Chapter 8 Potential Energy and Conservation of Energy Answers to Even Numbered Conceptual Questions 2 As water vapor rises there is an increase in the gravitational potential energy of the system Part of this potential energy is released as snow falls onto the mountain If an avalanche occurs the snow on the mountain accelerates down slope converting more gravitational potential energy to kinetic energy 4 The initial mechanical energy of the system is the gravitational potential energy of the mass Earth system As the mass moves downward the gravitational potential energy of the system decreases At the same time the potential energy of the spring increases as it is compressed Initially the decrease in gravitational potential energy is greater than the increase in spring potential energy which means that the mass gains kinetic energy Eventually the increase in spring energy equals the decrease in gravitational energy and the mass comes to rest 6 The object s kinetic energy is a maximum when it is released and a minimum when it reaches its greatest height The gravitational potential of the system is a minimum when the object is released and a maximum when the object reaches its greatest height 8 The jumper s initial kinetic energy is largely converted to a compressional spring like potential energy as the pole bends The pole straightens out converting its potential energy into gravitational potential energy As the jumper falls the gravitational potential energy is converted into kinetic energy and finally the kinetic energy is converted to compressional potential energy as the cushioning pad on the ground is compressed 10 When the toy frog is pressed downward work is done to compress the spring This work is stored in the spring as potential energy Later when the suction cup releases the spring the stored potential energy is converted into enough kinetic energy to lift the frog into the air 12 The total mechanical energy E decreases with time if air resistance is present Solutions to Problems and Conceptual Exercises 1 Picture the Problem The work done by a conservative force is indicated at right for a variety of different paths connecting the points A and B Strategy The work done by a conservative force is independent of the path taken From the middle path we see that the work to go from A to B is 15 J Similarly the work to go from B to A must be 15 J Use this principle to determine the works done along paths 1 and 2 Solution 1 a For the total work along this path to be 15 J the work on Path 1 must be 3 J 2 b For the total work along this path to be 15 J the work done on Path 2 must be 4 J Insight If there were a nonconservative force such as friction the works would be path dependent and we would need more information in order to answer the question Copyright 2010 Pearson Education Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced in any form or by any means without permission in writing from the publisher 8 1 James S Walker Physics 4th Edition Chapter 8 Potential Energy and Conservation of Energy Copyright 2010 Pearson Education Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced in any form or by any means without permission in writing from the publisher 8 2 James S Walker Physics 4th Edition Chapter 8 Potential Energy and Conservation of Energy 2 Picture the Problem The three paths of the object are depicted at right Strategy Find the work done by gravity W mgy when the object is moved downward W mgy when it is moved upward and zero when it is moved horizontally Sum the work done by gravity for each segment of each path Solution 1 Calculate the work for path 1 W1 mg y1 0 y2 0 y3 mg 4 0 m 1 0 m 1 0 m W1 3 2 kg 9 81 m s2 2 0 m 63 J 2 Calculate W for path 2 W2 mg 0 y4 0 3 2 kg 9 81 m s2 2 0 m 63 J 3 Calculate W for path 3 W3 mg y5 0 y6 3 2 kg 9 81 m s2 1 0 m 3 0 m 63 J Insight The work is path independent because gravity is a conservative force 3 Picture the Problem The three paths of the sliding box are depicted at right Strategy The work done by friction is W k mgd where d is the distance the box is pushed irregardless of direction because the friction force always acts in a direction opposite the motion Sum the work done by friction for each segment of each path Solution 1 Calculate the work for path 1 W1 k mg d1 d2 d3 d4 d5 k mg 4 0 4 0 1 0 1 0 1 0 m W1 0 26 3 7 kg 9 81 m s2 11 0 m 100 J 2 Calculate W for path 2 W2 k mg d6 d7 d8 0 26 3 7 kg 9 81 m s2 2 0 m 2 0 m 1 0 m 47 J 3 Calculate the work for path 3 W3 k mg d9 d10 d11 0 21 3 2 kg 9 81 m s2 1 0 m 3 0 m 3 0 m 66 J Insight The amount of work done depends upon the path because friction is a nonconservative force 4 Picture the Problem The physical situation is depicted at right 2 Strategy Use equation W 12 kx equation 7 8 to find the work done by the spring but caution is in order This work is positive when the force exerted by the spring is in the same direction that the block is traveling but it is negative when they point in opposite directions One 2 2 way to keep track of that sign convention is to say that W 12 k xi xf That way the work will always be negative if you start out at xi 0 because the spring force will always be in the opposite direction from the stretch or compression Copyright 2010 Pearson Education Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced in any form or by any means without permission in writing from the publisher 8 3 James S Walker Physics 4th Edition Chapter 8 Potential Energy and Conservation of Energy Solution 1 a Sum the work done by the spring for each segment of path 1 W1 12 k x12 x22 x22 x32 2 2 12 550 N m 02 0 040 m 0 040 m 0 020 m2 W1 0 44 J 0 33 J 0 11 J 2 Sum the work done by the spring for each segment of path 2 W2 12 k x12 x42 x42 x32 2 2 12 550 N m 02 …
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