Big Ideas 1 Work 2 Kinetic 3 Power is force times distance. energy is one-half mass times velocity sqared. is the rate at which work is done. 7 Work and Kinetic Energy The work done by this cyclist can increase his energy of motion. The relationship between work (force times distance) and energy of motion (kinetic energy) is developed in detail in this chapter. In this chapter we introdce the idea that force times distance is an important new physical qantity, which we refer to as work. Closely related to work is the energy of motion, or kinetic energy. Together, work and kinetic energy set the stage for a deeper nderstanding of the physical world. 195
196 CHAPTER 7 Work and KinETic Energy 7-1 Work Done by a Constant Force In this section we define work in the physics sense of the word and apply or definition to a variety of physical sitations. We start with the simplest case namely, the work done when force and displacement are in the same direction. Force in the Direction of Displacement When we psh a shopping cart in a store or pll a sitcase throgh an airport, we do work. The greater the force, the greater the work; the greater the distance, the greater the work. These simple ideas form the basis for or definition of work. To be specific, sppose we psh a box with a constant force F, as shown in FIGURE 7-1. If we move the box in the direction of F throgh a displacement d, the work W we have done is Fd: Definition of Work, W, When a Constant Force Is in the Direction of Displacement W = Fd 7-1 SI nit: newton-meter (N # m) = jole, J Work is the prodct of two magnitdes, and hence it is a scalar. In addition, notice that a small force acting over a large distance gives the same work as a large force acting over a small distance. For example, W = 11 N21400 m2 = 1400 N211 m2. A constant force of magnitde F, c F cacting in the direction of a displacement of magnitde d, c d F cdoes work W = Fd on the object. FIGURE 7-1 Work: constant force in the direction of motion A constant force F pshes a box throgh a displacement d. In this special case, where the force and displacement are in the same direction, the work done on the box by the force is W = Fd. The dimensions of work are newtons (force) times meters (distance), or N m. This combination of dimensions is called the jole (rhymes with school, as commonly prononced) in honor of James Prescott Jole (1818 1889), a dedicated physicist who is said to have condcted physics experiments even while on his honeymoon. We define a jole as follows: Definition of the jole, J 1 jole = 1 J = 1 N # m = 11kg # ms 2 2 # m = 1 kg # m 2 s 2 7-2 To get a better feeling for work and the associated nits, sppose yo exert a force of 82.0 N on the box in Figre 7-1 and move it in the direction of the force throgh a distance of 3.00 m. The work yo have done is W = Fd = 182.0 N213.00 m2 = 246 N # m = 246 J Similarly, if yo do 5.00 J of work to lift a book throgh a vertical distance of 0.750 m, the force yo have exerted on the book is F = W d = 5.00 J 0.750 m = 5.00 N # m 0.750 m = 6.67 N
7-1 Work Done by a Constant FoRCE 197 EXERCISE 7-1 WORK DONE One species of Darwin s finch, Geospiza magnirostris, can exert a force of 217 N with its beak as it cracks open a Tribls seed case. If its beak moves throgh a distance of 0.42 cm dring this operation, how mch work does the finch do to get the seed? REASONING AND SOLUTION The work the finch does is force times distance. In this case, the force is 217 N and the distance is 0.42 cm = 0.0042 m. Ths, W = Fd = 1217 N210.0042 m2 = 0.91 J This is a relatively small amont of work for a hman, bt significant for a small bird. Jst how mch work is a jole, anyway? Well, yo do one jole of work when yo lift a gallon of milk throgh a height of abot an inch, or lift an apple a meter. One jole of work lights a 100-watt lightblb for 0.01 second or heats a glass of water 0.00125 degree Celsis. Clearly, a jole is a modest amont of work in everyday terms. Additional examples of work are listed in Table 7-1. TABLE 7-1 Typical Vales of Work Eqivalent Activity work (J) Annal U.S. energy se 8 * 10 19 Mt. St. Helens erption 10 18 Brning 1 gallon of gas 10 8 Hman food intake/day 10 7 Melting an ice cbe 10 4 Lighting a 100-W blb for 6000 1 minte Heartbeat 0.5 Trning a page of a book 10-3 Hop of a flea 10-7 Breaking a bond in DNA 10-20 EXAMPLE 7-2 HEADING FOR THE ER An intern pshes an 87-kg patient on an 18-kg grney, prodcing an acceleration of 0.55 ms 2. (a) How mch work does the intern do in pshing the patient and grney throgh a distance of 1.9 m? Assme the grney moves withot friction. (b) How far mst the intern psh the grney to do 140 J of work? PICTURE THE PROBLEM Or sketch shows the physical sitation for this problem. Notice that the force exerted by the intern is in the same direction as the displacement of the grney; therefore, we know that the work is W = Fd. REASONING AND STRATEGY We are not given the magnitde of the force, F, so we cannot apply Eqation 7-1 1W = Fd2 directly. However, we are given the mass and acceleration of the patient and grney, and from them we can calclate the force with F = ma. The work done by the intern is then W = Fd, where d = 1.9 m. Known Mass of patient, 87 kg; mass of grney, 18 kg; acceleration, a = 0.55 ms 2 ; (a) pshing distance, d = 1.9 m; (b) work, W = 140 J. Unknown (a) Work done, W =? (b) Pshing distance, d =? F d = 1.9 m a = 0.55 m/s 2 SOLUTION Part (a) 1. First, find the force F exerted by the intern: F = ma = 187 kg + 18 kg210.55 ms 2 2 = 58 N 2. The work done by the intern, W, is the force times the distance: W = Fd = 158 N211.9 m2 = 110 J Part (b) 3. Use W = Fd to solve for the distance d: d = W F = 140 J 58 N = 2.4 m INSIGHT Yo might wonder whether the work done by the intern depends on the speed of the grney. The answer is no. The work done on an object, W = Fd, doesn t depend on whether the object moves throgh the distance d qickly or slowly. What does depend on the speed of the grney is the rate at which work is done, which we discss in detail in Section 7-4. PRACTICE PROBLEM PREDICT/CALCULATE (a) If the total mass of the grney pls patient is halved and the acceleration is dobled, does the work done by the intern increase, decrease, or remain the same? Explain. (b) Determine the work in this case. [Answer: (a) The work remains the same becase the two changes offset one another; that is, F = ma = 1m2212a2. (b) The work is 110 J, as before.] Some related homework problems: Problem 3, Problem 4
198 CHAPTER 7 Work and KinETic Energy Zero Distance Implies Zero Work Before moving on, let s note an interesting point abot or definition of work. It s clear from Eqation 7-1 that the work W is zero if the distance d is zero and this is tre regardless of how great the force might be, as illstrated in FIGURE 7-2. For example, if yo psh against a solid wall, yo do no work on it, even thogh yo may become tired from yor efforts. Similarly, if yo stand in one place holding a 50-pond sitcase in yor hand, yo do no work on the sitcase, even thogh yo soon feel worn ot. The fact that we become tired when we psh against a wall or hold a heavy object is de to the repeated contraction and expansion of individal cells within or mscles. Ths, even when we are at rest, or mscles are doing mechanical work on the microscopic level. Force at an Angle to the Displacement In FIGURE 7-3 we see a person plling a sitcase on a level srface with a strap that makes an angle with the horizontal in this case the force is at an angle to the direction of motion. How do we calclate the work now? Well, instead of force times distance, we say that work is the component of force in the direction of displacement times the magnitde of the displacement. In Figre 7-3, the component of force in the direction of the displacement is F cos and the magnitde of the displacement is d. Therefore, the work is F cos times d: Definition of Work When the Angle Between a Constant Force and the Displacement Is U W = 1F cos 2d = Fd cos 7-3 SI nit: jole, J Of corse, in the case where the force is in the direction of motion, the angle is zero; then W = Fd cos 0 = Fd # 1 = Fd, in agreement with Eqation 7-1. The component of force in the direction of displacement is F cos. This is the only component of the force that does work. FIGURE 7-2 Visalizing Concepts Force, Work, and Displacement The person doing psh ps does positive work as he moves pward, bt does no work when he holds himself in place. The weightlifter does positive work as she raises the weights, bt no work is done on the weights as she holds them motionless above her head. F F cos F F cos F d cos d d sin FIGURE 7-4 Force at an angle to direction of motion: another look The displacement of the sitcase in Figre 7-3 is eqivalent to a displacement of magnitde d cos in the direction of the force F, pls a displacement of magnitde d sin perpendiclar to the force. Only the displacement parallel to the force reslts in nonzero work; hence the total work done is F1d cos 2 as expected. d The work in this case is W = (F cos )d. FIGURE 7-3 Work: force at an angle to the direction of motion A person plls a sitcase with a strap at an angle to the direction of motion. The component of force in the direction of motion is F cos, and the work done by the person is W = (F cos )d. Eqally interesting is a sitation in which the force and the displacement are at right angles to one another. In this case = 90 and the work done by the force F is zero; W = Fd cos 90 = 0. This reslt leads natrally to an alternative way to think abot the expression W = Fd cos. In FIGURE 7-4 we show the displacement and the force for the sitcase in Figre 7-3. Notice that the displacement is eqivalent to a displacement in the direction of the force of magnitde 1d cos 2 pls a displacement at right angles to the force of magnitde 1d sin 2. Since the displacement at right angles to the force corresponds to zero work and the displacement in the direction of the force corresponds to a work W = F1d cos 2, it follows that the work done in this case is Fd cos, as given in Eqation 7-3. Ths, the work done by a force can be thoght of in the following two eqivalent ways:
7-1 Work Done by a Constant FoRCE 199 (i) Work is the component of force in the direction of the displacement times the magnitde of the displacement. (ii) Work is the component of displacement in the direction of the force times the magnitde of the force. In both interpretations, the mathematical expression for work is exactly the same, W = Fd cos, where is the angle between the force vector and the displacement vector when they are placed tail-to-tail. This definition of is illstrated in Figre 7-4. Finally, we can also express work as the dot prodct of the vectors F and d ; that is, W = F # d = Fd cos. Notice that the dot prodct, which is always a scalar, is simply the magnitde of one vector times the magnitde of the second vector times the cosine of the angle between them. We discss the dot prodct in greater detail in Appendix A. EXAMPLE 7-3 GRAVITY ESCAPE SYSTEM Big Idea 1 Work is force times distance when force and displacement are in the same direction. More generally, work is the component of force in the direction of displacement times the distance. PHYSICS IN CONTEXT Looking Back Even thogh work is a scalar qantity, notice that vector components (Chapter 3) are sed in the definition of work. RWP* In a gravity escape system (GES), an enclosed lifeboat on a large ship is deployed by letting it slide down a ramp and then contine in free fall to the water below. Sppose a 4970-kg lifeboat slides a distance of 5.00 m on a ramp, dropping throgh a vertical height of 2.50 m. How mch work does gravity do on the boat? PICTURE THE PROBLEM From or sketch, we see that the force of gravity mg and the displacement d are at an angle relative to one another when placed tail-to-tail, and that is also the angle the ramp makes with the vertical. In addition, we note that the vertical height of the ramp is h = 2.50 m and the length of the ramp is d = 5.00 m. REASONING AND STRATEGY By definition, the work done on the lifeboat by gravity is W = Fd cos, where F = mg, d = 5.00 m, and is the angle between mg and d. We are not given in the problem statement, bt from the right triangle that forms the ramp we see that cos = hd. Once is determined from the geometry of or sketch, it is straightforward to calclate W. Known Mass of lifeboat, m = 4970 kg; sliding distance, d = 5.00 m; vertical height, h = 2.50 m. Unknown Work done by gravity, W =? h = 2.50 m mg mg d d SOLUTION 1. First, find the component of F = mg in the direction of motion: F cos = 1mg2a h d b = 14970 kg219.81 ms 2 2a 2.50 m 5.00 m b = 24,400 N 2. Mltiply by distance to find the work: W = 1F cos 2d = 124,400 N215.00 m2 = 122,000 J 3. Alternatively, cancel d algebraically before sbstitting nmerical vales: W = Fd cos = 1mg21d2a h d b = mgh = 14970 kg219.81 ms 2 212.50 m2 = 122,000 J INSIGHT The work is simply W = mgh, exactly the same as if the lifeboat had fallen straight down throgh the height h. Working the problem symbolically, as in Step 3, reslts in two distinct advantages. First, it makes for a simpler expression for the work. Second, and more important, it shows that the distance d cancels; hence the work depends on the height h bt not on the distance. Sch a reslt is not apparent when we work solely with nmbers, as in Steps 1 and 2. PRACTICE PROBLEM Sppose the lifeboat slides halfway to the water, gets stck for a moment, and then starts p again and contines to the end of the ramp. What is the work done by gravity in this case? [Answer: The work done by gravity is exactly the same, W = mgh, independent of how the boat moves down the ramp.] Some related homework problems: Problem 10, Problem 11 Next, we present a Conceptal Example that compares the work reqired to move an object along two different paths. *Real World Physics applications are denoted by the acronym RWP.
200 CHAPTER 7 Work and KinETic Energy CONCEPTUAL EXAMPLE 7-4 PATH DEPENDENCE OF WORK Yo want to load a box into the back of a trck, as shown in the sketch. One way is to lift it straight p with constant speed throgh a height h, as shown, doing a work W 1. Alternatively, yo can slide the box p a loading ramp with constant speed a distance L, doing a work W 2. Assming the box slides on the ramp withot friction, which of the following statements is correct: (a) W 1 6 W 2, (b) W 1 = W 2, (c) W 1 7 W 2? W 1 W 2 f L F 1 mg h f F 2 mg sin f mg f L mg cos f h REASONING AND DISCUSSION Yo might think that W 2 is less than W 1, becase the force needed to slide the box p the ramp, F 2, is less than the force needed to lift it straight p. On the other hand, the distance p the ramp, L, is greater than the vertical distance, h, so perhaps W 2 shold be greater than W 1. In fact, these two effects cancel exactly, giving W 1 = W 2. To see this, we first calclate W 1. The force needed to lift the box with constant speed is F 1 = mg, and the height is h; therefore W 1 = mgh. Next, the work to slide the box p the ramp with constant speed is W 2 = F 2 L, where F 2 is the force reqired to psh against the tangential component of gravity. In the sketch we see that F 2 = mg sin f. The sketch also shows that sin f = hl; ths W 2 = 1mg sin f2l = 1mg21hL2L = mgh = W 1 : The two works are identical! Clearly, the ramp is a sefl device it redces the force reqired to move the box pward from F 1 = mg to F 2 = mg1hl2. Even so, it doesn t decrease the amont of work we need to do. As we have seen, the redced force on the ramp is offset by the increased distance. ANSWER (b) W 1 = W 2 Negative Work and Total Work Work depends on the angle between the force, F -908 6 6 08, and the displacement (or direction of motion), d (a) Positive work. This dependence gives rise to three distinct possibilities, as shown in FIGURE 7-5: (i) Work is positive if the force has a component F in the direction of motion 1-90 6 6 90 2. (ii) Work is zero if the force has no component in the direction of motion 1 d F = {90 2. (iii) Work is negative if the force has a component -908 opposite 6 6 to 08the direction of motion d (90 6 6 270 ). (a) Positive work = {908 Ths, whenever we calclate work, we mst be carefl abot its sign and not jst (b) Zero work assme it is positive. F d F d -908 6 6 08 (a) Positive work F d = {908 (b) Zero work F d 908 6 6 2708 (c) Negative work FIGURE 7-5 Positive, negative, and zero work Work is positive when the force is in the same general direction as the displacement and is negative if the force is generally opposite to the displacement. Zero work is done if the force is at right angles to the displacement. F F d d 908 6 6 2708 = {908 (c) Negative work
7-1 Work Done by a Constant FoRCE 201 When more than one force acts on an object, the total work is the sm of the work done by each force separately. Ths, if force F 1 does work W 1, force F 2 does work W 2, and so on, the total work is W total = W 1 + W 2 + W 3 + g = a W 7-4 Eqivalently, the total work can be calclated by first performing a vector sm of all the forces acting on an object to obtain F total and then sing or basic definition of work: W total = 1F total cos 2d = F total d cos 7-5 In this expression, is the angle between F total and the displacement d. In the next two Examples we calclate the total work in each of these ways. PROBLEM-SOLVING NOTE Be Carefl Abot the Angle U In calclating W = Fd cos, be sre that the angle yo se in the cosine is the angle between the force and the displacement vectors when they are placed tail-to-tail. Sometimes may be sed to label a different angle in a given problem. For example, is often sed to label the angle of a slope, in which case it may have nothing to do with the angle between the force and the displacement. To smmarize: Jst becase an angle is labeled doesn t mean it s atomatically the correct angle to se in the work formla. EXAMPLE 7-5 A COASTING CAR I A car of mass m coasts down a hill inclined at an angle f below the horizontal. The car is acted on by three forces: (i) the normal force N exerted by the road, (ii) a force de to air resistance, F f F air, and (iii) the force of gravity, mg d air. Find the total work done on the car as it travels a distance d along the road. f mg PICTURE THE PROBLEM Becase f is the angle the slope makes with the horizontal, it is also the angle between mg and the downward normal direction, as was shown in Figre 5-22. It follows that the angle between mg and the displacement d f is = 90 - f. Or sketch also shows that the angle between N and d is = 90, and the angle between F air and d is = 180. d f F air N d N = 908 d N f mg = 908 - f f mg f F air = 1808 d REASONING AND STRATEGY For each force we calclate the work sing W = Fd cos, where is the angle between that particlar force and the displacement d. The total work is the sm of the d work done by each of the three forces. Known Mass of car, m; angle of incline, f; = distance, 908 - f d; forces acting on car, N, F air, and mg. N Unknown Work done on car, W =? f = 908 SOLUTION d mg 1. We start with the work done by the normal force, N. W N = Nd cos = Nd cos 90 = Nd102 = 0 From the figre we see that = 90 for this force: 2. For the force of air resistance, = 180 : W air = F air d cos 180 = F air d1-12 = -F air d 3. For gravity the angle = 1808 F is = 90 - f, as indicated in the sketch. W mg = mgd cos 190 - f2 = mgd sin f air d Recall that cos190 - f2 = sin f (see Appendix A ): 4. The total work is the sm of the individal works: W total = W N + W air + W mg = 0 - F air d + mgd sin f INSIGHT The normal force is perpendiclar to the motion of the car, and ths does no work. Air resistance points in a direction that opposes the motion, so it does negative work. On the other hand, gravity has a component in the direction of motion; therefore, its work is positive. The physical significance of positive, negative, and zero work will be discssed in detail in the next section. PRACTICE PROBLEM Calclate the total work done on a 1550-kg car as it coasts 20.4 m down a hill with f = 5.00. Let the force de to air resistance be 15.0 N. [Answer: W total = W N + W air + W mg = 0 - F air d + mgd sin f = 0-306 J + 2.70 * 10 4 J = 2.67 * 10 4 J] Some related homework problems: Problem 15, Problem 63
202 CHAPTER 7 Work and KinETic Energy In the next Example, we first sm the forces acting on the car to find F total. Once the total force is determined, we calclate the total work sing W total = F total d cos. EXAMPLE 7-6 A COASTING CAR II Consider the car described in Example 7-5. Calclate the total work done on the car sing W total = F total d cos. PICTURE THE PROBLEM First, we choose the x axis to point down the slope, and the y axis to be at right angles to the slope. With this choice, there is no acceleration in the y direction, which means that the total force in that direction mst be zero. As a reslt, the total force acting on the car is in the x direction. The magnitde of the total force is mg sin f - F air, as can be seen in or sketch. mg cos f REASONING AND STRATEGY f We begin by finding the x component of each force vector, and then we sm them to find the total force acting on the car. As can be seen from the sketch, the total force points in the positive mg x direction that is, in the same direction as the displacement. Therefore, the angle in W = F total d cos is zero. Known Mass of car, m; angle of incline, f; distance, d; forces acting on car, N, F air, and mg. Unknown Work done on car, W =? F air N mg sin f F air mg sin f x force components (Enlarged) F total = mg sin f - F air y x SOLUTION 1. Referring to the sketch, we see that the magnitde of the total F total = mg sin f - F air force is mg sin f mins F air : 2. The direction of F total is the same as the direction of d ; W total = F total d cos = 1mg sin f - F air 2d cos 0 ths = 0. We can now calclate W total : = mgd sin f - F air d INSIGHT Notice that we were carefl to calclate both the magnitde and the direction of the total force. The magnitde (which is always positive) gives F total and the direction gives = 0, allowing s to se W total = F total d cos. PRACTICE PROBLEM Sppose the total work done on a 1620-kg car as it coasts 25.0 m down a hill with f = 6.00 is W total = 3.75 * 10 4 J. Find the magnitde of the force de to air resistance. [Answer: F air d = -W total + mgd sin f = 4030 J; ths F air = 14030 J2d = 161 N] Some related homework problems: Problem 15, Problem 63 Enhance Yor Understanding (Answers given at the end of the chapter) 1. A block slides a distance d to the right on a horizontal srface, as indicated in FIGURE 7-6. Rank the for forces that act on the block 1F, N, mg, ƒ k 2 in order of the amont of work they do, from most negative to most positive. Indicate ties where appropriate. f k N mg d F PROBLEM-SOLVING NOTE Work Can Be Positive, Negative, or Zero When yo calclate work, be sre to keep track of whether it is positive or negative. The distinction is important, since positive work increases speed, whereas negative work decreases speed. Zero work, of corse, has no effect on speed. Section Review FIGURE 7-6 The work done by a force is W = 1F cos 2d = Fd cos, where F is the force, d is the distance, and is the angle between the force and displacement. CONTINUED
7-2 KinETic Energy and the Work Energy Theorem 203 Work is positive when the force and displacement are in the same direction, zero if the force is perpendiclar to the displacement, and negative if the force and displacement are in opposite directions. 7-2 Kinetic Energy and the Work Energy Theorem Sppose yo drop an apple. As it falls, gravity does positive work on it, as indicated in FIGURE 7-7, and its speed increases. If yo toss the apple pward, gravity does negative work on it, and the apple slows down. In general, whenever the total work done on an object is positive, its speed increases; whenever the total work is negative, its speed decreases. In this section we derive an important reslt, the work energy theorem, which makes this connection between work and change in speed precise. Work and Kinetic Energy To begin, consider an apple of mass m falling throgh the air, and sppose that two forces act on the apple gravity, mg, and the average force of air resistance, F air. The total force acting on the apple, F total, gives the apple a constant downward acceleration of magnitde a = F total m Since the total force is downward and the motion is downward, the work done on the apple is positive. Now, sppose that the initial speed of the apple is v i, and that after falling a distance d its speed increases to v f. The apple falls with constant acceleration a; hence constant-acceleration kinematics (Eqation 2 12) gives With a slight rearrangement we find v f 2 = v i 2 + 2ad 2ad = v 2 2 f - v i Next, we sbstitte a = F total m into this eqation: 2a F total m bd = v f 2 2 - v i Mltiplying both sides by m and dividing by 2 yields F total d = 1 2 mv 2 f - 1 2 mv 2 i Force is in the direction of displacement c cso positive work is done on the apple. This cases the apple to speed p. (a) d mg Apple falling: W + 0, speed increases Negative work done on the apple c d mg ccases it to slow down. In this expression, F total d is simply the total work done on the apple. Ths we find W total = 1 2 mv 2 f - 1 2 mv 2 i This shows that the total work is directly related to the change in speed. Notice that W total 7 0 means v f 7 v i, W total 6 0 means v f 6 v i, and W total = 0 means v f = v i. The qantity 1 2 mv2 in the eqation for W total has a special significance in physics. We call it the kinetic energy, K: Definition of Kinetic Energy, K K = 1 2 mv2 7-6 SI nit: kg # m 2 s 2 = jole, J In general, the kinetic energy of an object is the energy de to its motion. We measre kinetic energy in joles, the same nits as work, and both kinetic energy and work are scalars. Unlike work, however, kinetic energy is never negative. Instead, K is always greater than or eqal to zero, independent of the direction of motion or the direction of any forces. (b) Apple tossed pward: W * 0, speed decreases FIGURE 7-7 Gravitational work (a) The work done by gravity on an apple that moves downward is positive. If the apple is in free fall, this positive work will reslt in an increase in speed. (b) In contrast, the work done by gravity on an apple that moves pward is negative. If the apple is in free fall, the negative work done by gravity will reslt in a decrease of speed.
204 CHAPTER 7 Work and KinETic Energy PHYSICS IN CONTEXT Looking Back The kinematic eqations of motion for constant acceleration (Chapters 2 and 4) are sed in the derivation of kinetic energy. PHYSICS IN CONTEXT Looking Ahead In Chapter 8 we introdce the concept of potential energy. The combination of kinetic and potential energy is referred to as mechanical energy, which will play a central role in or discssion of the conservation of energy. TABLE 7-2 Typical Kinetic Energies Approximate kinetic Sorce energy (J) Jet aircraft at 500 mih 10 9 Car at 60 mih 10 6 Home-rn baseball 10 2 Person at walking speed 50 Hosefly in flight 10-4 To get a feeling for typical vales of kinetic energy, consider yor kinetic energy when jogging. If we assme a mass of abot 62 kg and a speed of 2.5 ms, yor kinetic energy is K = 1 162 kg212.5 ms22 = 190 J. Additional examples of kinetic energy are 2 given in Table 7-2. EXERCISE 7-7 KINETIC ENERGY TAKES OFF A small airplane moving along the rnway dring takeoff has a mass of 690 kg and a kinetic energy of 25,000 J. (a) What is the speed of the plane? (b) By what mltiplicative factor does the kinetic energy of the plane change if its speed is tripled? REASONING AND SOLUTION (a) The kinetic energy of an object is given by K = 1 2 mv2. We know that K = 25,000 J and m = 690 kg; therefore the speed is 2K v = B m = 2125,000 J2 = 8.5 ms B 690 kg (b) Kinetic energy depends on the speed sqared, and hence tripling the speed increases the kinetic energy by a factor of nine. The Work Energy Theorem In the preceding discssion we sed a calclation of work to derive an expression for the kinetic energy of an object. The precise connection we derived between these qantities is known as the work energy theorem. This theorem can be stated as follows: Work Energy Theorem The total work done on an object is eqal to the change in its kinetic energy: W total = K = 1 2 mv f 2-1 2 mv i 2 7-7 Big Idea 2 Kinetic energy is onehalf mass times velocity sqared. The total work done on an object is eqal to the change in its kinetic energy. Ths, the work energy theorem says that when a force acts on an object over a distance doing work on it the reslt is a change in the speed of the object, and hence a change in its energy of motion. Eqation 7-7 is the qantitative expression of this connection. We have derived the work energy theorem for a force that is constant in direction and magnitde, bt it is valid for any force. In fact, the work energy theorem is completely general, making it one of the most important and fndamental reslts in physics. It is also a very handy tool for problem solving, as we shall see many times throghot this text. PHYSICS IN CONTEXT Looking Ahead The kinetic energy before and after a collision is an important characterizing featre, as we shall see in Chapter 9. In addition, look for kinetic energy to reappear when we discss rotational motion in Chapter 10, and again when we stdy ideal gases in Chapter 17. EXERCISE 7-8 WORK TO ACCELERATE A 220-kg motorcycle is crising at 14 ms. What is the total work that mst be done on the motorcycle to increase its speed to 19 ms? REASONING AND SOLUTION From the work energy theorem, we know that the total work reqired to change an object s kinetic energy is W total = K = 1 2 mv f 2-1 2 mv i 2. Given that m = 220 kg, v i = 14 ms, and v f = 19 ms, we find W total = 1 2 mv f 2-1 2 mv 2 i = 1 2 1220 kg2119 ms22-1 2 1220 kg2114 ms22 = 18,000 J This mch energy cold lift a 90-kg person throgh a vertical distance of abot 20 m. We now present Examples showing how the work energy theorem is sed in practical sitations.
EXAMPLE 7-9 HIT THE BOOKS 7-2 KinETic Energy and the Work Energy Theorem 205 A 4.10-kg box of books is lifted vertically from rest a distance of 1.60 m with a constant, pward applied force of 52.7 N. Find (a) the work done by the applied force, (b) the work done by gravity, and (c) the final speed of the box. PICTURE THE PROBLEM Or sketch shows that the direction of motion of the box is pward. In addition, we see that the applied force, F app, is pward and the force of gravity, mg, is downward. Finally, the box is lifted from rest (v i = 0) throgh a distance y = 1.60 m. v f =? y REASONING AND STRATEGY The applied force is in the direction of motion, so the work it does, W app, is positive. Gravity is opposite in direction to the motion; ths its work, W mg, is negative. The total work is the sm of W app and W mg and the final speed of the box is fond by applying the work energy theorem, W total = K. Known Mass of box, m = 4.10 kg; vertical distance, y = 1.60 m; applied force, F app = 52.7 N. Unknown (a) Work done by applied force, W app =? (b) work done by gravity, W mg =? (c) Final speed of box, v f =? v i = 0 F app mg y = 1.60 m SOLUTION Part (a) 1. First we find the work done by the applied force. W app = F app cos 0 y = 152.7 N211211.60 m2 = 84.3 J In this case, = 0 and the distance is y = 1.60 m: Part (b) 2. Next, we calclate the work done by gravity. The W mg = mg cos 180 y distance is y = 1.60 m, as before, bt now = 180 : = 14.10 kg219.81 ms 2 21-1211.60 m2 = -64.4 J Part (c) 3. The total work done on the box, W total, is the W total = W app + W mg = 84.3 J - 64.4 J = 19.9 J sm of W app and W mg : 4. To find the final speed, v f, we apply the work energy theorem. Recall that the box started at rest; ths v i = 0: W total = 1 2 mv f 2-1 2 mv i 2 = 1 2 mv 2 f INSIGHT As a check on or reslt, we can find v f in a completely different way. First, calclate the acceleration of the box with the reslt a = 1F app - mg2m = 3.04 ms 2. Next, se this reslt in the kinematic eqation v 2 = v 0 2 + 2a y. With v 0 = 0 and y = 1.60 m, we find v = 3.12 ms, in agreement with the reslts sing the work energy theorem. PRACTICE PROBLEM PREDICT/CALCULATE (a) If the box is lifted only a qarter of the distance, y = 11.60 m24 = 0.400 m, is the final speed 18, 14, or 12 of the vale fond in this Example? Explain. (b) Determine v f in this case. [Answer: (a) Work depends linearly on y, and v f depends on the sqare root of the work. As a reslt, it follows that the final speed is 214 = 1 the vale fond in 2 Step 4. (b) We find v f = 1 13.12 ms2 = 1.56 ms.] 2 Some related homework problems: Problem 28, Problem 29, Problem 65 v f = B 2W total m = 2119.9 J2 = 3.12 ms B 4.10 kg In the previos Example the initial speed was zero. This is not always the case, of corse. The next Example considers a case with a nonzero initial speed. EXAMPLE 7-10 GOING FOR A STROLL When a father takes his daghter ot for a spin in a stroller, he exerts a force F with a magnitde of 44.0 N at an angle of 32.0 below the horizontal, as shown in the sketch. The stroller and child together have a mass m = 22.7 kg. (a) How mch work does the father do on the stroller when he pshes it throgh a horizontal distance d = 1.13 m? (b) If the initial speed of the stroller is v i = 1.37 ms, what is its final speed? (Assme the stroller rolls with negligible friction.) CONTINUED
206 CHAPTER 7 Work and KinETic Energy PICTURE THE PROBLEM Or sketch shows the direction of motion and the directions of each of the forces. The normal forces and the force de to gravity are vertical, whereas the displacement is horizontal. The force exerted by the father has a vertical component, -F sin 32.0, and a horizontal component, F cos 32.0, where F = 44.0 N. Components of F (expanded view) F cos 32.08 32.08 32.08 F -F sin 32.08 F REASONING AND STRATEGY a. The normal forces 1 N 1 and N 22 and the weight (mg d ) do no work becase they are at right angles to the horizontal displacement. The force exerted by the father, however, has a horizontal mg component that does positive work on the stroller. Therefore, N N 2 the total work is simply the work done by the father. 1 b. After calclating the work in part (a), we can find the final speed v f by applying the work energy theorem with v i = 1.37 ms. Known Force, F = 44.0 N; angle below horizontal, = 32.0 ; mass, m = 22.7 kg; distance, d = 1.13 m; initial speed, v i = 1.37 ms. Unknown (a) Work done by the father, W father =? (b) Final speed, v f =? SOLUTION Part (a) 1. The work done by the father is the horizontal component of his force times the distance, 1F cos 2d. This is also the total work done on the stroller: Part (b) 2. Use the work energy theorem to solve for the final speed: W father = 1F cos 2d = 144.0 N21cos 32.0 211.13 m2 = 42.2 J = W total W total = K = 1 2 mv f 2-1 2 mv 2 i 1 2 mv f 2 = W total + 1 2 mv 2 i 2142.2 J2 3. Sbstitte nmerical vales to get the desired v f = + 11.37 ms22 B 22.7 kg reslt: = 2.37 ms INSIGHT Notice that the speed of the stroller increased by 1.00 ms, from 1.37 ms to 2.37 ms. If the stroller had started from rest, wold its speed increase from zero to 1.00 ms? No. The work energy theorem depends on the sqare of the speeds, and hence speeds don t simply add or sbtract in an intitive linear way. The final speed in this case is greater than 1.00 ms, as we show in the following Practice Problem. PRACTICE PROBLEM Sppose the stroller starts at rest. What is its final speed in this case? [Answer: v f = 22W total m = 1.93 ms, almost twice what simple linear reasoning wold sggest] Some related homework problems: Problem 29, Problem 62 v f = B 2W total m + v 2 i PROBLEM-SOLVING NOTE Be Carefl Abot Linear Reasoning Thogh some relationships are linear if yo doble the mass, yo doble the kinetic energy others are not. For example, if yo doble the speed, yo qadrple the kinetic energy. Be carefl not to jmp to conclsions based on linear reasoning. The final speeds in the previos Examples cold have been fond sing Newton s laws and the constant-acceleration kinematics of Chapter 2, as indicated in the Insight in Example 7-9. The work energy theorem provides an alternative method of calclation that is often mch easier to apply than Newton s laws. We retrn to this point in Chapter 8.
CONCEPTUAL EXAMPLE 7-11 7-3 Work Done by a Variable FoRCE 207 COMPARE THE WORK PREDICT/EXPLAIN (a) To accelerate a certain car from rest to the speed v reqires the work W 1, as shown in the sketch. The work needed to accelerate the car from v to 2v is W 2. Which of the following statements is correct: W 2 = W 1, W 2 = 2W 1, W 2 = 3W 1, W 2 = 4W 1? (b) Which of the following is the best explanation for yor prediction? I. The increase in speed is the same, so the work is also the same. II. To doble the speed reqires doble the work. III. Kinetic energy depends on v 2, and hence it takes for times as mch work to increase the speed to 2v. IV. For times as mch work is reqired to go from 0 to 2v as to go from 0 to v. Therefore, the work reqired to increase the speed from v to 2v is three times the original work. REASONING AND DISCUSSION A common misconception is to reason that becase we increase the speed by the same amont in each case, the work reqired is the same. It is not, and the reason is that work depends on the speed sqared rather than on the speed itself. To see how this works, first calclate W 1, the work needed to go from rest to a speed v. From the work energy theorem, with v i = 0 and v f = v, we find W 1 = 1 2 mv f 2-1 2 mv i 2 = 1 2 mv2. Similarly, the work W 2 needed to go from v i = v to v f = 2v is W 2 = 1 2 m(2v)2-1 2 mv2 = 3A 1 2 mv2 B = 3W 1. ANSWER (a) The reqired work is W 2 = 3W 1. (b) The best explanation is IV. W 1 W 2 y = 0 y 2y Enhance Yor Understanding (Answers given at the end of the chapter) 2. An object has an initial kinetic energy of 100 J. A few mintes later, after a single external force has acted on the object, its kinetic energy is 200 J. Is the work done by the force positive, negative, or zero? Explain. Section Review The kinetic energy of an object is one-half its mass times its velocity sqared. The total work done on an object is eqal to the change in its kinetic energy. 7-3 Work Done by a Variable Force F d Ths far we have calclated work only for constant forces, yet most forces in natre vary with position. For example, the force exerted by a spring depends on how far the spring is stretched, and the force of gravity between planets depends on their separation. In this section we show how to calclate the work for a force that varies with position. A Graphical Interpretation of Work First, let s review briefly the case of a constant force, and develop a graphical interpretation of work. FIGURE 7-8 shows a constant force plotted verss position, x. If the force acts in the positive x direction and moves an object a distance d, from x 1 to x 2, the work it does is W = Fd = F1x 2 - x 1 2. Referring to the figre, we see that the work is eqal to the shaded area 1 between the force line and the x axis. Next, consider a force that has the vale F 1 from x = 0 to x = x 1 and a different vale F 2 from x = x 1 to x = x 2, as in FIGURE 7-9 (a). The work in this case is the sm of the works done by F 1 and F 2. Therefore, W = F 1 x 1 + F 2 1x 2 - x 1 2. This, again, is the area between the force lines and the x axis. Clearly, this type of calclation can be extended to a force with any nmber of different vales, as indicated in FIGURE 7-9 (b). If a force varies continosly with position, we can approximate it with a series of constant vales that follow the shape of the crve, as shown in FIGURE 7-10 (a). It follows that the work done by the continos force is approximately eqal to the area of the corresponding Force O Area = Fd = W x 1 x 2 Position FIGURE 7-8 Graphical representation of the work done by a constant force A constant force F acting throgh a distance d does the work W = Fd. Note that Fd is also eqal to the shaded area between the force line and the x axis. 1 Usally, area has the dimensions of (length) * (length), or length 2. In this case, however, the vertical axis is force and the horizontal axis is distance. As a reslt, the dimensions of area are (force) * (distance), which in SI nits is N # m = J.
F 1 x 1 208 CHAPTER 7 Work and KinETic Energy FIGURE 7-10 Work done by a continosly varying force (a) A continosly varying force can be approximated by a series of constant vales that follow the shape of the crve. (b) The work done by the continos force is approximately eqal to the area of the small rectangles corresponding to the constant vales of force shown in part (a). (c) In the limit of an infinite nmber of vanishingly small rectangles, we see that the work done by the force is eqal to the area between the force crve and the x axis. Force Force Force O x 1 x 2 Position (a) Approximating a continos force O x 1 x 2 Position (b) Approximating the work done by a continos force O x 1 x 2 Position (c) A better approximation Force F 2 F 1 Force F 1 x 1 rectangles, as FIGURE 7-10 (b) shows. The approximation can be made better by sing more rectangles, as illstrated in FIGURE 7-10 (c). In the limit of an infinite nmber of vanishingly small rectangles, the area of the rectangles becomes identical to the area nder the force crve. Hence this area is the work done by the continos force. To smmarize: O The work done Position by a force in moving an object from one position to another is eqal to the corresponding area between the force crve and the x axis. (b) Work for the Spring Force A case of particlar interest is a spring. If we recall that the force exerted by a spring is given by F x = -kx (Section 6-2), it follows that the force we mst exert to hold the spring at the position x is +kx. This is illstrated in FIGURE 7-11, where we also show that the corresponding force crve is a straight line extending from the origin. Therefore, the work we do in stretching a spring from x = 0 (eqilibrim) to the general position x is the shaded trianglar area shown in FIGURE 7-12. This area is eqal to 1 1base21height2, where in this case the base is x and the height is kx. As a reslt, 2 the work is 1 2 1x21kx2 = 1 2 kx2. Similar reasoning shows that the work needed to compress a spring a distance x is also 1 2 kx2. Therefore: Work to Stretch or Compress a Spring a Distance x from Eqilibrim W = 1 2 kx2 7-8 SI nit: jole, J Force of spring Applied force -kx +kx x = 0 x Eqilibrim position of spring O F 2 (x 2 - x 1 ) F = kx Applied force Position FIGURE 7-11 Stretching a spring The force we mst exert on a spring to stretch it a distance x is +kx. Ths, applied force verss position for a spring is a straight line of slope k. x Force O x 1 x 2 Position O x 1 x 2 O Position Position (a) (b) FIGURE 7-9 Work done by a nonconstant force (a) A force with a vale F 1 from 0 to x 1 and a vale F 2 from x 1 to x 2 does the work W = F 1 x 1 + F 2 (x 2 - x 1 ). This is simply the combined area of the two shaded rectangles. (b) If a force takes on a nmber of different vales, the work it does is still the total area between the force lines and the x axis, jst as in part (a). (a) O Force Area = W Position x kx FIGURE 7-12 Work needed to stretch a spring a distance x The work done is eqal to the shaded area, which is a right triangle. The area of the triangle is 1 2 1x21kx2 = 1 2 kx2.
7-3 Work Done by a Variable FoRCE 209 We can get a feeling for the amont of work reqired to compress a typical spring in the following Exercise. EXERCISE 7-12 SPRING WORK (a) A toy snake has a spring with a force constant of 230 Nm. How mch work is reqired to stretch this spring 2.5 cm? (b) The coiled sspension spring in a car has a force constant of 21,000 Nm. If 3.5 J of work is done to compress this spring, what is the compression distance? REASONING AND SOLUTION Work is related to the force constant and compression (or stretch) by W = 1 2 kx2. We can apply this relationship to both qestions. (a) Sbstitte k = 230 Nm and x = 0.025 m into W = 1 2 kx2 : W = 1 2 kx2 = 1 2 1230 Nm210.025 m22 = 0.072 J (b) Solve W = 1 2 kx2 for x, then sbstitte W = 3.5 J and k = 21,000 Nm: x = B 2W k = B 213.5 J2 21,000 Nm = 0.018 m The work done in compressing or expanding a spring varies with the second power of x, the displacement from eqilibrim. The conseqences of this dependence are explored throghot the rest of this section. Springs in Other Contexts Or reslts for a spring apply to more than jst the classic case of a helical coil of wire. In fact, any flexible strctre satisfies the relationships F x = -kx and W = 1 2 kx2, given the appropriate vale of the force constant, k, and small enogh displacements, x. Several examples were mentioned in Section 6-2. Here we consider an example from the field of nanotechnology namely, the cantilevers sed in atomic-force microscopy (AFM). As we show in Example 7-13, a typical atomic-force cantilever is basically a thin silicon bar abot 250 mm in length, spported at one end like a diving board, with a sharp, hanging point at the other end. When the point is plled across the srface of a material like an old-fashioned phonograph needle in the groove of a record individal atoms on the srface case the point to move p and down, deflecting the cantilever. These deflections, which can be measred by reflecting a laser beam from the top of the cantilever, are then converted into an atomic-level pictre of the srface, as shown in FIGURE 7-13. A typical force constant for an AFM cantilever is on the order of 1 Nm, mch smaller than the 100 500-Nm force constant of a common lab spring. The implications of this are discssed in the following Example. EXAMPLE 7-13 FLEXING AN AFM CANTILEVER The work reqired to deflect a typical AFM cantilever by 0.10 nm is 1.2 * 10-20 J. (a) What is the force constant of the cantilever, treating it as an ideal spring? (b) How mch work is reqired to increase the deflection of the cantilever from 0.10 nm to 0.20 nm? FIGURE 7-13 Hman chromosomes, as imaged by an atomic-force microscope. Silicon rod PICTURE THE PROBLEM The pper sketch shows the cantilever and its sharp point being dragged across the srface of a material. In the lower sketch, we show an exaggerated view of the cantilever s deflection, and indicate that it is eqivalent to the stretch of an effective ideal spring with a force constant k. REASONING AND STRATEGY a. Given that W = 1.2 * 10-20 J for a deflection of x = 0.10 nm, we can find the effective force constant k sing W = 1 2 kx2. b. To find the work reqired to deflect from x = 0.10 nm to x = 0.20 nm, W 1S2, we calclate the work to deflect from x = 0 to x = 0.20 nm, W 0S2, and then sbtract the work needed to deflect from x = 0 to x = 0.10 nm, W 0S1. (Notice that we cannot simply assme that the work to go from x = 0.10 nm to x = 0.20 nm is the same as the work to go from x = 0 to x = 0.10 nm.) k x = 0.10 nm Effective ideal spring CONTINUED
210 CHAPTER 7 Work and KinETic Energy Known Deflection distance, x = 0.10 nm = 0.10 * 10-9 m; work to deflect, W = 1.2 * 10-20 J. Unknown (a) Force constant, k =? (b) Work to increase deflection, W 1S2 =? SOLUTION Part (a) 1. Solve W = 1 2 kx2 for the force constant k: k = 2W = 211.2 * 10-20 J2 = 2.4 Nm x 2 10.10 * 10-9 2 m2 Part (b) 2. First, calclate the work needed to deflect the W 0S2 = 1 2 kx2 cantilever from x = 0 to x = 0.20 nm: = 1 2 12.4 Nm210.2 * 10-9 m2 2 = 4.8 * 10-20 J 3. Sbtract from the above reslt the work to deflect from x = 0 to x = 0.10 nm, which the problem statement gives as 1.2 * 10-20 J: W 1S2 = W 0S2 - W 0S1 = 4.8 * 10-20 J - 1.2 * 10-20 J = 3.6 * 10-20 J INSIGHT Or reslts show that more energy is needed to deflect the cantilever the second 0.10 nm than to deflect it the first 0.10 nm. Why? The reason is that the force of the cantilever increases with distance; ths, the average force over the second 0.10 nm is greater than the average force over the first 0.10 nm. In fact, we can see from the graph that the average force between 0.10 nm and 0.20 nm 10.36 nn2 is three times the average force between 0 and F k = 2.4 N/m 0.10 nm 10.12 nn2. It follows that the work reqired for the second 0.10 nm is three times the work reqired for the first 0.10 nm. 0.48 nn PRACTICE PROBLEM PREDICT/CALCULATE A second cantilever has half the force constant of the cantilever in this Example. (a) Is the work reqired to deflect the second cantilever by 0.20 nm greater than, less than, or eqal to the work reqired to deflect the cantilever in this Example by 0.10 nm? Explain. (b) Determine the work reqired to deflect the second cantilever 0.20 nm. [Answer: (a) Halving the force constant halves the work, bt dobling the deflection qadrples the work. The net effect is that the work increases by a factor of two. (b) The work reqired is 2.4 * 10-20 J.] 0.36 nn 0.24 nn 0.12 nn O 0.10 nm 0.20 nm x Some related homework problems: Problem 31, Problem 36 Force kx 1 F av = kx 2 O x Position FIGURE 7-14 Work done in stretching a spring: average force The average force to stretch a spring from x = 0 to x is F av = 1 kx, and the work done is 2 W = F av d = 1 2 kx(x) = 1 2 kx2. BIO Another example of the work reqired to stretch a flexible strctre is the work done inside yor eye. The hman eye can accommodate, or focs on objects at different distances, by sing ciliary mscles to alter the shape of the flexible lens behind the ppil of the eye. (The physiology and optical properties of the eye are explored in more detail in Chapter 27.) The shorter the distance yo attempt to focs yor eye, the harder yor ciliary mscles mst work to change the shape of the lens. This is why long periods of viewing close-p objects, sch as a book or a compter screen, can lead to asthenopia, or eye strain. Using Average Force to Calclate Work An eqivalent way to calclate the work for a variable force is to mltiply the average force, F av, by the distance, d: W = F av d 7-9 For a spring that is stretched a distance x from eqilibrim, the force varies linearly from 0 to kx. Ths, the average force is F av = 1 kx, as indicated in FIGURE 7-14. Therefore, 2 the work is W = 1 2 kx(x) = 1 2 kx2 As expected, or reslt agrees with Eqation 7-8. Finally, when yo stretch or compress a spring from its eqilibrim position, the work yo do is always positive. The work done by a spring, however, may be positive or negative, depending on the sitation. For example, consider a block sliding to the right with an initial speed v 0 on a smooth, horizontal srface, as shown in FIGURE 7-15 (a). When the block begins to compress the spring, as in FIGURE 7-15 (b), the spring exerts a force
7-3 Work Done by a Variable FoRCE 211 Initial speed of block is v 0. v Eqilibrim position 0 of spring Spring is doing negative work on the block force and displacement are in opposite directions. v F 1 v = 0 F 2 (a) (b) (c) Final speed of block is again v 0. v Eqilibrim position 0 of spring Spring is doing positive work on the block force and displacement are in the same direction. v v = 0 F 1 F 2 (f) (e) (d) FIGURE 7-15 The work done by a spring can be positive or negative (a) A block slides to the right on a frictionless srface with a speed v 0 ntil it enconters a spring. (b) The spring now exerts a force to the left opposite to the block s motion and hence it does negative work on the block. This cases the block s speed to decrease. (c) The negative work done by the spring eventally is eqal in magnitde to the block s initial kinetic energy, at which point the block comes to rest momentarily. As the spring expands, (d) and (e), it does positive work on the block and increases its speed. (f) When the block leaves the spring, its speed is again eqal to v 0. on the block to the left that is, opposite to the block s direction of motion. As a reslt, the spring does negative work on the block, which cases the block s speed to decrease. Eventally the negative work done by the spring, W = - 1 2 kx2, is eqal in magnitde to the initial kinetic energy of the block. At this point, FIGURE 7-15 (c), the block comes to rest momentarily, and W = K = K f - K i = 0 - K i = -K i = - 1 2 mv 0 2 = - 1 2 kx2. After the block comes to rest, the spring expands back to its eqilibrim position, as we see in FIGURE 7-15 (d) (f). Dring this expansion the force exerted by the spring is in the same direction as the block s motion, and hence it does positive work in the amont W = 1 2 kx2. As a reslt, the block leaves the spring with the same speed it had initially. QUICK EXAMPLE 7-14 A BLOCK COMPRESSES A SPRING A block with a mass of 1.5 kg and an initial speed of v i = 2.2 ms slides on a frictionless, horizontal srface. The block comes into contact with a spring that is in its eqilibrim position, and compresses it ntil the block comes to rest momentarily. Find the maximm compression of the spring, assming its force constant is 475 Nm. v i Eqilibrim position of spring x Maximm compression REASONING AND SOLUTION Or sketch shows the block jst as it comes into contact with the spring with an initial speed of v i = 2.2 ms. We also show the spring at maximm compression, x, when the block is momentarily at rest. As the block comes to rest, the force of the spring is opposite to the v = 0 F CONTINUED
212 CHAPTER 7 Work and KinETic Energy block s displacement, and hence the work done by the spring is negative. The (negative) work done by the spring, W = - 1 2 kx2, is eqal to the (negative) change in kinetic energy, ΔK. Ths, we can set - 1 2 kx2 = K and solve for x to find the maximm compression. 1. Calclate the initial and final kinetic K i = 1 2 mv 2 i energies of the block: = 1 2 11.5 kg212.2 ms22 = 3.6 J K f = 0 2. Calclate the change in kinetic K = K f - K i = -3.6 J energy of the block: 3. Set the negative work done by the - 1 2 kx2 = K = -3.6 J spring eqal to the negative change in kinetic energy of the block: -2 K 4. Solve for the compression, x, and x = B k sbstitte nmerical vales: = B -21-3.6 J2 475 Nm = 0.12 m When the spring brings the block to rest, the kinetic energy it had initially is not lost it is stored in the spring itself and can be released later. We discss sitations like this, and their connection with energy conservation, in Chapter 8. Enhance Yor Understanding (Answers given at the end of the chapter) 3. As an object moves along the positive x axis, the force shown in FIGURE 7-16 acts on it. Is the work done by the force from x = 0 to x = 4 m greater than, less than, or eqal to the work done by the force from x = 5 m to x = 9 m? Explain. Force, F (N) 10 9 8 7 6 5 4 3 2 1 O 1 2 3 4 5 6 7 8 9 10 Position x, (m) FIGURE 7-16 Section Review Work is eqal to the area between a force crve and the x axis. Work is also eqal to the average force times displacement. 7-4 Power Power is a measre of how qickly work is done. To be precise, sppose the work W is performed in the time t. The average power delivered dring this time is defined as follows: Definition of Average Power, P P = W t 7-10 SI nit: Js = watt, W
7-4 Power 213 For simplicity of notation we drop the sal sbscript av for an average qantity and simply nderstand that the power P refers to an average power nless stated otherwise. The Dimensions of Power The dimensions of power are joles (work) per second (time). We define one jole per second to be a watt (W), after James Watt (1736 1819), the Scottish engineer and inventor who played a key role in the development of practical steam engines: 1 watt = 1 W = 1 Js 7-11 Of corse, the watt is the nit of power sed to rate the otpt of lightblbs. Another common nit of power is the horsepower (hp), which is sed to rate the otpt of car engines. It is defined as follows: 1 horsepower = 1 hp = 746 W 7-12 Thogh it sonds like a horse shold be able to prodce one horsepower, in fact, a horse can generate only abot 23 hp for sstained periods. The reason for the discrepancy is that when James Watt defined the horsepower as a way to characterize the otpt of his steam engines he prposely chose a nit that was overly generos to the horse, so that potential investors coldn t complain he was overstating the capability of his engines. Hman Power To get a feel for the magnitde of the watt and the horsepower, consider the power yo might generate when walking p a flight of stairs. Sppose, for example, that an 80.0-kg person walks p a flight of stairs in 20.0 s, and that the altitde gain is 12.0 ft (3.66 m). Referring to Example 7-3 and Conceptal Example 7-4, we find that the work done by the person is W = mgh = 180.0 kg219.81 ms 2 213.66 m2 = 2870 J. To find the power, we simply divide the work by the time: P = Wt = (2870 J)(20.0 s) = 144 W = 0.193 hp. Ths, a leisrely stroll p the stairs reqires abot 15 hp or 150 W. Similarly, the power prodced by a sprinter bolting ot of the starting blocks is abot 1 hp, and the greatest power most people can prodce for sstained periods of time is roghly 13 to 12 hp. Frther examples of power are given in Table 7-3. Hman-powered flight is a feat jst barely within or capabilities, since the most efficient hman-powered airplanes reqire a steady power otpt of abot 13 hp. In 1979 the Gossamer Albatross became the first (and so far the only) hman-powered aircraft to fly across the English Channel. This 22.25-mile flight from Folkestone, England, to Cap Gris-Nez, France took 2 hors 49 mintes and reqired a total energy otpt roghly eqivalent to climbing to the top of the Empire State Bilding 10 times. The Gossamer Albatross is shown in midflight in FIGURE 7-17. RWP Power otpt is also an important factor in the performance of a car. For example, sppose it takes a certain amont of work, W, to accelerate a car from 0 to 60 mih. If the average power provided by the engine is P, then according to Eqation 7-10, the amont of time reqired to reach 60 mih is t = WP. Clearly, the greater the power P, the less the time reqired to accelerate. Ths, in a loose way of speaking, we can say that of how fast it can go fast. Big Idea 3 Power is the rate at which work is done. The more work that is done in a shorter time, the greater the power. TABLE 7-3 Typical Vales of Power Approximate Sorce power (W) Hoover Dam 1.34 * 10 9 Car moving at 40 mih 7 * 10 4 Home stove 1.2 * 10 4 Snlight falling on one 1380 sqare meter Refrigerator 615 Television 200 Person walking p stairs 150 Hman brain 20 FIGURE 7-17 The Gossamer Albatross Twice the aircraft toched the srface of the water, bt the pilot was able to maintain control. EXAMPLE 7-15 PASSING FANCY To pass a slow-moving trck, yo want yor fancy 1.30 * 10 3 @kg car to accelerate from 13.4 ms 130.0 mih2 to 17.9 ms 140.0 mih2 in 3.00 s. What is the minimm power reqired for this pass? PICTURE THE PROBLEM Or sketch shows the car accelerating from an initial speed of v i = 13.4 ms to a final speed of v f = 17.9 ms. We assme the road is level, so that no work is done against gravity, and that friction and air resistance may be ignored. CONTINUED
214 CHAPTER 7 Work and KinETic Energy REASONING AND STRATEGY Power is work divided by time, and work is eqal to the change in kinetic energy as the car accelerates. We can determine the change in kinetic energy from the given mass of the car and its initial and final speeds. With this information at hand, we can determine the power with the relationship P = W/t = Kt. Known Mass of car, m = 1.30 * 10 3 kg; initial speed, v i = 13.4 ms; final speed, v f = 17.9 ms; time, t = 3.00 s. Unknown Minimm power, P =? 13.4 m/s 17.9 m/s SOLUTION 1. First, calclate the change in kinetic energy: K = 1 2 mv 2 f - 1 2 mv 2 i = 1 2 11.30 * 103 kg2117.9 ms2 2-1 2 11.30 * 103 kg2113.4 ms2 2 2. Divide by time to find the minimm power (the actal P = W t power wold have to be greater to overcome frictional losses): = 9.16 * 10 4 J = K t = 9.16 * 104 J 3.00 s = 3.05 * 10 4 W = 40.9 hp INSIGHT Sppose that yor fancy car contines to prodce the same 3.05 * 10 4 W of power as it accelerates from v = 17.9 ms 140.0 mih2 to v = 22.4 ms 150.0 mih2. Is the time reqired more than, less than, or eqal to 3.00 s? It will take more than 3.00 s. The reason is that K is greater for a change in speed from 40.0 mih to 50.0 mih than for a change in speed from 30.0 mih to 40.0 mih, becase K depends on speed sqared. Since K is greater, the time t = KP is also greater. PRACTICE PROBLEM Find the time reqired to accelerate from 40.0 mih to 50.0 mih with 3.05 * 10 4 W of power. [Answer: First, K = 1.18 * 10 5 J. Next, P = Kt can be solved for time to give t = KP. Ths, t = 3.87 s.] Some related homework problems: Problem 43, Problem 56, Problem 72 Power and Speed Finally, consider a system in which a car, or some other object, is moving with a constant speed v. For example, a car might be traveling phill on a road inclined at an angle above the horizontal. To maintain a constant speed, the engine mst exert a constant force F eqal to the combined effects of friction, gravity, and air resistance, as indicated in FIGURE 7-18. Now, as the car travels a distance d, the work done by the engine is W = Fd, and the power it delivers is FIGURE 7-18 Driving p a hill A car traveling phill at constant speed reqires a constant force, F, of magnitde mg sin + F air res + F friction applied in the direction of motion. F friction F air res N mg F P = W t F air res = Fd t F friction N mg sin mg cos y F x mg