Chapter 6 CONSERVATION OF ENERGY Conceptual Questions 1. Assuming the object can be treated as a point particle, the total work done on it by external forces is equal to the change in its kinetic energy. An object moving in a circle may be changing its speed as it goes around, so the total work done on it is not necessarily zero. 2. The force exerted on the backpack by your back and shoulders is directed upward and is perpendicular to your horizontal displacement. Hence it does not do any work on the backpack. (1) Now there is a component of your displacement directed downward, anti-parallel to the force on the backpack, so the force does negative work on the backpack. (2) In this case the backpack’s kinetic energy is increasing as you gain speed. The force exerted on the backpack is no longer vertical, but has a horizontal component in the direction you are moving. Thus, it does positive work on the backpack. 3. When the roads leading up a mountain wind back and forth, the angle of inclination of the road is less than that of a road going straight up the mountain. This reduces the force necessary to drive the car up the road and the required power output of the engine as well. The length of road is increased, however, so the total work that must be done to reach the top of the mountain remains the same (or may be larger if frictional forces are taken into account). 4. During the fall, the force of gravity on the mango is parallel to the mango’s displacement, so it does positive work. The force of gravity on the Earth due to the mango is directed upward, toward the mango, and has the same magnitude as the force of gravity on the mango (Newton’s third law). The Earth moves upward by a very small (imperceptible) amount as the mango falls, so the mango’s gravitational field does positive work on the Earth. Wm and WE are both positive, but WE is a very small number, close to zero, so Wm >> WE . 5. Yes, static friction can do work. As an example, imagine a book on a conveyor belt that carries it up an incline. The force of static friction on the book is directed upward along the surface of the belt and has a component that is parallel to the book’s displacement. Thus, the force of static friction does positive work on the book. (At the same time, the work done on the book by gravity is negative and the total work done on the book is zero.) 6. Work is done on the roller coaster by a tow chain or some other mechanism designed to increase its height with respect to the ground and thus to increase its gravitational potential energy. At the apex of the first hill, the kinetic energy of the roller coaster is negligibly small so that its total initial energy is equal to its potential energy. Energy is dissipated along the trip around the track as a result of frictional effects and air resistance. Thus, unless additional energy is added to the system—via another tow chain for example—the energy available to the roller coaster to climb subsequent hills is less than the original total energy—the hills must therefore be shorter. 7. When the ball reaches the ground, the gravitational potential energy it originally possessed will have been converted into the kinetic energy of its motion. If the ball is a rigid point-like particle, its kinetic energy will be conserved during the bounce and the ball’s velocity will be the same immediately before and after rebounding. Most balls however are made of deformable materials like rubber that compress when bouncing. The deformation process changes the state of the molecules that make up the ball—increasing the ball’s internal energy. The energy required for this process must be obtained via a decrease in the kinetic energy of the ball after the rebound—the maximum height attained by the ball will therefore be lower.

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8. The total mechanical energy of the gymnast swinging in a vertical circle about a crossbar has the same constant value for each point along the path (ignoring the relatively small amount of work done by the gymnast’s muscles during the swing). The gravitational potential energy of the gymnast is lowest at the bottom of the path and greatest at the top. Because the gymnast’s total energy is constant, this implies that the kinetic energy of the gymnast must be lowest at the top of the circle and greatest at the bottom. Correspondingly, the gymnast’s velocity is a minimum at the top of the loop and a maximum at the bottom. 9. The bicyclist requires a minimum amount of energy to climb the hill. This quantity is independent of the means that the bicyclist employs to acquire the energy and is solely a function of the height of the hill (the energy required is also affected by the work done by frictional and drag forces—the magnitude of this effect is approximately equal for any method used by the bicyclist to climb the hill and therefore doesn’t affect our reasoning). After beginning the ascent, a component of the gravitational force acts in the direction opposite to the displacement thereby increasing the amount of negative work done on the rider with respect to the amount done while riding on flat land. Therefore, the rate at which the rider must do work to acquire the necessary energy is greater when pedaling uphill than when on flat land. It is thus advantageous to acquire as much energy as possible before the ascent when the amount of kinetic energy gained per amount of work done by the rider is greatest. 10. When pushing the crate with a force parallel to the ground, the force of friction acting to impede its motion is proportional to the normal force acting on the crate—in this situation, the normal force is equal to the crate’s weight. When pulling the crate with a rope angled above the horizontal, the normal force on the crate is less than its weight—the force of friction is therefore reduced. To keep the crate moving across the floor, the applied force in the parallel direction must be greater than or equal to the force of friction—pulling on the rope therefore requires a smaller parallel applied force. The work done in moving an object is equal to the product of the displacement through which it has been moved and the force component parallel to the direction of motion. The applied force component parallel to the ground is smaller when pulling the crate with the rope—thus, the work done to move the crate with the rope must be less, regardless of the weight of the crate or the displacement. 11. Such animals have larger than average leg muscles located predominantly inside the body such that they don’t have to move with the legs. As a result, the legs of these animals are thinner than the legs of slower animals of similar size. The less massive legs require less work to accelerate and decelerate—more of the animal’s energy can therefore go into increasing its kinetic energy and thus its speed. 12. Because an ideal spring has zero mass, Newton’s second law implies that the net force exerted on it must be zero (provided it has a finite acceleration). The forces exerted by a spring on objects attached to its ends are equal and opposite to the forces exerted by those objects on the spring, according to Newton’s third law. The forces exerted by the spring must therefore be equal in magnitude and opposite in direction so that the spring experiences a zero net force. The work done on the two attached objects is not necessarily the same, because the distances they move can be different. Consider for example a spring attached to a heavy lead ball on one end and a Ping-Pong ball on the other. If the spring were initially in a stretched position and then released, the end attached to the lead ball would hardly move at all compared to the other end. The work done by the spring on the Ping-Pong ball would therefore be greater than that done on the heavy ball. 13. Zorba is correct. You get to a top speed sooner on the first slide, so it takes less time to get to the bottom, but the

final speeds are the same from mgh = 12 mv 2 .

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Problems 1. Strategy Use Eq. (6-1). Solution Find the work done by Denise dragging her basket of laundry.

30.0 N

W = F ∆r cos θ = (30.0 N)(5.0 m) cos 60.0° = 75 J

60.0° 5.0 m

x

2. Strategy The distance is equal to the speed times the time interval. Use Eq. (6-1). Solution Find the work done by the rope on the sled.

W = F ∆r cos θ = Tv∆t cos θ = (240 N)(1.5 m s)(10.0 s) cos 30.0° = 3.1 kJ 3. Strategy and Solution Since the book undergoes no displacement, no work is done on the book by Hilda. 4. Strategy The angle between the tension and the displacement is zero. Use Eq. (6-1). Solution Find the work done by the towrope on the water-skier. W = F ∆r cos θ = (240 N)(54 m) cos 0° = 13 kJ 5. Strategy Use Newton’s second law and Eq. (6-2). Solution Find the net force on the barge. ΣFy = T sin θ − T sin θ = 0 and ΣFx = T cos θ + T cos θ = Fx . Find the work done on the barge. W = Fx ∆x = (2T cos θ )∆x = 2(1.0 kN) cos 45°(150 m) = 210 kJ

y 1.0 kN 45° 45°

x 150 m

1.0 kN

6. (a) Strategy The force is equal to the weight of the pile driver. Use Eq. (6-1). Solution Find the work done to raise the pile driver. W = F ∆r cos θ = mg ∆r cos 0° = (402 kg)(9.80 N kg)(12 m) = 47 kJ (b) Strategy and Solution The work done by gravity is negative the work done to raise the pile driver since the

force of gravity is opposite the driver’s motion, so Wgravity, up = −Wdriver = −47 kJ . (c) Strategy and Solution The motion of the driver is in the same direction as the force of gravity, so the work

done by gravity is opposite that found in part (b); therefore, Wgravity, down = −Wgravity, up = 47 kJ . 7. (a) Strategy Consider the work done on the carton by Jennifer and the work done of the carton by gravity. Solution Jennifer does positive work on the carton because the carton moves up in the direction of the force applied by Jennifer. Gravity does negative work on the carton because the carton moves in the direction opposite the force due to gravity. The absolute value of each amount of work done is the same, so the total work done on the carton is zero. (b) Strategy Use Eq. (6-1). Let the +y-axis point downward. Solution The force of gravity is parallel to the displacement of the litter. W = F ∆r cos θ = mg ∆r cos θ = (1.2 kg)(9.80 m s 2 )(0.75 m) cos 0° = 8.8 J 342

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Chapter 6: Conservation of Energy

8. Strategy Use Eq. (6-2). Let the x-axis point in the direction of motion. Solution The force of friction is opposite the motion of the box. Dirk’s horizontal force is in the direction of motion.

66.0 N 4.80 N

x

W = Fx ∆x = ( F − f k )∆x = (66.0 N − 4.80 N)(2.50 m) = 153 J 9. Strategy Use Eq. (6-2). Let the x-axis point in the direction of motion. Solution The force of friction is opposite the motion of the box, and according to Newton’s second law, it is equal to f k = µk N = µk mg . Juana’s horizontal force is in the direction of motion. Solve for the displacement. W = Fx ∆x, so W W W 74.4 J ∆x = = = = = 1.3 m . Fx F − f k F − µk mg 124 N − 0.120(56.8 kg)(9.80 m s 2 )

N x F fk mg

10. Strategy Use Eq. (6-6). Solution Compute the kinetic energy of the automobile. 1 1 K = mv 2 = (1600 kg)(30.0 m s)2 = 720 kJ 2 2 11. Strategy The work done on the briefcase by the executive is equal to the change in kinetic energy of the briefcase. Use Eqs. (6-6) and (6-7). Solution Find the work done by the executive on the briefcase. 1 1 W = ∆K = m(vf 2 − vi 2 ) = (5.00 kg) ⎡ (2.50 m s)2 − 0 ⎤ = 15.6 J ⎣ ⎦ 2 2 12. Strategy Use Eq. (6-6) for each case. Solution Compute the kinetic energies.

1 2 1 mv = (70.5 kg)(27.8 m s) 2 = 27.2 kJ . 2 2 1 2 1 For Howard and his bike, the kinetic energy was K = mv = (70.5 kg)(68.04 m s) 2 = 163 kJ . 2 2

The kinetic energy of Murphy and his bike was K =

13. Strategy The kinetic energy of the sack is equal to the work done on it by Sam. Use Eqs. (6-2), (6-6), and (6-7). Solution (a) Compute the kinetic energy of the sack. 1 1 ∆K = m(vf 2 − vi 2 ) = mv 2 − 0 = K = W = Fx ∆x, so K = (2.0 N)(0.35 m) = 0.70 J . 2 2 (b) Solve for the speed of the sack. 1 2K 2(0.70 J) K = mv 2 , so v = = = 0.37 m s . 2 m 10.0 kg

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Physics

14. Strategy Use Eqs. (6-2), (6-6), and (6-7). Solution Find the magnitude of the force. 1 1 mv 2 (12 kg)(0.40 m s) 2 W = Fx ∆x = ∆K = m(vf 2 − vi 2 ) = mv 2 − 0, so Fx = = = 0.12 N . 2 2 2∆x 2(8.0 m) 15. Strategy Use Eq. (6-6) for the initial and final kinetic energies. Solution Compute the change in the kinetic energy of the ball. 1 1 ∆K = m(vf 2 − vi 2 ) = (0.10 kg) ⎡(2.0 m s) 2 − (2.0 m s) 2 ⎤ = 0 ⎣ ⎦ 2 2 Since the ball bounced back with the same speed, its kinetic energy did not change. 16. Strategy The sum of the work done on Jim and his skateboard by gravity and that of friction is equal to the change in kinetic energy of Jim and his skateboard. Use Eqs. (6-6) and (6-7). Solution Find the work done by friction on Jim and his skateboard. Let the y-axis point upward. 1 1 Wtotal = Wgravity + Wfriction = − mg ∆y + Wfriction = ∆K = K f − Ki = mv 2 − 0 = mv 2 , so 2 2 1 2 2 2 Wfriction = mv + mg ∆y = (65.0 kg)[(9.00 m s) 2 + (9.80 m s )(0 − 5.00 m)] = −550 J . 2 17. Strategy Use Eqs. (6-6) and (6-7). Solution Compute the work done by the wall on the skater. 1 1 Wtotal = ∆K = K f − Ki = 0 − mv 2 = − (69.0 kg)(11.0 m s)2 = −4.17 kJ 2 2 18. (a) Strategy Use Eqs. (6-6) and (6-7). The weight is equal to W = mg. Solution Calculate the work done on the plane by the cables. 1 mg 220 × 103 N W = ∆K = m(vf 2 − vi 2 ) = (0 − vi 2 ) = − (67 m s) 2 = −50 MJ 2 2g 2(9.80 N kg)

The work done on the plane by the cables is −50 MJ . (b) Strategy The force due to the cables is opposite the direction of motion. Use Eq. (6-1). Solution Find the force exerted on the plane by the cables. 5.0 × 107 J W =− = − 600 kN. W = F ∆r cos180°, so F = − ∆r 84 m The force exerted on the plane by the cables is 600 kN opposite the plane’s direction of motion.

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Chapter 6: Conservation of Energy

19. Strategy Use Eq. (6-6) to compute the kinetic energies; then form a ratio to compare. Solution Compute the kinetic energies of the car and the meteoroid. 1 1 K meteoroid = mv 2 = (0.0050 kg)(48 × 103 m s)2 = 5.8 MJ 2 2 1 K car = (1100 kg)(29 m s)2 = 0.46 MJ 2 Form the ratio. K meteoroid 5.8 MJ = > 12 0.46 MJ K car

The meteoroid has more than 12 times the kinetic energy of the car. 20. Strategy Since U = 0 at ground level, the potential energy of Sean and the parachute at the top of the tower is equal to the negative of the work done by gravity as Sean climbed the tower. Solution Find the potential energy of Sean and the parachute at the top of the tower.

U = mghtower = (68.0 kg)(9.80 m s 2 )(82.3 m) = 54.8 kJ . 21. (a) Strategy and Solution Since the floor is level, the motion of the desk is perpendicular to the force due to gravity; therefore, the change in the desk’s gravitational potential energy is zero. (b) Strategy The motion of the desk is in the direction of the applied constant force. Use Eq. (6-2). Solution Compute the work done by Justin.

W = Fx ∆x = (340 N)(10.0 m) = 3.4 kJ (c) Strategy and Solution Justin did work against friction, not gravity, so the energy has been dissipated as heat by friction between the bottom of the desk and the floor. 22. (a) Strategy The energy saved is equal to the potential energy that the paint would have had had it been lifted by the plane to cruising altitude. Use Eq. (6-9), where the ground level is zero and the height of the plane is h. Solution Find the energy saved.

U = mgh = (100 kg)(9.80 m s 2 )(12, 000 m) = 12 MJ (b) Strategy Use the work-kinetic energy theorem. Solution Find the energy saved. 1 1 W = ∆K = mvf 2 − 0 = (100 kg)(250 m s)2 = 3.1 MJ 2 2 23. Strategy Use Eq. (6-9). Solution (a) Since the orange returns to its original position (∆y = 0) and air resistance is ignored, the change in its

potential energy is 0 . (b) Let the y-axis point upward and the initial position be y = 0.

∆U grav = mg ∆y = (0.30 kg)(9.80 m s 2 )(−1.0 m − 0) = −2.9 J 345

Chapter 6: Conservation of Energy

Physics

24. (a) Strategy Find the change in potential energy from the change in height. Use this to find the change in kinetic energy and, thus, the speed. Solution Find the change in height. ∆h = A sin θ = (2.00 m) sin 30.0° = 1.00 m Find the speed of the brick. 1 ∆U = mg ∆h = ∆K = mv 2 , so v = 2 g ∆h = 2(9.80 m s 2 )(1.00 m) = 4.43 m s . 2 (b) Strategy Friction does negative work on the brick, slowing it. Solution Find the work done by friction. Wf = F ∆x = ( µ mg cos θ )A = µ mg A cos θ Find the speed of the brick. 1 ∆K = mv 2 = mg ∆h − µ mg A cos θ = mg A sin θ − µ mg A cos θ , so 2 v = 2 g A(sin θ − µ cos θ ) = 2(9.80 m s 2 )(2.00 m)(sin 30.0° − 0.10 cos 30.0°) = 4.03 m s . 25. (a) Strategy and Solution Since there are two pulleys, only half the force is required to move the mass (but

twice the length of rope must be pulled), so the pulley system multiplies the force exerted by a factor of 2 . (b) Strategy Use ∆U = mg ∆h. Solution Find the change in potential energy of the weight.

∆U = mg ∆h = (48.0 kg)(9.80 m s 2 )(4.00 m) = 1.88 kJ (c) Strategy and Solution By conservation of energy, the work done to lift the mass is equal to its change in

potential energy, so W = ∆U = 1.88 kJ . (d) Strategy and Solution Twice the length of rope must be pulled to do a given amount of work while applying half the force, so the length of rope pulled is 8.00 m. 26. Strategy Use Newton’s second law and Eq. (6-10). Solution The total work is given by Wtotal = Wvs. friction + Wvs. gravity . Find the work done against friction. ΣFy = N − mg cos φ = 0, so f = µ N = µ mg cos φ .

Wvs. friction = f ∆x = µ mg cos φ L = µ mg

2

L2 − h 2 ⎛L⎞ L = µ mgh ⎜ ⎟ − 1 and the work done against gravity is L ⎝h⎠

2 2 ⎛ ⎞ ⎛ ⎞ ⎛L⎞ ⎛ 4.0 m ⎞ ⎟ = 2.5 kJ . Wvs. gravity = mgh. So, Wtotal = mgh ⎜ µ ⎜ ⎟ − 1 + 1⎟ = (1400 N)(1.0 m) ⎜ 0.20 ⎜ 1 1 − + ⎟ ⎜ ⎝h⎠ ⎟ ⎜ ⎟ ⎝ 1.0 m ⎠ ⎝ ⎠ ⎝ ⎠

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27. (a) Strategy Use conservation of energy. Solution Find the speed of the cart as it passes point 3. 1 1 1 1 E1 = mv12 if y1 = 0 and E3 = mv32 + mgy3 . Ef = E3 = mv32 + mgy3 = Ei = E1 = mv12 , so 2 2 2 2

v3 = v12 − 2 gy3 = (20.0 m s) 2 − 2(9.81 m s 2 )(10.0 m) = 14.3 m s . (b) Strategy Use the result of part (a), replacing 3 with 4. If the result is real—the argument of the square root is nonnegative—the cart will reach position 4. Solution Compute the speed of the cart at position 4.

v4 = v12 − 2 gy4 = (20.0 m s) 2 − 2(9.81 m s 2 )(20.0 m) = 3 m s The answer is yes; the cart will reach position 4. 28. Strategy Use conservation of energy. Solution Find a general expression for the speed of the cart. 1 1 E4 = K 4 + U 4 = mv42 + mgy4 = Ei , so En = K n + U n = mvn 2 + mgyn , where n = 1, 2, or 3. 2 2 Solve for vn .

1 1 mv 2 + mgyn = Ei = mv42 + mgy4 , so vn = v42 + 2 g ( y4 − yn ). 2 n 2 Compute the speed at each position. Ef =

v1 = (15 m s)2 + 2(9.80 m s 2 )(20.0 m − 0) = 25 m s v2 = (15 m s)2 + 2(9.80 m s 2 )(20.0 m − 15.0 m) = 18 m s v3 = (15 m s) 2 + 2(9.80 m s 2 )(20.0 m − 10.0 m) = 21 m s 29. Strategy The initial height of the rope is l cos θ where l is the length of the rope and θ is the angle it makes with the vertical. Then ∆y = l cos θ − l = l (cos θ − 1). Use conservation of energy. Solution Find Bruce’s speed at the bottom of the swing. 1 1 ∆K = mv 2 − 0 = mv 2 = −∆U = − mg ∆y = mgl (1 − cos θ ), so 2 2 v = 2 gl (1 − cos θ ) = 2(9.80 m s 2 )(20.0 m)(1 − cos 35.0°) = 8.42 m s .

θ

l cos θ

l − l cos θ

30. Strategy Use conservation of energy. Solution Find the maximum height achieved by the swinging child. 1 K f − Ki = mv 2 − 0 = U i − U f = mgytop − mgybottom , so 2 v2 (4.9 m s) 2 + ybottom = + 0.70 m = 1.9 m . ytop = 2g 2(9.80 m s 2 )

347

l

Chapter 6: Conservation of Energy

Physics

31. Strategy Use Eq. (6-10) to find the nonconservative work. Solution Calculate the work done by friction and air resistance during the run. 1 Wtotal = Wc + Wnc = ∆K = mvf 2 , so 2 1 1 1 2 Wnc = mvf − Wc = mvf 2 − mgh = (75 kg)(12 m s)2 − (75 kg)(9.80 m s 2 )(78 m) = −52 kJ . 2 2 2 32. Strategy Assume frictional forces and air resistance are negligible. Use conservation of energy. Solution Find h, the highest position the car reaches above the bottom of the hill.

∆U = mgh − mghi = −∆K =

1 mv 2 − 0, so h = 2 i

vi 2 2g

+ hi =

(20.0 m s)2 2(9.80 m s 2 )

+ 5.0 m = 25 m .

33. Strategy Use conservation of energy. Solution (a) Solve for the final speed of the ball. 1 1 ∆K = mvf 2 − mv 2 = −∆U = mgh, so vf = 2 2

v 2 + 2 gh .

(b) By inspection of the equation found in part (a), we find that the final speed is independent of the angle. 34. Strategy Since energy is conserved and nonconservative forces do no work, ∆K = −∆U . Let the y-axis point upward. Solution The initial speeds are zero and the final speeds are the same (due to the rope). Since m1 < m2 , block 1

moves up the incline and block 2 falls. Let d be the distance block 1 moves along the incline, then ∆r1 = d sin θ and ∆r2 = − d .

1 1 m v 2 + m2 v 2 = −∆U = −∆U1 − ∆U 2 = − m1 g ∆r1 − m2 g ∆r2 = − m1 gd sin θ − m2 g (− d ), so 2 1 2 2 gd (−m1 sin θ + m2 ) 2(9.80 m s 2 )(1.4 m)[−(12.4 kg) sin 36.9° + 16.3 kg] = = 2.9 m s . m1 + m2 12.4 kg + 16.3 kg

∆K = ∆K1 + ∆K 2 = v=

35. Strategy Plot the force on the vertical axis and the spring length on the horizontal axis. Solution The graph is shown at the right.

(b) Find the y-intercept. F = 1.00 N = kx + b = (0.286 N cm)(14.5 cm) + b, so b = −3.15 N. The force on the spring is zero when the spring is relaxed. Set F = 0. b −3.15 N 0 = kx0 + b, so x0 = − = − = 11.0 cm . k 0.286 N cm

348

5.00 Force, F (N)

(a) Determine the slope of the line to find k, since F = kx. 5.00 N − 1.00 N 4.00 N k= = = 0.286 N cm 28.5 cm − 14.5 cm 14.0 cm

4.00 3.00 2.00 1.00 0.00 0.0 10.0 20.0 30.0 Spring length, x (cm)

Physics

Chapter 6: Conservation of Energy

36. (a) Strategy Since the gravitational field is uniform, the work done by gravity is Wgrav = Fy ∆y = − mg ∆y, where

the y-axis points up. Solution Note that the slope is inclined at 15.0° to the horizontal.

Wgrav = − mg ∆y = −(75.0 kg)(9.80 m

s 2 )[0 − (32.0

m) sin15.0°]

32.0

15.0°

m

(32.0 m) sin 15.0°

= 6.09 kJ The normal force is perpendicular to the motion of the skier, so the work done by the normal force is 0 J. (b) Strategy Refer to part (a). Use conservation of energy and Newton’s second law. Solution The work done by gravity is the same as found in part (a), 6.09 kJ. As before, the normal force is 0 J. The total work done on the skier is equal to the sum of the work done by gravity and the work done by friction. We use Eqs. (6-6) and (6-7) to find the work done by friction. 1 1 Wtotal = Wgravity + Wfriction = − mg ∆y + Wfriction = ∆K = K f − Ki = mv 2 − 0 = mv 2 , so 2 2 1 2 Wfriction = mv + mg ∆y 2 1 = (75.0 kg)(10.0 m s) 2 + (75.0 kg)(9.80 m s 2 )[0 − (32.0 m) sin15.0°] = −2.34 kJ . 2 Now that we know the work done by friction, we use Eq. (6-2) to find the force of friction. W −2337 J W = Fx ∆x = f k ∆x = Wfriction , so f k = friction = = −73.0 N. 32.0 m ∆x The force of friction is 73.0 N opposite the direction of motion. To find the coefficient of kinetic friction, we draw a diagram and use Newton’s second law. N

y x

mg

15.0°

15.0° mg cos 15.0°

ΣFy = N − mg cos15.0° = 0, so N = mg cos15.0°. Since f k = µk N , the coefficient of kinetic friction is f fk 73.0 N µk = k = = = 0.103 . N mg cos15.0° (75.0 kg)(9.80 m s 2 ) cos15.0° 37. Strategy Use the result for escape speed found in Example 6.8. Solution Replace the values for Earth with those for the Moon.

v=

2GM Moon RMoon

=

2(6.674 × 10−11 N ⋅ m 2 kg 2 )(7.35 × 1022 kg) 1.74 × 106 m

= 2.37 km s

38. Strategy Use the result for escape speed found in Example 6.8. Solution The magnitude of the gravitational field is given by GM R 2 = 30.0 m s 2 . Find the escape speed.

vesc =

2GM ⎛ GM = 2⎜ R ⎝ R2

⎞ 2 7 ⎟ R = 2(30.0 m s )(6.00 × 10 m) = 60.0 km s ⎠

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Chapter 6: Conservation of Energy

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39. Strategy Use conservation of energy and the result for escape speed found in Example 6.8. Solution Replacing the values for Earth with those for the Zoroaster, we find that the escape speed for Zoroaster is given by vesc = 2GM Z RZ . Find the speed of the meteor when it hits the surface of the planet.

∆K =

GM Z m 1 1 mvf 2 − mvi 2 = −∆U = , so 2 2 RZ

vf = vi 2 +

2GM Z RZ

= vi 2 + vesc 2 = (5.0 km s) 2 + (12.0 km s) 2 = 13.0 km s .

40. Strategy In the equation for the escape speed found in Example 6.8, replace the values for Earth with appropriate values for the fictional planet. Use proportional reasoning and the relationship between the volume of a sphere and its radius to relate the mass and radius of the planet with those of Earth. Solution Find the escape speed. Earth: 2GM E ME vesc = and ρ E = density = . 4πR 3 RE E 3

Planet: vesc = =

2GM = R

2G ρ V = R E

2G 2 RE

⎛ M ⎞ E ⎟ ⎡ 4 π (2 R )3 ⎤ = 8GM E ⎜ E ⎥ ⎜ 4 π RE3 ⎟ ⎢⎣ 3 RE ⎦ ⎝3 ⎠

8(6.674 × 10−11 N ⋅ m 2 kg 2 )(5.974 × 1024 kg) 6.37 × 106 m

= 22.4 km s

41. Strategy Use Eq. (6-14) and form a ratio. Solution Find the ratio of the potential energies at perigee and apogee.

U perigee U apogee

GmM

=

− 2R E E GmM − 4R E E

=

4 = 2 2

42. Strategy The initial kinetic and potential energies are zero. Neglect the drag force on the meteor due to the atmosphere. Use conservation of energy and Eq. (6-14). Solution Let M be the mass of Earth, m be the mass of the meteor, R be the radius of Earth, and h be the height of the stratosphere. Find the minimum speed of the meteor when it reaches the stratosphere. 1 GMm Ei = 0 = Ef = mvf 2 − , so 2 R+h 2GM 2(6.674 × 10−11 N ⋅ m 2 kg 2 )(5.974 × 1024 kg) = = 11.2 km s . vf = R+h 6.371× 106 m + 40 × 103 m

350

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Chapter 6: Conservation of Energy

43. Strategy Ignore air resistance. Use conservation of energy. Solution Find the required initial speed of the projectile. GM E m GM E m 1 Ei = mvi 2 − , so = Ef = 0 − RE 2 5 RE

vi =

8GM E 5 RE

=

8(6.674 × 10−11 N ⋅ m 2 kg 2 )(5.974 × 1024 kg) 5(6.37 × 106 m)

= 10.0 km s .

44. Strategy Use conservation of energy. Solution Find the comet’s speed at perihelion. 1 GmM 1 GmM = K a + U a = mva 2 − K p + U p = mvp 2 − , so 2 rp 2 ra

⎛1 1⎞ vp = va 2 + 2GM ⎜ − ⎟ ⎜ rp ra ⎟ ⎝ ⎠ 1 1 ⎛ ⎞ = (10.0 × 103 m s) 2 + 2(6.674 × 10−11 N ⋅ m 2 kg 2 )(1.987 × 1030 kg) ⎜ − ⎟ 10 12 ⎝ 8.9 × 10 m 5.3 × 10 m ⎠ = 55 km s 45. Strategy Use Newton’s second law and law of universal gravitation. Solution Calculate the orbital speed. v 2 GM E m ∑ Fr = = mar = m orb , so vorb = 4.0 RE (4.0 RE )2

GM E 4.0 RE

.

Calculate the escape speed. GM E m 1 Ki + U i = mvesc 2 − = K f + U f = 0 + 0, so vesc = 2 4.0 RE

GM E 2.0 RE

.

Find the change in speed. GM E GM E ⎛ 1 1 ⎞ GM E ∆v = vesc − vorb = − =⎜ − ⎟ 2.0 RE 4.0 RE ⎝ 2.0 2.0 ⎠ RE

1 ⎞ (6.674 × 10−11 N ⋅ m 2 kg 2 )(5.974 × 1024 kg) ⎛ 1 =⎜ − = 1.6 km s ⎟ 6.371× 106 m ⎝ 2.0 2.0 ⎠

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46. Strategy The small velocity toward Earth is just enough to upset the equilibrium of the rock and send it toward Earth, but we can neglect it in the calculation of the rock’s final speed and thus neglect the initial kinetic energy of the rock as well. Use conservation of energy. Solution Let the distances from the equilibrium point to the centers’ of Earth and the Moon be d E and d M ,

respectively; and let the distance between the centers of the Earth and the Moon be R. Let h = 700, 000 m, and let mr , m, and M be the masses of the rock, Moon, and Earth, respectively. Find the speed of the rock when it encounters Earth’s atmosphere. Gmr M Gmr m Gmr M Gmr m 1 Ki + U i = 0 + U i = − + = K f + U f = mr v 2 − + , so dE dM 2 h R−h ⎛ M m M m ⎞ v = 2G ⎜⎜ − + + − ⎟⎟ . ⎝ dE dM h R − h ⎠ We need to find d E and d M . At the equilibrium point, the forces due to Earth and the Moon on the rock are equal; so according to Newton’s law of universal gravitation, −

Gmr M dE

2

=−

Gmr m dM

2

, or

dE = dM

M . m

M m R R + dM = dE + dE , so d M = and d E = . m M 1+ M m 1+ m M Substitute these into the equation for v. Now, R = d E + d M = d M

⎡ M (1 + m M ) m(1 + M m ) M m ⎤ m ⎞ ⎛m−M M v = 2G ⎢ − + + − + − ⎥ = 2G ⎜ ⎟ R R h R − h ⎥⎦ h R−h⎠ ⎝ R ⎢⎣ G = 6.674 × 10−11 N ⋅ m 2 kg 2 , m = 7.349 × 1022 kg, M = 5.974 × 1024 kg, R = 3.845 × 108 m, and h = 700, 000 m + 6.371× 106 m. Substituting these values into the equation for v gives a speed for the rock of 10,500 m s . 47. Strategy Use proportional reasoning. Solution The force is linear with respect to the displacement of the string. If the string is pulled back half as far (20.0 cm) as in Example 6.9 (40.0 cm), the average force is only half that as in Example 6.9. Therefore, the work ⎛ 1 ⎞⎛ 1 ⎞ done is ⎜ ⎟ ⎜ ⎟ (32 J) = 8 J . ⎝ 2 ⎠⎝ 2 ⎠ 48. Strategy The work done on the spring is negative the work done by the spring. Use the relationship between work and the extension or compression of a spring. Solution Find the work done to stretch the spring. 1 1 W = kx 2 = (20.0 N m)(0.40 m)2 = 1.6 J 2 2 49. Strategy The work done by the hammer on the object is represented by the area between the curve and the x-axis. Solution Compute the work done in driving the nail.

W = (50 N)(0.012 m) + (120 N)(0.050 m − 0.012 m) = 5.2 J

352

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Chapter 6: Conservation of Energy

50. (a) Strategy The increase in the force is F and the length that the tendon increases is x if the tendon is modeled as a spring and Hooke’s law is used. Solution Compute the spring constant. F 4800 N − 3200 N 1600 N k= = = = 3200 N cm x 0.50 cm 0.50 cm (b) Strategy The triangular area under a force of the muscle vs. the stretch of the tendon graph is equal to the work done by the muscle. Solution Compute the work done by the muscle in stretching the tendon. 1 1 W = Fx = (4800 N − 3200 N)(0.0050 m) = 4.0 J 2 2 51. (a) Strategy Use Hooke’s law and form a proportion. Solution F k= 1= x1

Form the proportion. F2 x2

Solve for x2 to find the amount that the spring stretches. x2 =

F2 7.0 N x = (3.5 cm) = 4.9 cm F1 1 5.0 N

(b) Strategy Substitute known values for F1 and x1 to find k. Solution Compute the spring constant. F 5.0 N k= 1= = 1.4 N cm x1 3.5 cm (c) Strategy The triangular area under a forces on the spring vs. the stretch of the spring graph is equal to the work done by the forces. Solution Compute the work done by the forces on the spring. 1 1 W = Fx = (5.0 N)(0.035 m) = 88 mJ 2 2 52. (a) Strategy Solve for k in Hooke’s law. Solution Compute the spring constant. F 120 N ⎛ 109 nm ⎞ k= = ⎜ ⎟ = 6.0 × 1010 N m x 2.0 nm ⎜⎝ 1 m ⎟⎠ (b) Strategy Solve for x in Hooke’s law and use the value for k found in part (a). Solution Compute the compression of the block. F 480 N x= = = 8.0 nm k 6.0 × 1010 N m

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Chapter 6: Conservation of Energy

Physics

(c) Strategy Since the forces due to the block are opposite to the directions of compression, the block does negative work during the compression. The work done by the applied forces is positive and equal to the negative of the work done by the block. Solution Compute the work done. 1 1 W = −Wblock = kx 2 = (6.0 × 1010 N m)(8.0 × 10−9 m) 2 = 1.9 µJ 2 2 53. (a) Strategy Set the weight of the mass equal to the force in Hooke’s law. Solution Compute the spring constant. mg (1.4 kg) ( 9.80 N kg ) W = mg = F = kx, so k = = = 1.9 N cm . x 7.2 cm (b) Strategy Use Eq. (6-24). Solution Compute the elastic potential energy stored in the spring. 1 1 ⎡ (1.4 kg) ( 9.80 N kg ) ⎤ 2 U elastic = kx 2 = ⎢ ⎥ (0.072 m) = 0.49 J 2 2⎣ 0.072 m ⎦ (c) Strategy Solve for m in the equation for k found in part (a). Solution Compute the second mass. ⎛m g⎞x kx x 12.2 cm m2 = 2 = ⎜⎜ 1 ⎟⎟ 2 = 2 m1 = (1.4 kg) = 2.4 kg g x g x 7.2 cm 1 ⎝ 1 ⎠ 54. Strategy W = Fx ∆x and the work is represented by the area under the curve, A = (1 2)bh. Solution Find the work done in each situation.

(a) W =

1 (0.20 m)(15 N) = 1.5 J 2

(b) W =

1 1 (0.20 m)(15 N) − (0.10 m)(7.5 N) = 1.1 J 2 2

55. Strategy and Solution Since the displacement of the model airplane is zero, zero work has been done on it by the string. 56. Strategy The work done by the force on the object is represented by the area between the curve and the x-axis. The area under the axis represents negative work done. Solution Compute the work done by the force. 1 1 W = (2.0 N)(1.0 m) + (1.0 N)(1.0 m) + (−1.0 N)(1.0 m) = 0.5 J 2 2 57. Strategy and Solution Let E be the elastic energy stored in the legs. This energy is converted into gravitational potential energy, mgh, when the kangaroo jumps. Since only one leg is used, and since h ∝ E , the kangaroo can

only jump half as high, or (0.70 m) 2 = 0.35 m .

354

Physics

Chapter 6: Conservation of Energy

58. Strategy The mechanical energy is constant, so we set the elastic potential energy of the spring on the toy gun equal to the gravitational potential energy of the rubber ball. Assume x  h. Solution mgh = (1 2)kx 2 , so h ∝ x 2 . Thus, the height reached by the ball is proportional to the square of the compression of the spring. Form a proportion.

h2 h1

=

2 x 2 ⎛ 2x ⎞ , so h2 = 2 h1 = ⎜ ⎟ h = 4h . ⎝ x ⎠ x12 x12

x22

59. Strategy The elastic potential energy of the catapult, (1 2)kx 2 , is converted into gravitational potential energy of the pebble, mgh. Solution Find the maximum height achieved by the pebble. 1 kx 2 (320 N m)(0.20 m) 2 mgh = kx 2 , so h = = = 13 m . 2 2mg 2(0.051 kg)(9.80 m s 2 ) 60. Strategy Use conservation of energy. Solution (a) Compute the speed of the block as it passes through the equilibrium point.

1 1 k Ki + U i = 0 + kd 2 = K f + U f = mv 2 + 0, so v = d . 2 2 m (b) Find the maximum distance below the equilibrium point that the block will reach. 1 1 m k m =d = d . Ki + U i = mv 2 + 0 = K f + U f = 0 + kx 2 , so x = v 2 2 k m k 61. Strategy Take the surface of the unstretched trampoline to be y = 0. Use conservation of energy and Newton’s second law. Solution Find the spring constant from the gravitational potential energy. 1 2mgh mgh = kymin 2 , so k = . 2 ymin 2

Use Newton’s second law for the situation where the gymnast is at rest. ∑ F = ky − mg = 0, so ⎛y 2⎞ y 2 mg (−0.75 m)2 y= = mg ⎜ min ⎟ = min = = 8.7 cm . ⎜ 2mgh ⎟ k 2h 2(2.5 m + 0.75 m) ⎝ ⎠ 62. Strategy The stretched length of the bungee cord (and the distance George falls) must be no more than 55.0 m − 2.00 m = 53.0 m. The gravitational potential energy decrease of George as he falls must equal the elastic potential energy increase of the bungee cord. Solution To find the spring constant, set Eqs. (6-24) and (6-13) equal and solve for k. 2mgyfall 2(75.0 kg)(9.80 m s 2 )(53.0 m) 1 kystretch 2 = mgyfall , so k = = = 115 N m . 2 ystretch 2 (53.0 m − 27.0 m)2

355

Chapter 6: Conservation of Energy

Physics

63. (a) Strategy The increase in kinetic energy of the block is equal to the decrease in its potential energy. Let the potential energy be zero at y = 0.25 m. Solution To find the speed of the block, set Eqs. (6-6) and (6-13) equal and solve for v. 1 2 mv = mgy, so v = 2 gy = 2(9.80 m s 2 )(0.25 m) = 2.2 m s . 2 (b) Strategy The elastic potential energy increase of the spring is equal to the decrease in gravitational potential energy of the block. Let the potential energy be zero at y = 0 m. Solution To find the compression of the spring, set Eqs. (6-24) and (6-13) equal and solve for x.

1 2 kx = mgy, so x = 2

2mgy = k

2(2.0 kg)(9.80 m s 2 )(0.50 m) = 0.21 m . 450 N m

(c) Strategy and Solution Since the surface is frictionless, no nonconservative forces do work on the block. So, the block will return to its previous height, or 0.50 m. 64. Strategy As Lars climbs the stairs, he increases his gravitational potential energy. The rate of potential energy increase must be equal to the rate he does work. Solution To find the time for Lars to climb the stairs, use Eqs. (6-26) and (6-9) and solve for ∆t.

Pav =

∆E ∆U mg ∆y mg ∆y (82.4 kg)(9.80 m s 2 )(12.0 m − 0) = = , so ∆t = = = 13.0 s . ∆t ∆t ∆t Pav 746 W

65. Strategy Watts are joules per second and there are 3600 seconds in 1 hour. Solution Show that 1 kW ⋅ h = 3.6 MJ. J 3600 s 1 kW ⋅ h = 103 ⋅ h ⋅ = 3.6 × 106 J = 3.6 MJ s 1h 66. Strategy Use the definition of average power and the potential energy in a uniform gravitational field. Solution Find the minimum time required for the man to lift the boxes. ∆E Pav = and ∆E = ∆U = mtotal gh, so ∆t ∆E 50mgh 50(10.0 kg)(9.80 m s 2 )(2.00 m) ⎛ 1 min ⎞ ∆t = = = ⎜ ⎟ = 4.08 min . Pav Pav 40.0 W ⎝ 60 s ⎠ 67. Strategy As Rosie lifts the trunk, she does work on it and increases its gravitational potential energy. Solution To find the average rate of work Rosie does on the trunk (that is, the power she supplies), use Eqs. (6-26) and (6-13). ∆E ∆W mgh (220 N)(4.0 m) = = = = 22 W Pav = ∆t ∆t ∆t 40 s

356

Physics

Chapter 6: Conservation of Energy

68. Strategy Assume that friction is negligible. Use Eq. (6-27). Solution The rate at which gravity does work on the bicycle and rider is G G P = mgv cos θ , where θ is the angle between v and g, or θ = 90° + φ . Find φ .

v

φ

θ φ

g

5.0 m

5.0 m 100 m , so φ = tan −1 0.050. 100 m The power output of the rider is equal to the rate of change of potential energy, which equals −P. Therefore, tan φ =

Prider = − P = −(75 kg)(9.80 m s 2 )(4.0 m s) cos(90° + tan −1 0.050) = 150 W .

69. Strategy Use Eq. (6-27). Solution (a) Find the force exerted on the cyclist by the air. G ∆U G Pa + Pc = 0 since = 0. v is antiparallel to Fa . ∆t Pa −120 W Pa = Fa v cos θ , so Fa = = = 20 N . v cos θ (6.0 m s) cos180° (b) Find the speed of the cyclist. Pc 120 W v= = = 6.7 m s Fc cos θ (−18 N) cos180° 70. (a) Strategy Use the definition of average power. The change in energy is equal to the change in kinetic energy. Solution Find the average mechanical power output. 2 2 2 1 ⎡ ⎤ ∆E ∆K 2 m(vf − vi ) (1000.0 kg) ⎣ (40.0 m s) − 0 ⎦ Pav = = = = = 80.0 kW 2(10.0 s) ∆t ∆t ∆t (b) Strategy Relate the mechanical energy required to the chemical energy provided per liter of gasoline. Solution Find the volume of gasoline consumed. mechanical energy required K = = chemical energy provided (efficiency)(46 MJ L) per liter

1 (1000.0 2

kg)(40.0 m s)2

(46 × 106 J L)(0.22)

= 0.079 L

71. Strategy Use the definition of average power. The change in energy is equal to the change in kinetic energy. Solution Determine the average mechanical power the engine must supply. 2 2⎤ 2 2 1 ⎡ ∆E 2 m(vf − vi ) (1200 kg) ⎣(30.0 m s) − (20.0 m s) ⎦ = = = 60 kW Pav = ∆t ∆t 2(5.0 s)

357

Chapter 6: Conservation of Energy

Physics

72. (a) Strategy Use the definition of average power. The change in energy is equal to the change in gravitational potential energy. Solution Find the woman’s average power output. ∆E ∆U mgh (62 kg)(9.80 m s 2 )(5.0 m) Pav = = = = = 510 W 6.0 s ∆t ∆t ∆t (b) Strategy and Solution The body would have to be 100% efficient for the answer to part (a) to be equal to the average power input. So, the answer is no. 73. Strategy Relate the change in gravitational potential energy of the person to the energy provided by the carbohydrate. Solution Find the mass of carbohydrate required for the person to climb the stairs. energy mgh (74 kg)(9.80 m s 2 )(15 m) = = = 6.2 g energy available per gram 0.100(energy per gram) 0.100(17.6 × 103 J g) The other 90% of the energy is dissipated as heat.

74. Strategy The instantaneous power is given by Eq. (6-27), where θ = 0°. Obtain the necessary values of the force from the graph. Solution (a) Compute the instantaneous power.

Pav = Fv cos θ = Fx vx = (800 N)(11 m s) = 8.8 kW (b) As in part (a), we have Pav = Fv cos θ = Fx vx = (400 N)(16 m s) = 6.4 kW . 75. Strategy Use the definition of average power. The change in energy is equal to the change in kinetic energy. Solution Find the engine’s average power output. 2 2 2 1 ⎡ ⎤ ∆E ∆K 2 m(vf − vi ) (500.0 kg) ⎣(125 m s) − 0 ⎦ = = = = 930 kW Pav = ∆t ∆t ∆t 2(4.2 s) 76. (a) Strategy Use Eq. (6-9). Solution Calculate the change in gravitational potential energy of the water. ∆U = mg ( yf − yi ) = (1 kg)(9.80 N kg)(0 − 50 m) = −500 J

(b) Strategy The mass flow rate times the potential energy change per unit mass gives the rate at which gravitational potential energy is lost by the river. Solution Compute the rate. P = (5.5 × 106 kg s)(500 J kg) = 3 GW

(c) Strategy The electrical power output divided by the power per household gives the total number of households supplied. Solution Compute the number of households. 0.10(3 × 109 W) = 300, 000 households 1× 103 W household 358

Physics

Chapter 6: Conservation of Energy

77. Strategy Use conservation of energy. Solution Compute the required speed of the high jumper. 1 1 Ki + U i = mv 2 + 0 = mv 2 = K f + U f = 0 + mgh = mgh, so 2 2 v = 2 gh = 2(9.80 m s 2 )(1.2 m) = 4.8 m s . 78. Strategy Use conservation of energy. Solution Find the maximum height of the pole-vaulter’s center of gravity. 1 1 Ki + U i = mv 2 + mghi = K f + U f = 0 + mghf , so mv 2 = mg (hf − hi ), or 2 2 v2 (10.0 m s) 2 hf = +h = + 1.0 m = 6.1 m . 2 g i 2(9.80 m s 2 ) 79. Strategy Use conservation of energy. Neglect drag. Solution Find the speed of the hang glider. 1 1 ∆K = mvf 2 − mvi 2 = −∆U = mgh, so vf = 2 gh + vi 2 = 2(9.80 m s 2 )(8.2 m) + (9.5 m s) 2 = 16 m s . 2 2 80. Strategy Use the work-kinetic energy theorem. The work done by friction is equal to the force of friction, f k = µk N = µk mg , times the distance the car skidded d. Solution Relate the speed to the distance. 1 1 ∆K = mvf 2 − 0 = mvf 2 = W = f k d = µk mgd 2 2 Let the first case be represented by the subscript 1, and the second by 2: 1 1 mv 2 = µ mgd1 and mv22 = µ mgd 2 . Form a proportion to find the distance of the skid. 2 1 2 1 mv 2 1 2 1 mv 2 2 2

2

=

2

⎛v ⎞ µ mgd1 ⎛ 60 mi h ⎞ , so d 2 = ⎜⎜ 2 ⎟⎟ d1 = ⎜ ⎟ (50 ft) = 200 ft . v µ mgd 2 ⎝ 30 mi h ⎠ ⎝ 1⎠

81. Strategy Use Newton’s second law and law of universal gravitation. Solution Prove that U = −2 K for any gravitational circular orbit. ∑ Fr = m

v2 r

mv 2 1 2 mv 2 K U

=

GMm r2 GMm

= mar =

mv 2 , so r

r2 GMm = r 1 ⎛ GMm ⎞ = ⎜ ⎟ 2⎝ r ⎠ 1 =− U 2 = −2 K

359

Chapter 6: Conservation of Energy

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82. Strategy Use conservation of energy. Solution v y can be found from the kinetic energy gained due to gravity. 1 mv 2 = mgh, so v y 2 = 2 gh. 2 y vx can be found from the kinetic energy gained from the spring’s elastic potential energy. 1 1 k ( xi 2 − xf 2 ) mvx 2 = k ( xi 2 − xf 2 ), so vx 2 = . 2 2 m Compute the speed of the ball when it hits the gound.

v = vx 2 + v y 2 =

k ( xi 2 − xf 2 ) (28 N m)[(0.18 m)2 − (0.12 m)2 ] + 2 gh = + 2(9.80 m s 2 )(1.4 m) m 0.056 kg

= 6.0 m s . 83. Strategy Use Hooke’s law and Newton’s laws. Solution (a) The mass connected to the lower spring exerts a force on the lower spring equal to its weight, W. The spring stretches an amount x = F k = W k . The lower spring exerts a force on the upper spring equal to F = W , and causes it to stretch by x = F k = W k . Thinking of the two springs as a single spring:

2x =

F F 2F k + = = x′, so F = x′ = k ′x′. Therefore, 2 k k k

k = k ′, the effective spring constant. 2

(b) Sum the forces on the mass. F + F − W = kx + kx − W = 2kx − W = 0, so W = 2kx = k ′x.

Therefore, 2k = k ′, the effective spring constant. 84. (a) Strategy Use conservation of energy. Solution Find the speed of the car at the top of the loop. 1 1 1 ∆K = mvf 2 − mvi 2 = mvf 2 − 0 = −∆U = − mg (hf − hi ) = mg (hi − hf ), so 2 2 2 vf = 2 g (hi − hf ) = 2(9.80 m s 2 )(40.0 m − 20.0 m) = 19.8 m s . (b) Strategy Use Newton’s second law. Solution Find the normal force on the car exerted by the track. ⎛ v2 ⎞ ΣFr = mar = N + mg , so N = m(ar − g ) = m ⎜ − g ⎟ . ⎜ r ⎟ ⎝ ⎠ ⎡ 2 g (hi − hf ) ⎤ ⎛h −h 1⎞ ⎛ 40.0 m − 20.0 m 1 ⎞ − g ⎥ = 2mg ⎜ i f − ⎟ = 2(988 kg)(9.80 m s 2 ) ⎜ − ⎟ = 29.0 kN N = m⎢ 2⎠ 10.0 m 2⎠ r ⎝ ⎣ ⎦ ⎝ r (c) Strategy Use Newton’s second law. Solution Set the normal force on the roller coaster car equal to zero at the top to find the minimum height. 10.0 m v2 r N = 0 = m − mg , so gr = v 2 = 2 g (hi − hf ). Thus, hi = + hf = + 20.0 m = 25.0 m . 2 2 r 360

Physics

Chapter 6: Conservation of Energy

85. Strategy Use Eqs. (6-10) and (6-11) for the work and energy solution. Use Newton’s second law and Eq. (2-13) for the force solution. Let d = 8.0 m, d1 = 5.0 m, and d 2 = 8.0 m − 5.0 m = 3.0 m. Solution First method: Find the constant tension using work and energy. Wtotal = Wc + Wnc = ∆K = 0, since the speeds at the top and bottom of the incline

.0 m

d=8 15°

d sin 15°

are zero. Also, Wc = −∆U and Wnc = Td 2 . Wnc = Td 2 = −Wc = ∆U , so

∆U mg ∆y mg (0 − d sin θ ) (4.0 kg)(9.80 m s 2 )(8.0 m) sin15° mgd sin θ = = =− =− = −27 N. 3.0 m d2 d2 d2 d2 The sign is negative because the work done by the tension is opposite the box’s motion. So, the magnitude of the tension is 27 N. T=

Second method: Find the speed of the block just before the person grasps the cord using Newton’s second law. ΣFy = N − mg cos θ = 0 and ΣFx = mg sin θ = ma. vf 2 − vi 2 = vf 2 − 0 = 2 g sin θ d1

N T

y x mg

Let vf = v. Find the tension.

15° mg cos 15° mg sin 15°

T ΣFx = −T + mg sin θ = ma x , so a x = − + g sin θ . m ⎛ T ⎞ vf 2 − vi 2 = 0 − v 2 = −2 g sin θ d1 = 2a x ∆x = 2 ⎜ − + g sin θ ⎟ d 2 , so ⎝ m ⎠ ⎛ d1 + d 2 ⎞ d ⎛ 8.0 m ⎞ 2 T = mg sin θ ⎜⎜ ⎟⎟ = mg sin θ d = (4.0 kg)(9.80 m s ) sin15° ⎜⎝ 3.0 m ⎟⎠ = 27 N. d ⎝ ⎠ 2 2

86. (a) Strategy The spring does work against gravity. Use Eq. (6-20). Solution 1 2mg ∆y 2(780 N)(68 m − 182 m) Wspring = − kx 2 = −Wg = mg ∆y, so k = − =− = 25 N m . 2 2 x (182 m − 68 m − 30.0 m)2

(b) Strategy The kinetic energy gained is the sum of the positive work done by gravity (which increases the kinetic energy during the fall) and the negative work done by the cord (which decreases the kinetic energy). Solution Find the speed of the jumper. 1 2 1 mv = − mg ∆y − kx 2 , so 2 2 k 2 25.2 N m v = −2 g ∆y − x = −2(9.80 m s 2 )(92 m − 182 m) − (182 m − 92 m − 30.0 m)2 = 25 m s . 780 N m 2 9.80 m s

87. Strategy Use conservation of energy. Solution The elastic potential energy of the spring is converted to gravitational potential energy, so 1 2 kx = mgh = mg (l sin θ ) where l is the distance the object travels up the incline. 2 kx 2 (40.0 N m)(0.20 m) 2 Thus, l = = = 0.33 m . 2mg sin θ 2(0.50 kg)(9.80 N kg) sin 30.0° 361

Chapter 6: Conservation of Energy

Physics

88. (a) Strategy According to the work-kinetic energy theorem, the total work done on the stunt woman is equal to her change in kinetic energy. Solution Wtotal = ∆K = K f − Ki =

1 1 mv 2 − 0 = (62.5 kg)(10.5 m s)2 = 3.45 kJ 2 f 2

(b) Strategy The work done by gravity is negative the change in gravitational potential energy of the stunt woman. Solution Wgrav = −∆U = U i − U f = − mg ∆y = −(62.5 kg)(9.80 m s 2 )(−8.10 m) = 4.96 kJ

(c) Strategy Use Eq. (6-10) to find the nonconservative work done by air resistance. Solution Wtotal = Wcons + Wnc = Wgrav + Wair = ∆K , so Wair = ∆K − Wgrav =

1 ⎡1 ⎤ mv 2 + mg ∆y = (62.5 kg) ⎢ (10.5 m s) 2 + (9.80 m s 2 )(−8.10 m) ⎥ = −1.52 kJ . 2 f 2 ⎣ ⎦

(d) Strategy Use Eq. (6-2) to find the magnitude of the average constant force of air resistance during the fall. Solution Let the +y-axis point upward. W −1516 J = 187 N Wair = Fair ∆y, so Fair = air = ∆y −8.10 m Thus, the magnitude of the average force of air resistance is 187 N. 89. Strategy Use conservation of energy and Newton’s second law. Solution Find k. mg ΣFy = kx1 − mg = 0, so k = . Find vmax . x1 1 1 Ki + K f = 0 + mvmax 2 = U i + U f = kxmax 2 + 0, so 2 2 k g 9.80 m s 2 = xmax = (0.100 m − 0.050 m) = 1.6 m s . vmax = xmax m x1 0.060 m − 0.050 m

F = kx1 0.20 kg

mg

90. (a) Strategy Use the definition of average power and the fact that the change in energy is equal to the person’s increase in potential energy. Solution ∆E ∆U mg ∆y (70 kg)(9.80 m s 2 )(740 m) ⎛ 1 h ⎞ Pav = = = = ⎜ ⎟ = 94 W 1.5 h ∆t ∆t ∆t ⎝ 3600 s ⎠ (b) Strategy Since the human body is only 25% efficient, it takes 4.0 units of chemical energy for every single unit of potential energy gained. Solution Find the amount of chemical energy used in the hike. Echem = 4.0∆U grav = 4.0mg ∆y = 4.0(70 kg)(9.80 m s 2 )(740 m) = 2.0 MJ

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Chapter 6: Conservation of Energy

(c) Strategy and Solution Using the given conversion factor, the number of Calories of food energy used for ⎛ 1 Calorie ⎞ the hike is 4(70 kg)(9.80 m s 2 )(740 m) ⎜ ⎟ = 490 Calories . ⎝ 4.186 × 103 J ⎠ 91. (a) Strategy Let +x be up the incline. Use Newton’s second law and Eq. (6-27). Solution Compute the power the engine must deliver. ΣFx = Fair − mg sin φ = 0 at terminal speed. The rate at which air resistance dissipates energy is Pair = Fair v cos180° = − Fair v = − mg sin φ v. (We use cos 180° since the force of air resistance is opposite the car’s velocity.) The power the engine must deliver to drive the car on level ground is Pengine = − Pair = mg sin φ v = (1500 kg)(9.80 m s 2 )(20.0 m s) sin 2.0° = 10 kW .

(b) Strategy The power available to climb the hill is the power delivered by the engine minus the dissipating power of air resistance. Solution From part (a), for a slope of φ : P = mgv sin φ , so φ = sin −1

P 40.0 × 103 W − 10.26 × 103 W = sin −1 = 5.8° . mgv (1500 kg)(9.80 m s 2 )(20.0 m s)

92. Strategy Use Hooke’s law and Eq. (6-24). Solution Since F = kx, compressing a spring a certain distance x0 requires a force F0. After cutting the spring in half, compressing the spring x0 is equivalent to compressing the original spring 2x0, which would require a force of 2F0. This is equivalent to the short spring having a spring constant twice as large as the original spring. Find Ushort. 1 2 1 kx and U short = (2k ) x 2 . Form a ratio. 2 2 1 (2k ) x 2 = 2 = 2, so U short = 2U long = 2(10.0 J) = 20.0 J . 1 kx 2 2

U long = U short U long

93. (a) Strategy Use the definition of average power. The change in energy is equal to the gravitational potential energy. Solution Find the average power the motor must deliver. ∆E mgh (1202 kg − 801 kg)(9.80 m s 2 )(40.0 m) = = = 2.62 kW Pav = ∆t ∆t 60.0 s (b) Strategy Without the counterweight, the motor must deliver more power. Solution Find the average power the motor must deliver. (1202 kg)(9.80 m s 2 )(40.0 m) Pav = = 7.85 kW 60.0 s The answer is significantly larger.

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94. (a) Strategy Use the work-kinetic energy theorem. Solution Compute the work done by the pitcher. 1 1 2 W = ∆K = mvf 2 − 0 = (0.153 kg) ( 40.2 m s ) = 124 J 2 2 (b) Strategy Divide the energy available to do work by the energy required to throw a fastball. Solution Compute the number of fastballs required to “burn off” the meal. energy available to do work 0.200(1520 × 103 cal) ⎛ 4.186 J ⎞ = ⎜ ⎟ = 10,300 fastballs J energy per fastball thrown 123.6 ⎝ cal ⎠ fastball

95. Strategy The basal metabolic rate is equal to the number of kilocalories per day required by a person resting under standard conditions. Solution (a) Compute Jermaine’s basal metabolic rate. ⎛ 1 kcal ⎞⎛ 0.015 mol ⎞ ⎛ 1440 min ⎞ BMR = ⎜ ⎟ = 2200 kcal day ⎟⎜ ⎟⎜ ⎝ 0.010 mol ⎠⎝ min ⎠ ⎝ day ⎠ (b) Find the mass of fat lost. 2160 kcal day ⎛ 2.2 lb ⎞ ⎜ ⎟ = 0.51 lb day 9.3 kcal g ⎜⎝ 103 g ⎟⎠ Since Jermaine is not resting the entire time, he loses more than 0.51 lb. 96. Strategy Tarzan’s kinetic energy must be great enough that his gravitational potential energy can increase by mgh, where h = 1.7 m. Solution If Tarzan just makes it across the gully, his final kinetic energy is zero. Let his initial potential energy be zero. Set his initial kinetic energy equal to his final potential energy and solve for his initial speed. 1 mv 2 = mgh, so vi = 2 gh = 2(9.80 m s 2 )(1.7 m) = 5.8 m s . 2 i 97. (a) Strategy Draw a diagram and use trigonometry. Solution Referring to the diagram, we see that when Jane is at the lowest point of her swing, L = h + L cos 20°. Solving for h, we find that h = L − L cos 20°.

L 20°

L cos 20°

h

(b) Strategy We assume that no nonconservative forces act (significantly) on Jane. Thus, ∆E = 0. Solution Use conservation of energy to find Jane’s speed at the lowest point of her swing. 1 1 1 1 1 1 ∆E = 0 = ∆K + ∆U = mvf 2 − mvi 2 + mg ∆y = mvf 2 − mvi 2 + mg (0 − h), so vf 2 = vi 2 + gh, or 2 2 2 2 2 2 vf = vi 2 + 2 gh = vi 2 + 2 gL(1 − cos 20°) = (4.0 m s)2 + 2(9.80 m s 2 )(7.0 m)(1 − cos 20°) = 4.9 m s .

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(c) Strategy When Jane’s entire initial kinetic energy is converted into gravitational potential energy, she will have reached her maximum height. Solution Use conservation of energy to find how high Jane can swing (with respect to her lowest point). 1 1 ∆E = 0 = ∆K + ∆U = 0 − mvi 2 + mg ∆y = − mvi 2 + mg (hmax − hmin ), so 2 2 2 vi 2 (4.0 m s) hmax = + L(1 − cos 20°) = + (7.0 m)(1 − cos 20°) = 1.24 m . 2g 2(9.80 m s 2 ) 98. Strategy Ignore drag due to the atmosphere and gravitational forces due to the Moon and the other planets. Use conservation of energy and Eq. (6-14) for the potential energy. Solution Find the minimum speed. RE-S = Earth-Sun distance Ei =

GM E m GM S m 1 mvi 2 − − = Ef = 0 + 0, so RE RE-S 2

⎛ 5.974 × 1024 kg 1.987 × 1030 kg ⎞ ⎛M M ⎞ vi = 2G ⎜⎜ E + S ⎟⎟ = 2(6.674 × 10−11 N ⋅ m 2 kg 2 ) ⎜ + ⎟ = 43.5 km s . 6 ⎜ 1.50 × 1011 m ⎟⎠ ⎝ RE RE-S ⎠ ⎝ 6.371× 10 m

99. Strategy Draw a diagram. Then, use Newton’s second law and conservation of energy. Let the positive direction be away from the slope. N

Solution According to Newton’s second law, ΣFr = N − mg cos θ = mar = − mv 2 R . When the normal force becomes zero, we have mg cos θ = mv 2 R , or mgR cos θ = mv 2 . When this condition is true, the skier leaves the surface of the ice. Note from the figure that h = R cos θ , thus, the condition becomes mgh = mv 2 . Now, mgh is the final gravitational

potential energy (U i = mgR) and mv 2 is twice the final kinetic energy ( Ki = 0).

mg cos θ

mg h

θ

So, the condition becomes U f = 2 K f , or K f = U f 2. Use conservation of energy to find h in terms of R. 1 3 3 3 0 = ∆K + ∆U = K f − Ki + U f − U i = U f − 0 + U f − U i = U f − U i , so mgh − mgR = h − R = 0, or 2 2 2 2 2 h= R . 3

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100. (a) Strategy Use Hooke’s law and Newton’s laws of motion. Solution According to Hooke’s law, F1 = k1 x1 and F2 = k2 x2 . Imagine that one spring (1) is suspended from a ceiling and the other (2), attached to the bottom of the first, has a mass m attached to its bottom end. Assume that the masses of the springs are negligible, and that the system is in equilibrium. The mass connected to the lower spring exerts a force on the lower spring equal to its weight, W. The spring stretches an amount x2 = F2 k2 = W k2 . The lower spring exerts a force on the upper spring equal to F2 = W , and causes it to stretch by x1 = F1 k1 = F2 k1 = W k1. So, F1 = F2 , thus k1 x1 = k2 x2 . Let F1 = F2 = F and x = x1 + x2 , and imagine the two springs in series as only one spring which stretches an amount x in response to a force F. Find the effective spring constant, k.

x = x1 + x2 =

F1 k1

+

F2 k2

=

⎛1 1 F F + = F ⎜⎜ + k1 k2 ⎝ k1 k2

The effective spring constant is k =

k1k2 k1 + k2

⎞ ⎛1 1 ⎟⎟ , so F = ⎜⎜ + ⎠ ⎝ k1 k2

⎞ ⎟⎟ ⎠

−1

×x=

k1k2 k1 + k2

x = kx.

.

(b) Strategy Use the result from part (a) and Eq. (6-24). Solution Compute the potential energy stored in the spring. 1 1 k1k2 2 (500 N m)(300 N m) U = kx 2 = x = (0.040 m) 2 = 0.15 J 2 2 k1 + k2 2(500 N m + 300 N m) 101. (a) Strategy Use Hooke’s law and Newton’s laws of motion. Solution According to Hooke’s law, F1 = k1 x1 and F2 = k2 x2 . Imagine that the springs are suspended from a ceiling such that the bottom of each is at the same height. Then a mass m is attached to the bottom of both, the springs stretch, and the system comes to equilibrium. Assume that the masses of the springs are negligible. Sum the vertical forces. F1 + F2 − W = 0, so W = F1 + F2 = k1 x1 + k2 x2 . Assuming the springs are attached to the same point on the top of the mass, x1 = x2 = x. W = k1 x1 + k2 x2 = k1 x + k2 x = (k1 + k2 ) x = kx = W So, in response to a force that stretches the springs (W, in this case), the springs act like one spring with a spring constant k = k1 + k2 .

(b) Strategy Use the result from part (a) and Eq. (6-24). Solution Compute the potential energy stored in the spring. 1 1 1 U = kx 2 = (k1 + k2 ) x 2 = (500 N m + 300 N m)(0.020 m)2 = 0.16 J 2 2 2

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102. (a) Strategy Use conservation of energy and the relationship between radial acceleration and tangential speed. Solution The total kinetic energy required for the bob to travel the full circle is equal to the gravitational potential energy difference between the bottom and the top of the circle, mgh = mg[2( L − d )] = 2mg ( L − d ), plus the kinetic energy required for the bob to have enough speed at the top of the circle to complete it. The radial acceleration must be equal to that due to gravity at the top of the circle for the bob to just complete the circle. So, ar = vtop 2 / r = g , or vtop = gr = g ( L − d ). Find the total kinetic energy of the bob at the bottom of the circle. 1 1 1 K total = mv 2 = 2mg ( L − d ) + mvtop 2 = 2mg ( L − d ) + mg ( L − d ), so v = 2 2 2

5g (L − d ) .

(b) Strategy h = L(1 − cos θ ) since θ = 0° gives h = L(1 − 1) = 0 and θ = 90° gives h = L(1 − 0) = L. Use conservation of energy. Solution Find the minimum angle. U i + Ki = U f + K f 1 mgh + 0 = 0 + mv 2 2 1 mgL(1 − cos θ ) = m[5 g ( L − d )] 2 5⎛ L−d ⎞ 5⎛ d ⎞ 1 − cos θ = ⎜ ⎟ = ⎜1 − ⎟ 2⎝ L ⎠ 2⎝ L⎠ 5 5d 5d 3 − cos θ = − −1 = − + 2 2L 2L 2 ⎛ 5d 3 ⎞ θ = cos −1 ⎜ − ⎟ ⎝ 2L 2 ⎠ 103. Strategy Use Newton’s second law, Hooke’s law, and Eq. (6-24). Solution Find the distance that the tendon stretches. T 4.7 kN ∑ F = T − kx = 0, so x = = = 1.3 cm . k 350 kN m Find the stored elastic energy. 1 1 T2 T2 (4.7 × 103 N) 2 U = kx 2 = k = = = 32 J 2 2 k 2 2k 2(350 × 103 N m) 104. (a) Strategy The work is represented by the area under the curve. Estimate the work done during stretching and contraction; then, find the total work done. Solution Estimate the work. 1 1 Wstretch ≈ (0.34 m)(18 N) = 3.1 J and Wcontract ≈ (0.34 m − 0.02 m)(16 N) = 2.6 J. 2 2 Therefore, the total work = 3.1 J − 2.6 J = 0.5 J (b) Strategy and Solution Hooke’s Law is a conservative force, so the total work to stretch and contract the rubber band would be zero. (c) Strategy and Solution The work done on the rubber band does not all go into increasing its elastic potential energy; some of the energy is dissipated as heat. 367

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105. (a) Strategy Use Newton’s second law and Eq. (6-10). Solution Find the speed at the bottom of the incline. 1 2 fd ∆K = mv 2 − 0 = Wc + Wnc = mgh − fd so v = 2 gh − . 2 m Use Newton’s second law with +y perpendicular to the incline and +x down the incline. ΣFy = N − mg cos θ = 0, so N = mg cos θ .

N f

mg cos θ

θ

mg sinθ

mg

Now, d = 0.85 m, h = d sin θ , and f = µ N = µ mg cos θ . Substitute. 2 µ mgd cos θ = 2 gd (sin θ − µ cos θ ) v = 2 gd sin θ − m Find the maximum compression. 1 1 K f + U f = 0 + kx 2 = Ki + U i = mv 2 + 0, so 2 2 x=v

m = k

2mgd (sin θ − µ cos θ ) = k

2(0.50 kg)(9.80 m s 2 )(0.85 m) (sin 30.0° − 0.25cos 30.0°) 35 N m

= 26 cm . (b) Strategy When the block is accelerated by the spring, it attains its previous kinetic energy and speed. Solution Find the distance along the incline, d ′. 1 ∆K = 0 − mv 2 = Wc + Wnc = − mgh − fd ′ = − mgd ′ sin θ − µ mgd ′ cos θ = − d ′[ mg (sin θ + µ cos θ )], so 2 v2 2 gd (sin θ − µ cos θ ) sin 30.0° − 0.25cos 30.0° d′ = = = (85 cm) = 34 cm . 2 g (sin θ + µ cos θ ) 2 g (sin θ + µ cos θ ) sin 30.0° + 0.25cos 30.0°

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106. (a) Strategy The blades sweep out a circle of radius L. The air moves though the circular area at a speed v; therefore, the distance the air moves in a time ∆t is d = v∆t. The volume of the air is equal to the area of the circle swept out by the blades times the distance d. Solution Find the volume of air. V = Ad = π L2 v∆t = π (4.0 m)2 (10 m s)(1.0 s) = 500 m3

(b) Strategy The mass of air is equal to its volume times its density. Solution Find the mass of the air.

m = V ρ = (500 m3 )(1.2 kg m3 ) = 600 kg (c) Strategy Use Eq. (6-6). Solution Find the translational kinetic energy of the air. 1 1 K = mv 2 = (600 kg)(10 m s)2 = 30 kJ 2 2 (d) Strategy Use the definition of average power. The power output is 40% of the kinetic energy per unit time. Solution Find the electrical power output. ∆E 0.40(30 kJ) P= = = 12 kW ∆t 1.0 s (e) Strategy Use the results of parts (a) through (d). Solution Form a proportion. 3 2 1 1 ρVv 2 1 3 P5 ⎛ 2 v ⎞ ∆E 2 mv 1 1 ρπ L2 v∆tv 2 ρπ L2 3 2 3 ⎟ = ⎛⎜ ⎞⎟ = . P= v ∝ v , so = = = = =⎜ P10 ⎜ v ⎟ 2 ∆t 2 8 ∆t ∆t ∆t ⎝2⎠ ⎝ ⎠

So, the power output would decrease to 1 8 of its previous value. The power production of wind turbines is inconsistant, since modest changes in wind speed produce large changes in power output.

107. Strategy and Solution The kinetic energy of a volume of wind passing through the circular area swept out by the rotor blades in time ∆t is 12 mv 2 , where m = ρV = ρ Ad = ρ (π L2 )(v∆t ) and v is the speed of the wind; therefore, ∆E ε ∆K = , where ε is the kinetic energy is given by K = 12 ρπ L2 ∆tv3 . The average power generated is Pav = ∆t ∆t the efficiency of the energy conversion from kinetic energy to electrical energy. ε ∆K ερπ L2 3 Therefore, Pav = = v ∝ v3 . ∆t 2

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108. Strategy Use the method outlined in the problem statement, the relationship between mass, density, and volume, and Eq. (6-9). Solution Find how the speed with which animals of similar shape can run up a hill depends upon the size of the animals. ∆U mg ∆y ∆U Pmax ∝ L2 and = = mgv ∝ mv. If ρ is the mass density, then m = ρV ≈ ρ L3 , so ∝ L3v. ∆t ∆t ∆t ∆U , then L3v ∝ L2 , or v ∝ 1 L . Thus, if Pmax = ∆t 109. Strategy Use conservation of energy. Solution According to the graph, the potential energy in the region under consideration is 300 J. So, U f = 300 J.

Initially, the kinetic energy is 200 J and the potential energy is 0, so Ki = 200 J and U i = 0. Compute the final kinetic energy; that is, the kinetic energy in the region under consideration. E = ∆K + ∆U = K f − Ki + U f − U i = K f − 200 J + 300 J − 0 = K f + 100 J = 0, so K f = −100 J, which is impossible, since kinetic energy cannot be negative. Therefore, the answer is no, the particle cannot enter the region 3 cm < x < 8 cm. Since the particle cannot enter the region, it must remain in the region x < 3 cm. 110. Strategy Use conservation of energy. Solution According to the graph, the potential energy in the region under consideration is 300 J. So, U f = 300 J.

Initially, the kinetic energy is 400 J and the potential energy is 0, so Ki = 400 J and U i = 0. Compute the final kinetic energy; that is, the kinetic energy in the region under consideration. E = ∆K + ∆U = K f − Ki + U f − U i = K f − 400 J + 300 J − 0 = K f − 100 J = 0, so K f = 100 J . Since the final kinetic energy is positive, the answer is yes, the particle can enter the region 3 cm < x < 8 cm.

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