A P T E R 9 QUANTITATIVE TOOLS CONCEPTS

Transcription

1 Ch09n10v3.26_Ch09n10v2.34 9/24/10 5:14 PM Page 1 C H A P T E R 9 Work CONCEPTS 9.1 Force displacement 9.2 Positive and negative work 9.3 Energy diagrams 9.4 Choice of system QUANTITATIVE TOOLS 9.5 Work done on a single particle 9.6 Work done on a many-particle system 9.7 Variable and distributed forces 9.8 Power 2010 by Eric Mazur Text in progress for Addison Wesley, Inc. A Division of Pearson Education San Francisco, CA All rights reserved. No part of this chapter may be reproduced, in any form or by any means, electronic or mechanical, without permission in writing from the author or from Pearson. Sample uncorrected material. Not for resale or distribution.

2 Ch09n10v3.26_Ch09n10v2.34 9/24/10 5:14 PM Page 2 2 CHAPTER 9 WORK Concepts As you learned in Chapter 8, a system subject to external forces that is to say, forces exerted by objects outside the system on objects inside the system is not isolated. The relationship between the momentum of a nonisolated system and the vector sum of the forces exerted on that system is simple: the vector sum of the forces is equal to the time rate of change in the system s momentum (Eq. 8.4). In addition to affecting a system s momentum, external forces also can change the energy of the system. In the simplest case, a single force exerted on a system consisting of a single object causes that object to accelerate, and so the force changes the object s kinetic energy. Forces can also change the physical state of objects (think of compressing or stretching a spring or crumpling a piece of paper by exerting a force on it). Because the physical state of an object is related to the object s internal energy, we see that forces can change not only an object s kinetic energy but its internal energy as well. To describe these changes in energy, physicists use the concept of work: Work is the change in the energy of a system due to external forces. F Æ ext When an external force causes a change in the energy of a system, physicists refer to this change either as the work done by the force F Æ ext on the system or as the work done by the object or person exerting force F Æ ext on the system. Because work is a change in energy, the unit for work is the same as that for energy the joule. Note that this definition of work is much more limited than the meaning of work in ordinary language, which refers to any physical or mental effort or activity. You may be doing work for your physics class while reading this book, but there is no work involved in the physics sense. 9.1 Force displacement Work amounts to a mechanical transfer of energy either from a system to its environment or from the environment to the system. For instance, if you (part of the environment ) push a friend on a bicycle (system is rider plus bicycle) and your friend accelerates, the force exerted by you on your friend transfers energy from your body to hers, increasing her kinetic energy. If you compress a spring, you transfer energy from your body to the spring, where it is stored in the form of elastic potential energy. In each case, the change in the system s energy is caused by an external force a force exerted by you on the system. If you wonder why a force has to be external to do work on a system that is, to change the energy of the system remember that forces represent interactions between objects. Any interaction between objects within the system rearranges the energy between these objects but doesn t change the overall amount of energy in the system. Only an interaction between an object inside the system and an object outside it an external interaction can transfer energy across the system boundary. Do external forces always cause a change in the energy of a system? To answer this question, it is helpful to consider an example. 9.1 Imagine pushing against a brick wall as shown in Figure 9.1a. (a) Considering the wall as the system, is the force you exert on it internal or external? (b) Does this force accelerate the wall? Change its shape? Raise its temperature? (c) Does the energy of the wall change as a result of the force you exert on it? (d) Does the force you exert on the wall do work on the wall? As this checkpoint illustrates, external forces exerted on a system do not always do work on the system. The other two examples in Figure 9.1, however, do involve changes in energy, and hence in these cases work is done by your external force on the system. If, as shown in Figure 9.1b, you push against something that has wheels and so can accelerate, its kinetic energy changes. If you push against a wall made of some deformable material, the wall changes shape, and so its potential energy changes. Why is the work zero in Figure 9.1a but nonzero in b and c? The key to answering this question lies in looking at the point of application of each force (the point at which the force is exerted on the system, as indicated by the black dots in Figure 9.1). When you push against the brick wall, it neither moves nor deforms, and so the point of application of the force does not move. When you push the cart, you must move along with the cart to accelerate it; in other words, the point of application of the force is displaced. Similarly, when your force changes the shape of the deformable wall, the point of application of the force is displaced.

3 Ch09n10v3.26_Ch09n10v2.34 9/24/10 5:14 PM Page Positive and negative work 3 (a) r F = 0 (b) (c) Force displacement r F Point of application of force F pw F pc r F F pc mattress F pm Point of application does not move, so force does no work on wall. Point of application moves, so force does work on cart. Point of application moves, so force does work on mattress. Figure 9.1 An external force does work on a system only if it causes the point at which it is applied (represented by a black dot) to move. In order for a force to do work, the point of application of the force must undergo a displacement. Whenever an external force is exerted on a system and the displacement of the point of application of the force is zero, no work is done by the force on the system. Conversely, in one dimension, when the force displacement (shorthand for the displacement of the point of application of the force ) is nonzero, the external force does work on the system, and so the energy of the system changes. (In more than one dimension, the angle between force and force displacement also plays a role. We shall discuss this point in Chapter 10.) * To distinguish the force displacement from other displacements, we add a subscript F to the symbol for displacement: r Æ F. of the force is at the hand, which moves to compress the spring. (b) The point of application of the force of gravity exerted by Earth on the ball is at the ball, which moves. (c) The point of application is on the ground, which doesn t move. (d) The point of application is on the floor of the elevator, which moves. 9.2 You throw a ball straight up in the air. Which of the following forces do work on the ball while you throw it? Consider the interval from the instant the ball is at rest in your hand to the instant it leaves your hand at speed v. (a) The force of gravity exerted by Earth on the ball. (b) The contact force exerted by your hand on the ball. Exercise 9.1: Displaced forces For which of the following forces is the force displacement nonzero? (a) The force exerted by a hand compressing a spring. (b) The force exerted by Earth on a ball thrown up. (c) The force exerted by the ground on you at the instant you jump up. (d) The force exerted by the floor of an elevator on you as the elevator moves down at constant speed. Solution: (a), (b) and (d). (a) The point of application * Two additional external forces are exerted on the cart in Figure 9.1 that have nonzero force displacements: the force of gravity and the contact force exerted by the ground. However, these two forces are equal in magnitude and opposite in direction and so they cancel out : their net effect is the same as that of no force. Therefore no work is done by these forces on the cart. 9.2 Positive and negative work When the energy of a system increases as a result of an external force exerted on the system, the change in energy is positive, and so the work done by that external force on the system is said to be positive. An external force can also decrease the energy of a system, however. For instance, imagine slowing down a friend coasting on a bicycle. In this case, the kinetic energy of the system decreases, the change in energy is negative, and so the work done by the external force (that is, by you on your friend) on the system is said to be negative.

4 Ch09n10v3.26_Ch09n10v2.34 9/24/10 5:14 PM Page 4 4 CHAPTER 9 WORK Concepts Force and force displacement point in same direction v = 0 v F pc r F (b) Cart slows down, so negative work is done on it Force and force displacement point in opposite directions r F v = 0 Figure 9.2 Work done on a system is positive if the system gains kinetic energy and negative if the system loses kinetic energy. Also, the work done by a force is positive when the force and force-displacement vectors point in the same direction and negative when they point in opposite directions. 9.3 A ball is thrown vertically upward. (a) As it moves upward, it slows down under the influence of gravity. Considering the changes in energy of the ball, is the work done by Earth on the ball positive or negative? (b) After reaching its highest position, the ball moves downward, gaining speed. Is the work done by the gravitational force exerted on the ball during this motion positive or negative? Examples of positive and negative work are illustrated in Figure 9.2. In both cases the person exerts a force F Æ pc on the cart. What then is the difference between these two situations? The difference lies in the direction in which the force is applied. To speed the cart up, the person must push in the direction of travel of the cart; the force vector F Æ and the force displacement r Æ pc F point in the same direction. To slow the cart down, the force must be applied in the direction opposite the cart s direction of travel; the force F Æ pc and the force displacement r Æ F point in opposite directions. The work done by a force on a system is positive v F pc when the force and the force displacement point in the same direction and negative when they point in opposite directions. 9.4 Go back to Checkpoint 9.3 and answer the same two questions in terms of the directions of the applied forces and the force displacements. Do your answers agree with those you gave in Checkpoint 9.3? Let us next consider a situation involving changes in potential energy. Figure 9.3a shows a spring being compressed by two moving blocks. There are several ways to look at this problem. We can consider the spring and the blocks as a closed system whose energy must remain constant, or we can consider just the spring as a system on which the blocks do work. For the closed blocks-spring system, the potential energy of the spring increases during the compression by an amount equal to the decrease in the blocks kinetic energy. Because no external forces are exerted on the system, no work is involved. Considering just the spring as our system in Figure 9.3a, we note that the increase in the spring s potential energy results from the work done by the blocks on it. Because the spring s potential energy increases, the work done on the system must be positive. Indeed, the forces exerted by the blocks on the spring point in the same direction as the force displacements. Figure 9.3b shows the reverse situation: two blocks are held against a compressed spring and then released, so that they accelerate away from each other. For the closed blocks-spring system, the energy doesn t change, and we can account for the increase in kinetic energy of the blocks by a decrease in the potential energy of the expanding spring. If, however, we consider the spring by itself as our system, the decrease in energy of the system implies that the work done by the blocks on the spring is negative. Let us see if we reach the same conclusion for the * Careful! You may think the spring exerts a force on the blocks and not the other way around. Don t forget, however, that the force exerted by the spring on each block is just one part of an interaction pair. If you are having trouble picturing the force exerted by the blocks on the spring, imagine the spring expanding without having to push against the blocks: the expansion would be easier because there would be no resistance from the blocks. So, in a sense, the forces exerted by the blocks on the spring slows down the expansion of the spring.

5 Ch09n10v3.26_Ch09n10v2.34 9/24/10 5:14 PM Page energy diagrams 5 System = spring + blocks Forces are internal to system, so they do no work on system, and system s energy does not change. System = spring Forces are external to system, so work they do on system changes system s energy. (a) Moving blocks compress spring (b) Spring expands, pushing blocks apart 1 1 F 1s F 1s F 1s F 2s F 2s 1 2 F 2s 2 2 E E E F 1s 1 2 r F rf 1 2 r F F 1s F 1s F 2s F 2s rf F 2s 1 2 Force and force displacement in same direction: work done on system is positive, system energy increases. Force and force displacement in opposite directions: work done on system is negative, system energy decreases. Figure 9.3 Only forces that are external to a system can do work on the system and change the system s energy sign of the work in Figure 9.3b when we look at the directions of the forces and the force displacements. Because we are interested in the work done on the spring, we need to consider the forces exerted by the blocks on it. These forces still point toward the spring, as shown in Figure 9.3b. * The force displacements, however, now point in the opposite directions, away from the spring. Because each force and its force displacement point in opposite directions, the work done by the blocks on the spring is negative. Everything is consistent: both energy considerations and the directions of the forces and force displacements tell us that the work done by the blocks on the spring is negative as the spring expands. Exercise 9.2: Positive and negative work Is the work done by the following forces considered in Exercise 9.1 positive, negative, or zero? In each case the system is the object on which the force is exerted. (a) The force exerted by a hand compressing a spring. (b) The force exerted by Earth on a ball thrown up. (c) The force exerted by the ground on you at the instant you jump up. (d) The force exerted by the floor of an elevator on you as the elevator moves down at constant speed. Solution: (a) Positive. To compress a spring, I must move my hand in the same direction as I push. (b) Negative. The force exerted by Earth points down; the point of application moves up. (c) Zero, because the point of application is on the ground, which doesn t move. (d) Negative. The force exerted by the elevator floor points up; the point of application moves down. 9.5 Suppose that instead of the two moving blocks in Figure 9.3a, just one block is used to compress the spring while the other end of the spring is held against a wall. (a) Is the system comprising the block and the spring closed? (b) When the system is defined as being only the spring, is the work done by the block on the spring positive, negative, or zero? How can you tell? (c) Is the work done by the wall on the spring positive, negative, or zero?

6 Ch09n10v3.26_Ch09n10v2.34 9/24/10 5:14 PM Page 6 6 CHAPTER 9 WORK Concepts 9.3 Energy diagrams In analyzing situations involving work, we extend the bar diagrams introduced in Chapter 5 to include work done on the system. Consider, for example, the situation in Figure 9.4a. The energy of the system represented by this diagram increases from 10 J to 15 J. In order for this 5-J increase to occur, an amount of work equal to 5 J must be done by external forces on the system. The length of the vertical bar representing the final energy is equal to the sum of the lengths of the bars representing the initial energy and the work done by the external forces on the system. Figure 9.4b represents a system whose energy decreases. In order for the energy to decrease, the work done by external forces on the system must be negative, and so the bar representing this work extends below the baseline. Because any of the four kinds of energy discussed in Chapter 7 (kinetic energy, potential energy, source energy, thermal energy) can change in situations involving work, our bar diagrams need more detail in order to show in what form the energy transferred to a system appears when some external force does work on it. The additional bars make these diagrams more cumbersome to draw, however, and therefore we introduce the following simplification. Instead of drawing one set of bars for the initial individual energies and another set for the final individual energies, we draw just one set representing the change in each category of energy and add a fifth bar, denoted by W, representing the work done by external forces on the (a) initial + work = energy 10 J 5 J = 15 J final energy Positive work done on system increases system s energy. (b) initial energy 10 J + work = 6 J 4 J = final energy Negative work done on system decreases system s energy. Figure 9.4 Bar diagrams show how the energy of a system changes when work is done on the system. If the energy of the system increases, the work done on the system is positive. If the energy of the system decreases, the work done on the system is negative. The system s final energy is the sum of the initial energy and the work done on the system. (a) (b) (c) External work by person changes system s kinetic and potential energy. F pc v Earth initial energy change in energy system. We call these diagrams energy diagrams. Energy diagrams are a graphic representation of conservation of energy: they show that the energy of a system can only change due to transfer of energy into or out of the system. Just as with free-body diagrams, the first step in drawing an energy diagram is to define the system. In addition we should show any external force whose point of application undergoes a nonzero displacement these are the forces that do work on the system. Consider, for example, the situation shown in Figure 9.5a: a person pushing a pair of carts connectfinal energy K U E s E th K U E s E th K U E s E th W K F pc U v We can represent the changes in energy by initial and final bar diagrams or by a single energy diagram. Earth changes in system s energy equal work done on system Figure 9.5 Changes in a system s energy can be shown (a) by initial and final bar diagrams, or more compactly, (b) by a single energy diagram, which shows the changes in energy in each category and a column showing work done on the system.

7 Ch09n10v3.26_Ch09n10v2.34 9/24/10 5:14 PM Page Choice of system 7 PROCEDURE: Drawing energy diagrams 1. Specify the system under consideration by listing the components inside the system. 2. Sketch the system in its initial and final states. (The initial and final states may be defined for you by the problem, or you may have to choose yourself the most helpful states to examine.). Include in your sketch any external forces exerted on the system that undergo a nonzero force displacement and show a dot at the point of application of those forces. 3. Determine any nonzero changes in energy for each of the four categories of energy, taking into account the four basic energy conversion processes illustrated in Figure 7.13: a. Did the speed of any components of the system change? If so, determine whether the system s kinetic energy increased or decreased and draw a bar representing K for the system. For positive K, the bar extends above the baseline; for negative K, it extends below the baseline. (For some problems you may wish to draw separate K bars for different objects in the system; if so, be sure to specify clearly the system component corresponding to each bar and verify that the entire system is represented by the sum of the components.) b. Did the configuration of the system change in a reversible way? If so, draw a bar representing the change in potential energy U for the system. If necessary, draw separate bars for the changes in separate types of potential energy, such as changes in elastic and gravitational potential energy. c. Was any source energy consumed? If so, draw a bar showing E s. Source energy usually decreases, making E s negative, and so the bar extends below the baseline. Keep in mind that conversion of source energy is always accompanied by generation of thermal energy (Figure 7.13c and d). d. Does friction occur within the system, or is any source energy consumed? If so, draw a bar showing E th. In nearly all cases we consider, the amount of thermal energy increases, and so is positive. E th 4. Determine whether or not any work W is done by external forces on the system. Determine whether this work is negative or positive. Draw a bar representing this work, making the length of the bar equal to the sum of the lengths of the other bars in the diagram. If no work is done by external forces on the system, leave the bar for work blank, then go back and adjust the lengths of the other bars so that their sum is zero. ed by a spring. The pushing accelerates the carts and compresses the spring. Let us chose a system comprising the pair of carts and the spring; Figure 9.5a shows its initial and final states. We also draw a vector representing the external force F Æ pc exerted by the person on the left cart and indicate the position of the point of application of that force with a dot. Because the carts speed up, the final kinetic energy of the system is larger than the initial kinetic energy and the change in kinetic energy K K f K i is therefore positive. This change in kinetic energy corresponds to the upward pointing arrow marked K in Figure 9.5b, which is a set of the bar diagrams we used before now (no work values shown). The compression of the spring causes the elastic potential energy stored in the spring to increase, and so the change in potential energy U U f U i is also positive, as shown by the upward pointing arrow in Figure 9.5b. To represent this same information in an energy diagram, shown in Figure 9.5c, we draw bars to represent the changes K and U. The heights of the bars, which are equal to the heights of the the arrows in Figure 9.5b, now represents the changes in each energy category. The E s and E th bars are empty because there are no changes in these energies. The change in the system s energy comes from the work done by external forces on the system. The length of the bar representing the work done on the system in Figure 9.5c must therefore equal the sum of the K and U bars. Note that these energy arguments tell us that the work done by external forces on the system must be positive. We can verify this sign by checking the directions of the force exerted on the system and its

8 Ch09n10v3.26_Ch09n10v2.34 9/24/10 5:14 PM Page 8 8 CHAPTER 9 WORK Concepts force displacement. From Figure 9.5a, we see both are to the right and so the work done by external forces on the system is indeed positive. The general procedure for making energy diagrams is summarized in the Procedure box on page. Exercise 9.3: Cart launch A cart is at rest on a low-friction track. A person gives the cart a shove, and after moving down the track the cart hits a spring, which slows down the cart as the spring compresses. Draw an energy diagram for the system comprising the person and the cart over the time interval from the instant the cart is at rest until it has begun to slow down. Solution: Following the steps in the Procedure box, I begin by listing the objects making up the system: cart and person. Then I sketch the system in its initial and final states (Figure 9.6). The external forces exerted on the system are: force of gravity on person and cart, contact force exerted by ground on person, contact force exerted by track on cart, and contact force exerted by spring on cart. The vertical forces add up to zero, and so the work done by them on the system is zero. The force exerted by the spring on the cart, however, undergoes a force displacement, so I include that force in my diagram and indicate its point of application, as shown in Figure 9.6. Figure 9.6 Next I determine whether there are any changes in energy. Kinetic energy: the cart begins at rest and ends with nonzero speed, which means a gain in kinetic energy (Figure 9.7). Potential energy: neither the configuration of the cart nor that of the person change in a reversible way, and so the potential energy does not change. (The configuration of the spring does change in a reversible way, but the spring is not part of the system.) Source energy: the person has to exert effort to give the cart a shove, and so source energy is consumed. The source energy of the system therefore decreases. Thermal energy: no friction is involved, but the conversion of source energy is always accompanied by the generation of thermal energy, and so the thermal energy increases. (The person s muscles heat up.) In Figure 9.6 the force exerted by the spring on the cart points to the left and the force displacement is to the right. This means the work done by the spring on the cart is negative, and the W bar of my energy diagram must extend below the baseline. I adjust the length of the bars such that the length of the W bar is equal to the sum of the lengths of the other bars, yielding the energy diagram shown in Figure 9.7. Figure Draw an energy diagram for the cart in Figure 9.2b. Exercise 9.4: Compressing a spring A block initially at rest is released on an inclined surface. The block slides down, compressing a spring at the bottom of the incline; there is friction between the surface and the block. Consider the time interval from a little after release, when the block is moving at some initial speed v, until it comes to rest against the spring. Draw an energy diagram for the system comprising the block, surface, and Earth. Solution: I begin by listing the objects making up the system: block, surface, and Earth. Then I sketch the initial and final states of the system (Figure 9.8). The spring exerts external forces on the system. The bottom end of the spring exerts a force on the surface edge but this force has a force displacement of zero. The top end of the spring exerts a force F Æ c sb on the block. Because this force undergoes a nonzero force displacement, I include it in my diagram and show a dot at its point of application.

9 Ch09n10v3.26_Ch09n10v2.34 9/24/10 5:14 PM Page Dissipation of energy at a system boundary 9 Figure 9.8 Next I determine whether there are any energy changes. Kinetic energy: The block s kinetic energy goes to zero, and the kinetic energies of the surface and Earth do not change. Thus the kinetic energy of the system decreases, and K is negative. Potential energy: As the block moves down, the gravitational potential energy of the block-earth system decreases, and so U is negative. (Because the spring gets compressed, its elastic potential energy changes, but the spring is not part of the system.) Source energy: None (no fuel, food, or other source of energy is converted in this problem). Thermal energy: As the block slides, energy is dissipated by the friction between surface and block and so E th is positive. To determine the work done on the system, I look at the external forces exerted on it. The point of application of the external force F Æ c sb exerted by the spring on the block undergoes a force displacement opposite the direction of the force, and so that force does negative work on the system. Thus the work done on the system by the external forces on the system is negative, and the W bar extends below the baseline (Figure 9.9). I adjust the length of the bars such that the length of the W bar is equal to the sum of the lengths of the other three bars, yielding the energy diagram shown in Figure 9.9. Figure Draw an energy diagram for the situation presented in Exercise 9.4, but choose the system comprising block, spring, surface, and Earth. 9.4 Choice of system For a given situation, different choices of system yield different energy diagrams, as you just saw if you did Checkpoint 9.7. To examine this point further, consider the situation shown in Figure Using a long rope, a person lowers a basket from a balcony, smoothly putting the basket down on the ground. The basket initially moves downward at speed v i, but friction between the person s hands and the rope slows the basket so that it ends up at rest on the ground. For simplicity we assume no source energy is consumed the person just lets the rope slide, requiring no physical effort. Let us begin by choosing the basket and Earth as our system, a choice that leads to the energy diagram shown in Figure 9.10a. Because the basket comes to rest, its kinetic energy decreases. As the distance between Earth and basket decreases, the potential energy of the Earth-basket system decreases. There is no conversion of source energy nor any dissipation of energy within the system, and so we leave the columns for E s and E th blank. Next we examine the external forces exerted on the system. The person exerts forces both on Earth and, via the rope, on the basket. The latter involves a nonzero force displacement (the knot, which is the point of application for the force exerted by the rope on the basket, moves downward), and so work is done on the system. This work is negative because the force exerted by the rope on the basket is upward and the force displacement is downward. In order to make the work equal to the change in the system s energy, the length of the bar representing work must be made equal to the sum of the lengths of the K and U bars. What this energy diagram tells us is that the energy of the basket-earth system decreases as energy is transferred out of the system. Because energy is transferred out of the system, the work done by the person on the system is negative. What happens if we include the person and the rope in the system, as is Figure 9.10b? The changes in kinetic and potential energy are the same as before, but now the person no longer exerts external forces and therefore cannot do any work on the system. In fact, because we have included everything in the system, there are no external forces, and so there is no work done by external forces on the system the system is closed. For this choice of system, we have to take into account dissipation of energy because of friction

10 Ch09n10v3.26_Ch09n10v2.34 9/24/10 5:14 PM Page CHAPTER 9 Work Concepts (a) (b) (c) System = basket + Earth System = person + rope + basket + Earth System = person + rope + basket F c rb v Friction converts kinetic to thermal energy. v E E v F c rb F c rb v = 0 v = 0 v = 0 Earth Earth Earth Earth Earth F G Eb K U E s E th W K U E s E th W K U E s E th W Negative work done by rope on basket equals decrease in system s kinetic and potential energy. No work done on system; kinetic and potential energy transformed to thermal energy in system. Work done by Earth on basket equals decrease in system s kinetic energy plus increase in system s thermal energy. Figure 9.10 Different choices of system yield different energy diagrams. between the person s hands and the rope, and so we have a positive E th. The resulting energy diagram tells us the following: the initial kinetic energy of the basket and the initial potential energy of the system are converted to thermal energy (the rope and the person s hands heat up). No work is done by external forces on the system because the system is closed. This conclusion is entirely consistent with the one we drew on the basis of Figure 9.10a, but in the process we ve obtained more information on where the energy went. Finally we consider the same situation for a system comprising only basket, person, and rope, as in Figure 9.10c. The changes in kinetic and thermal energy are as before. For this choice of system, however, there is no change in gravitational potential energy (remember that gravitational potential energy refers to the configuration of Earth and the basket together, and Earth is no longer part of the system). What was counted as gravitational potential energy in Figure 9.10a and b appears as work done by the gravitational force on the system here. Earth exerts a downward external force on the basket as the basket moves down, and so Earth does positive work on the basket. (Earth also exerts a downward force on the person, but this force involves no force displacement.) In Section 9.5, I shall show that the positive work done by the gravitational force on the system in Figure 9.10c is exactly equal to the negative of the change in gravitational potential energy in Figure 9.10a and b. As this example shows, in order to avoid double-counting, it is important always to remember the following point: Gravitational potential energy always refers to the relative position of various parts within a system, never to the relative position of one component of the system and its environment. Whenever you are looking at the energy associated with the gravitational interaction between an object inside your system and an object outside the system, do not include that energy in the U bar of your energy diagram. That energy must be accounted for in the W part of the diagram instead. The energy diagram of Figure 9.10c shows how the initial kinetic energy of the basket and the positive

11 Ch09n10v3.26_Ch09n10v2.34 9/24/10 5:14 PM Page 11 [(H1F)] 11 work done by Earth on the system end up as thermal energy. All three choices of system lead to a valid conclusion. By looking at a situation from various points of view, you gain a better understanding of what energy conver sions and transfers take place. Also, by looking at a situation in various ways and verifying that your answers are consistent with one another, you can convince yourself that you have not overlooked something. 9.8 Draw an energy diagram for just the basket in Figure While you are generally free to chose whichever system suits you best, there are certain choices of system that lead to complications. In particular, When drawing an energy diagram, do not chose a system for which friction occurs at the boundary of the system. The reason for this is simple: wherever friction occurs, thermal energy is generated. The problem is that this thermal energy appears on both sides of the surface where friction occurs. If you pull on a rope to draw a large box toward yourself, as in Figure 9.11, friction between floor and box heats up both box and floor. The thermal energy generated ends up partly in the floor and partly in the box. How much ends up where depends, among other things, on the materials rubbing against each other and is not something that can be determined easily. So, if you chose just the box as your system, an unknown amount of energy flows out of the system, and it is no longer possible to do the energy accounting for the system. There is another reason friction at a system boundary leads to problems. Whenever friction occurs at a boundary, the frictional force is external to the system (it is exerted by the environment on the system) and hence has to be taken into account in the computation of the work done on the system. The force of friction, however, is not applied at a single well-defined location. Instead, it is distributed over the surfaces that move relative to each other. Consequently it is not possible to determine the force displacement, which is the other piece of information necessary to determine the work done on the system. These problems with friction at a system boundary don t mean we can never treat any situations involving friction. Both our analysis of Figure 9.10 and Exercise 9.4 involve friction, but for all choices of system we considered, the two objects r ubbing against each other were kept together either inside or outside the system. (a) System = person + box (b) System = person + box + Earth Friction generates thermal energy in Earth and box but we don t know how much ends up in Earth & how much in box. If system also includes Earth, all thermal energy remains in system v v v E v E Earth Earth Earth Earth So, if system excludes Earth, we can t do energy accounting. so energy accounting is easy.?? K U E s E th W K U E s E th W Figure 9.11 If you need to do energy accounting, don t choose a system for which friction occurs at the system boundary.

12 Ch09n10v3.26_Ch09n10v2.34 9/24/10 5:14 PM Page CHAPTER 9 Work Concepts Example 9.5: Puck and spring A puck sliding on a horizontal surface hits the free end of a spring that is held fixed at the other end. There is friction between puck and surface. Consider the interval from the instant before the puck comes in contact with the spring until the instant the puck has zero velocity and the spring reaches maximum compression. Draw an energy diagram for a system that contains the puck and for which the work done on the system is nonzero. (Your system may include objects in addition to the puck if you like.) 1 Getting started: I begin by making situation sketches at the beginning and end of the time interval of interest. The items in my sketches are the puck, the surface, and the spring (Figure 9.12). 3 Execute plan: My energy diagram is shown in Figure Because the puck slows to a stop, its kinetic energy decreases. Because there are no reversible changes in the system configuration and because no source energy is consumed, my bars for U and E s must be zero. The friction causes dissipation of energy, and so the E th bar shows an increase. Because the external force exerted by the spring on the system points in the direction opposite the direction of the force displacement, the work done by the spring on the system is negative. I adjust the length of the bars such that the length of the W bar is equal to the sum of the lengths of the other bars. Figure 9.13 Figure 9.12 If I include all three items in my system, the system is closed, and so the work done by external forces on the system is zero. Because there is friction between surface and puck, I must keep the two together. Given that the puck must be part of the system, I have no choice but to let my system be the puck plus the surface, and so I draw my system boundary around these two objects. The only external force exerted on this system is the contact force exerted by the spring on the puck, and so I add that force to my diagram and indicate its point of application. 2 Devise plan: All I have to do is follow the procedure for drawing an energy diagram for my system. 4 Evaluate result: The puck slows down under the combined action of the frictional force and the force exerted by the spring. So part of the puck s initial kinetic energy is converted to thermal energy, and the rest goes into compressing the spring. Because the spring is not part of the system, that energy leaves the system as (negative) work. 9.9 (a) Draw an energy diagram for the situation shown in Figure 9.11 for the system comprising Earth, the box, and the floor. Assume the box keeps moving at constant velocity, and consider the rope to be part of the person. (b) Explain each bar in the energy diagram of Figure 9.11a.

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14 Ch09n10v3.26_Ch09n10v2.34 9/24/10 5:14 PM Page CHAPTER 9 WORK Quantitative Tools SE L F - Q U I Z 9.5 Dissipation at the boundary of a system 1. A stunt performer falls from the roof of a two-story building onto a mattress on the ground. The mattress compresses, bringing the performer to rest without his getting hurt. Is the work done by the mattress on the performer positive, negative, or zero? Is the work done by the performer on the mattress positive, negative, or zero? 2. Consider a weightlifter holding a barbell motionless above his head. (a) Is the sum of the forces exerted on the barbell zero? (b) Is the weightlifter exerting a force on the barbell? (c) If yes, does this force do any work on the barbell? (d) Does the energy of the barbell change? (e) Are your answers to (c) and (d) consistent in light of the relationship between work and energy? 3. Do any of the systems in Figure 9.14 undergo a change in potential energy? If yes, is the change positive or negative? Ignore any friction. (a) Figure 9.14 (b) (c) (d) 14 v v = 0 v = 0 v v = 0 v v v = 0 v Earth Earth Earth Earth 4. On which systems in Figure 9.14 does the gravitational force do work? 5. A person pushes a cart up a small hill, gaining speed along the way. Friction in the cart s wheels is negligible. Name the components of the system depicted in each of the energy diagrams shown in Figure (a) (b) (c) Figure 9.15 Earth Earth Earth Earth K U E s E th W K U E s E th W K U E s E th W ANSWERS 1. Because the mattress brings the performer to rest, F Æ c mp must point upward. The point of application of this force moves downward as the mattress compresses, meaning the work done by the mattress on the performer is negative. The work done by the performer on the mattress is positive because F Æ c pm is downward and its point of application moves downward. 2. (a) Yes, because the barbell remains motionless. (b) Yes, he exerts an upward force to counter the downward gravitational force. (c) No, because the point at which the lifter exerts force on the barbell is not displaced. (d) No. (e) They are consistent. If no work is done on a system, the energy of the system does not change. 3. (a), (c) and (d). Because none of the situations involves friction or conversion of source energy, the only things you need to consider are changes in kinetic and potential energy and work done by external forces on the system. The system in (a) is closed, and so the decrease in kinetic energy causes an increase in gravitational potential energy: U 0. In (b), Earth is not included in the system, and so there can be no potential energy component of the system s energy. In (c), the system is closed, and so the increase in kinetic energy must be due to a decrease in gravitational potential energy: U 0. In (d), Earth does negative work on the system, and so the energy of the system must decrease. Because the kinetic energy increases, the elastic potential energy of the spring must decrease: U (b) and (d) because Earth is external to these systems. When Earth is part of the system, any energy associated with the gravitational interaction with Earth is included as potential energy. 5. If there is a change in potential energy, the cart and Earth must be in the system; if there is a change in source energy, the person must be in the system. (a) cart and Earth. The person does positive work on the system. (b) person, cart, and Earth. This system is closed; no work is done by external forces on it. Source energy is converted to increase the kinetic and potential energies (and some of it is dissipated). There is more kinetic and potential energy in this system than in (a) because the person is included in the system. (c) person and cart. Earth does negative work on the system.

15 Ch09n10v3.26_Ch09n10v2.34 9/24/10 5:14 PM Page Work done on a single particle Work done on a single particle In the first part of this chapter we defined work as the change in the energy of a system due to external forces and represented it by the letter W. The SI unit of work is the same as that of energy, J. When work is done by external forces on a system, the energy of the system changes by an amount E W. (9.1) This equation is called the energy law and embodies conservation of energy: because energy cannot be created or destroyed, the energy of a system can change only if energy is transferred into or out of the system. * For a system that is closed, W = 0 and so the energy law takes on the form we encountered in Chapter 5: E 0. To calculate the work done by an external force on a system, we start by considering a simple case: a single particle of inertia m subject to a constant force F Æ. The term particle refers to any object having no internal structure and no extent in space. Because it lacks extent and internal structure, a particle cannot change shape and therefore any internal energy is fixed ( E int 0 ). Only the kinetic energy of a particle can change, and so E K(particle). (9.2) To calculate what happens to the kinetic energy of a particle subject to a constant force F Æ, we consider the particle s motion over a time interval Æ t t f t i for the case where the particle s velocity at instant t i is vi. The constant force gives the particle a constant acceleration given by the equation of motion (Eq. 8.8): a x F x. (9.3) m F x m At the end of the time interval, the x component of the particle s velocity is given by Eq. 3.4, v x,f v x,i a x t, (9.4) and the x component of the particle s displacement is given by Eq. 3.7, x v x,i t 1 2 a x( t) 2. (9.5) Because we are considering a particle, the displacement of the particle is also the force displacement. Using these expressions, we can evaluate the change in the particle s kinetic energy: K K f K i 1 2 m(v2 f v 2 i ). (9.6) * For now, we consider only mechanical transfers of energy and ignore any other type of energy transfer. Later we ll also take into account transfer of energy by heating and radiation.

16 Ch09n10v3.26_Ch09n10v2.34 9/24/10 5:14 PM Page CHAPTER 9 WORK Quantitative Tools Substituting Eq. 9.4 into 9.6 and keeping in mind that v 2 2 v x and a 2 2 a x, we get K 1 2 m[(v x,i a x t) 2 v 2 x,i] 1 2 m[2v x,ia x t a 2 x ( t) 2 ] ma x [v x,i t 1 2 a x( t) 2 ] ma x x F F x x F, (9.7) where the substitution v x,i t 1 2 a x( t) 2 x F comes from Eq. 9.5, the substitution ma x =F x comes from Eq. 9.3, and where x F carries a subscript F to remind us that this is the force displacement (the displacement of the point of application of the force). Having found this result K F x x F, we can use Eq. 9.2, E K, to say E F x x F and then Eq. 9.1, E W, to finally say W F x x F (constant force exerted on particle, one dimension). (9.8) In words: For motion in one dimension the work done by a constant force exerted on a particle equals the product of the x components of the force and the force displacement. If the external force F Æ and the force displacement r Æ F point in the same direction, F x and x F are either both positive or both negative, and so the product F x x F is positive. Consequently, the kinetic energy increases. When the force and the particle s displacement point in opposite directions, F x and x F have opposite algebraic signs. The work done by the force on the particle is then negative, and so the kinetic energy of the particle decreases. If more than one force is exerted on a particle, the x component of its acceleration is given by Eq. 8.8: a x = S F x m. Substituting this expression for the x component of the acceleration into Eq. 9.7 gives K F x x F. Moving on to Eq. 9.8, we obtain the work done by these forces on the particle: W ( F x ) x F (constant forces exerted on particle, one dimension). (9.9) Equation 9.9 is what we call the work equation for a particle. Together with the energy law (Eq. 9.1), this equation allows us to deal with systems that are not closed. Notice the parallel between our treatments of momentum/impulse and energy/work (Figure 9.16). Conservation of momentum gives rise to the momentum Æ law (Eq. 4.18, p Æ Æ J, where J is the impulse delivered to the system). For an isolated system, the impulse is zero and so the momentum of the system does not change. If the system is not isolated, we can use the impulse equation, Eq. 8.25, Æ J F Æ t, to calculate the change in the momentum of the system. Conservation of energy gives rise to the energy law (Eq. 9.1). The work done on a closed system is zero, and so the energy of the system does not change. If the system is not closed, we can use the work equation, Eq. 9.9, to calculate the change in the energy of the system. When solving problems, it is possible to

17 Ch09n10v3.26_Ch09n10v2.34 9/24/10 5:14 PM Page Work done on a single particle 17 conservation of momenum conservation of energy p J (momentum law) E W (energy law) system isolated system not isolated system closed system not closed J 0 J ( F ) t (impulse equation) W 0 W ( F x ) x F (work equation) Figure 9.16 Energy and momentum changes for systems. choose either a closed system or a system that is not closed. As the next example shows, both choices lead to the same answer. If you are unable to find a solution via one approach, try the other. Although we derived Eqs. 9.8 and 9.9 for a particle, they can also be used for any rigid object as long as only the kinetic energy of the object changes. In that case, we can use Eqs. 9.2 to calculate the change in the energy of the object. Because all particles making up the object undergo the same displacement r Æ, Eq. 9.8 gives the work done by a constant force exerted on the object. Example 9.6: Work done by gravity A ball of inertia m b is released from rest and falls vertically. What is the ball s final kinetic energy after a displacement x x f x i? 1 Getting started: I begin by making a sketch of the initial and final conditions and drawing an energy diagram for the ball (Figure 9.17). I choose an x axis pointing up. Because the ball s internal energy doesn t change as it falls (its shape and temperature do not change), I can treat the ball as a particle. Therefore only its kinetic energy changes. I can also assume air resistance is small enough to be ignored, so that the only external force exerted on the ball is a constant gravitational force. This force has a nonzero force displacement and so does work on the ball. I therefore include this force in my diagram. 2 Devise plan: If I treat the ball as a particle, Eqs. 9.1 and 9.2 tell me that the change in the ball s kinetic energy is equal to the work done on it by the constant force of gravity, the x component of which is given by Eq. 8.17: F G Ebx m b g. (The minus sign shows that the force points in the negative x direction.) To calculate the work done by this force on the ball, I use Eq Execute plan: Substituting the x component of the gravitational force exerted on the ball and the force displacement x f x i into Eq. 9.8, I get initial final W F G Ebx x F m b g(x f x i ). Figure 9.17