount of energy to be expended, the greater the power. W Fd W Fd and V q q (0.065 N)(0.25 m) 6 3710 C 4.4102 V Physics: Principles and Problems Solutions Manual 437 Chapter 21 continued 89. Lasers Lasers are used to try to produce controlled fusion reactions. These lasers require brief pulses of energy that are stored in large rooms filled with capacitors. One such room has a capacitance of 61103 F charged to a potential difference of 10.0 kV. between them is 300 V, what is the capacitance of the system? 0.060 C 1 2 a. Given that W CV 2, find the energy V 300 V ■ 2 V 1 (61103 F)(1.00104 V)2 2 3.1106 J b. The capacitors are discharged in 10 ns (1.0108 s). What power is produced? 3.1106 J W 3.11014 W P t 1.0108 s c. If the capacitors are charged by a generator with a power capacity of 1.0 kW, how many seconds will be required to charge the capacitors? 1.010 W Figure 21-25 q C V q CV (4.7108 F)(120 V) 5.6106 C 5.6 C page 587 Level 1 90. How much work does it take to move 0.25 C between two parallel plates that are 0.40 cm apart if the field between the plates is 6400 N/C? V E W qV qEd (2.5107 C)(6400 N/C)(4.0103 m) d 120 V 4.8104 V/m 3 2.510 95. An electron is placed between the plates of the capacitor in Problem 93 above, as in Figure 21-26. What force is exerted on that electron? 6.4106 J 91. How much charge is stored on a 0.22-F parallel plate capacitor if the plates are 1.2 cm apart and the electric field between them is 2400 N/C? q CV CEd m) 6.3 C m V 120 V C 0.047 F 0.25 cm ■ Figure 21-26 92. Two identical small spheres, 25 cm apart, carry equal but opposite charges of 0.060 C, as in Figure 21-25. If the potential difference 438 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. V Ed N/C)(1.2102 300 V 93. The plates of a 0.047 F capacitor are 0.25 cm apart and are charged to a potential difference of 120 V. How much charge is stored on the capacitor? Mixed Review F)(2400 94. What is the strength of the electric field between the plates of the capacitor in Problem 93 above? 3.1106 J W t 3.1103 s 3 q 6.0108 C C 21010 F 1 W CV 2 (2.2107 25 cm stored in the capacitors. P 0.060 C Chapter 21 continued 1 2 F E W area bh q F Eq (4.8104 V/m)(1.61019 C) 7.71015 N 96. How much work would it take to move an additional 0.010 C between the plates at 120 V in Problem 93? W V q W qV (1.0108 C)(120 V) 1.2106 J Level 2 97. The graph in Figure 21-27 represents the charge stored in a capacitor as the charging potential increases. What does the slope of the line represent? Charge Stored on Capacitor (25 V)(12.5 C) 1 2 160 J 101. The work found in Problem 100 above is not equal to qV. Why not? The potential difference is not constant as the capacitor is charged. Therefore, the area under the graph must be used to find work, not just simple multiplication. 102. Graph the electric field strength near a positive point charge as a function of distance from it. E Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. q (C) 15 d 10 103. Where is the field of a point charge equal to zero? 5 0 10 20 30 V (V) ■ Figure 21-27 capacitance of the capacitor 98. What is the capacitance of the capacitor represented by Figure 21-27? C slope 0.50 F 99. What does the area under the graph line in Figure 21-27 represent? work done to charge the capacitor 100. How much work is required to charge the capacitor in problem 98 to a potential difference of 25 V? Physics: Principles and Problems Nowhere, or at an infinite distance from the point charge. 104. What is the electric field strength at a distance of zero meters from a point charge? Is there such a thing as a true point charge? Infinite. No. Thinking Critically pages 587–588 105. Apply Concepts Although a lightning rod is designed to carry charge safely to the ground, its primary purpose is to prevent lightning from striking in the first place. How does it do that? The sharp point on the end of the rod leaks charge into the atmosphere before it has the chance to build up enough potential difference to cause a lightning strike. Solutions Manual 439 Chapter 21 continued 106. Analyze and Conclude In an early set of experiments in 1911, Millikan observed that the following measured charges could appear on a single oil drop. What value of elementary charge can be deduced from these data? a. 6.5631019 C f. 18.081019 C b. 8.2041019 C g. 19.711019 C c. 11.501019 C h. 22.891019 C d. 13.131019 C i. 26.131019 C e. 16.481019 C 1.631019 C. Subtracting adjacent values, b a, c b, d c, etc. yields 1.6411019 C, 3.301019 C, 1.631019 C, 3.351019 C, 1.601019 C, 1.631019 C, 3.181019 C, 3.241019 C. There are two numbers, approximately 1.6311019 C and 3.21019 C, that are common. Averaging each similar group produces one charge of 1.631019 C and one charge of 3.271019 C (which is two times 1.6411019 C). Dividing 1.631019 C into each piece of data yields nearly whole-number quotients, indicating it is the value of an elementary charge. 3.00106 C 5.00106 C A B 0.800 m Draw the spheres and vectors representing the fields due to each charge at the given point. ■ Figure 21-28 EA ER EB A 0.800 m B Now do the math: FA KqA (9.0109 Nm2/C2)(3.00106 C) EA 4.22104 N/C 2 2 q d (0.800 m) FB KqB (9.0109 Nm2/C2)(5.00106 C) EB 7.03104 N/C 2 2 q d (0.800 m) EAx EA cos 60.0° (4.22104 N/C)(cos 60.0°) 2.11104 N/C EAy EA sin 60.0° (4.22104 N/C)(sin 60.0°) 3.65104 N/C 440 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 107. Analyze and Conclude Two small spheres, A and B, lie on the x-axis, as in Figure 21-28. Sphere A has a charge of 3.00106 C. Sphere B is 0.800 m to the right of sphere A and has a charge of 5.00106 C. Find the magnitude and direction of the electric field strength at a point above the x-axis that would form the apex of an equilateral triangle with spheres A and B. Chapter 21 continued EBx EB cos (60.0°) (7.03104 N/C)(cos 60.0°) 3.52104 N/C EBy EB sin (60.0°) (7.03104 N/C)(sin 60.0°) 6.09104 N/C Ex EAx E Bx (2.11104 N/C) (3.52104 N/C) 5.63104 N/C Ey EAy E By (3.65104 N/C) (6.09104 N/C) 2.44104 N/C Ex2 Ey2 6.14104 N/C ER Ey tan Ex Ey tan1 Ex 2.44104 N/C 5.6310 N/C tan1 4 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 23.4° 108. Analyze and Conclude In an ink-jet printer, drops of ink are given a certain amount of charge before they move between two large, parallel plates. The purpose of the plates is to deflect the charges so that they are stopped by a gutter and do not reach the paper. This is shown in Figure 21-29. The plates are 1.5-cm long and have an electric field of E 1.2106 N/C between them. Drops with a mass m 0.10 ng, and a charge q 1.01016 C, are moving horizontally at a speed, v 15 m/s, parallel to the plates. What is the vertical displacement of the drops when they leave the plates? To answer this question, complete the following steps. 1.5 cm q m Gutter v E 1.2106 N/C ■ Figure 21-29 a. What is the vertical force on the drops? F Eq (1.01016 C)(1.2106 N/C) 1.21010 N b. What is their vertical acceleration? F 1.21010 N 1.2103 m/s2 a 13 m 1.010 kg c. How long are they between the plates? L 1.5102 m t 1.0103 s 15 m/s v d. How far are they displaced? 1 y at 2 2 1 (1.2103 m/s2)(1.0103 s)2 2 6.0104 m 0.60 mm Physics: Principles and Problems Solutions Manual 441 Chapter 21 continued 109. Apply Concepts Suppose the Moon had a net negative charge equal to q, and Earth had a net positive charge equal to 10q. What value of q would yield the same magnitude of force that you now attribute to gravity? Equate the expressions for gravitational force and Coulombic force between Earth and the Moon: Gm m d 10Kq 2 d Kq q d E M E M F 2 2 2 where q is the net negative charge of the Moon and qE, the net positive charge of Earth, is 10q. Solve symbolically before substituting numbers. q 10K GmEmM (6.671011 Nm2/kg2)(6.001024 kg)(7.311022 kg) (10)(9.0109 Nm2/C2) 1.81013C Writing in Physics page 588 110. Choose the name of an electric unit, such as coulomb, volt, or farad, and research the life and work of the scientist for whom it was named. Write a brief essay on this person and include a discussion of the work that justified the honor of having a unit named for him. Student answers will vary. Some examples of scientists they could choose are Volta, Coulomb, Ohm, and Ampère. Cumulative Review a. How long does it take light to travel the distance to the mountain and back? (35 km/trip)(2 trips)(1000 m/1 km)/ 3.00108 m/s 2.3104 s b. Assume that Michelson used a rotating octagon with a mirror on each face of the octagon. Also assume that the light reflects from one mirror, travels to the other mountain, reflects off of a fixed mirror on that mountain, and returns to the rotating mirrors. If the rotating mirror has advanced so that when the light returns, it reflects off of the next mirror in the rotation, how fast is the mirror rotating? 4 2.310 s (8 mirrors/rev) 1.8103 s/rev T 1 mirror 1 1 f 5.6102 rev/s 3 T 1.810 s/rev Note that if students carry extra digits from part a to prevent rounding errors, they will get an answer of 5.4102 rev/s. 442 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. page 588 111. Michelson measured the speed of light by sending a beam of light to a mirror on a mountain 35 km away. (Chapter 16) Chapter 21 continued c. If each mirror has a mass of 1.0101 g and rotates in a circle with an average radius of 1.0101 cm, what is the approximate centripetal force needed to hold the mirror while is it rotating? h you h mountain d you d mountain h youd mountain h mountain d you (2 m)(50,000 m) Fc 4 2mf 2r 50 m 4 2(0.010 kg)(5.6102 (0.10 m) 1.210 4 N Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Note that the answer should be 1.2104 N regardless of whether the students use 5.4102 rev/s or 5.6104 rev/s for f. 112. Mountain Scene You can see an image of a distant mountain in a smooth lake just as you can see a mountain biker next to the lake because light from each strikes the surface of the lake at about the same angle of incidence and is reflected to your eyes. If the lake is about 100 m in diameter, the reflection of the top of the mountain is about in the middle of the lake, the mountain is about 50 km away from the lake, and you are about 2 m tall, then approximately how high above the lake does the top of the mountain reach? (Chapter 17) Since the angle of incidence of the light from the top of the mountain is equal to its angle of reflection from the lake, you and the reflection of the top of the mountain form a triangle that is similar to a triangle formed by the mountain and the top of its reflection in the lake. Your height makes up one side, h you 2 m and the top of the mountain is halfway across the lake, d you 50 m. The mountain is a distance d mountain 50,000 m from its reflection. Find h mountain by equating the ratios of the sides of the two similar triangles. Physics: Principles and Problems 2000 m rev/s)2 113. A converging lens has a focal length of 38.0 cm. If it is placed 60.0 cm from an object, at what distance from the lens will the image be? (Chapter 18) 1 1 1 di do f d f do f o di (60.0 cm)(38.0 cm) 60.0 cm 38.0 cm 104 cm The image is 104 cm from the lens. 114. A force, F, is measured between two charges, Q and q, separated by a distance, r. What would the new force be for each of the following? (Chapter 20) a. r is tripled F/9 b. Q is tripled 3F c. both r and Q are tripled F/3 d. both r and Q are doubled F/2 e. all three, r, Q, and q, are tripled F Solutions Manual 443 Chapter 21 continued Challenge Problem page 579 The plates of a capacitor attract each other because they carry opposite charges. A capacitor consisting of two parallel plates that are separated by a distance, d, has capacitance, C. 1. Derive an expression for the force between the two plates when the capacitor has charge, q. Combine the following equations: q V F Eq, E , and V C d d q V F Eq q d C q2 q d Cd 2. What charge must be stored on a 22-F capacitor to have a force of 2.0 N between the plates if they are separated by 1.5 mm? q2 Cd F so q FCd (2.0 N )(2.2 105 F )(1.5 103 m) 2.6104 C Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 444 Solutions Manual Physics: Principles and Problems CHAPTER 22 Current Electricity Practice Problems 22.1 Current and Circuits pages 591–600 page 594 1. The current through a lightbulb connected across the terminals of a 125-V outlet is 0.50 A. At what rate does the bulb convert electric energy to light? (Assume 100 percent efficiency.) P IV (0.50 A)(125 V) 63 J/s 63 W 2. A car battery causes a current of 2.0 A through a lamp and produces 12 V across it. What is the power used by the lamp? P IV (2.0 A)(12 V) 24 W 3. What is the current through a 75-W lightbulb that is connected to a 125-V outlet? 6. An automobile panel lamp with a resistance of 33 is placed across a 12-V battery. What is the current through the circuit? 12 V V I 0.36 A 33 R 7. A motor with an operating resistance of 32 is connected to a voltage source. The current in the circuit is 3.8 A. What is the voltage of the source? V IR (3.8 A)(32 ) 1.2102 V 8. A sensor uses 2.0104 A of current when it is operated by a 3.0-V battery. What is the resistance of the sensor circuit? V 3.0 V R 1.5104 4 P IV I P 75 W I 0.60 A V 125 V Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. page 598 For all problems, assume that the battery voltage and lamp resistances are constant, no matter what current is present. 4. The current through the starter motor of a car is 210 A. If the battery maintains 12 V across the motor, how much electric energy is delivered to the starter in 10.0 s? P IV and E Pt Thus, E IVt (210 A)(12 V)(10.0 s) 2.5104 2.010 A 9. A lamp draws a current of 0.50 A when it is connected to a 120-V source. a. What is the resistance of the lamp? 120 V V R 2.4102 I 0.50 A b. What is the power consumption of the lamp? P IV (0.50 A)(120 V) 6.0101 W J 10. A 75-W lamp is connected to 125 V. 5. A flashlight bulb is rated at 0.90 W. If the lightbulb drops 3.0 V, how much current goes through it? P IV P 0.90 W I 0.30 A V 3.0 V Physics: Principles and Problems a. What is the current through the lamp? 75 W P I 0.60 A V 125 V b. What is the resistance of the lamp? V 125 V R 2.1102 I 0.60 A Solutions Manual 445 Chapter 22 continued 11. A resistor is added to the lamp in the previous problem to reduce the current to half of its original value. A 85 mA a. What is the potential difference across the lamp? 53 4.5 V I The new value of the current is V 4.5 V R 53 0.60 A 0.30 A 2 I V IR (0.30 A)(2.1102 ) 6.3101 V b. How much resistance was added to the circuit? The total resistance of the circuit is now V 125 V Rtotal 4.2102 I 0.30 A 0.085 A 14. Add a voltmeter to measure the potential difference across the resistors in problems 12 and 13 and repeat the problems. Both circuits will take the following form. A V I Therefore, Rres Rtotal RIamp 4.2102 2.1102 2.1102 c. How much power is now dissipated in the lamp? P IV (0.30 A)(6.3101 V) 19 W 15. Draw a circuit using a battery, a lamp, a potentiometer to adjust the lamp’s brightness, and an on-off switch. Lamp Battery Switch Potentiometer A 4.80 A 12.5 60.0 V 16. Repeat the previous problem, adding an ammeter and a voltmeter across the lamp. I V 60.0 V I 4.80 A R 12.5 13. Draw a series-circuit diagram showing a 4.5-V battery, a resistor, and an ammeter that reads 85 mA. Determine the resistance and label the resistor. Choose a direction for the conventional current and indicate the positive terminal of the battery. 446 Solutions Manual V A Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. page 600 12. Draw a circuit diagram to include a 60.0-V battery, an ammeter, and a resistance of 12.5 in series. Indicate the ammeter reading and the direction of the current. Because the ammeter resistance is assumed zero, the voltmeter readings will be 60.0 V for Practice Problem 12 and 4.5 V for Practice Problem 13. Chapter 22 continued Section Review P1 V 2/R1 (12 V)2/12 12 W 22.1 Current and Circuits pages 591–600 P P2 P1 16 W 12 W 4.0 W page 600 17. Schematic Draw a schematic diagram of a circuit that contains a battery and a lightbulb. Make sure the lightbulb will light in this circuit. P2 V 2/R2 (12 V)2/9.0 16 W 4.0 W increase 21. Energy A circuit converts 2.2103 J of energy when it is operated for 3.0 min. Determine the amount of energy it will convert when it is operated for 1 h. 3 2.210 J E (60.0 min) 3.0 min 4.4104 J 22. Critical Thinking We say that power is “dissipated” in a resistor. To dissipate is to use, to waste, or to squander. What is “used” when charge flows through a resistor? 18. Resistance Joe states that because R V/I, if he increases the voltage, the resistance will increase. Is Joe correct? Explain. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. No, resistance depends on the device. When V increases, so will I. 19. Resistance You want to measure the resistance of a long piece of wire. Show how you would construct a circuit with a battery, a voltmeter, an ammeter, and the wire to be tested to make the measurement. Specify what you would measure and how you would compute the resistance. The potential energy of the charges decreases as they flow through the resistor. This decrease in potential energy is used to produce heat in the resistor. Practice Problems 22.2 Using Electric Energy pages 601–605 page 603 23. A 15- electric heater operates on a 120-V outlet. a. What is the current through the heater? A V 120 V I 8.0 A R V 15 b. How much energy is used by the heater in 30.0 s? E I 2Rt (8.0 A)2(15 )(30.0 s) 2.9104 J Measure the current through the wire and the potential difference across it. Divide the potential difference by the current to obtain the wire resistance. 20. Power A circuit has 12 of resistance and is connected to a 12-V battery. Determine the change in power if the resistance decreases to 9.0 . c. How much thermal energy is liberated in this time? 2.9104 J, because all electric energy is converted to thermal energy. 24. A 39- resistor is connected across a 45-V battery. a. What is the current in the circuit? V 45 V I 1.2 A R Physics: Principles and Problems 39 Solutions Manual 447 Chapter 22 continued b. How much energy is used by the resistor in 5.0 min? 2 V E t R 2 (45 V) (5.0 min)(60 s/min) (39 ) 1.6104 J 25. A 100.0-W lightbulb is 22 percent efficient. This means that 22 percent of the electric energy is converted to light energy. a. How many joules does the lightbulb convert into light each minute it is in operation? E Pt (0.22)(100.0 J/s)(1.0 min) (60 s/min) 1.3103 J b. How many joules of thermal energy does the lightbulb produce each minute? E Pt 4.7103 J t 2 For a given amount of energy, doubling the voltage will divide the time by 2. 2.2 h t 1.1 h 2 page 605 28. An electric space heater draws 15.0 A from a 120-V source. It is operated, on the average, for 5.0 h each day. a. How much power does the heater use? P IV (15.0 A)(120 V) 1800 W 1.8 kW b. How much energy in kWh does it consume in 30 days? a. If 220 V are applied across it, what is the current through the stove element? 220 V V I 2.0101 A 11 b. How much energy does the element convert to thermal energy in 30.0 s? E I 2Rt (2.0101 A)2(11 )(30.0 s) 1.3105 J c. The element is used to heat a kettle containing 1.20 kg of water. Assume that 65 percent of the heat is absorbed by the water. What is the water’s increase in temperature during the 30.0 s? Q mCT with Q 0.65E 0.65E (0.65)(1.3105 J) T mC (1.20 kg)(4180 J/kg°C) 17°C Solutions Manual 270 kWh c. At $0.12 per kWh, how much does it cost to operate the heater for 30 days? Cost ($0.12/kWh)(270 kWh) $32.40 29. A digital clock has a resistance of 12,000 and is plugged into a 115-V outlet. a. How much current does it draw? V 115 V I 9.6103 A R 12,000 b. How much power does it use? P VI (115 V)(9.6103 A) 1.1 W c. If the owner of the clock pays $0.12 per kWh, how much does it cost to operate the clock for 30 days? Cost (1.1103 kWh)($0.12/kWh) (30 days)(24 h/day) $0.10 30. An automotive battery can deliver 55 A at 12 V for 1.0 h and requires 1.3 times as much energy for recharge due to its lessthan-perfect efficiency. How long will it Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 26. The resistance of an electric stove element at operating temperature is 11 . 448 E IVt I(2V ) E Pt (1.8 kW)(5.0 h/day)(30 days) (0.78)(100.0 J/s)(1.0 min) (60.0 s/min) R 27. A 120-V water heater takes 2.2 h to heat a given volume of water to a certain temperature. How long would a 240-V unit operating with the same current take to accomplish the same task? Chapter 22 continued take to charge the battery using a current of 7.5 A? Assume that the charging voltage is the same as the discharging voltage. Echarge (1.3)IVt (1.3)(55 A)(12 V)(1.0 h) 858 Wh 858 Wh E t 9.5 h IV (7.5 A)(12 V) 31. Rework the previous problem by assuming that the battery requires the application of 14 V when it is recharging. Echarge (1.3)IVt (1.3)(55 A)(12 V)(1.0 h) 858 Wh E 858 Wh t 8.2 h IV (7.5 A)(14 V) Section Review Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 22.2 Using Electric Energy pages 601–605 page 605 32. Energy A car engine drives a generator, which produces and stores electric charge in the car’s battery. The headlamps use the electric charge stored in the car battery. List the forms of energy in these three operations. Mechanical energy from the engine converted to electric energy in the generator; electric energy stored as chemical energy in the battery; chemical energy converted to electric energy in the battery and distributed to the headlamps; electric energy converted to light and thermal energy in headlamps. V 22/R (0.5V1)2/R P1 0.25 P2 V 12/R V 12 35. Efficiency Evaluate the impact of research to improve power transmission lines on society and the environment. Research to improve power transmission lines would benefit society in cost of electricity. Also, if less power was lost during transmission, less coal and other power-producing resources would have to be used, which would improve the quality of our environment. 36. Voltage Why would an electric range and an electric hot-water heater be connected to a 240-V circuit rather than a 120-V circuit? For the same power, at twice the voltage, the current would be halved. The I 2R loss in the circuit wiring would be dramatically reduced because it is proportional to the square of the current. 37. Critical Thinking When demand for electric power is high, power companies sometimes reduce the voltage, thereby producing a “brown-out.” What is being saved? Power, not energy; most devices will have to run longer. Chapter Assessment Concept Mapping page 610 38. Complete the concept map using the following terms: watt, current, resistance. Electricity 33. Resistance A hair dryer operating from 120 V has two settings, hot and warm. In which setting is the resistance likely to be smaller? Why? Hot draws more power, P IV, so the fixed voltage current is larger. Because I V/R the resistance is smaller. 34. Power Determine the power change in a circuit if the applied voltage is decreased by one-half. Physics: Principles and Problems rate of flow rate of conversion opposition to flow current power resistance ampere watt ohm Solutions Manual 449 Chapter 22 continued Mastering Concepts page 610 39. Define the unit of electric current in terms of fundamental MKS units. (22.1) 1 A 1 C/1 s b. Which device converts chemical energy to electric energy? 1 c. Which device turns the circuit on and off? 2 40. How should a voltmeter be connected in Figure 22-12 to measure the motor’s voltage? (22.1) d. Which device provides a way to adjust speed? 3 44. Describe the energy conversions that occur in each of the following devices. (22.1) 4 electric energy to heat and light a. an incandescent lightbulb 3 1 b. a clothes dryer electric energy to heat and kinetic energy c. a digital clock radio 2 ■ Figure 22-12 The positive voltmeter lead connects to the left-hand motor lead, and the negative voltmeter lead connects to the right-hand motor lead. Break the circuit between the battery and the motor. Then connect the positive ammeter lead to the positive side of the break (the side connected to the positive battery terminal) and the negative ammeter lead to the negative side nearest the motor. 42. What is the direction of the conventional motor current in Figure 22-12? (22.1) from left to right through the motor 43. Refer to Figure 22-12 to answer the following questions. (22.1) 45. Which wire conducts electricity with the least resistance: one with a large cross-sectional diameter or one with a small cross-sectional diameter? (22.1) A larger-diameter wire has a smaller resistance because there are more electrons to carry the charge. 46. A simple circuit consists of a resistor, a battery, and connecting wires. (22.1) a. Draw a circuit schematic of this simple circuit. I I b. How must an ammeter be connected in a circuit for the current to be correctly read? The ammeter must be connected in series. a. Which device converts electric energy to mechanical energy? 4 450 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 41. How should an ammeter be connected in Figure 22-12 to measure the motor’s current? (22.1) electric energy to light and sound Chapter 22 continued Applying Concepts V I pages 610–611 51. Batteries When a battery is connected to a complete circuit, charges flow in the circuit almost instantaneously. Explain. I A c. How must a voltmeter be connected to a resistor for the potential difference across it to be read? The voltmeter must be connected in parallel. V 52. Explain why a cow experiences a mild shock when it touches an electric fence. I By touching the fence and the ground, the cow encounters a difference in potential and conducts current, thus receiving a shock. I A 47. Why do lightbulbs burn out more frequently just as they are switched on rather than while they are operating? (22.2) Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. The low resistance of the cold filament allows a high current initially and a greater change in temperature, subjecting the filament to greater stress. 48. If a battery is short-circuited by a heavy copper wire being connected from one terminal to the other, the temperature of the copper wire rises. Why does this happen? (22.2) The short circuit produces a high current, which causes more electrons to collide with the atoms of the wire. This raises the atoms’ kinetic energies and the temperature of the wire. 49. What electric quantities must be kept small to transmit electric energy economically over long distances? (22.2) the resistance of the wire and the current in the wire 50. Define the unit of power in terms of fundamental MKS units. (22.2) m2 s kg 2 J kgm2 C J W 3 s C s A potential difference is felt over the entire circuit as soon as the battery is connected to the circuit. The potential difference causes the charges to begin to flow. Note: The charges flow slowly compared to the change in potential difference. s Physics: Principles and Problems 53. Power Lines Why can birds perch on highvoltage lines without being injured? No potential difference exists along the wires, so there is no current through the birds’ bodies. 54. Describe two ways to increase the current in a circuit. Either increase the voltage or decrease the resistance. 55. Lightbulbs Two lightbulbs work on a 120-V circuit. One is 50 W and the other is 100 W. Which bulb has a higher resistance? Explain. 50-W bulb V2 V2 P , so R R P Therefore, the lower P is caused by a higher R. 56. If the voltage across a circuit is kept constant and the resistance is doubled, what effect does this have on the circuit’s current? If the resistance is doubled, the current is halved. s Solutions Manual 451 Chapter 22 continued 57. What is the effect on the current in a circuit if both the voltage and the resistance are doubled? Explain. Motor 1.5 A No effect. V IR, so I V /R, and if the voltage and the resistance both are doubled, the current will not change. 12 V 58. Ohm’s Law Sue finds a device that looks like a resistor. When she connects it to a 1.5-V battery, she measures only 45106 A, but when she uses a 3.0-V battery, she measures 25103 A. Does the device obey Ohm’s law? No. V IR, so R V/I. At 1.5 V, 1.5 V 4 R 6 3.310 4510 3.0 V At 3.0 V, R 120 3 2510 A A device that obeys Ohm’s law has a resistance that is independent of the applied voltage. ■ Figure 22-13 a. How much power is delivered to the motor? P VI (12 V)(1.5 A) 18 W b. How much energy is converted if the motor runs for 15 min? E Pt (18 W)(15 min)(60 s/min) 1.6104 J 62. Refer to Figure 22-14 to answer the following questions. A 59. If the ammeter in Figure 22-4a on page 596 were moved to the bottom of the diagram, would the ammeter have the same reading? Explain. 18 27 V V I 60. Two wires can be placed across the terminals of a 6.0-V battery. One has a high resistance, and the other has a low resistance. Which wire will produce thermal energy at a faster rate? Why? a. What should the ammeter reading be? the wire with the smaller resistance b. What should the voltmeter reading be? V2 P R Smaller R produces larger power P dissipated in the wire, which produces thermal energy at a faster rate. Mastering Problems 22.1 Current and Circuits pages 611–612 Level 1 61. A motor is connected to a 12-V battery, as shown in Figure 22-13. 452 Solutions Manual ■ Figure 22-14 27 V I V /R 1.5 A 18 27 V c. How much power is delivered to the resistor? P VI (27 V)(1.5 A) 41 W d. How much energy is delivered to the resistor per hour? E Pt (41 W)(3600 s) 1.5105 J Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Yes, because the current is the same everywhere in this circuit. Chapter 22 continued 63. Refer to Figure 22-15 to answer the following questions. E Pt (4.5 W)(3600 s) 1.6104 J A 65. Toasters The current through a toaster that is connected to a 120-V source is 8.0 A. What power is dissipated by the toaster? 9.0 27 V V P IV (8.0 A)(120 V) 9.6102 W I ■ d. How much energy is delivered to the resistor per hour? Figure 22-15 a. What should the ammeter reading be? 27 V I V /R 3.0 A 9.0 b. What should the voltmeter reading be? 27 V 67. A lamp draws 0.50 A from a 120-V generator. a. How much power is delivered? b. How much energy is converted in 5.0 min? P VI (27 V)(3.0 A) 81 W E Pt J (6.0101 W) 5.0 min 1 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 64. Refer to Figure 22-16 to answer the following questions. 18 V 18,000 J 1.8104 J a. How many joules of energy does the battery deliver to the motor each second? P IV (210 A)(12 V) 2500 J/s or 2.5103 J/s I ■ 60 s min 68. A 12-V automobile battery is connected to an electric starter motor. The current through the motor is 210 A. A 9.0 V E t The definition of power is P , so d. How much energy is delivered to the resistor per hour? E Pt (81 W)(3600 s) P IV (1.2 A)(120 V) 1.4102 W P IV (0.50 A)(120 V) 6.0101 W c. How much power is delivered to the resistor? 2.9105 66. Lightbulbs A current of 1.2 A is measured through a lightbulb when it is connected across a 120-V source. What power is dissipated by the bulb? Figure 22-16 a. What should the ammeter reading be? 9.0 V I V /R 0.50 A 18 b. What should the voltmeter reading be? 9.0 V c. How much power is delivered to the resistor? b. What power, in watts, does the motor use? P 2.5103 W 69. Dryers A 4200-W clothes dryer is connected to a 220-V circuit. How much current does the dryer draw? P IV P 4200 W I 19 A V 220 V P VI (9.0 V)(0.50 A) 4.5 W Physics: Principles and Problems Solutions Manual 453 Chapter 22 continued 70. Flashlights A flashlight bulb is connected across a 3.0-V potential difference. The current through the bulb is 1.5 A. Table 22-2 Voltage, V Current, I Resistance, R V/I a. What is the power rating of the bulb? P IV (1.5 A)(3.0 V) 4.5 W b. How much electric energy does the bulb convert in 11 min? E The definition of power is P , so t E Pt (4.5 W)(11 min) 60 s min 3.0103 J 71. Batteries A resistor of 60.0 has a current of 0.40 A through it when it is connected to the terminals of a battery. What is the voltage of the battery? R ______________ ______________ ______________ ______________ ______________ ______________ ______________ ______________ ______________ ______________ 143 154 143 161 , R 148 , R 150 , , R 159 , R 143 , , R 154 , R 157 , 0.04 I (amps) 0.02 12.00 8.00 4.00 0.00 4.00 8.00 12.00 0.04 0.06 15 V (volts) Level 2 75. Some students connected a length of nichrome wire to a variable power supply to produce between 0.00 V and 10.00 V across the wire. They then measured the current through the wire for several voltages. The students recorded the data for the voltages used and the currents measured, as shown in Table 22-2. 454 Solutions Manual c. Does the nichrome wire obey Ohm’s law? If not, for all the voltages, specify the voltage range for which Ohm’s law holds. Ohm’s law is obeyed when the resistance of a device is constant and independent of the potential difference. The resistance of the nichrome wire increases somewhat as the magnitude of the voltage increases, so the wire does not quite obey Ohm’s law. Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. V 75 V I 5.0 A 0.0140 0.0270 0.0400 0.0520 0.0630 0.0140 0.0280 0.0390 0.0510 0.0620 0.06 73. What voltage is placed across a motor with a 15- operating resistance if there is 8.0 A of current? V IR 2.00 4.00 6.00 8.00 10.00 2.00 4.00 6.00 8.00 10.00 b. Graph I versus V. V IR (1.5 A)(4.0 ) 6.0 V 74. A voltage of 75 V is placed across a 15- resistor. What is the current through the resistor? (amps) R R R R 72. What voltage is applied to a 4.0- resistor if the current is 1.5 A? V IR (8.0 A)(15 ) (amps) a. For each measurement, calculate the resistance. V IR (0.40 A)(60.0 ) 24 V 1.2102 V (volts) Chapter 22 continued 76. Draw a series circuit diagram to include a 16- resistor, a battery, and an ammeter that reads 1.75 A. Indicate the positive terminal and the voltage of the battery, the positive terminal of the ammeter, and the direction of conventional current. V 28 V a. What is the lamp’s resistance when it is on? V IR 120 V V R 3.0102 I A 0.40 A b. When the lamp is cold, its resistance is I 1.75 A 79. The current through a lamp connected across 120 V is 0.40 A when the lamp is on. 1 as great as it is when the lamp is hot. 5 16 What is the lamp’s cold resistance? I V IR (1.75 A)(16 ) 28 V 77. A lamp draws a 66-mA current when connected to a 6.0-V battery. When a 9.0-V battery is used, the lamp draws 75 mA. a. Does the lamp obey Ohm’s law? No. The voltage is increased by a 9.0 factor of 1.5, but the current 6.0 75 is increased by a factor of 1.1 66 b. How much power does the lamp dissipate when it is connected to the 6.0-V battery? 15(3.0102 ) 6.0101 c. What is the current through the lamp as it is turned on if it is connected to a potential difference of 120 V? V IR 120 V V I 2.0 A 1 R 6.010 Level 3 80. The graph in Figure 22-17 shows the current through a device called a silicon diode. Current in a Diode c. How much power does it dissipate at 9.0 V? P IV (75103 A)(9.0 V) 0.68 W 78. Lightbulbs How much energy does a 60.0-W lightbulb use in half an hour? If the lightbulb converts 12 percent of electric energy to light energy, how much thermal energy does it generate during the half hour? E P t E Pt (60.0 W)(1800 s) 1.08105 J If the bulb is 12 percent efficient, 88 percent of the energy is lost to heat, so Q (0.88)(1.08105 J) 9.5104 J Current (A) Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. P IV (66103 A)(6.0 V) 0.40 W 0.02 0.01 0 0.2 0.4 0.6 0.8 Voltage (V) ■ Figure 22-17 a. A potential difference of 0.70 V is placed across the diode. What is the resistance of the diode? From the graph, I 22 m/A, and V IR, so 0.70 V V R 32 2 I 2.210 A b. What is the diode’s resistance when a 0.60-V potential difference is used? 0.60 V V R 1.2102 3 I 5.210 A c. Does the diode obey Ohm’s law? No. Resistance depends on voltage. Physics: Principles and Problems Solutions Manual 455 Chapter 22 continued 81. Draw a schematic diagram to show a circuit including a 90-V battery, an ammeter, and a resistance of 45 connected in series. What is the ammeter reading? Draw arrows showing the direction of conventional current. 40.0 V A 2A I 45 90 V I ■ a. the maximum safe current V IR P I 2R 90 V V I 2 A 45 R I 22.2 Using Electric Energy pages 612–613 Level 1 82. Batteries A 9.0-V battery costs $3.00 and will deliver 0.0250 A for 26.0 h before it must be replaced. Calculate the cost per kWh. E IVt (0.0250 A)(9.0 V)(26.0 h) 5.9 Wh 5.9103 kWh $3.00 cost Rate 3 E 5.910 kWh 5.010 W 1 A RP 40.0 1 b. the maximum safe voltage P V 2/R 101 W)(40.0 ) V PR (5.0 45 V 86. Utilities Figure 22-19 represents an electric furnace. Calculate the monthly (30-day) heating bill if electricity costs $0.10 per kWh and the thermostat is on one-fourth of the time. $510/kWh P I 2R 0.15 A R 220 P 5.0 W 84. A 110-V electric iron draws 3.0 A of current. How much thermal energy is developed in an hour? Q E VIt (110 V)(3.0 A)(1.0 h)(3600 s/h) 1.2106 J Level 2 85. For the circuit shown in Figure 22-18, the maximum safe power is 5.0101 W. Use the figure to find the following: 456 Solutions Manual 4.80 240.0 V ■ Figure 22-19 V2 E (t ) R (240.0 V)2 4.80 (30 d)(24 h/d)(0.25) 2160 kWh Cost (2160 kWh)($0.100/kWh) $216 87. Appliances A window air conditioner is estimated to have a cost of operation of $50 per 30 days. This is based on the assumption that the air conditioner will run half of the time Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Thermostat 83. What is the maximum current allowed in a 5.0-W, 220- resistor? I Figure 22-18 Chapter 22 continued and that electricity costs $0.090 per kWh. Determine how much current the air conditioner will take from a 120-V outlet. Q E I 2Rt Cost (E )(rate) $50 Cost E $0.090/kWh rate E IVt (556 kWh)(1000 W/kW) E I (120 V)(30 d)(24 h/d)(0.5) Vt 12.9 A 60 s min 91. A 6.0- resistor is connected to a 15-V battery. a. What is the current in the circuit? V IR 88. Radios A transistor radio operates by means of a 9.0-V battery that supplies it with a 50.0-mA current. a. If the cost of the battery is $2.49 and it lasts for 300.0 h, what is the cost per kWh to operate the radio in this manner? P IV (0.050 A)(9.0 V) 0.45 W 4.5104 kW $2.49 Cost 4 (4.510 kW)(300.0 h) $18.00/kWh Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. (1.2 A)2(50.0 )(5.0 min) 2.2104 J 556 kWh b. The same radio, by means of a converter, is plugged into a household circuit by a homeowner who pays $0.12 per kWh. What does it now cost to operate the radio for 300.0 h? Cost ($0.12/kWh) (4.5104 kW)(300 h) $0.02 Mixed Review page 613 Level 1 89. If a person has $5, how long could he or she play a 200 W stereo if electricity costs $0.15 per kWh? Cost Rate E Pt Cost t (Rate)(P) $5 90. A current of 1.2 A is measured through a 50.0- resistor for 5.0 min. How much heat is generated by the resistor? 1 kW ($0.15/kWh)(200 W) 1000 W 200 h Physics: Principles and Problems V 15 V I 2.5 A R 6.0 b. How much thermal energy is produced in 10.0 min? Q E I 2Rt (2.5 A)2(6.0 )(10.0 min) 60 s min 2.3104 J Level 2 92. Lightbulbs An incandescent lightbulb with a resistance of 10.0 when it is not lit and a resistance of 40.0 when it is lit has 120 V placed across it. a. What is the current draw when the bulb is lit? V 120 V I 3.0 A R 40.0 b. What is the current draw at the instant the bulb is turned on? V 120 V I 12 A R 10.0 c. When does the lightbulb use the most power? the instant it is turned on 93. A 12-V electric motor’s speed is controlled by a potentiometer. At the motor’s slowest setting, it uses 0.02 A. At its highest setting, the motor uses 1.2 A. What is the range of the potentiometer? The slowest speed’s resistance is R V/I 12 V/0.02 A 600 . The fastest speed’s resistance is R V/I 12 V/1.2 A 1.0101 . The range is 1.0101 to 600 . Solutions Manual 457 Chapter 22 continued Level 3 94. An electric motor operates a pump that irrigates a farmer’s crop by pumping 1.0104 L of water a vertical distance of 8.0 m into a field each hour. The motor has an operating resistance of 22.0 and is connected across a 110-V source. a. What current does the motor draw? V IR Q T mC 6 1.110 J (20.0 kg)(4180 J/kgC°) 13°C d. At $0.08 per kWh, how much does it cost to operate the heating coil 30 min per day for 30 days? 1.1106 J 5 min Cost (30 days) 110 V V I 5.0 A R 22.0 $0.08 1 kWh 3.6106 J kWh b. How efficient is the motor? EW mgd $4.40 (1104 kg)(9.80 m/s2)(8.0 m) 8105 J Em IVt (5.0 A)(110 V)(3600 s) 2.0106 J 5 2.010 J 40% 95. A heating coil has a resistance of 4.0 and operates on 120 V. b. The heater is being used to heat a room containing 50 kg of air. If the specific heat of air is 1.10 kJ/kg°C, and 50 percent of the thermal energy heats the air in the room, what is the change in air temperature in half an hour? Q mCT Q T mC V IR 5 (0.5)(910 J) 120 V V I 3.0101 A (50.0 kg)(1100 J/kgC°) 4.0 b. What energy is supplied to the coil in 5.0 min? I 2Rt (3.0101 A)2(4.0 )(5.0 min) 60 s min 1.1106 J c. If the coil is immersed in an insulated container holding 20.0 kg of water, what will be the increase in the temperature of the water? Assume 100 percent of the heat is absorbed by the water. Q mCT Solutions Manual 8°C c. At $0.08 per kWh, how much does it cost to run the heater 6.0 h per day for 30 days? 500 J 6.0 h 3600 s Cost s day h 1kWH $0.08 (30 days) 6 3.610 J kWh $7 Thinking Critically page 614 97. Formulate Models How much energy is stored in a capacitor? The energy needed to increase the potential difference of a charge, q, Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. a. What is the current in the coil while it is operating? 458 a. How much energy is delivered to the heater in half an hour? 9105 J 810 J 100 6 E 96. Appliances An electric heater is rated at 500 W. E Pt (5102 W)(1800 s) Ew Efficiency 100 Em R 30 min day Chapter 22 continued Potential difference (V) 4.0 → Microwave converter 1440 J/s → Heat converter 1080 J/s Wasted 360 J/s Wasted 270 J/s → Useful output 810 J/s b. Derive an equation for the rate of temperature increase ( T/s) from the information presented in Chapter 12. Solve for the rate of temperature rise given the rate of energy input, the mass, and the specific heat of a substance. 1 Q T mC t t 3.0 2.0 1.0 0.0 1.0 2.0 3.0 4.0 5.0 Charge (C) q 5.0 C Voltage V 5.0 V 1.0 F C Energy E area under curve Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Input 1440 J/s → 5.0 Figure 22-2b on page 593. Label the function of each block according to total joules per second. → is represented by E qV. But in a capacitor, V q/C. Thus, as charge is added, the potential difference increases. As more charge is added, however, it takes more energy to add the additional charge. Consider a 1.0-F “supercap” used as an energy storage device in a personal computer. Plot a graph of V as the capacitor is charged by adding 5.0 C to it. What is the voltage across the capacitor? The area under the curve is the energy stored in the capacitor. Find the energy in joules. Is it equal to the total charge times the final potential difference? Explain. 1 (5.0 V)(5.0 C) 2 13 J No. Graphically, total change times final potential difference is exactly twice the area under the curve. Physically, it means that each coulomb would require the same maximum amount of energy to place it on the capacitor. Actually, the amount of energy needed to add each charge increases as charge accumulates on the capacitor. 98. Apply Concepts A microwave oven operates at 120 V and requires 12 A of current. Its electric efficiency (converting AC to microwave radiation) is 75 percent, and its conversion efficiency from microwave radiation to heating water is also 75 percent. a. Draw a block power diagram similar to the energy diagram shown in Physics: Principles and Problems c. Use your equation to solve for the rate of temperature rise in degrees Celsius per second when using this oven to heat 250 g of water above room temperature. 1 Q T mC t t 810 J/s (0.25 kg)(4180 J/kg°C) 0.78°C/s d. Review your calculations carefully for the units used and discuss why your answer is in the correct form. The kg unit cancels and the J unit cancels, leaving °C/s. e. Discuss, in general terms, different ways in which you could increase the efficiency of microwave heating. The efficiency of conversion from electric energy to microwave energy is 75 percent. It might be possible to find a way to convert electric energy to radiation using a different approach that would be more efficient. The efficiency of conversion from microwave radiation to thermal energy in water is 75 percent. It might be possible to use a different frequency of electromagnetic radiation to improve this rating. Or, it might be possible to find a new Solutions Manual 459 Chapter 22 continued f. Discuss, in efficiency terms, why microwave ovens are not useful for heating everything. The conversion efficiency from microwave energy to thermal energy is good for water. It’s not as good for other materials. The containers and dishes designed for use with microwave ovens convert little of the energy. g. Discuss, in general terms, why it is not a good idea to run microwave ovens when they are empty. The empty oven means that the microwave energy has to be dissipated in the oven. This can lead to overheating of the oven components and to their failure. 99. Analyze and Conclude A salesclerk in an appliance store states that microwave ovens are the most electrically efficient means of heating objects. The physical size of a resistor is determined by its power rating. Resistors rated at 100 W are much larger than those rated at 1 W. 101. Make and Use Graphs The diode graph shown in Figure 22-17 on page 612 is more useful than a similar graph for a resistor that obeys Ohm’s law. Explain. The volt–ampere graph for a resistor obeying Ohm’s law is a straight line and is seldom necessary. 102. Make and Use Graphs Based on what you have learned in this chapter, identify and prepare two parabolic graphs. voltage–power and current–power In the case of heating a cup of water, an immersion heater uses only resistance for energy conversion and is nearly 100 percent efficient. A microwave oven uses two energy conversions (electricity to microwave radiation to heat) and is typically around 50 percent efficient. In the case of heating a potato, a microwave oven heats mostly the potato and is more efficient than an electric oven or skillet, which also heats the air, cabinets, racks, etc. c. Formulate a diplomatic reply to the clerk. “It can be true, but it depends on the specific application.” 460 Solutions Manual 30 20 10 0 5 10 15 20 Voltage across a 10- resistor 2 PV R 40 Watts dissipated b. Formulate an argument to support the clerk’s claim. Hint: Think about heating a specific object. P I 2R 40 30 20 10 0 1 2 Amperes through a 10- resistor Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. a. Formulate an argument to refute the clerk’s claim. Hint: Think about heating a specific object. 100. Apply Concepts The sizes of 10- resistors range from a pinhead to a soup can. Explain. Watts dissipated geometry of radiating objects to be heated to improve the efficiency. Chapter 22 continued Writing in Physics page 614 103. There are three kinds of equations encountered in science: (1) definitions, (2) laws, and (3) derivations. Examples of these are: (1) an ampere is equal to one coulomb per second, (2) force is equal to mass times acceleration, (3) power is equal to voltage squared divided by resistance. Write a one-page explanation of where “resistance is equal to voltage divided by current” fits. Before you begin to write, first research the three categories given above. The student’s answer should include the idea (1) that, for devices obeying Ohm’s law, the voltage drop is proportional to current through the device and (2) that the formula R V/I, the definition of resistance, is a derivation from Ohm’s law. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 104. In Chapter 13, you learned that matter expands when it is heated. Research the relationship between thermal expansion and high-voltage transmission lines. Answers will vary, but students should determine that transmission lines can become hot enough to expand and sag when they have high currents. Sagging lines can be dangerous if they touch objects beneath them, such as trees or other power lines. S (20 kg)(3.34105 J/kg)/(273 K) 2.4104 J/K 106. When you go up the elevator of a tall building, your ears might pop because of the rapid change in pressure. What is the pressure change caused by riding in an elevator up a 30-story building (150 m)? The density of air is about 1.3 kg/m3 at sea level. (Chapter 13) P gh (1.3 kg/m3)(9.80 m/s2)(150 m) 1.9 kPa or about 2/100 of the total air pressure 107. What is the wavelength in air of a 17-kHz sound wave, which is at the upper end of the frequency range of human hearing? (Chapter 15) v f 343 m/s v 0.020 m 2.0 cm f 17,000 Hz 108. Light of wavelength 478 nm falls on a double slit. First-order bright bands appear 3.00 mm from the central bright band. The screen is 0.91 m from the slits. How far apart are the slits? (Chapter 19) xd L L x d (478109 m)(0.91 m) (3.0010 m) Cumulative Review 3 page 614 105. A person burns energy at the rate of about 8.4106 J per day. How much does she increase the entropy of the universe in that day? How does this compare to the entropy increase caused by melting 20 kg of ice? (Chapter 12) 1.4104 m S Q/T where T is the body temperature of 310 K. S (8.4106 J)/(310 K) 2.7104 J/K For melting ice Physics: Principles and Problems 109. A charge of 3.0106 C is 2.0 m from a second charge of 6.0105 C. What is the magnitude of the force between them? (Chapter 20) q q d A B F K 2 (9.0109 Nm2/C2) (3.0106 C)(6.0105 C) (2.0 m)2 0.41 N Solutions Manual 461 Chapter 22 continued Challenge Problem page 604 Use the figure to the right to help you answer the questions below. 1. Initially, the capacitor is uncharged. Switch 1 is closed, and Switch 2 remains open. What is the voltage across the capacitor? Switch 2 Switch 1 1.5 F 15 V 1200 15 V 2. Switch 1 is now opened, and Switch 2 remains open. What is the voltage across the capacitor? Why? It remains 15 V because there is no path for the charge to be removed. 3. Next, Switch 2 is closed, while Switch 1 remains open. What is the voltage across the capacitor and the current through the resistor immediately after Switch 2 is closed? 15 V and 13 mA 4. As time goes on, what happens to the voltage across the capacitor and the current through the resistor? 462 Solutions Manual Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. The capacitor voltage remains at 15 V because there is no path to discharge the capacitor; the current remains at 13 mA because the battery voltage is constant at 15 V. However, if the battery and capacitor were real components instead of ideal circuit components, the capacitor voltage eventually would become zero due to leakage, and the current eventually would become zero due to battery depletion. Physics: Principles and Problems CHAPTER 23 Series and Parallel Circuits Practice Problems 23.1 Simple Circuits pages 617–626 page 619 1. Three 20- resistors are connected in series across a 120-V generator. What is the equivalent resistance of the circuit? What is the current in the circuit? R R1 R2 R3 20 20 20 60 120 V V I 2A 60 R 2. A 10-, 15-, and 5- resistor are connected in a series circuit with a 90-V battery. What is the equivalent resistance of the circuit? What is the current in the circuit? R R1 R2 R3 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 10 15 5 30 V 90 V I 3A 30 R 3. A 9-V battery is in a circuit with three resistors connected in series. a. If the resistance of one of the resistors increases, how will the equivalent resistance change? It will increase. b. What will happen to the current? V I , so it will decrease. R c. Will there be any change in the battery voltage? No. It does not depend on the resistance. 4. A string of holiday lights has ten bulbs with equal resistances connected in series. When the string of lights is connected to a 120-V outlet, the current through the bulbs is 0.06 A. Physics: Principles and Problems a. What is the equivalent resistance of the circuit? 120 V V R 2103 I 0.06 A b. What is the resistance of each bulb? R 2103 Rbulb 2102 10 10 5. Calculate the voltage drops across the three resistors in problem 2, and verify that their sum equals the voltage of the battery. V1 IR1 (3 A)(10 ) 30 V V2 IR2 (3 A)(15 ) 45 V V3 IR3 (3 A)(5 ) 15 V V1 V2 V3 30 V 45 V 15 V 90 V voltage of battery page 622 6. The circuit shown in Example Problem 1 is producing these symptoms: the ammeter reads 0 A, VA reads 0 V, and VB reads 45 V. What has happened? RB has failed. It has infinite resistance, and the battery voltage appears across it. 7. Suppose the circuit shown in Example Problem 1 has these values: RA 255 , RB 292 , and VA 17.0 V. No other information is available. a. What is the current in the circuit? V 17.0 V I 66.7 mA R 255.0 b. What is the battery voltage? First, find the total resistance, then solve for voltage. R RA RB 255 292 547 V IR (66.7 mA)(547 ) 36.5 V Solutions Manual 463 Chapter 23 continued c. What are the total power dissipation and the individual power dissipations? P IV (66.7 mA)(36.5 V) 2.43 W PA I 2RA (66.7 mA)2(255 ) 1.13 W PB I 2RB (66.7 mA)2(292 ) 1.30 W d. Does the sum of the individual power dissipations in the circuit equal the total power dissipation in the circuit? Explain. Yes. The law of conservation of energy states that energy cannot be created or destroyed; therefore, the rate at which energy is converted, or power dissipated, will equal the sum of all parts. 8. Holiday lights often are connected in series and use special lamps that short out when the voltage across a lamp increases to the line voltage. Explain why. Also explain why these light sets might blow their fuses after many bulbs have failed. 9. The circuit in Example Problem 1 has unequal resistors. Explain why the resistor with the lower resistance will operate at a lower temperature. The resistor with the lower resistance will dissipate less power, and thus will be cooler. 10. A series circuit is made up of a 12.0-V battery and three resistors. The voltage across one resistor is 1.21 V, and the voltage across another resistor is 3.33 V. What is the voltage across the third resistor? VC Vsource (VA VB) 12.0 V (1.21 V 3.33 V) 7.46 V page 623 11. A 22- resistor and a 33- resistor are connected in series and placed across a 120-V potential difference. a. What is the equivalent resistance of the circuit? R R1 R2 22 33 55 b. What is the current in the circuit? 120 V V I 2.2 A 55 R c. What is the voltage drop across each resistor? V1 IR1 R1 V R (22 ) 120 V 55 48 V V2 IR2 (33 ) 72 V 120 V 55 d. What is the voltage drop across the two resistors together? V 48 V 72 V 1.20102 V 12. Three resistors of 3.3 k, 4.7 k, and 3.9 k are connected in series across a 12-V battery. a. What is the equivalent resistance? R 3.3 k 4.7 k 3.9 k 1.2101 k b. What is the current through the resistors? V 12 V I 4 R 1210 1.0 mA 1.0103 A c. What is the voltage drop across each resistor? V IR V1 (3.3 k)(1.0103 A) 3.3 V V2 (4.7 k)(1.0103 A) 4.7 V 464 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. If not for the shorting mechanism, the entire set would go out when one lamp burns out. After several lamps fail and then short, the total resistance of the remaining working lamps results in an increased current that is sufficient to blow the fuse. Vsource VA VB VC Chapter 23 continued V3 (3.9 k)(1.0103 A) 3.9 V so V 3.3 V, 4.7 V, and 3.9 V d. Find the total voltage drop across the three resistors. V 3.3 V 4.7 V 3.9 V 11.9 V 13. A student makes a voltage divider from a 45-V battery, a 475-k resistor, and a 235-k resistor. The output is measured across the smaller resistor. What is the voltage? VR RA RB (45 V)(235 k) B VB 15 V 475 k 235 k 14. Select a resistor to be used as part of a voltage divider along with a 1.2-k resistor. The drop across the 1.2-k resistor is to be 2.2 V when the supply is 12 V. 1 1 1 120.0 60.0 40.0 R 20.0 b. What is the current through the entire circuit? V 12.0 V I 0.600 A R 20.0 c. What is the current through each branch of the circuit? 120.0 V 12.0 V I2 0.200 A VR VB R2 (12.0 V)(1.2 k) 1.2 k 2.2 V 60.0 V 12.0 V I3 0.300 A R3 40.0 17. Suppose that one of the 15.0- resistors in problem 15 is replaced by a 10.0- resistor. 5.3 k Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1 1 1 1 R3 R1 R2 R R1 B RB RA page 626 15. Three 15.0- resistors are connected in parallel and placed across a 30.0-V battery. a. What is the equivalent resistance of the parallel circuit? 1 1 1 1 R2 R1 R3 R 1 1 1 15.0 15.0 15.0 3 15.0 R 5.00 b. What is the current through the entire circuit? V 30.0 V I 6.00 A 5.00 c. What is the current through each branch of the circuit? V a. What is the equivalent resistance of the parallel circuit? V 12.0 V I1 0.100 A VR RA RB B VB R 16. A 120.0- resistor, a 60.0- resistor, and a 40.0- resistor are connected in parallel and placed across a 12.0-V battery. 30.0 V I 2.00 A R1 15.0 Physics: Principles and Problems a. Does the equivalent resistance change? If so, how? Yes, it gets smaller. b. Does the amount of current through the entire circuit change? If so, in what way? Yes, it gets larger. c. Does the amount of current through the other 15.0- resistors change? If so, how? No, it remains the same. Currents are independent. 18. A 150- branch in a circuit must be reduced to 93 . A resistor will be added to this branch of the circuit to make this change. What value of resistance should be used and how must the resistor be connected? A parallel resistor will be required to reduce the resistance. 1 1 1 RA RB R Solutions Manual 465 Chapter 23 continued 1 1 1 1 1 RA RB R 93 150 RA 2.4102 2.4102 in parallel with the 150- resistance 19. A 12-, 2-W resistor is connected in parallel with a 6.0-, 4-W resistor. Which will become hotter if the voltage across them keeps increasing? Neither. They both reach maximum dissipation at the same voltage. I T I1 I 2 I 3 I4 120 mA 250 mA 380 mA 2.1 A 0.12 A 0.25 A 0.38 A 2.1 A 2.9 A 22. Total Current A series circuit has four resistors. The current through one resistor is 810 mA. How much current is supplied by the source? 810 mA. Current is the same everywhere in a series circuit. 2 V P R V PR The voltage is equal across parallel resistors, so: V P1R1 P2R2 (2 W)( 12 ) (4 W)( 6.0 ) 5 V maximum Section Review page 626 20. Circuit Types Compare and contrast the voltages and the currents in series and parallel circuits. The student’s answer should include the following ideas: (1) In a series circuit, the current in each of the devices is the same, and the sum of the device voltage drops equals the source voltage. (2) In a parallel circuit, the voltage drop across each device is the same and the sum of the currents through each loop equals the source current. 21. Total Current A parallel circuit has four branch currents: 120 mA, 250 mA, 380 mA, and 2.1 A. How much current is supplied by the source? 466 Solutions Manual a. What is the potential difference across the switch when it is closed (turned on)? 0 V; V IR with R 0 b. What is the potential difference across the switch if another 75-W bulb is added in series? 0 V; V IR with R 0 24. Critical Thinking The circuit in Figure 23-8 has four identical resistors. Suppose that a wire is added to connect points A and B. Answer the following questions, and explain your reasoning. R R A B R R ■ Figure 23-8 a. What is the current through the wire? 0 A; the potentials of points A and B are the same. b. What happens to the current through each resistor? nothing c. What happens to the current drawn from the battery? nothing Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 23.1 Simple Circuits pages 617–626 23. Circuits A switch is connected in series with a 75-W bulb to a source of 120 V. Chapter 23 continued d. What happens to the potential difference across each resistor? nothing Practice Problems 23.2 Applications of Circuits pages 627–631 page 630 25. A series-parallel circuit has three resistors: one dissipates 2.0 W, the second 3.0 W, and the third 1.5 W. How much current does the circuit require from a 12-V battery? By conservation of energy (and power): PT P1 P2 P3 Section Review 23.2 Applications of Circuits pages 627–631 page 631 Refer to Figure 23-13 for questions 29–33, and 35. The bulbs in the circuit are identical. A I3 1 V 12 V Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 26. There are 11 lights in series, and they are in series with two lights in parallel. If the 13 lights are identical, which of them will burn brightest? The 11 lights in series will burn brighter. The parallel lights each will conduct half of the current of the series lights and they will burn at one-fourth the intensity of the series lights since P I 2R. 27. What will happen to the circuit in problem 26 if one of the parallel lights burns out? Then, all of the working lights are in series. The 12 working lights will burn with equal intensity. 28. What will happen to the circuit in problem 26 if one of the parallel lights shorts out? Then, the shorted light will reduce the voltage across itself and its parallel companion to 0. The 11 series lights will burn with equal, but increased, intensity and the two parallel lights will go out. Physics: Principles and Problems 2 C V 3 ■ 6.5 W PT 6.5 W I 0.54 A I2 A I1 A 2.0 W 3.0 W 1.5 W PT IV B A Figure 23-13 29. Brightness How do the bulb brightnesses compare? Bulbs 2 and 3 are equal in brightness but dimmer than bulb 1. 30. Current If I3 measures 1.7 A and I1 measures 1.1 A, how much current is flowing in bulb 2? I3 I1 I2 I2 I3 I1 1.7 A 1.1 A 0.6 A 31. Circuits in Series The wire at point C is broken and a small resistor is inserted in series with bulbs 2 and 3. What happens to the brightnesses of the two bulbs? Explain. Both dim equally. The current in each is reduced by the same amount. 32. Battery Voltage A voltmeter connected across bulb 2 measures 3.8 V, and a voltmeter connected across bulb 3 measures 4.2 V. What is the battery voltage? These bulbs are in series, so: VT V1 V2 3.8 V 4.2 V 8.0 V Solutions Manual 467 Chapter 23 continued 33. Circuits Using the information from problem 32, determine if bulbs 2 and 3 are identical. No. Identical bulbs in series would have identical voltage drops since their currents are the same. 34. Circuit Protection Describe three common safety devices associated with household wiring. fuses, circuit breakers, ground-fault circuit interrupters 35. Critical Thinking Is there a way to make the three bulbs in Figure 23-13 burn with equal intensity without using any additional resistors? Explain. Yes. Because intensity is proportional to power, it would be necessary to use a bulb at location 1 that has four times the operating resistance of each of those at locations 2 and 3. V 2/4R (V/2)2/R Chapter Assessment Concept Mapping Resistors in circuits series circuit Type of circuit parallel circuit constant current Principle constant potential R R1 R2 R3 . . . 468 Resistance Solutions Manual 1 1 1 . . . R R1 R2 page 636 37. Why is it frustrating when one bulb burns out on a string of holiday tree lights connected in series? (23.1) When one bulb burns out, the circuit is open and all the bulbs go out. 38. Why does the equivalent resistance decrease as more resistors are added to a parallel circuit? (23.1) Each new resistor provides an additional path for the current. 39. Several resistors with different values are connected in parallel. How do the values of the individual resistors compare with the equivalent resistance? (23.1) The equivalent resistance will be less than that of any of the resistors. 40. Why is household wiring constructed in parallel instead of in series? (23.1) Appliances in parallel can be run independently of one another. 41. Why is there a difference in equivalent resistance between three 60- resistors connected in series and three 60- resistors connected in parallel? (23.1) In a series circuit, the current is opposed by each resistance in turn. The total resistance is the sum of the resistors. In a parallel circuit, each resistance provides an additional path for current. The result is a decrease in total resistance. 42. Compare the amount of current entering a junction in a parallel circuit with that leaving the junction. (A junction is a point where three or more conductors are joined.) (23.1) The amount of current entering a junction is equal to the amount of current leaving. Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. page 636 36. Complete the concept map using the following terms: series circuit, R R1 R2 R3, constant current, parallel circuit, constant potential. Mastering Concepts Chapter 23 continued 43. Explain how a fuse functions to protect an electric circuit. (23.2) The purpose of a fuse is to prevent conductors from being overloaded with current, causing fires due to overheating. A fuse is simply a short length of wire that will melt from the heating effect if the current exceeds a certain maximum. 44. What is a short circuit? Why is a short circuit dangerous? (23.2) A short circuit is a circuit that has extremely low resistance. A short circuit is dangerous because any potential difference will produce a large current. The heating effect of the current can cause a fire. 45. Why is an ammeter designed to have a very low resistance? (23.2) Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. An ammeter must have low resistance because it is placed in series in the circuit. If its resistance were high, it would significantly change the total resistance of the circuit and thus serve to reduce the current in the circuit, thereby changing the current it is meant to measure. 46. Why is a voltmeter designed to have a very high resistance? (23.2) A voltmeter is placed in parallel with the portion of the circuit whose difference in potential is to be measured. A voltmeter must have very high resistance for the same reason that an ammeter has low resistance. If the voltmeter had low resistance, it would lower the resistance of the portion of the circuit it is across and increase the current in the circuit. This would produce a higher voltage drop across the part of the circuit where the voltmeter is located, changing the voltage it is measuring. 47. How does the way in which an ammeter is connected in a circuit differ from the way in which a voltmeter is connected? (23.2) An ammeter is connected in series; a voltmeter is connected in parallel. Physics: Principles and Problems Applying Concepts page 636 48. What happens to the current in the other two lamps if one lamp in a three-lamp series circuit burns out? If one of the lamp filaments burns out, the current will cease and all the lamps will go out. 49. Suppose the resistor, RA, in the voltage divider in Figure 23-4 is made to be a variable resistor. What happens to the voltage output, VB, of the voltage divider if the resistance of the variable resistor is increased? VB VRB/(RA RB). As RA increases, VB will decrease. 50. Circuit A contains three 60- resistors in series. Circuit B contains three 60- resistors in parallel. How does the current in the second 60- resistor of each circuit change if a switch cuts off the current to the first 60- resistor? Circuit A: There will be no current in the resistor. Circuit B: The current in the resistor will remain the same 51. What happens to the current in the other two lamps if one lamp in a three-lamp parallel circuit burns out? If one of the filaments burns out, the resistance and the potential difference across the other lamps will not change; therefore, their currents will remain the same. 52. An engineer needs a 10- resistor and a 15- resistor, but there are only 30- resistors in stock. Must new resistors be purchased? Explain. No, the 30- resistors can be used in parallel. Three 30- resistors in parallel will give a 10- resistance. Two 30- resistors in parallel will give a 15- resistance. Solutions Manual 469 Chapter 23 continued 53. If you have a 6-V battery and many 1.5-V bulbs, how could you connect them so that they light but do not have more than 1.5 V across each bulb? Connect four of the bulbs in series. The voltage drop across each will be (6.0 V)/4 1.5 V. 54. Two lamps have different resistances, one larger than the other. a. If the lamps are connected in parallel, which is brighter (dissipates more power)? The lamp with the lower resistance: P IV and I V/R, so P V 2/R. Because the voltage drop is the same across both lamps, the smaller R means larger P, and thus will be brighter. b. When the lamps are connected in series, which lamp is brighter? The lamp with the higher resistance; P IV and V IR, so P I 2R. Because the current is the same in both lamps, the larger R means larger P, and thus will be brighter. Adding a resistor to the circuit increases the total resistance. series g. If the current through one resistor in the circuit goes to zero, there is no current in the entire circuit. series h. If the current through one resistor in the circuit goes to zero, the current through all other resistors remains the same. parallel i. This form is suitable for house wiring. parallel 56. Household Fuses Why is it dangerous to replace the 15-A fuse used to protect a household circuit with a fuse that is rated at 30 A? The 30-A fuse allows more current to flow through the circuit, generating more heat in the wires, which can be dangerous. Mastering Problems 23.1 Simple Circuits pages 637–638 Level 1 57. Ammeter 1 in Figure 23-14 reads 0.20 A. a. The current is the same everywhere throughout the entire circuit. A1 22 series b. The total resistance is equal to the sum of the individual resistances. series c. The voltage drop across each resistor in the circuit is the same. parallel d. The voltage drop in the circuit is proportional to the resistance. series e. Adding a resistor to the circuit decreases the total resistance. parallel 470 Solutions Manual A2 V 15 A3 ■ Figure 23-14 a. What should ammeter 2 indicate? 0.20 A, because current is constant in a series circuit. b. What should ammeter 3 indicate? 0.20 A, because current is constant in a series circuit. Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 55. For each of the following, write the form of circuit that applies: series or parallel. f. Chapter 23 continued 58. Calculate the equivalent resistance of these series-connected resistors: 680 , 1.1 k, and 10 k. 64. A 22- lamp and a 4.5- lamp are connected in series and placed across a potential difference of 45 V as shown in Figure 23-15. R 680 1100 10,000 12 k 22 59. Calculate the equivalent resistance of these parallel-connected resistors: 680 , 1.1 k, and 10.2 k. 4.5 45 V 1 1 1 1 R1 R2 R3 R R ■ 1 1 1 1 0.68 k 1.1 k 10.2 k 0.40 k V 5.50 V 6.90 V 12.4 V 61. A parallel circuit has two branch currents: 3.45 A and 1.00 A. What is the current in the energy source? I 3.45 A 1.00 A 4.45 A Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. a. What is the equivalent resistance of the circuit? 22 4.5 26 b. What is the current in the circuit? 60. A series circuit has two voltage drops: 5.50 V and 6.90 V. What is the supply voltage? Level 2 62. Ammeter 1 in Figure 23-14 reads 0.20 A. a. What is the total resistance of the circuit? V 45 V I 1.7 A R 27 c. What is the voltage drop across each lamp? V IR (1.7 A)(22 ) 37 V V IR (1.7 A)(4.5 ) 7.7 V d. What is the power dissipated in each lamp? P IV (1.7 A)(37 V) 63 W P IV (1.7 A)(7.7 V) 13 W 65. Refer to Figure 23-16 to answer the following questions. R R1 R2 15 22 37 b. What is the battery voltage? V1 V IR (0.20 A)(37 ) 7.4 V c. How much power is delivered to the 22- resistor? P I 2R (0.20 Figure 23-15 A)2(22 ) 0.88 W d. How much power is supplied by the battery? A 35 10.0 V P IV (0.20 A)(7.4 V) 1.5 W 63. Ammeter 2 in Figure 23-14 reads 0.50 A. a. Find the voltage across the 22- resistor. V IR (0.50 A)(22 ) 11 V b. Find the voltage across the 15- resistor. V IR (0.50 A)(15 ) 7.5 V c. What is the battery voltage? V V1 V2 (11 V) (7.5 V) 19 V Physics: Principles and Problems 15 ■ V2 Figure 23-16 a. What should the ammeter read? R R1 R2 35 15 I V/R (10.0 V)/(35 15 ) 0.20 A b. What should voltmeter 1 read? V IR (0.20 A)(35 ) 7.0 V Solutions Manual 471 Chapter 23 continued c. What should voltmeter 2 read? 67. For Figure 23-18, the battery develops 110 V. V IR (0.20 A)(15 ) 3.0 V d. How much energy is supplied by the battery per minute? E Pt VIt A2 V 20.0 (10.0 V)(0.20 A)(1 min)(60 s/min) 10.0 50.0 A1 120 J ■ 66. For Figure 23-17, the voltmeter reads 70.0 V. Figure 23-18 a. Which resistor is the hottest? 10.0 . Since P V 2/R and V is constant in a parallel circuit, the smallest resistor will dissipate the most power. V A 35 b. Which resistor is the coolest? 15 V 50.0 . Since P V 2/R and V is constant in a parallel circuit, the largest resistor will dissipate the least power. 50 ■ A4 A3 Figure 23-17 c. What will ammeter 1 read? a. Which resistor is the hottest? 1 1 1 1 R1 R2 R3 R 50 . Since P and I is constant in a series circuit, the largest value of resistance will produce the most power. I 2R 1 R 1 1 1 R1 R2 R3 20.0 50.0 10.0 15 . Since P I 2R and I is constant in a series circuit, the smallest value of resistance will produce the least power. c. What will the ammeter read? Use Ohm’s law: I V/R 5.88 1.1102 V V I 19A R d. What will ammeter 2 read? V 1.1102 V I 5.5 A (70.0 V)/(35 ) 2.0 A R R R1 R2 R3 35 15 50 0.1 k P I 2R (2.0 A)2(0.1 k)(1000 /k) 20.0 e. What will ammeter 3 read? d. What is the power supplied by the battery? First, find the total resistance: 5.88 V 1.1102 V I 2.2 A R f. 50.0 What will ammeter 4 read? V 1.1102 V I 11 A R 10.0 68. For Figure 23-18, ammeter 3 reads 0.40 A. a. What is the battery voltage? V IR (0.40 A)(50.0 ) 2.0101 V 4102 W 472 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1 1 1 1 b. Which resistor is the coolest? Chapter 23 continued b. What will ammeter 1 read? Find the equivalent resistance: 1 1 1 1 R1 R2 R3 R R11 R12 R13 c. What power is dissipated by each light? 1 1 1 1 50.0 10.0 1 0.17 5.88 2.0101 V V I 3.4 A 5.88 R c. What will ammeter 2 read? V 2.0101 V I 1.0 A 20.0 R d. What will ammeter 4 read? V 2.0101 V I 2.0 A R 10.0 69. What is the direction of the conventional current in the 50.0- resistor in Figure 23-18? down Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. R is the sum of the resistances of 18 lamps, so each resistance is 2.3102 13 18 1 R 20.0 b. What is the resistance of a single light? 64 W 3.6 W 18 72. One of the lights in problem 71 burns out. The light shorts out the bulb filament when it burns out. This drops the resistance of the lamp to zero. a. What is the resistance of the light string now? There are now 17 lamps in series instead of 18 lamps. The resistance 17 is (2.3102 ) 2.2102 18 b. Find the power dissipated by the string. 2 2 V (120 V) P 65 W 2 2.210 R c. Did the power increase or decrease when the bulb burned out? It increased. 70. The load across a battery consists of two resistors, with values of 15 and 47 , connected in series. a. What is the total resistance of the load? R R1 R2 15 47 62 b. What is the voltage of the battery if the current in the circuit is 97 mA? V IR (97 mA)(62 ) 6.0 V 71. Holiday Lights A string of 18 identical holiday tree lights is connected in series to a 120-V source. The string dissipates 64 W. a. What is the equivalent resistance of the light string? V2 P a. Compute the equivalent resistance of the parallel circuit. 1 1 1 R2 R1 R 1 R 1 1 R1 R2 1 1 1 16.0 20.0 8.89 b. What is the total current in the circuit? V 40.0 V I 4.50 A R Req V2 (120 V)2 Req 2.3102 P 73. A 16.0- and a 20.0- resistor are connected in parallel. A difference in potential of 40.0 V is applied to the combination. 64 W Physics: Principles and Problems 8.89 c. What is the current in the 16.0- resistor? V 40.0 V I1 2.50 A R1 16.0 Solutions Manual 473 Chapter 23 continued 74. Amy needs 5.0 V for an integrated-circuit experiment. She uses a 6.0-V battery and two resistors to make a voltage divider. One resistor is 330 . She decides to make the other resistor smaller. What value should it have? VR2 V2 R1 R2 1 R 1 1 1 1 1 VR V2 52 12 (6.0 V)(330 ) 330 66 5.0 V 75. Pete is designing a voltage divider using a 12-V battery and a 82- resistor as RB. What resistor should be used as RA if the output voltage across RB is to be 4.0 V? VR RA RB VR VB B RA RB VR VB B RA RB (12 V)(82 ) 82 4.0 V d. Find the voltage drop across the television and the hair dryer. VRA (120 V)(9.8 ) V1 96 V RA RB 9.8 2.5 a. Find the resistance of the television. V2 V P IV and I , so P , or R R (120 V)2 R 52 P 275 W b. The television and 2.5- wires connecting the outlet to the fuse form a series circuit that works like a voltage divider. Find the voltage drop across the television. VR RA RB A VA (120 V)(52 ) 52 2.5 IA IB IC ■ Figure 23-19 The parallel combination of the two 30.0- resistors has an equivalent resistance of 15.0 . So R 30.0 15.0 45.0 78. Refer to Figure 23-19 and assume that each resistor dissipates 120 mW. Find the total dissipation. P 3(120 mW) 360 mW 79. Refer to Figure 23-19 and assume that IA 13 mA and IB 1.7 mA. Find IC. IC IA IB 13 mA 1.7 mA 11 mA 110 V 474 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Level 3 76. Television A typical television dissipates 275 W when it is plugged into a 120-V outlet. V2 9.8 23.2 Applications of Circuits pages 638–639 Level 1 77. Refer to Figure 23-19 and assume that all the resistors are 30.0 . Find the equivalent resistance. B VB 1 1 1 RA RB R RA RB 2 R1 R2 1.6102 c. A 12- hair dryer is plugged into the same outlet. Find the equivalent resistance of the two appliances. Chapter 23 continued 80. Refer to Figure 23-19 and assume that IB 13 mA and IC 1.7 mA. Find IA. IA IB IC 13 mA 1.7 mA Half the total current is in each parallel branch because the sum of the resistances in each branch are equal. P I 2R (0.25 A)2(30.0 ) 1.9 W 15 mA P I 2R (0.25 A)2(20.0 ) 1.2 W Level 2 81. Refer to Figure 23-20 to answer the following questions. P I 2R (0.25 A)2(10.0 ) 0.62 W P I 2R (0.25 A)2(40.0 ) 2.5 W The 25.0- resistor is the hottest. 25.0 The 10.0- resistor is the coolest. 30.0 10.0 25 V ■ 20.0 40.0 Figure 23-20 a. Determine the total resistance. The 30.0- and 20.0- resistors are in series. R1 30.0 20.0 50.0 82. A circuit contains six 60-W lamps with a resistance of 240- each and a 10.0- heater connected in parallel. The voltage across the circuit is 120 V. Find the current in the circuit for the following situations. a. Four lamps are turned on. 1 1 1 1 1 R1 R2 R3 R4 R 1 1 1 240 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 240 1 240 The 10.0- and 40.0- resistors are in series. R2 10.0 40.0 50.0 240 4 240 R1 and R2 are in parallel. 240 R 0.060 k 1 1 1 R1 R2 R V 120 V I 2.0 A 1 R 1 1 R1 R 0.060 k b. All of the lamps are turned on. R2 1 1 1 50.0 50.0 25.0 and is in series with the 25.0- resistor R Total 25.0 25.0 50.0 b. Determine the current through the 25- resistor. use Ohm’s law and R Total V 4 25 V I 0.50 A RTotal 50.0 c. Which resistor is the hottest? Coolest? P I 2R (0.50 A)2(25.0 ) 6.25 W Physics: Principles and Problems 1 6 R 240 240 R 0.040 k 6 120 V V I 3.0 A R 0.040 k c. Six lamps and the heater are operating. 1 1 1 R 0.040 k 10.0 5 1 4.010 4.010 R 8.0 1 5 V 120 V I 15 A R 8.0 Solutions Manual 475 Chapter 23 continued 83. If the circuit in problem 82 has a 12-A fuse, will the fuse melt if all the lamps and the heater are on? Yes. The 15-A current will melt the 12-A fuse. 84. During a laboratory exercise, you are supplied with a battery of potential difference V, two heating elements of low resistance that can be placed in water, an ammeter of very small resistance, a voltmeter of extremely high resistance, wires of negligible resistance, a beaker that is well insulated and has negligible heat capacity, and 0.10 kg of water at 25°C. By means of a diagram and standard symbols, show how these components should be connected to heat the water as rapidly as possible. A Heating element #1 V Heating element #2 V 86. Home Circuit A typical home circuit is shown in Figure 23-21. The wires to the kitchen lamp each have a resistance of 0.25 . The lamp has a resistance of 0.24 k. Although the circuit is parallel, the lead lines are in series with each of the components of the circuit. 0.25 Kitchen light 240 Switch box 120 V Power saw Wall outlets 0.25 ■ Figure 23-21 a. Compute the equivalent resistance of the circuit consisting of just the lamp and the lead lines to and from the lamp. R 0.25 0.25 0.24 k 0.24 k b. Find the current to the lamp. Q mCT (0.10 kg)(4.2 kJ/kg°C)(75°C) 32 kJ (energy needed to raise temperature of water to 100°C) Q mHv (0.10 kg)(2.3106 J/kg) 2.3102 kJ (energy needed to vaporize the water) Qtotal 32 kJ 2.3102 kJ 2.6102 kJ (total energy needed) Energy is provided at the rate of P IV (5.0 A)(45 V) 0.23 kJ/s. V 120 V I 0.50 A R 0.24 k c. Find the power dissipated in the lamp. P IV (0.50 A)(120 V) 6.0101 W Mixed Review page 639 Level 1 87. A series circuit has two voltage drops: 3.50 V and 4.90 V. What is the supply voltage? V 3.50 V 4.90 V 8.40 V 88. A parallel circuit has two branch currents: 1.45 A and 1.00 A. What is the current in the energy source? I 1.45 A 1.00 A 2.45 A 89. A s