* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download CHAPTER 6 WORK AND ENERGY
Survey
Document related concepts
Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup
Eigenstate thermalization hypothesis wikipedia , lookup
Internal energy wikipedia , lookup
Relativistic quantum mechanics wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Relativistic mechanics wikipedia , lookup
Kinetic energy wikipedia , lookup
Newton's laws of motion wikipedia , lookup
Hunting oscillation wikipedia , lookup
Spinodal decomposition wikipedia , lookup
Centripetal force wikipedia , lookup
Transcript
CHAPTER 6 WORK AND ENERGY PROBLEMS ______________________________________________________________________________ 1. SSM REASONING AND SOLUTION We will assume that the tug-of-war rope remains parallel to the ground, so that the force that moves team B is in the same direction as the displacement. According to Equation 6.1, the work done by team A is 3 W = ( F cosθ )s = (1100 N)(cos 0 °)(2.0 m) = 2.2 × 10 J ______________________________________________________________________________ 2. REASONING The work done by a constant force is given by Equation 6.1 as W = ( F cos θ ) s , where F is the magnitude of the force, s is the magnitude of the F displacement, and θ is the angle between the force and the displacement. When the amplifier is being lowered at a constant velocity, it is in equilibrium so the net force Amplifier in the vertical direction is zero. Thus, the magnitude of the force F exerted by the student is equal to the magnitude of the amplifier’s weight mg. Since the amplifier is moving down, the angle between the force F and the displacement s is θ = 180°. mg SOLUTION a. The work done by the student in lowering the amplifier is W = (F cos θ)s = (mg cos θ)s = (10.5 kg)(9.80 m/s2)(cos 180°)(1.82 m − 0.75 m) = −1.10 × 102 J b. The work done by the gravitational force (the weight mg of the amplifier) is W = (mg cos θ )s = (10.5 kg)(9.80 m/s2)(cos 0°)(1.82 m − 0.75 m) = +1.10 × 102 J ______________________________________________________________________________ 3. REASONING AND SOLUTION According to Equation 6.1, W = Fs cos θ, the work is a. W = (94.0 N)(35.0 m) cos 25.0° = 2980 J b. W = (94.0 N)(35.0 m) cos 0° = 3290 J ______________________________________________________________________________ s Chapter 6 Problems 4. REASONING AND SOLUTION 207 Each locomotive does work W = Ts cos θ = (5.00 × 103 N)(2.00 × 103 m) cos 20.0° = 9.40 × 106 J The net work is then WT = 2W = 1.88 × 10 7 J ______________________________________________________________________________ 5. SSM REASONING AND SOLUTION Solving Equation 6.1 for the angle θ, we obtain W 1.10 × 103 J −1 θ = cos = cos = 42.8° Fs (30.0 N)(50.0 m) ______________________________________________________________________________ −1 6. REASONING AND SOLUTION a. In both cases, the lift force L is perpendicular to the displacement of the plane, and, therefore, does no work. As shown in the drawing in the text, when the plane is in the dive, there is a component of the weight W that points in the direction of the displacement of the plane. When the plane is climbing, there is a component of the weight that points opposite to the displacement of the plane. Thus, since the thrust T is the same for both cases, the net force in the direction of the displacement is greater for the case where the plane is diving. Since the displacement s is the same in both cases, more net work is done during the dive . b. The work done during the dive is Wdive = (T + W cos 75° ) s , while the work done during the climb is Wclimb = (T + W cos 115° ) s . Therefore, the difference between the net work done during the dive and the climb is Wdive – Wclimb = ( T + W cos 75 °) s – (T + W cos 115°) s = Ws (cos 75 ° – cos 115° ) 4 3 7 = (5.9 × 10 N)(1.7 × 10 m)(cos 75° – cos 115° ) = 6.8 × 10 J ______________________________________________________________________________ 7. REASONING The drawing shows three of the forces +y 29.0° that act on the cart: F is the pushing force that the F shopper exerts, f is the frictional force that opposes the f +x motion of the cart, and mg is its weight. The displacement s of the cart is also shown. Since the cart moves at a constant velocity along the +x direction, it is in mg s equilibrium. The net force acting on it in this direction is zero, ΣFx = 0. This relation can be used to find the magnitude of the pushing force. The work done 208 WORK AND ENERGY by a constant force is given by Equation 6.1 as W = ( F cos θ ) s , where F is the magnitude of the force, s is the magnitude of the displacement, and θ is the angle between the force and the displacement. SOLUTION a. The x-component of the net force is zero, ΣFx = 0, so that F cos29.0° − f = 0 144 42444 3 (4.9a) ΣFx The magnitude of the force that the shopper exerts is F = f 48.0 N = = 54.9 N . cos 29.0° cos 29.0° b. The work done by the pushing force F is W = ( F cos θ ) s = ( 54.9N )( cos29.0° )( 22.0m ) = 1060J (6.1) c. The angle between the frictional force and the displacement is 180°, so the work done by the frictional force f is W = ( f cos θ ) s = ( 48.0N ) ( cos180.0° ) ( 22.0m ) = −1060J d. The angle between the weight of the cart and the displacement is 90°, so the work done by the weight mg is ( ) W = ( mg cos θ ) s = (16.0kg ) 9.80m/s 2 ( cos90 °) ( 22.0m ) = 0 J ______________________________________________________________________________ 8. REASONING AND SOLUTION a. The work done by the applied force is W = Fs cos θ = (2.40 × 102 N)(8.00 m) cos 20.0° = 1.80 × 103 J b. The work done by the frictional force is Wf = f k s cos θ, where f k = µ s (mg − F sin θ ) = (0.200) (85.0 kg)(9.80 m/s2 ) − (2.40 × 102 N)(sin 20.0°) = 1.50 × 102 N Now Wf = (1.50 × 102 N)(8.00 m) cos 180° = − 1.20 × 103 J Chapter 6 Problems 209 ______________________________________________________________________________ 9. SSM REASONING AND SOLUTION According to Equation 6.1, the work done by the husband and wife are, respectively, [ Husband] [ Wife] WH = ( FH cos θH )s WW = ( FW cos θW )s Since both the husband and the wife do the same amount of work, ( FH cos θH )s = ( FW cos θW )s Since the displacement has the same magnitude s in both cases, the magnitude of the force exerted by the wife is cos θH cos 58° = (67N) = 45N cos θW cos 38° ______________________________________________________________________________ FW = FH 10. REASONING AND SOLUTION The work done by the force P is WP = Ps cos 30.0° = 0.866 P s Similarly, the work done by the frictional force is Wf = f k s cos 180° = – f k s where f k = µk (mg – P sin 30.0°) Then Wf = – (0.200)[(1.00 × 102 kg)(9.80 m/s2) – 0.500P]s = (0.100P – 196)s The net work is zero if WP + Wf = 0 or 0.866 Ps + (0.100P – 196)s = 0 Solving for P gives P = 203 N . ______________________________________________________________________________ 210 WORK AND ENERGY 11. REASONING AND SOLUTION The net work done on the car is WT = WF + Wf + Wg + WN WT = Fs cos 0.0° + f s cos 180° – mgs sin 5.0° + FNs cos 90° Rearranging this result gives F= WT s + f + mg sin5.0° ( ) 150 × 103 J + 524 N + (1200kg ) 9.80m/s 2 sin5.0° = 2.07 × 103 N 290m ______________________________________________________________________________ = 12. REASONING AND SOLUTION The work done on the arrow by the bow is given by W = Fs cos 0° = Fs This work is converted into kinetic energy according to the work energy theorem. W = 12 mvf2 − 12 mv02 Solving for v f, we find that vf = 2W + v02 = m 2 ( 65 N ) ( 0.90 m ) –3 + ( 0 m/s ) = 39 m/s 2 75 × 10 kg ______________________________________________________________________________ 13. SSM REASONING The work done to launch either object can be found from Equation 6.3, 1 2 1 2 the work-energy theorem, W = KEf − KE 0 = mvf2 − mv02 . SOLUTION a. The work required to launch the hammer is ( ) W = 12 mvf2 − 12 mv02 = 12 m vf2 − v02 = 12 (7.3 kg) (29 m/s)2 − ( 0 m/s ) = 3.1× 103 J 2 b. Similarly, the work required to launch the bullet is Chapter 6 Problems ( ) 211 W = 12 m vf2 − v02 = 12 (0.0026 kg) (410 m/s) 2 − ( 0 m/s ) = 2.2 ×10 2 J 2 ______________________________________________________________________________ 14. REASONING The average kinetic energy is given by Equation 6.2 as K E = 12 mv 2 . The mass of the aircraft carrier is known, and its average speed v can be obtained from Equation 2.1 as the distance traveled divided by the elapsed time. SOLUTION The average speed of the aircraft carrier is equal to the distance s it travels divided by the elapsed time t, or v = s / t . Therefore, the average kinetic energy is 2 s KE = mv = m t 1 2 2 1 2 2 208 × 10 3 m 10 1 = 2 8.35 × 10 kg = 1.14 × 10 J ( 3.50h) 3600s 1 h ______________________________________________________________________________ ( 15. SSM WWW 7 ) REASONING AND SOLUTION The work required to bring each car up 1 1 to speed is, from the work-energy theorem, W = KEf − KE 0 = mvf2 − mv02 . Therefore, 2 2 ( ) 2 ( ) 2 WB = 1 m vf2 − v02 = 1 (1.20 × 103 kg) (40.0 m/s)2 − ( 0 m/s )2 = 9.60 × 105 J 2 WB = 1 m vf2 − v02 = 1 (2.00 × 103 kg) (40.0 m/s)2 − ( 0 m/s ) 2 = 1.60 ×106 J 2 The additional work required to bring car B up to speed is, therefore, 6 5 5 WB − WA = (1.6 × 10 J) – (9.60 × 10 J) = 6.4 × 10 J ______________________________________________________________________________ 16. REASONING AND SOLUTION From the work-energy theorem, Equation 6.3, 1 2 1 2 1 2 2 W = 2 mvf − 2 mv0 = 2 m vf − v0 1 a. W = 2 (7420kg) (8450m/s) 2 –(2820m/s) 2 = 2.35 ×1011 J 212 WORK AND ENERGY b. W = 2 (7420 kg) (2820 m/s) 2 − (8450 m/s) 2 = –2.35 × 1011 J ______________________________________________________________________________ 1 17. REASONING AND SOLUTION a. The work-energy theorem gives W = (1/2)mv f2 – (1/2)mv o2 = (1/2)(45 × 10–3 kg)(41 m/s)2 = 38 J b. From the definition of work W = Fs cos 0° so F = W/s = (38 J)/(1.0 × 10–2 m) = 3.8 × 103 N ______________________________________________________________________________ 18. REASONING AND SOLUTION The work energy theorem, W = 12 mvf2 − 12 mv02 , gives vf = 2W 2 + v0 m where W = Fs cos 180° = − (4.0 × 105 N)(2500 × 103 m) = − 1.0 × 1012 J Now vf = ( 2 −1.0 × 1012 J ) + (11000m/s) 2 = 9 × 103 m/s 5.0 × 104 kg ______________________________________________________________________________ 19. SSM REASONING According to the work-energy theorem, the kinetic energy of the sled increases in each case because work is done on the sled. The work-energy theorem is given by Equation 6.3: W = KEf − KE 0 = 1 mv2f − 1 mv20 . The work done on the sled is given by 2 2 Equation 6.1: W = ( F cosθ )s . The work done in each case can, therefore, be expressed as W1 = (F cos 0°)s = 1 2 1 2 mv2f − mv02 = ∆KE1 and W2 = ( F cos 62 °)s = 1 mvf2 − 1 mv02 = ∆KE2 2 2 The fractional increase in the kinetic energy of the sled when θ = 0° is Chapter 6 Problems 213 ∆KE1 ( F cos 0°)s = = 0.38 KE0 KE0 Therefore, Fs = (0.38) KE0 (1) The fractional increase in the kinetic energy of the sled when θ = 62° is ∆KE2 (F cos 62°)s Fs = = (cos 62 °) KE 0 KE0 KE0 (2) Equation (1) can be used to substitute in Equation (2) for Fs. SOLUTION Combining Equations (1) and (2), we have ∆KE2 Fs (0.38) KE0 = (cos 62°) = (cos 62°) = (0.38)(cos 62°) = 0.18 KE 0 KE0 KE0 Thus, the sled's kinetic energy would increase by 18 % . ______________________________________________________________________________ 20. REASONING Since the person has an upward acceleration, there must be a net force acting in the upward direction. The net force ΣFy is related to the acceleration ay by Newton’s second law, Σ Fy = may , where m is the mass of the person. This relation will allow us to determine the tension in the cable. The work done by the tension and the person’s weight can be found directly from the definition of work, Equation 6.1. SOLUTION a. The free-body diagram at the right shows the two forces that act on the person. Applying Newton’s second law, we have T +y s T424 − mg 1 3 = ma y Σ Fy Solving for the magnitude of the tension in the cable yields mg T = m(ay + g) = (79 kg)(0.70 m/s2 + 9.80 m/s2) = 8.3 × 102 N b. The work done by the tension in the cable is WT = ( T cos θ ) s = (8.3 × 102 N) (cos 0°) (11 m) = 9.1 × 103 J (6.1) 214 WORK AND ENERGY c. The work done by the person’s weight is WW = ( mg cos θ ) s = (79kg) ( 9.8m/s 2 ) (cos 180°) (11 m) = −8.5 × 103 J (6.1) d. The work-energy theorem relates the work done by the two forces to the change in the kinetic energy of the person. The work done by the two forces is W = WT + WW : WT + WW = 12 mv 2f − 12 mv 20 14 243 (6.3) W Solving this equation for the final speed of the person gives vf = v02 + ( 2 W + WW m T ) 2 ( 9.1 × 103 J − 8.5 × 103 J ) = 4 m / s 79kg ______________________________________________________________________________ = ( 0 m/s ) 2 + 21. REASONING It is useful to divide this problem into two parts. The first part involves the skier moving on the snow. We can use the work-energy theorem to find her speed when she comes to the edge of the cliff. In the second part she leaves the snow and falls freely toward the ground. We can again employ the work-energy theorem to find her speed just before she lands. SOLUTION The drawing at the right shows the three forces that act on the skier as she glides on the snow. The forces are: her weight mg, the normal force FN, and the kinetic frictional force fk. Her displacement is labeled as s. The work-energy theorem, Equation 6.3, is FN +y fk s 65.0° 25.0° W = mv − mv 1 2 2 f 1 2 2 0 mg where W is the work done by the net external force that acts on the skier. The work done by each force is given by Equation 6.1, W = ( F cos θ ) s , so the work-energy theorem becomes ( mg cos65.0o ) s + ( f k42444444444 cos180o ) s + ( FN cos90o ) s = 12 mvf2 − 12 mv02 1444444444 4 3 W Chapter 6 Problems 215 Since cos 90° = 0, the third term on the left side can be eliminated. The magnitude f k of the kinetic frictional force is given by Equation 4.8 as f k = µk FN . The magnitude FN of the normal force can be determined by noting that the skier does not leave the surface of the slope, so ay = 0 m/s2. Thus, we have that ΣFy = 0, so FN − mg cos25.0° = 0 14442444 3 or FN = mg cos 25.0° Σ Fy The magnitude of the kinetic frictional force becomes f k = µk FN = µk mg cos25.0 o . Substituting this result into the work-energy theorem, we find that ( mg cos65.0o ) s + ( µk mg cos25.0o ) ( cos180o )3s = 12 mvf2 − 12 mv02 14444444442444444444 W Algebraically eliminating the mass m of the skier from every term, setting cos 180° = –1 and v 0 = 0 m/s, and solving for the final speed v f, gives vf = 2 gs ( cos65.0o − µk cos 25.0o ) = 2 ( 9.80m/s 2 ) (10.4m ) cos65.0o − ( 0.200) cos25.0o = 7.01m/s The drawing at the right shows her displacement s during free fall. Note that the displacement is a vector that starts where she leaves the slope and ends where she touches the ground. The only force acting on her during the free fall is her weight mg. The work-energy theorem, Equation 6.3, is 3.50 m θ s mg W = 12 mvf2 − 12 mv02 The work W is that done by her weight, so the work-energy theorem becomes s = 12 mv f2 − 12 mv02 ( mg cos θ)3 144244 W In this expression θ is the angle between her weight (which points vertically downward) and her displacement. Note from the drawing that s cos θ = 3.50 m. Algebraically eliminating the mass m of the skier from every term in the equation above and solving for the final speed v f gives 216 WORK AND ENERGY vf = v02 + 2 g ( s cos θ ) = ( 7.01m/s ) 2 + 2 ( 9.80m/s 2 ) ( 3.50m ) = 10.9m/s ______________________________________________________________________________ 22. REASONING AND SOLUTION Since the boat travels with constant speed, the net force on the boat is zero. Therefore, the drive force must be equal in magnitude and opposite in direction to the resistive force due to the water, FR. Since the force due to the water is resistive, it always points opposite to the direction of motion of the boat and θ = 180°. After the engine is shut off, the resistive force due to the water does negative work on the boat thereby bringing it to a halt. a. From the work-energy theorem, W = 12 mvf2 − 12 mv02 where W is the work done by the resistive force due to the water. Since the boat coasts to a halt, v f = 0 m/s, and we have ( FR cos θ )s = 12 m ( 0m/s ) − 2 1 mv02 2 Solving for s gives mv02 (480 kg)(8.9 m/s)2 s=− =− = 25 m 2FR cos θ 2(760 N)(cos 180°) b. The time required for the boat to come to a halt can be found from Equation 2.4: v = v0 + at where v = 0 m/s since the final speed of the boat is zero. Solving for t yields t= ( 0 m/s) − v0 a = − v0 F/m = − v0m F = − (8.9 m/s)(480 kg) = 5.6 s ( − 760 N) Notice that the value substituted for F is negative. This is because F and v0 point in opposite directions, and the value substituted for v 0 is positive. ______________________________________________________________________________ 23. SSM WWW REASONING When the satellite goes from the first to the second orbit, its kinetic energy changes. The net work that the external force must do to change the orbit can be 1 1 found from the work-energy theorem: W = KEf − KE 0 = mvf2 − mv02 . The speeds vf and 2 2 Chapter 6 Problems 217 v0 can be obtained from Equation 5.5 for the speed of a satellite in a circular orbit of radius r. Given the speeds, the work energy theorem can be used to obtain the work. SOLUTION According to Equation 5.5, v = GME r . Substituting into the work-energy theorem, we have W = 1 2 mvf2 − 1 mv02 2 = m 1 2 ( vf2 − v02 ) GM 2 GM E E = m − rf r0 1 2 2 = GM E m 1 − 1 2 rf r0 Therefore, W= (6.67×10 –11 N ⋅ m 2 /kg 2 )(5.98×10 24 kg)(6200 kg) 2 1 1 = 1.4 × 1011 J × − 6 7 7.0 × 10 m 3.3 × 10 m ______________________________________________________________________________ 24. REASONING AND SOLUTION work-energy theorem: The net work done on the plane can be found from the 2 2 2 2 W = 12 mvf − 12 mv0 = 12 m vf − v0 The tension in the guideline provides the centripetal force required to keep the plane moving in a circular path. Since the tension in the guideline becomes four times greater, Tf = 4 T0 or mvf2 mv02 = 4 rf r0 Solving for vf2 gives vf2 = 4v02 rf r0 Substitution of this expression into the work-energy theorem gives 4v2 r W = m 0 f − v02 = r0 1 2 1 mv 2 0 2 4 rf r 0 − 1 218 WORK AND ENERGY Therefore, 1 2 14 m 2 W = 2 (0.90 kg)(22 m/s) 4 − 1 = 5.4 ×10 J 16 m ______________________________________________________________________________ 25. REASONING AND SOLUTION PE = mgh = (55.0 kg)(9.80 m/s2)(443m) = 2.39 × 105 J (6.5) ______________________________________________________________________________ 26. REASONING The work done by the weight of the basketball is given by Equation 6.1 as W = ( F cos θ ) s , where F = mg is the magnitude of the weight, s θ is the angle between the weight and the displacement, and s is the magnitude of the displacement. The drawing shows that the weight and displacement are parallel, so that θ = 0°. The potential energy of the basketball is given by Equation 6.5 as mg PE = mgh, where h is the height of the ball above the ground. SOLUTION a. The work done by the weight of the basketball is W = ( F cos θ ) s = mg (cos 0°)(h0 − hf) = (0.60 kg)(9.80 m/s2)(6.1 m − 1.5 m) = 2 7 J b. The potential energy of the ball, relative to the ground, when it is released is PE0 = mgh0 = (0.60 kg)(9.80 m/s2)(6.1 m) = 3 6 J (6.5) c. The potential energy of the ball, relative to the ground, when it is caught is PEf = mghf = (0.60 kg)(9.80 m/s2)(1.5 m) = 8.8J (6.5) d. The change in the ball’s gravitational potential energy is ∆PE = PEf − PE0 = 8.8 J – 36 J = −27 J We see that the change in the gravitational potential energy is equal to –27 J = −W , where W is the work done by the weight of the ball (see part a). ______________________________________________________________________________ 27. SSM REASONING During each portion of the trip, the work done by the resistive force is given by Equation 6.1, W = ( F cosθ )s . Since the resistive force always points opposite to the Chapter 6 Problems 219 displacement of the bicyclist, θ = 180°; hence, on each part of the trip, W = ( F cos 180° )s = −Fs . The work done by the resistive force during the round trip is the algebraic sum of the work done during each portion of the trip. SOLUTION a. The work done during the round trip is, therefore, Wtotal = W1 + W2 = −F1 s1 − F2 s2 = −(3.0 N)(5.0 × 103 m) − (3.0 N)(5.0 × 103 m) = –3.0 × 10 4 J b. Since the work done by the resistive force over the closed path is not zero, we can conclude that the resistive force is not a conservative force . ______________________________________________________________________________ 28. REASONING AND SOLUTION a. From the definition of work, W = Fs cos θ. For upward motion θ = 180° so W = – mgs = – (71.1 N)(2.13 m – 1.52 m) = –43 J b. The change in potential energy is ∆PE = mghf – mgh0 = (71.1 N)(2.13 m – 1.52 m) = +43 J ______________________________________________________________________________ 29. REASONING AND SOLUTION The vertical height of the skier is h = s sin 14.6° so s ∆PE = mgs sin 14.6° h 14.6° = (75.0 kg)(9.80 m/s2)(2830 m) sin 14.6° = 5.24 × 105 J ______________________________________________________________________________ 30. REASONING AND SOLUTION The change in the pole-vaulter's potential energy is PEf – PE0 = mghf – mgh0 Let hf = 0 m, so that h0 = 5.80 m. Then, 220 WORK AND ENERGY PEf − PE0 0 J – 3.70 × 103 J = = 6.38 × 102 N hf − h0 0 m – 5.80 m ______________________________________________________________________________ mg = 31. SSM REASONING The only nonconservative force that acts on Rocket man is the force generated by the propulsion unit. Thus, Equation 6.7b can be used to determine the net work done by this force. SOLUTION From Equation 6.7b, we have Wnc = ∆KE + ∆PE = ( 1 2 ) mvf2 − 12 mv02 + ( mghf − mgh0 ) Since rocket man starts from rest, v 0 = 0 m/s. If we take h0 = 0 m on the ground, we have Wnc = 1 mvf2 + mghf = m 2 ( 1 v2 2 f + ghf ) = (136 kg) 12 (5.0 m/s) 2 + (9.80 m/s 2 )(16 m) = 2.3 ×10 4 J ______________________________________________________________________________ 32. REASONING The only two forces that act on the gymnast are his weight and the force exerted on his hands by the high bar. The latter is the (non-conservative) reaction force to the force exerted on the bar by the gymnast, as predicted by Newton's third law. This force, however, does no work because it points perpendicular to the circular path of motion. Thus, Wnc = 0 J, and we can apply the principle of conservation of mechanical energy. SOLUTION The conservation principle gives 1 mv f2 + mghf 214 4244 3 Ef 1 = 2 mv 02 + mgh0 144244 3 E0 Since the gymnast's speed is momentarily zero at the top of the swing, v0 = 0 m/s. If we take hf = 0 m at the bottom of the swing, then h0 = 2r , where r is the radius of the circular path followed by the gymnast's waist. Making these substitutions in the above expression and solving for vf , we obtain vf = 2gh0 = 2g(2r) = 2(9.80 m/s2 )(2 × 1.1 m) = 6.6 m/s ______________________________________________________________________________ Chapter 6 Problems 221 33. REASONING Since air resistance is being neglected, the only force that acts on the falling water is the conservative gravitational force (its weight). Since the height of the falls and the speed of the water at the bottom are known, we may use the conservation of mechanical energy to find the speed of the water at the top of the falls. SOLUTION The conservation of mechanical energy, as expressed by Equation 6.9b, states that 1 mv 2 + mgh = 1 mv 2 + mgh f 0 2 2 144244 3f 144244 30 Ef Ef The mass m can be eliminated algebraically from this equation, since it appears as a factor in every term. Solving for v 0 gives v0 = = v f2 + 2 g ( hf − h0 ) ( 25.8m/s )2 + 2 ( 9.80m/s 2 ) ( 0 m − 33.2m ) = 3.9m/s ______________________________________________________________________________ 34. REASONING AND SOLUTION 1 mv 2 f 2 The conservation of energy gives + mghf = 1 mv 2 0 2 + mgh0 Rearranging gives hf − h0 2 2 14.0m/s ) − (13.0m/s ) ( = = 1.4m 2 9.80m/s2 ______________________________________________________________________________ 35. ( ) SSM REASONING No forces other than gravity act on the rock since air resistance is being ignored. Thus, the net work done by nonconservative forces is zero, Wnc = 0 J. Consequently, the principle of conservation of mechanical energy holds, so the total mechanical energy remains constant as the rock falls. If we take h = 0 m at ground level, the gravitational potential energy at any height h is, according to 1 Equation 6.5, PE = mgh . The kinetic energy of the rock is given by Equation 6.2: KE = mv 2 . 2 In order to use Equation 6.2, we must have a value for v 2 at each desired height h. The quantity v 2 can be found from Equation 2.9 with v0 = 0 m/s, since the rock is released from rest. At a height h, the rock has fallen through a distance (20.0 m) – h , and according to Equation 2.9, v 2 = 2ay = 2a[(20.0 m) – h] . Therefore, the kinetic energy at any height h is given by 222 WORK AND ENERGY KE = ma[ (20.0 m) – h ] . The total energy of the rock at any height h is the sum of the kinetic energy and potential energy at the particular height. SOLUTION The calculations are performed below for h = 10.0 m . The table that follows also shows the results for h = 20.0 m and h = 0 m. PE = mgh = ( 2.00 kg)(9.80 m/s 2 )(10.0 m) = 196 J KE = ma[ (20.0 m) – h ] = (2.00 kg )(9.80 m/s 2 )(20.0 m – 10.0 m) = 196 J E = KE + PE = 196 J + 196 J = h (m) 20.0 10.0 0 KE (J) 0 196 392 392 J PE (J) 392 196 0 E (J) 392 392 392 Note that in each case, the value of E is the same, because mechanical energy is conserved. ______________________________________________________________________________ 36. REASONING Since air resistance is being neglected, the only force that acts on the golf ball is the conservative gravitational force (its weight). Since the maximum height of the trajectory and the initial speed of the ball are known, the conservation of mechanical energy can be used to find the kinetic energy of the ball at the top of the highest point. The conservation of mechanical energy can also be used to find the speed of the ball when it is 8.0 m below its highest point. SOLUTION a. The conservation of mechanical energy, Equation 6.9b, states that 1 mv 2 f 2 123 + mghf = 12 mv02 + mgh0 KE f Solving this equation for the final kinetic energy, KEf, yields KE f = 12 mv02 + mg ( h0 − hf = b. 1 2 ) ( 0.0470kg ) ( 52.0m/s ) 2 + ( 0.0470kg ) ( 9.80m/s 2 ) ( 0 m − 24.6m ) = The conservation of mechanical energy, Equation 6.9b, states that 52.2 J Chapter 6 Problems 1 mv 2 + mgh f 2 144244 3f = 223 1 mv 2 + mgh 0 2 144244 30 Ef Ef The mass m can be eliminated algebraically from this equation, since it appears as a factor in every term. Solving for v f, and noting that the final height of the ball is hf = 24.6 m – 8.0 m = 16.6 m, we have that vf = v 02 + 2 g ( h0 − hf ) = ( 52.0m/s ) 2 ( ) + 2 9.80m/s 2 ( 0 m − 16.6m ) = 4 8 . 8 m / s ______________________________________________________________________________ 37. REASONING AND SOLUTION Mechanical energy is conserved if no friction acts. Ef = E0 or 1 mv 2 f 2 + mgh = vf =? h = 5.0 m 1 mv 2 0 2 Taking h0 = 0 m and rearranging yields vf = v02 − 2 gh = v o = 11.0 m/s (11.0m/s )2 − 2 ( 9.80m/s 2 ) (5.0m ) = 4.8m/s ______________________________________________________________________________ 38. REASONING AND SOLUTION The hook is let go when it is h0 = 1.5 m above the ground and traveling vertically upward. Its initial speed is v0 = 2π r 2π (1.5 m ) = = 28 m/s 1s T 3 The conservation of energy requires 1 mv 2 f 2 + mghf = 12 mv02 + mgh0 At maximum height v f = 0 m/s, so v02 ( 28 m/s ) + 1.5 m = 42 m hf = + h0 = 2g 2 9.80 m/s2 2 ( ) ______________________________________________________________________________ 224 WORK AND ENERGY 39. SSM WWW REASONING Gravity is the only force acting on the vaulters, since friction and air resistance are being ignored. Therefore, the net work done by the nonconservative forces is zero, and the principle of conservation of mechanical energy holds. SOLUTION Let E2f represent the total mechanical energy of the second vaulter at the ground, and E20 represent the total mechanical energy of the second vaulter at the bar. Then, the principle of mechanical energy is written for the second vaulter as 1 mv22f + mgh2f 2144 2443 E2f 1 2 = mv20 + mgh20 21442443 E20 Since the mass m of the vaulter appears in every term of the equation, m can be eliminated algebraically. The quantity h20 = h , where h is the height of the bar. Furthermore, when the vaulter is at ground level, h2 f = 0 m. Solving for v20 we have v 20 = v 22 f − 2 gh (1) In order to use Equation (1), we must first determine the height h of the bar. The height h can be determined by applying the principle of conservation of mechanical energy to the first vaulter on the ground and at the bar. Using notation similar to that above, we have 1 2 mv1f + mgh1f 2144 2443 E1f 1 2 = mv10 + mgh10 2 1442443 E10 where E10 corresponds to the total mechanical energy of the first vaulter at the bar. The height of the bar is, therefore, v2 − v2 (8.90 m/s)2 − (1.00 m/s) 2 h = h10 = 1f 10 = = 3.99 m 2g 2(9.80 m/s 2 ) The speed at which the second vaulter clears the bar is, from Equation (1), 2 v20 = v 2f − 2 gh = (9.00 m/s) 2 − 2(9.80 m/s 2 )(3.99 m) = 1.7 m/s ______________________________________________________________________________ 40. REASONING AND SOLUTION Since energy is conserved, the ball will have the same speed at the bottom of its swing whether it is moving toward or away from the crane. As the cable swings to an angle of 32.0°, the ball will rise a distance h = L (1 – cos 32.0°) Chapter 6 Problems 225 where L is the length of the cable. The conservation of energy then gives 1 mv 2 0 2 = 12 mvf2 + mgh Solving for v f gives vf = v02 − 2 gL (1 − cos32.0° ) = ( 6.50m/s )2 − 2 ( 9.80m/s2 ) (10.0m )( 1 − cos32.0°) = 3.53m/s ______________________________________________________________________________ 41. SSM REASONING Friction and air resistance are being ignored. The normal force from the slide is perpendicular to the motion, so it does no work. Thus, no net work is done by nonconservative forces, and the principle of conservation of mechanical energy applies. SOLUTION Applying the principle of conservation of mechanical energy to the swimmer at the top and the bottom of the slide, we have 1 mv2f + mghf 2 4 1 4244 3 Ef 1 = mv20 + mgh0 2 4 1 4244 3 E0 If we let h be the height of the bottom of the slide above the water, hf = h , and h0 = H . Since the swimmer starts from rest, v0 = 0 m/s, and the above expression becomes 1 2 v 2 f + gh = gH Solving for H, we obtain v2 H = h+ f 2g Before we can calculate H, we must find vf and h. Since the velocity in the horizontal direction is constant, ∆x 5.00 m vf = = = 10.0 m/s ∆t 0.500 s The vertical displacement of the swimmer after leaving the slide is, from Equation 3.5b (with down being negative), y = 2 ay t 2 = 2 (–9.80 m/s 2 )( 0.500 s) 2 = −1.23 m 1 1 Therefore, h = 1.23 m. Using these values of vf and h in the above expression for H, we find 226 WORK AND ENERGY v2 (10.0 m/s) 2 H = h + f = 1.23 m + = 6.33 m 2g 2(9.80 m/s 2 ) ______________________________________________________________________________ 42. REASONING AND SOLUTION If air resistance is ignored, the only nonconservative force that acts on the skier is the normal force exerted on the skier by the snow. Since this force is always perpendicular to the direction of the displacement, the work done by the normal force is zero. We can conclude, therefore, that mechanical energy is conserved. 1 mv 20 + 2 1 mgh 0 = mv 2f + mgh f 2 Since the skier starts from rest v0 = 0 m/s. Let hf define the zero level for heights, then the final gravitational potential energy is zero. This gives mgh 0 = 1 2 mv 2f (1) At the crest of the second hill, the two forces that act on the skier are the normal force and the weight of the skier. The resultant of these two forces provides the necessary centripetal force to keep the skier moving along the circular arc of the hill. When the skier just loses contact with the snow, the normal force is zero and the weight of the skier must provide the necessary centripetal force. mg = mv 2f r v 2f =gr so that F N mg (2) Substituting this expression for v 2f into Equation (1) gives mgh 0 = 1 2 mgr Solving for h0 gives r 36 m = = 18 m 2 2 ______________________________________________________________________________ h0 = 43. REASONING AND SOLUTION At the bottom of the circular path of the swing, the centripetal force is provided by the tension in the rope less the weight of the swing and rider. That is, mv 2 = T − mg r { FC Chapter 6 Problems 227 Solving for the mass yields m= T v2 +g r The energy of the swing is conserved if friction is ignored. The initial energy, E0, when the swing is released is completely potential energy and is E0 = mgh0, where h0 = r (1 – cos 60.0°) = 12 r is the vertical height of the swing. At the bottom of the path the swing's energy is entirely kinetic and is Ef = 12 mv2 Setting Eo = Ef and solving for v gives v = 2gh0 = gr The expression for the mass now becomes 8.00 × 102 N T = = 40.8kg 2 g 2 9.80m/s 2 ______________________________________________________________________________ m= ( ) 44. REASONING AND SOLUTION If air resistance is ignored, the only nonconservative force that acts on the person is the normal force exerted on the person by the surface. Since this force is always perpendicular to the direction of the displacement, the work done by the normal force is zero. We can conclude, therefore, that mechanical energy is conserved. 1 mv 20 + 2 1 2 mgh 0 = mv 2f + mgh f (1) 228 WORK AND ENERGY where the final state pertains to the position where the person leaves the surface. Since the person starts from rest v0 = 0 m/s. Since the radius of the surface is r, h0 = r, and hf = r cos θf where θf is the angle at which the person leaves the surface. r θf r cos θf r Equation (1) becomes 1 mgr = mv 2f + mg( r cos θf ) 2 (2) In general, as the person slides down the surface, the two forces that act on him are the normal force FN and the weight mg. The centripetal force required to keep the person moving in the circular path is the resultant of FN and the radial component of the weight, mg cos θ. When the person leaves the surface, the normal force is zero, and the radial component of the weight provides the centripetal force. mg cos θf = mv 2f r F N mg cos θ θ θ θ mg v2f = gr cos θf (3) ⇒ Substituting this expression for v 2f into Equation (2) gives mgr = 12 mg (r cos θf ) + mg (r cos θf ) Solving for θf gives 2 θf = cos −1 = 48° 3 ______________________________________________________________________________ 45. SSM REASONING AND SOLUTION The work-energy theorem can be used to determine the change in kinetic energy of the car according to Equation 6.8: Wnc = ( mv + mgh ) − ( mv + mgh ) 1442443 1442443 1 2 2 f f Ef 1 2 2 0 0 E0 The nonconservative forces are the forces of friction and the force due to the chain mechanism. Thus, we can rewrite Equation 6.8 as Chapter 6 Problems 229 Wfriction + Wchain = ( KEf + mghf ) − ( KE 0 + mgh0 ) We will measure the heights from ground level. Solving for the change in kinetic energy of the car, we have KEf − KE0 = Wfriction + Wchain − mghf + mgh0 = − 2.00×104 J + 3.00×10 4 J − (375 kg)(9.80 m/s 2 )(20.0 m) + (375 kg)(9.80 m/s 2 )(5.00 m)= –4.51×104 J ______________________________________________________________________________ 46. REASONING AND SOLUTION The work done by air resistance is equal to the energy lost by the ball. 2.00 m Wnc = Ef – E0 Wnc = Wnc = ( 1 2 ) ( 2 mvf + mghf − 1 (0.600 2 1 2 2 mv0 + mgh0 3.10 m ) kg)(4.20 m/s)2 + (0.600 kg)(9.80 m/s2)(3.10 m) – 12 (0.600 kg)(7.20 m/s)2 – (0.600 kg)(9.80 m/s2)(2.00 m) = −3.8J ______________________________________________________________________________ 47. REASONING AND SOLUTION a. The work done by non-conservative forces is given by Equation 6.7b as Wnc = ∆KE + ∆PE ∆PE = Wnc – ∆KE so Now ∆KE = 1 2 mvf2 − 1 2 mv02 = 1 2 (55.0 kg)[(6.00 m/s)2 − (1.80 m/s) 2 ] = 901 J and ∆PE = 80.0 J – 265 J – 901 J = −1086 J b. ∆PE = mg (h – h0) so h – h0 = ∆ PE –1086 J = = –2.01 m mg (55.0 kg ) 9.80 m/s 2 ( ) 230 WORK AND ENERGY Thus, the skater’s vertical position has changed by 2.01 m , and the skater is below the starting point . ______________________________________________________________________________ 48. REASONING a. Since there is no air friction, the only force that acts on the projectile is the conservative gravitational force (its weight). The initial and final speeds of the ball are known, so the conservation of mechanical energy can be used to find the maximum height that the projectile attains. b. When air resistance, a nonconservative force, is present, it does negative work on the projectile and slows it down. Consequently, the projectile does not rise as high as when there is no air resistance. The work-energy theorem, in the form of Equation 6.6, may be used to find the work done by air friction. Then, using the definition of work, Equation 6.1, the average force due to air resistance can be found. SOLUTION a. The conservation of mechanical energy, as expressed by Equation 6.9b, states that 1 mv2 + mgh = 1 mv 2 + mgh f 0 2 2 144244 3f 144244 30 Ef E0 The mass m can be eliminated algebraically from this equation since it appears as a factor in every term. Solving for the final height hf gives hf = 1 2 ( v02 − vf2 ) + h 0 g Setting h0 = 0 m and v f = 0 m/s, the final height, in the absence of air resistance, is hf = vo2 − vf2 2g = (18.0m/s ) 2 − ( 0 m/s ) 2 ( 2 9.80m/s2 ) = 16.5m b. The work-energy theorem is Wnc = ( 1 mv 2 f 2 2 − 12 mv0 ) + ( mgh f − mgh0 ) (6.6) Chapter 6 Problems 231 where Wnc is the nonconservative work done by air resistance. According to Equation 6.1, the work can be written as Wnc = ( FR cos180° ) s , where FR is the average force of air resistance. As the projectile moves upward, the force of air resistance is directed downward, so the angle between the two vectors is θ = 180° and cos θ = –1. The magnitude s of the displacement is the difference between the final and initial heights, s = hf – h0 = 11.8 m. With these substitutions, the work-energy theorem becomes ( ) − FR s = 12 m vf2 − vo2 + mg ( hf − h0 ) Solving for FR gives FR = = 1 2 ( ) m vf2 − vo2 + mg ( hf − h0 ) −s 1 2 ( 0.750kg ) ( 0m/s ) 2 − (18.0m/s ) 2 + ( 0.750kg ) ( 9.80m/s 2 ) (11.8m) = − (11.8m) 2.9N ______________________________________________________________________________ 49. REASONING AND SOLUTION The force exerted by the bat on the ball is the only nonconservative force acting. The work due to this force is Wnc = mv f2 – mv02 + mg ( hf – h0 ) 1 1 2 2 Taking h0 = 0 m at the level of the bat, v 0 = 40.0 m/s just before the bat strikes the ball and v f to be the speed of the ball at hf = 25.0 m, we have vf = 2Wnc m + v02 – 2ghf 2(70.0 J) 2 + ( 40.0 m/s ) –2(9.80 m/s2 )(25.0 m) = 45.9 m/s 0.140 kg ______________________________________________________________________________ = 50. REASONING a. Because the kinetic frictional force, a nonconservative force, is present, it does negative work on the skier. The work-energy theorem, in the form of Equation 6.6, may be used to find the work done by this force. 232 WORK AND ENERGY b. Once the work done by the kinetic frictional force is known, the magnitude of the kinetic frictional force can be determined by using the definition of work, Equation 6.1, since the magnitude of the skier’s displacement is known. SOLUTION a. The work Wnc done by the kinetic frictional force is related to the object’s kinetic and potential energies by Equation 6.6: Wnc = ( 1 2 mv2f − 12 mv20 ) + ( mgh − mgh0 ) f The initial height of the skier at the bottom of the hill is h0 = 0 m, and the final height is hf = s sin 25° (see the drawing). Thus, the work is Wnc = ( = 1 2 1 mv 2 f 2 − 12 mv02 s hf 25° ) + mg ( s sin 25° − h ) 0 ( 63kg ) ( 4 . 4 m / s ) 2 − 12 ( 63kg )( 6 . 6 m / s ) 2 ( ) + ( 63kg ) 9.80m/s 2 (1.9m ) sin25° − 0 m = −2 7 0 J b. The work done by the kinetic frictional force is, according to Equation 6.1, Wnc = ( f k cos180° ) s , where f k is the magnitude of the kinetic frictional force, and s is the magnitude of the skier’s displacement. The displacement of the skier is up the hill and the kinetic frictional force is directed down the hill, so the angle between the two vectors is θ = 180° and cos θ = –1. Solving the equation above for f k, we have Wnc −270J = = 140N −s −1.9m ______________________________________________________________________________ fk = 51. REASONING The work-energy theorem, as expressed in Equation 6.7b, is W nc = ∆ KE + ∆ PE and can be applied to method 1 and method 2. In either method, the speed is constant, so that ∆KE = 0 J. In addition, we take the height at ground level to be h0 = 0 m, so that hf = h and ∆ PE = mghf – mgh0 = mgh . Thus, the work-energy theorem becomes Wnc = mgh (1) Chapter 6 Problems 233 SOLUTION In method 1, the nonconservative work that you do is the only nonconservative work present, so that Equation (1) becomes Wnc = W you, 1 = mgh In method 2, there are two contributions to the nonconservative work, so that Equation (1) becomes Wnc = –485 J + Wyou, 2 = mgh Since Wyou, 2 = 2Wyou, 1 = 2mgh, we have –485 J + 2 mgh = mgh 485 J = mgh or But h = (3.00 m) sin θ, so it follows that 485 J = mg(3.00 m ) sin θ Therefore, 485 J 485 J –1 = 21.5° θ = sin = sin 2 mg ( 3.00 m ) ( 45.0kg ) 9.80 m/s ( 3.00 m ) ______________________________________________________________________________ –1 ( ) 52. REASONING AND SOLUTION a. The lost energy is Elost = E0 – Ef The ball dropped from rest, so its initial energy is purely potential. The ball is momentarily at rest at the highest point in its rebound, so its final energy is also purely potential. Then ( ) 2 Elost = mgh0 – mghf = ( 0.60kg ) 9.80 m/s (1.22 m ) – ( 0.69 m ) = 3.1 J b. The work done by the player must compensate for this loss of energy. Elost 3.1 J = 24 N (cos θ) s (cos 0°)(0.13 m) ______________________________________________________________________________ Elost = (F cos θ) s 53. ⇒ F= = SSM REASONING AND SOLUTION According to the work-energy theorem as given in Equation 6.8, we have 234 WORK AND ENERGY Wnc = ( 1 mvf2 2 + mghf )−( 1 2 mv02 + mgh0 ) The metal piece starts at rest and is at rest just as it barely strikes the bell, so that vf = v0 = 0 m/s. ( 2 ) In addition, hf = h and h0 = 0 m, while Wnc = 0.25 12 M v , where M and v are the mass and speed of the hammer. Thus, the work-energy theorem becomes 0.25 ( 1 M v2 2 ) = mgh Solving for the speed of the hammer, we find 2mgh 2(0.400 kg)(9.80 m/s 2 )(5.00 m) = = 4.17 m/s 0.25M 0.25 (9.00 kg) ______________________________________________________________________________ v = 54. REASONING AND SOLUTION For the actual motion of the rocket ∆KE = Wnc – ∆PE = Wnc – mgh (6.7b) ∆KE = –8.00 × 102 J – (3.00 kg)(9.80 m/s2)(1.00 × 102 m) = –3740 J Since ∆KE = KEf – KE0 = – KE0 , the initial kinetic energy of the rocket is then KE0 = 3740 J If the rocket were launched with this initial kinetic energy and no air resistance, the potential energy of the rocket at the top of its trajectory would be PE = mgh = 3740 J Hence, h= 3740J ( 3.00kg ) ( 9.80m/s2 ) = 127m ______________________________________________________________________________ 55. REASONING We can find the power produced by the hero by using Equation 6.10a, P = W / t . In order to use this expression, we must first determine the work done by the hero. In order to calculate the work, we must first find the net force. The net force can be found from Newton's second law. SOLUTION There are two forces that act on the villain while he is being lifted upward. They are the upward vertical force F exerted on the villain by the hero, and the downward weight mg. Chapter 6 Problems 235 With upward taken as positive, Newton’s second law gives F − mg = ma = 0 , since the villain moves straight upward at constant speed. Therefore, F = mg , and the work done by the hero in lifting the villain straight up through a vertical distance h is W = Fh = mgh . Therefore, the power produced by the hero is W mgh (91 kg)(9.80 m/s 2 )(1.2 m) = = = 2.1 × 103 W t t 0.51 s ______________________________________________________________________________ P= 56. REASONING AND SOLUTION The work done by the person in turning the crank one revolution is W = F (2πr) = (22 N)(2π)(0.28 m) = 39 J The power expended is then W 39J = = 3.0 × 101 W (6.10a) t 1.3s ______________________________________________________________________________ P= 57. SSM REASONING AND SOLUTION One kilowatt ⋅ hour is the amount of work or energy generated when one kilowatt of power is supplied for a time of one hour. From Equation 6.10a, we know that W = P t . Using the fact that 1 kW = 1.0 × 10 3 J/s and that 1h = 3600 s, we have 3 3 6 1.0 kWh = (1.0 × 10 J/s)(1 h) = (1.0 × 10 J/s)(3600 s) = 3.6 × 10 J ______________________________________________________________________________ 58. REASONING AND SOLUTION gravity, W = mgh, so The work done by the crane is the work done against ( )( ) 3.00 × 102 kg 9.80m/s2 (10.0m) mgh t= = = 73.5s P 4.00 × 102 W ______________________________________________________________________________ 59. REASONING The average power developed by the cheetah is equal to the work done by the cheetah divided by the elapsed time (Equation 6.10a). The work, on the other hand, can be related to the change in the cheetah’s kinetic energy by the work-energy theorem, Equation 6.3. SOLUTION a. The average power is P= W t (6.10a) 236 WORK AND ENERGY where W is the work done by the cheetah. This work is related to the change in the cheetah’s kinetic energy by Equation 6.3, W = 12 mvf2 − 12 mv02 , so the average power is W P= = t = mvf2 − 12 mv02 t 1 (110kg )( 2 7 m / s )2 − 2 1 2 1 2 ( 110kg ) ( 0 m / s )2 4.0s = 1.0 × 104 W b. The power, in units of horsepower (hp), is ( ) 1hp 4 P = 1.0 × 10 W = 13hp 745.7W ______________________________________________________________________________ 60. REASONING AND SOLUTION a. The power developed by the engine is P = Fv = (2.00 × 102 N)(20.0 m/s) = 4.00 × 103 W b. The force required of the engine in order to maintain a constant speed up the slope is F = Fa + mg sin 37.0° The power developed by the engine is then P = Fv = (Fa + mg sin 37.0°)v (6.11) P = [2.00 × 102 N + (2.50 × 102 kg)(9.80 m/s2)sin 37.0°](20.0 m/s) = 3.35 × 10 4 W ______________________________________________________________________________ 61. SSM REASONING AND SOLUTION In the drawings below, the positive direction is to the right. When the boat is not pulling a skier, the engine must provide a force F1 to overcome the resistive force of the water FR. Since the boat travels with constant speed, these forces must be equal in magnitude and opposite in direction. FR – FR + F1 = ma = 0 or FR = F1 (1) F1 Chapter 6 Problems 237 When the boat is pulling a skier, the engine must provide a force F2 to balance the resistive force of the water FR and the tension in the tow rope T. T F – FR + F2 – T = ma = 0 R F 2 or F2 = FR + T (2) The average power is given by P = Fv , according to Equation 6.11 in the text. Since the boat moves with the same speed in both cases, v 1 = v 2, and we have P1 P 2 = F1 F2 or F2 = F1 P2 P1 Using Equations (1) and (2), this becomes F1 + T = F1 P2 P1 Solving for T gives P T = F1 2 − 1 P1 The force F1 can be determined from P1 = F1v1 , thereby giving 7.50 × 104 W 8.30 × 104 W P1 P 2 − 1 − 1 = 6.7 × 102 N = 4 v1 P1 12m/s 7.50 × 10 W ______________________________________________________________________________ T= 62. REASONING AND SOLUTION The following drawings show the free-body diagrams for the car in going both up and down the hill. The force FR is the combined force of air resistance and friction, and the forces FU and FD are the forces supplied by the engine in going uphill and downhill respectively. 238 WORK AND ENERGY Going up the hill FN Going down the hill FN FU FR FD FR mg sin θ mg cos θ mg sin θ mg cos θ Writing Newton's second law in the direction of motion for the car as it goes uphill, taking uphill as the positive direction, we have FU − FR − mg sin θ = ma = 0 Solving for FU , we have FU = FR + mg sin θ Similarly, when the car is going downhill, Newton's second law in the direction of motion gives FR − FD − mg sin θ = ma = 0 so that FD = FR − mg sin θ Since the car needs 47 hp more to sustain the constant uphill velocity than the constant downhill velocity, we can write P U = P D + ∆P where PU is the power needed to sustain the constant uphill velocity, PD is the power needed to sustain the constant downhill velocity, and ∆ P = 47 hp . In terms of SI units, 746 W ∆P = ( 47 hp ) = 3.51 ×10 4 W 1 hp Using Equation 6.11 ( P = Fv ), the equation PU = P D + ∆P can be written as FU v = FD v + ∆P Using the expressions for FU and FD , we have Solving for θ, we find ( FR + mg sin θ ) v = ( FR − mg sin θ ) v + ∆ P Chapter 6 Problems 239 ∆P 3.51 ×10 4 W –1 θ = sin = 2.0° = sin 2 2 mgv 2(1900 kg)(9.80 m/s )(27 m/s) ______________________________________________________________________________ –1 63. SSM REASONING The area under the force-versus-displacement graph over any displacement interval is equal to the work done over that displacement interval. SOLUTION a. Since the area under the force-versus-displacement graph from s = 0 to 0.50 m is greater for bow 1, it follows that bow 1 requires more work to draw the bow fully . b. The work required to draw bow 1 is equal to the area of the triangular region under the forceversus-displacement curve for bow 1. Since the area of a triangle is equal one-half times the base of the triangle times the height of the triangle, we have W1 = 1 (0.50m)(350 N) = 88 J 2 For bow 2, we note that each small square under the force-versus-displacement graph has an area of (0.050 m)(40.0 N) = 2.0 J We estimate that there are approximately 31.3 squares under the force-versus-displacement graph for bow 2; therefore, the total work done is 2.0 J (31.3 squares) = 63 J square Therefore, the additional work required to stretch bow 1 as compared to bow 2 is 88 J – 63 J = 25 J ______________________________________________________________________________ 64. REASONING AND SOLUTION The work done during each interval is equal to the area under the force vs. displacement curve over that interval. a. Since the area under the curve from 0 to 1.0 m is triangular, the area under the curve over this interval is (1/2) × height × base: W01 = 1 ( +6.0N)(1.0m − 0 m ) = 3.0 J 2 240 WORK AND ENERGY b. Since there is no area under the curve, the work done is zero . This is reasonable since there is no net force acting on the object during this interval. c. In the interval from 2.0 to 4.0 m, the area under the curve has two distinct regions. From 2.0 to 3.0 m, the area is triangular in shape, while from 3.0 to 4.0 m the region is rectangular. The total work done from 2.0 to 4.0 m is the sum of the areas of these two regions. 1 W 23 = 2 (–6.0 N)(3.0 m – 2.0 m) = – 3.0 J and W 34 = (–6.0 N)(4.0 m – 3.0 m) = – 6.0 J Thus, the total work done in the interval from 2.0 to 4.0 m is W24 = W23 + W34 = (–3.0 J) + (–6.0 J) = –9.0 J Notice in the interval from 2.0 to 4.0 m, the area under the curve (and hence the work done) is negative since the force is negative in that interval. ______________________________________________________________________________ 65. REASONING AND SOLUTION a. The work done is equal to the colored area under the graph. Therefore, the work done by the net force on the skater from 0 to 3.0 m is Area under graph = ( 31 N − 0 N ) ( 3.0 m − 0 m ) = 93 J from 0 to 3.0 m b. From 3.0 m to 6.0 m, the net force component F cos θ is zero, and there is no area under the curve. Therefore, no work is done . c. The speed of the skater will increase from 1.5 m/s when s = 0 m to a final value vf at s = 3.0 m . Since the force drops to zero at s = 3.0 m , the speed remains constant at the value vf from s = 3.0 m to s = 6.0 m ; therefore, the speed at s = 6.0 m is vf . We can find vf from the work-energy theorem (Equation 6.3). According to the work-energy theorem, the work done on the skater is given by 1 2 1 2 W = mvf2 − mv02 Solving for vf , we find 2W 2(93 J) + v02 = + (1.5 m/s) 2 = 2.3 m/s m 65 kg ______________________________________________________________________________ vf = Chapter 6 Problems 241 66. REASONING The area under the force-versus-displacement graph over any displacement interval is equal to the work done over that interval. From Example 16, we know that the total work done in drawing back the string of the bow is 60.5 J, corresponding to a total area under the curve of 242 squares (each square represents 0.250 J of work). The percentage of the total work done over any displacement interval is, therefore, Work done during interval Number of squares under curve during interval ×100 = × 100 Total work done during all intervals Total number of squares under entire curve SOLUTION a. We estimate that there are 130 squares under the curve from s = 0 to 0.306 m ; therefore, the percentage of work done during the interval in question is 130 squares ×100 = 54% 242 squares b. Similarly, we estimate that there are 112 squares under the curve in the interval from s = 0.306 m to 0.500 m ; therefore, the percentage of work done during the interval in question is 112 squares × 100 = 46% 242 squares ______________________________________________________________________________ 67. REASONING AND SOLUTION a. The net work done is equal to the area under the curve. For the first 10 m, the area is triangular and the area under the curve is equal to (1/2) × height × base; therefore W0,10 = 12 (10.0N)(10.0m − 0 m )= 5.00× 101 J From 10 to 20 m the region is rectangular and the area under the curve is W10,20 = (10.0N)(20.0m − 10.0m) = 1.00 ×102 J Thus, the total work done by the force is W0,10 + W10,20 = (5.00 × 101 J) + (1.00 × 102 J) = 1.50 × 102 J b. From the work-energy theorem, W = 12 mvf2 − 12 mv02 Since the object is initially at rest, v 0 = 0 m/s. Solving for v f gives 242 WORK AND ENERGY 2W 2(1.50 ×10 2 J) vf = = = 7.07 m/s m 6.00 kg ______________________________________________________________________________ 68. REASONING The work done by the tension in the cable is given by Equation 6.1 as W = ( T cos θ ) s . Since the elevator is moving upward at a constant velocity, it is in equilibrium, and the magnitude T of the tension must be equal to the magnitude mg of the elevator’s weight; T = mg. SOLUTION a. The tension and the displacement vectors point in the same direction (upward), so the angle between them is θ = 0°. The work done by the tension is W = ( T cos θ ) s = ( mg cos θ ) s (6.1) = (1200kg ) (9.80m/s 2 ) cos0° ( 3 5 m ) = 4.1 × 10 5 J b. The weight and the displacement vectors point in opposite directions, so the angle between them is θ = 180°. The work done by the weight is W = ( mg cos θ ) s = (1200kg ) ( 9.80m/s 2 ) cos180° ( 3 5 m ) = −4.1 × 10 5 J (6.1) ______________________________________________________________________________ 69. SSM REASONING AND SOLUTION Since we are ignoring friction and air resistance, the net work done by the nonconservative forces is zero, and the principle of conservation of mechanical energy holds. Let E0 be the total mechanical energy of the vaulter at take-off and Ef be the total mechanical energy as he clears the bar. The principle of conservation of mechanical energy gives 1 mv2f + mghf = 1 mv20 + mgh0 214 2 4 4244 3 1 4244 3 Ef E0 For maximum height hf , the vaulter just clears the bar with zero speed, vf = 0 m/s. If we let h0 = 0 m at ground level, we have 1 2 mghf = mv20 Solving for hf , we find v02 (9.00 m/s) 2 hf = = = 4.13 m 2 g 2(9.80 m/s 2 ) ______________________________________________________________________________ Chapter 6 Problems 70. REASONING AND SOLUTION 243 The work done by the pitcher is Wnc = F (π r) Assuming that only the pitcher does work on the ball Wnc = 1 2 mvf2 − 12 mv02 + mghf − mgh0 Then, vf = = 2Wnc m + v02 – 2 g ( hf – h0 ) 2(28 N) π ( 0.51 m ) + (12 m/s ) – 2(9.80 m/s2 )(–1.02 m) = 23 m/s 2 0.25 kg ______________________________________________________________________________ 71. SSM REASONING AND SOLUTION The work done by the retarding force is given by Equation 6.1:W = ( F cosθ )s . Since the force is a retarding force, it must point opposite to the direction of the displacement, so that θ = 180°. Thus, we have 3 6 W = ( F cosθ )s = (3.0 × 10 N)(cos 180 °)(850 m) = –2.6 × 10 J The work done by this force is negative , because the retarding force is directed opposite to the direction of the displacement of the truck. ______________________________________________________________________________ 72. REASONING AND SOLUTION The work done by nonconservative forces on the balloon is Wnc = ∆KE + ∆PE = +9.70 × 104 J The change in kinetic energy of the balloon is ∆KE = 1 mv 2 f 2 = 1 (5.00 2 × 102 kg)(8.00 m/s)2 = +1.60 × 104 J The change in the potential energy is ∆PE = mgh = (5.00 × 102 kg)(9.80 m/s2)h = (4.90 × 103 kg⋅m/s2)h Hence, h= 9.70 × 104 J − 1.60 × 104 J = 16.5m 4.90 × 103 kg ⋅ m/s 2 ______________________________________________________________________________ 244 WORK AND ENERGY 73. REASONING AND SOLUTION The applied force does work WP = Ps cos 0° = (150 N)(7.0 m) = 1.0 × 103 J The frictional force does work Wf = f ks cos 180° = – µkFNs where FN = mg, so Wf = – (0.25)(55 kg)(9.80 m/s2)(7.0 m) = –940 J The normal force and gravity do no work , since they both act at a 90° angle to the displacement. ______________________________________________________________________________ 74. REASONING The average power generated by the tension in the cable is equal to the work done by the tension divided by the elapsed time (Equation 6.10a). The work can be related to the change in the lift’s kinetic and potential energies via the work-energy theorem. SOLUTION a. The average power is P= Wnc t (6.10a) where Wnc is the work done by the nonconservative tension force. This work is related to the lift’s kinetic and potential energies by Equation 6.6, Wnc = ( 1 2 ) mvf2 − 12 mv02 + ( mghf − mgh0 ) , so the average power is W P = nc = t ( 1 2 mvf2 − 12 mv02 ) + (mgh f − mgh0 ) t Since the lift moves at a constant speed, v 0 = v f, the average power becomes ( 4skiers ) 65 ( ) kg 9.80m/s 2 (1 4 0 m) mg ( hf − h0 ) skier P= = = 3.0 × 103 W t 60s ( 2 min) 1min ______________________________________________________________________________ Chapter 6 Problems 75. 245 SSM REASONING The work-energy theorem can be used to determine the net work done on the car by the nonconservative forces of friction and air resistance. The work-energy theorem is, according to Equation 6.8, Wnc = ( 1 mv 2 f 2 ) ( + mghf − 1 2 2 mv0 + mgh0 ) The nonconservative forces are the force of friction, the force of air resistance, and the force provided by the engine. Thus, we can rewrite Equation 6.8 as Wfriction + Wair + Wengine = ( 1 2 ) ( 2 mvf + mghf − 2 mv0 + mgh0 1 2 ) This expression can be solved for Wfriction + Wair . SOLUTION We will measure the heights from sea level, where h0 = 0 m. Since the car starts from rest, v0 = 0 m/s. Thus, we have Wfriction + Wair = m ( 1 v2 2 f ) + ghf − Wengine = (1.50 × 103 kg) 12 (27.0 m/s)2 + (9.80 m/s2 )(2.00× 102 m) – (4.70 × 10 6 J) = –1.21 ×10 6 J ______________________________________________________________________________ 76. REASONING AND SOLUTION Relative to her final position in the water, the diver initially had energy mgh0. All of this energy was dissipated as work in traveling through 3.00 m of water. Wnc = 1 mv 2 f 2 – 1 mv 2 0 2 + mghf – mgh0 Thus, Wnc = Fs cos 180° = mghf – mgh0 Then, ( 55.0kg ) ( 9.80 m/s 2 ) ( –13.0 m ) F= = ( 3.00 m ) cos180° 2340 N ______________________________________________________________________________ 77. REASONING AND SOLUTION The conservation of energy applied between point A and the top of the trajectory gives 246 WORK AND ENERGY KEA + mghA = mgh where h = 4.00 m. Rearranging, we find KEA = mg(h – hA ) or ( ) vA = 2g ( h − hA ) = 2 9.80m/s 2 ( 4.00m − 3.00m ) = 4.43m/s ______________________________________________________________________________ 78. REASONING AND SOLUTION consists of the full weight of the car. When the car is at the top of the track the centripetal force mv2/r = mg Applying the conservation of energy between the bottom and the top of the track gives (1/2)mv2 + mg(2r) = (1/2)mv02 Using both of the above equations v02 = 5gr so r = v02/(5g) = (4.00 m/s)2/(49.0 m/s2) = 0.327 m ______________________________________________________________________________ 79. SSM WWW REASONING After the wheels lock, the only nonconservative force acting on the truck is friction. The work done by this conservative force can be determined from the work-energy theorem. According to Equation 6.8, Wnc = E f − E0 = ( 1 2 2 ) ( mvf + mghf − 1 2 2 mv0 + mgh0 ) (1) where Wnc is the work done by friction. According to Equation 6.1, W = ( F cosθ )s ; since the force of kinetic friction points opposite to the displacement, θ = 180°. According to Equation 4.8, the kinetic frictional force has a magnitude of f k = µk FN , where µk is the coefficient of kinetic friction and FN is the magnitude of the normal force. Thus, Wnc = ( f k cosθ ) s = µk FN (cos 180°) s = −µk FN s (2) Chapter 6 Problems Since the truck is sliding down an incline, we refer to the freebody diagram in order to determine the magnitude of the normal force FN . The free-body diagram at the right shows the three forces that act on the truck. They are the normal force FN , the force of kinetic friction f k , and the weight mg . The weight has been resolved into its components, and these vectors are shown as dashed arrows. From the free body diagram and the fact that the truck does not accelerate perpendicular to the incline, we can see that the magnitude of the normal force is given by 247 fk FN mg 15° mg cos 15° mg sin 15° FN = mg cos 15° Therefore, Equation (2) becomes Wnc = − µk (mg cos 15°) s (3) Combining Equations (1), (2), and (3), we have − µk ( mg cos 15° ) s = ( 1 2 2 mvf + mghf )−( 1 2 2 mv0 + mgh0 ) (4) This expression may be solved for s. SOLUTION When the car comes to a stop, vf = 0 m/s. If we take hf = 0 m at ground level, then, from the drawing at the right, we see that h0 = s sin 15° . Equation (4) then gives s h0 15° v20 (11.1 m/s) 2 = = 13.5 m 2g(µk cos 15° – sin 15 ° ) 2(9.80 m/s 2 )[(0.750)(cos 15 ° ) – sin 15 °] ______________________________________________________________________________ s= 80. CONCEPT QUESTIONS a. Since the force exerted by the elevator and the displacement are in the same direction on the upward part of the trip, the angle between them is 0°, and so the work done by the force is positive. b. Since the force exerted by the elevator and the displacement are in opposite directions on the downward part of the trip, the angle between them is 180°, and so the work done by the force is negative. SOLUTION FN +y W W 248 WORK AND ENERGY a. The free-body diagram at the right shows the three forces that act on you: W is your weight, Wbelongings is the weight of your belongings, and FN is the normal force exerted on you by the floor of the elevator. Since you are moving upward at a constant velocity, you are in equilibrium and the net force in the y direction must be zero: FN − W − Wbelongings = 0 14442444 3 ΣFy (4.9b) Therefore, the magnitude of the normal force is FN = W + Wbelongings .The work done by the normal force is W = ( FN cos θ ) s = (W + Wbelongings ) ( cos θ ) s (6.1) = ( 685N + 915N )( cos0° ) (15.2m ) = 2 4 3 0 0 J b. During the downward trip, you are still in equilibrium since the elevator is moving with a constant velocity. The magnitude of the normal force is now FN = W. The work done by the normal force is W = ( FN cos θ ) s = ( W cos θ ) s = ( 685N )( cos180° )(15.2m ) = −1 0 4 0 0 J (6.1) ______________________________________________________________________________ 81. CONCEPT QUESTIONS a. Since the asteroid is slowing down, it is decelerating. The force must point in a direction that is opposite to that of the displacement. b. The direction of the force is opposite to the displacement of the asteroid, so the angle between them is 180°. Therefore, the force does negative work. c. The asteroid is slowing down, so its speed is decreasing. Thus, its kinetic energy is changing. d. The work done by the force is related to the change in the asteroid’s kinetic energy by the work-energy theorem: W = 12 mv 2f − 12 mv02 , Equation 6.3. SOLUTION a. The work-energy theorem states that the work done by the force acting on the asteroid is equal to its final kinetic energy minus its initial kinetic energy: Chapter 6 Problems W = 12 mvf2 − = 1 2 1 2 249 mv02 (6.3) ( 4.5 × 10 kg ) ( 5500m/s) 4 2 − 1 2 ( 4.5 × 10 kg ) ( 7100m/s) 4 2 = −4.5 × 1011 J b. The work done by the force acting on the asteroid is given by Equation 6.1 as W = ( F cos θ ) s . Since the force acts to slow down the asteroid, the force and the displacement vectors point in opposite directions, so the angle between them is θ = 180°. The magnitude of the force is F= W −4.5 × 1011 J = = 2.5 × 105 N 6 s cosθ 1.8 × 10 m cos180° ( ) (6.1) ______________________________________________________________________________ 82. CONCEPT QUESTIONS a. Since there is no friction, there are only two forces acting on each box, the gravitational force (its weight) and the normal force. The normal force, being perpendicular to the displacement of a box, does no work. The gravitational force does work. However, it is a conservative force, so the conservation of mechanical energy applies to the motion of each box. This principle states that the final speed depends only on the initial speed, and the initial and final heights. The final speed does not depend on the mass of the box or on the steepness of the slope. Thus, both boxes reach B with the same speed. b. The kinetic energy is given by the expression K E = 12 mv2 . Both boxes have the same speed v when they reach B. However, the heavier box has the greater mass. Therefore, the heavier box has the greater kinetic energy. SOLUTION a. The conservation of mechanical energy states that 1 mv 2 B 2 + mghB = 1 mv 2 A 2 + mghA (6.9b) Solving for the speed v B at B gives vB = vA2 + 2 g ( hA − hB ) = ( 0 m /s ) 2 ( ) + 2 9.80m/s 2 ( 4.5m − 1.5m ) = b. The speed of the heavier box is the same as the lighter box: 7.7m/s . c. The ratio of the kinetic energies is 7.7m/s 250 WORK AND ENERGY ( KE B ) heavierbox ( KE B ) lighterbox ( = ( ) mv ) 1 2 mvB2 1 2 2 B heavier box = mheavier box mlighterbox = 44kg = 11kg 4.0 lighterbox ______________________________________________________________________________ 83. CONCEPT QUESTIONS a. The kinetic frictional force does negative work, because the force is directed opposite to the displacement of the student. b. Since the kinetic frictional force does negative work, the final total mechanical energy must be less than the initial total mechanical energy. Therefore, the total mechanical energy decreases. c. The work done by the kinetic frictional force is related to the change in the student’s total mechanical energy by the work-energy theorem, Wnc = E f − E 0 , Equation 6.8. SOLUTION The work-energy theorem states that ( ) Wnc = 12 mv f2 + mghf − 12 mv02 + mgh0 144244 3 144 42444 3 Ef E0 Solving for the final speed gives vf = = (6.8) 2Wnc + v02 + 2 g ( h0 − hf ) m ( 2 −6.50 × 103 J ) + (0 m / s ) 2 ( ) + 2 9.80m/s 2 (11.8m ) = 8 . 6 m / s 83.0kg ______________________________________________________________________________ 84. CONCEPT QUESTIONS a. The two types of energy that are changing are the kinetic energy and the gravitational potential energy. The kinetic energy is increasing, because the speed of the helicopter is increasing. The gravitational potential energy is increasing, because the height of the helicopter is increasing. b. The work done by the lifting force is related to the kinetic and potential energies by the workenergy theorem, Equation 6.6. c. The average power is defined as the work divided by the time, Equation 6.10a, so the time must be known in addition to the work. SOLUTION Chapter 6 Problems 251 The average power generated by the lifting force is equal to the work done by the force divided by the elapsed time (Equation 6.10a). The work can be related to the change in the helicopter’s total mechanical energy, which can be determined. The average power is P= Wnc t (6.10a) where Wnc is the work done by the nonconservative lifting force. This work is related to the helicopter’s kinetic and potential energies by Equation 6.6, Wnc = ( 1 2 ) mvf2 − 12 mv02 + ( mghf − mgh0 ) , so the average power is W P = nc = t ( 1 mv 2 f 2 − 12 mv02 ) + (mgh f − mgh0 ) t ( = 1m 2 (v 2 f − v02 ) + mg ( h f − h0 ) t ) 2 − ( 0 m /s ) + ( 810kg ) 9.80m/s 2 [8.2m − 0 m ] P= = 2.4 × 104 W 3.5s ______________________________________________________________________________ 1 2 ( 810kg ) ( 7.0m/s ) 2 85. CONCEPT QUESTIONS a. The kinetic energy is changing because the skier’s speed in changing. b. The work is positive. Because the speed of the skier is increasing, the net external force must be in the same direction as the displacement of the skier. c. The work done by the net external force is equal to the change in the kinetic energy of the skier. SOLUTION The work done by the net external force acting on the skier can be found from the work-energy theorem in the form of Equation 6.3, because the final and initial kinetic energies of the skier can be determined: W = 12 mvf2 − = 1 2 1 2 mv02 ( 70.3kg ) (11.3m/s ) 2 − 12 ( 70.3kg )( 6.10m/s )2 = 3.2 × 10 3 J ______________________________________________________________________________