Solutions to Physics: Principles with Applications, 5/E, Giancoli Chapter 6

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1 CHAPTER 6 1. Because there is no acceleration, the contact orce must have the same magnitude as the weight. The displacement in the direction o this orce is the vertical displacement. Thus, W = F Æy = () Æy = (75.0 kg)(9.80 m/s )(10.0 m) = J.. (a) Because there is no acceleration, the horizontal applied orce must have the same magnitude as the riction orce. Thus, W = F Æx = (180 N)(6.0 m) = J. (b) Because there is no acceleration, the vertical applied orce must have the same magnitude as the weight. Thus, W = F y = y = (900 N)(6.0 m) = J. 3. Because there is no acceleration, rom the orce diagram we see that F N =, and F = F r = µ k. Thus, W = F x cos 0 = µ k x cos 0 = (0.70)(150 kg)(9.80 m/s )(1.3 m)(1) = J. F F N N y x r ² x 4. Because there is no acceleration, the net work is zero, that is, the (positive) work o the car and the (negative) work done by the average retarding orce must add to zero. Thus, W net = W car + F r Æx cos 180 = 0, or F r = W car /Æx cos 180 = ( J)/( m)( 1) = 5 N. F N F y x r ² x 5. Because the speed is zero beore the throw and when the rock reaches the highest point, the positive work o the throw and the (negative) work done by the (downward) weight must add to zero. Thus, W net = W throw + h cos 180 = 0, or h = W throw / cos 180 = (115 J)/(0.35 kg)(9.80 m/s )( 1) = 36.1 m. 6. The maximum amount o work that the hammer can do is the work that was done by the weight as the hammer ell: W max = h cos 0 = (.0 kg)(9.80 m/s )(0.40 m)(1) = 7.8 J. People add their own orce to the hammer as it alls in order that additional work is done beore the hammer hits the nail, and thus more work can be done on the nail. Page 6 1

2 7. The minimum work is needed when there is no acceleration. (a) From the orce diagram, we write ΣF = ma: y-component: F N cos θ = 0; x-component: F min sin θ = 0. For a distance d along the incline, we have W min = F min d cos 0 = d sin θ (1) = (1000 kg)(9.80 m/s )(300 m) sin 17.5 = J. (b) When there is riction, we have x-component: F min sin θ µ k F N = 0, or F min = sin θ + µ k cos θ = 0, For a distance d along the incline, we have W min = F min d cos 0 = d (sin θ + µ k cos θ)(1) = (1000 kg)(9.80 m/s )(300 m)(sin cos 17.5 ) = J. y r x F N θ F d 8. Because the motion is in the x-direction, we see that the weight and normal orces do no work: W FN = W = 0. From the orce diagram, we write ΣF = ma: x-component: F cos θ F r = 0, or F r = F cos θ. For the work by these two orces, we have W F = F Æx cos θ = (1 N)(15 m) cos 0 = J. θ F F r N F N y x W r = F cos θ Æx cos 180 = (1 N) cos 0 (15 m)( 1) = J. As expected, the total work is zero: W F = W r = J. 9. Because the net work must be zero, the work to stack the books will have the same magnitude as the work done by gravity. For each book the work is times the distance the center is raised (zero or the irst book, one book-height or the second book, etc.). W 1 = 0, W = h, W 3 = h;. Thus or eight books, we have W = W 1 + W + W W 8 = h( ) = (1.8 kg)(9.80 m/s )(0.046 m)(8) = 3 J. Page 6

3 10. (a) From the orce diagram, we write ΣF = ma: y-component: F N cos θ = 0; x-component: F µ k F N + sin θ = 0. Thus we have F = µ k F N + sin θ = µ k cos θ + sin θ θ = (sin θ µ k cos θ) = (80 kg)(9.80 m/s )(sin cos 30 ) = N. (b) Because the piano is sliding down the incline, we have W F = F d cos 180 = (4. 10 J)(4.3 m)( 1) = J. (c) For the riction orce, we have W r = µ k cos θ d cos 180 = (0.40)(80 kg)(9.80 m/s )(cos 30 )(4.3 m)( 1) = J. (d) For the orce o gravity, we have W grav = d cos 60 = (80 kg)(9.80 m/s )(4.3 m)(cos 60 ) = J. (e) Because the normal orce does no work, we have W net = W grav + W F + W r + W N = J J J + 0 = 0. v F r θ F N F y x 11. (a) To ind the required orce, we use the orce diagram to write ΣF y = ma y : F Mg = Ma, so we have F = M(a + g) = M(0.10g + g) = 1.10Mg. (b) For the work, we have W F = Fh cos 0 = 1.10Mgh. F +y a Mg 1. From the graph we obtain the orces at the two ends: at d A = 10.0 m, (F cos θ) A = 150 N; at d B = 35.0 m, (F cos θ) B = 50 N. The work done in moving the object is the area under the F cos θ vs. x graph. I we assume the graph is a straight line, we have W =!(150 N + 50 N)(35.0 m 10.0 m) = J. 13. The work done in moving the object is the area F x (N) under the F cos θ vs. x graph. (a) For the motion rom 0.0 m to 10.0 m, we ind 400 the area o two triangles and one rectangle: W =!(400 N)(3.0 m 0.0 m) + TZ 100 (400 N)(7.0 m 3.0 m) + 0!(400 N)(10.0 m 7.0 m) = J. 00 (b) For the motion rom 0.0 m to 15.0 m, we add the negative area o two triangles and one rectangle: W = J!(00 N)1.0 m 10.0 m) (00 N)(13.5 m 1.0 m)!(00 N)(15.0 m 13.5 m) = J. x (m) Page 6 3

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5 14. We obtain the orces at the beginning and end o the motion: at x 1 = m, F 1 = kx 1 = (88 N/m)(0.038 m) = 3.34 N; at x = m, F = kx = (88 N/m)(0.058 m) = 5.10 N. From the graph the work done in stretching the object is the area under the F vs. x graph: W = =!(3.34 N N)(0.058 m m) = J. F(x) 0 x 1 x x 15. The work done in moving the object is the area under the F x vs. x graph. For the motion rom 0.0 m to 11.0 m, we ind the area o two triangles and one rectangle: W =!(4.0 N)(3.0 m 0.0 m) + (4.0 N)(8.0 m 3.0 m) +!(4.0 N)(11.0 m 8.0 m) = J F x (N ) x(m) 16. We obtain the orces at the beginning and end o the motion: at r E + h = m m = m, F = GM E m/r = ( N m /kg )( kg)(1300 kg)/ [( m)] = N. at r E = m, F 1 = GM E m/r E = ( N m /kg )( kg)(1300 kg)/ 0 [( m)] = N. From the graph the work done in stretching the object is the area under the F vs. r graph, which we approximate as a straight line: W =![F 1 + F ]h =!( N N)( m) = J. This will be a slight overestimate. F(r) r E r E + h r 17. We ind the speed rom KE = ; J =!( kg)v, which gives v = 484 m/s. 18. (a) KE = = KE 1 = ( 1 ), which gives v = v 1 Ã, so the speed increases by a actor o Ã. (b) KE = =!m(v 1 ) = 4( 1 ) = 4 KE 1, so the kinetic energy increases by a actor o The work done on the electron decreases its kinetic energy: W = ÆKE = 0 = 0!( kg)( m/s) = J. Page 6 5

6 0. The work done on the car decreases its kinetic energy: W = ÆKE = 0 = 0!(1000 kg)[(110 km/h)/(3.6 ks/h)] = J. Page 6 6

7 1. The percent increase in the kinetic energy is % increase = [( 1 )/ 1 ](100%) = (v v 1 )(100%)/v 1 = [(100 km/h) (90 km/h) ](100%)/(90 km/h) = 3%.. The work done on the arrow increases its kinetic energy: W = Fd = ÆKE = 0 ; (95 N)(0.80 m) =!(0.080 kg)v 0, which gives v = 44 m/s. 3. The work done by the orce o the glove decreases the kinetic energy o the ball: W = Fd = ÆKE = 0 ; F(0.5 m) = 0!(0.140 kg)(35 m/s), which gives F = N. The orce by the ball on the glove is the reaction to this orce: N in the direction o the motion o the ball. 4. The work done by the braking orce decreases the kinetic energy o the car: W = ÆKE; Fd = 0 = 0 0. Assuming the same braking orce, we orm the ratio: d /d 1 = (v 0 /v 01 ) = (1.5) = On a level road, the normal orce is, so the kinetic riction orce is µ k. Because it is the (negative) work o the riction orce that stops the car, we have W = ÆKE; µ k d = 0 ; (0.4)m(9.80 m/s )(88 m) = 0, which gives v 0 = 7 m/s (97 km/h or 60 mi/h). Because every term contains the mass, it cancels. 6. The work done by the air resistance decreases the kinetic energy o the ball: W = F air d = ÆKE = 0 =!m(0.90v 0 ) 0 = 0 [(0.90) 1]; F air (15 m) =!(0.5 kg)[(95 km/h)/(3.6 ks/h)] [(0.90) 1], which gives F air = 1.1 N. 7. With m 1 = m, or the initial condition we have KE 1 =!KE ;!m 1 v 1 =!(!m v ), or m v 1 =!m v, which gives v 1 =!v. Ater a speed increase o Æv, we have KE 1 = KE ;!m 1 (v 1 + Æv) =!m (v + Æv) ; m (!v m/s) = m (v m/s). When we take the square root o both sides, we get Ã(!v m/s) = ± (v m/s), which gives a positive result o v = For the other speed we have v 1 =!v = 3.5 m/s. 7.1 m/s. Page 6 7

8 8. (a) From the orce diagram we write ΣF y = ma y : F T = ma; F T (0 kg)(9.80 m/s ) = (0 kg)(0.150)(9.80 m/s ), which gives F T = N. (b) The net work is done by the net orce: W net = F net h = (F T )h = [ N (0 kg)(9.80 m/s )](1.0 m) = J. (c) The work is done by the cable is W cable = F T h = ( N)(1.0 m) = J. (d) The work is done by gravity is W grav = h = (0 kg)(9.80 m/s )(1.0 m) = J. Note that W net = W cable + W grav. (e) The net work done on the load increases its kinetic energy: W net = ÆKE = 0 ; J =!(0 kg)v 0, which gives v = 7.86 m/s. F T +y a 9. The potential energy o the spring is zero when the spring is not stretched (x = 0). For the stored potential energy, we have PE =!kx 0; 5 J =!(440 N/m)x 0, which gives x = 0.34 m. 30. For the potential energy change we have ÆPE = Æy = (6.0 kg)(9.80 m/s )(1. m) = 71 J. 31. For the potential energy change we have ÆPE = Æy = (64 kg)(9.80 m/s )(4.0 m) = J. 3. (a) With the reerence level at the ground, or the potential energy we have PE a = y a = (.10 kg)(9.80 m/s )(.0 m) = 45.3 J. (b) With the reerence level at the top o the head, or the potential energy we have PE b = (y b h)= (.10 kg)(9.80 m/s )(.0 m 1.60 m) = 1.3 J. (c) Because the person lited the book rom the reerence level in part (a), the potential energy is equal to the work done: 45.3 J. In part (b) the initial potential energy was negative, so the inal potential energy is not the work done. 33. (a) With the reerence level at the ground, or the potential energy change we have ÆPE = Æy = (55 kg)(9.80 m/s )(3100 m 1600 m) = J. (b) The minimum work would be equal to the change in potential energy: W min = ÆPE = J. (c) Yes, the actual work will be more than this. There will be additional work required or any kinetic energy change, and to overcome retarding orces, such as air resistance and ground deormation. Page 6 8

9 34. The potential energy o the spring is zero when the spring is not stretched or compressed (x= 0). (a) For the change in potential energy, we have ÆPE =!kx!kx 0 =!k(x x 0 ). (b) I we call compressing positive, we have ÆPE compression =!k(+ x 0 ) 0 =!kx 0 ; ÆPE stretching =!k( x 0 ) 0 =!kx 0. The change in potential energy is the same. 35. We choose the potential energy to be zero at the ground (y = 0). Because the tension in the vine does no work, energy is conserved, so we have E = KE i + PE i = KE + PE ; i + y i = + y ;!m(5.6 m/s) + m(9.80 m/s )(0) =!m(0) + m(9.80 m/s )h, which gives h = 1.6 m. No, the length o the vine does not aect the height; it aects the angle. L θ h y = 0 v We choose the potential energy to be zero at the bottom (y = 0). Because there is no riction and the normal orce does no work, energy is conserved, so we have E = KE i + PE i = KE + PE ; i + y i = + y ;!m(0) + m(9.80 m/s )(15 m) = + m(9.80 m/s )(0), which gives v = 49.5 m/s. This is 180 km/h! It is a good thing there is riction on the ski slopes. 37. We choose the potential energy to be zero at the bottom (y = 0). Because there is no riction and the normal orce does no work, energy is conserved, so we have E = KE i + PE i = KE + PE ; i + y i = + y ; i + m(9.80 m/s )(0) =!m(0) + m(9.80 m/s )(1.35 m), which gives v i = 5.14 m/s. 38. We choose the potential energy to be zero at the ground (y = 0). We ind the minimum speed by ignoring any rictional orces. Energy is conserved, so we have E = KE i + PE i = KE + PE ; i + y i = + y ; i + m(9.80 m/s )(0) =!m(0.70 m/s) + m(9.80 m/s )(.10 m), which gives v i = 6.5 m/s. Note that the initial velocity will not be horizontal, but will have a horizontal component o 0.70 m/s. Page 6 9

10 39. We choose y = 0 at the level o the trampoline. (a) We apply conservation o energy or the jump rom the top o the platorm to the trampoline: E = KE i + PE i = KE + PE ; 0 + H = 1 + 0;!m(5.0 m/s) + m(9.80 m/s )(3.0 m) = 1, +y H which gives v 1 = 9. m/s. (b) We apply conservation o energy rom the landing on the trampoline to the maximum depression o the trampoline. I we ignore the small y = 0 change in gravitational potential energy, we have E = KE i + PE i = KE + PE ; = 0 +!kx ;!(75 kg)(9. m/s) =!( N/m)x, which gives x = 0.35 m. This will increase slightly i the gravitational potential energy is taken into account. v 0 v We choose y = 0 at point B. With no riction, energy is conserved. The initial (and constant) energy is E = E A = y A + A = m(9.8 m/s )(30 m) + 0 = (94 J/kg)m. At point B we have E = y B + B ; (94 J/kg)m = m(9.8 m/s )(0) + B, which gives v B = 4 m/s. At point C we have E = y C + C ; (94 J/kg)m = m(9.8 m/s )(5 m) + C, which gives v C = 9.9 m/s. At point D we have E = y D + D ; (94 J/kg)m = m(9.8 m/s )(1 m) + D, which gives v D = 19 m/s. A h A C h C D B hd 41. We choose the potential energy to be zero at the ground (y = 0). Energy is conserved, so we have E = KE i + PE i = KE + PE ; i + y i = + y ;!m(185 m/s) + m(9.80 m/s )(65 m) = + m(9.80 m/s )(0), which gives v = 199 m/s. Note that we have not ound the direction o the velocity. Page 6 10

11 L 0 y = 0 +y 4. (a) For the motion rom the bridge to the lowest point, we use energy conservation: KE i + PE gravi + PE cordi = KE + PE grav + PE cord ; = 0 + ( h) +!k(h L 0 ) ; 0 = (60 kg)(9.80 m/s )(31 m) +!k(31 m 1 m), which gives k = N/m. (b) The maximum acceleration will occur at the lowest point, where the upward restoring orce in the cord is maximum: kx max = ma max ; ( N/m)(31 m 1 m) (60 kg)(9.80 m/s ) = (60 kg)a max, which gives a max = m/s. h x v = We choose the potential energy to be zero at the compressed position (y = 0). (a) For the motion rom the release point to where the ball leaves the spring, we use energy conservation: KE i + PE gravi + PE springi = KE + PE grav + PE spring ; !kx = + x + 0;!(900 N/m)(0.150 m) =!(0.300 kg)v + (0.300 kg)(9.80 m/s )(0.150 m), which gives v = 8.03 m/s. (b) For the motion rom the release point to the highest point, we use energy conservation: KE i + PE gravi + PE springi = KE + PE grav + PE spring ; !kx = 0 + h + 0; !(900 N/m)(0.150 m) = (0.300 kg)(9.80 m/s )h, which gives h = 3.44 m. x v = 0 v h +y y = With y = 0 at the bottom o the circle, we call the start point A, the bottom o the circle B, and the top o the circle C. At the top o the circle we have the orces and F N, both downward, that provide the centripetal acceleration: + F N = mv C /r. The minimum value o F N is zero, so the minimum speed at C is ound rom v C min = gr. From energy conservation or the motion rom A to C we have KE A + PE A = KE C + PE C ; 0 + h = C + (r), thus the minimum height is ound rom gh =!v C min + gr =!gr + gr, which gives h =.5r. h A F N B C r 45. The potential energy is zero at x = 0. For the motion rom the release point, we use energy conservation: E = KE i + PE i = KE + PE ; E = 0 +!kx 0 = +!kx, which gives E =!kx 0 = +!kx Page 6 11

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13 46. The maximum acceleration will occur at the lowest point, where the upward restoring orce in the spring is maximum: kx max Mg = Ma max = M(5.0g), which gives x max = 6.0Mg/k. With y = 0 at the initial position o the top o the spring, or the motion rom the break point to the maximum compression o the spring, we use energy conservation: KE i + PE gravi + PE springi = KE + PE grav + PE spring ; h v = 0 +y 0 + Mgh + 0 = 0 + Mg( x max ) +!kx max. When we use the previous result, we get Mgh = [6.0(Mg) /k] +!k(6.0mg/k), which gives k = 1Mg/h. y = 0 x 47. The maximum acceleration will occur at the maximum compression o the spring: kx max = Ma max = M(5.0g), which gives x max = 5.0Mg/k. For the motion rom reaching the spring to the maximum compression o the spring, we use energy conservation: KE i + PE springi = KE + PE spring ;!Mv + 0 = 0 +!kx max. When we use the previous result, we get!mv =!k(5.0mg/k), which gives k = 5Mg /v = 5(100 kg)(9.80 m/s ) /[(100 km/h)/(3.6 ks/h)] = N/m. 48. (a) The work done against gravity is the increase in the potential energy: W = h = (75.0 kg)(9.80 m/s )(10 m) = J. (b) I this work is done by the orce on the pedals, we need to ind the distance that the orce acts over one revolution o the pedals and the number o revolutions to climb the hill. We ind the number o revolutions rom the distance along the incline: N = (h/ sin θ)/(5.10 m/revolution) = [(10 m)/ sin 7.50 ]/(5.10 m/revolution) = 180 revolutions. Because the orce is always tangent to the circular path, in each revolution the orce acts over a distance equal to the circumerence o the path: ¹D. Thus we have W = NF¹D; J = (180 revolutions)f¹(0.360 m), which gives F = 433 N. 49. The thermal energy is equal to the loss in kinetic energy: E thermal = ÆKE = i =!()(6500 kg)[(95 km/h)/(3.6 ks/h)] 0 = J. 50. We choose the bottom o the slide or the gravitational potential energy reerence level. The thermal energy is the negative o the change in kinetic and potential energy: E thermal = (ÆKE + ÆPE) = i + (h i h ) = 0!(17 kg)(.5 m/s) + (17 kg)(9.80 m/s )(3.5 m 0) = J. Page 6 13

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15 51. (a) We ind the normal orce rom the orce diagram or the ski: y-component: F N1 = cos θ; which gives the riction orce: F r1 = µ k cos θ. For the work-energy principle, we have W NC = ÆKE + ÆPE = ( i ) + (h h i ); F N1 L µ k cos θ L = ( 0) + (0 L sin θ); (0.090)(9.80 m/s ) cos 0 (100 m) =!v (9.80 m/s )(100 m) sin 0, which gives v = m/s. (b) On the level the normal orce is F N =, so the riction orce is F r = µ k. For the work-energy principle, we have W NC = ÆKE + ÆPE = ( i ) + (h h i ); µ k D = (0 i ) + (0 0); (0.090)(9.80 m/s )D =!( m/s), which gives D =.9 10 m. F r1 y = 0 θ F r F N 5. On the level the normal orce is F N =, so the riction orce is F r = µ k. For the work-energy principle, we have W NC = ÆKE + ÆPE = ( i ) + (h h i ); F(L 1 + L ) µ k L = ( 0) + (0 0); (350 N)(15 m + 15 m) (0.30)(90 kg)(9.80 m/s )(15 m) =!(90 kg)v, which gives v = 1 m/s. 53. We choose y = 0 at point B. For the work-energy principle applied to the motion rom A to B, we have W NC = ÆKE + ÆPE = ( B A ) + (h B h A ); 0.L = ( B ) + (0 h A ); 0.(9.80 m/s )(45.0 m) =!v B! (1.70 m/s) (9.80 m/s )(30 m), which gives v B = 0 m/s. A C h A h C D B hd 54. We ind the normal orce rom the orce diagram or the skier: y-component: F N = cos θ; which gives the riction orce: F r = µ k cos θ. For the work-energy principle or the motion up the incline, we have W NC = ÆKE + ÆPE = ( i ) + (h h i ); µ k cos θ L = (0 i ) + (L sin θ 0); µ k (9.80 m/s ) cos 18 (1. m) =!(1.0 m/s) + (9.80 m/s )(1. m) sin 18, which gives µ k = F r y = 0 L F N θ Page 6 15

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17 55. On the level the normal orce is F N =, so the riction orce is F r = µ k. The block is at rest at the release point and where it momentarily stops beore turning back. For the work-energy principle, we have W NC = ÆKE + ÆPE = ( i ) + (!kx!kx i ); µ k L = (0 0) +!k(x x i ); µ k (0.50 kg)(9.80 m/s )(0.050 m m) =!(180 N/m)[(0.03 m) ( m) ], which gives µ k = We ind the spring constant rom the orce required to compress the spring: k = F/x i = ( 0 N)/( 0.18 m) = 111 N/m. On the level the normal orce is F N =, so the riction orce is F r = µ k. The block is at rest at the release point and where it momentarily stops beore turning back. For the work-energy principle, we have W NC = ÆKE + ÆPE = ( i ) + (!kx!kx i ); µ k L = (0 0) +!k(x x i ); (0.30)(0.180 kg)(9.80 m/s )(0.18 m + x ) =!(111 N/m)[x ( 0.18 m) ]. This reduces to the quadratic equation 55.5x x 1.70 = 0, which has the solutions x = 0.17 m, 0.18 m. The negative solution corresponds to no motion, so the physical result is x = 0.17 m. 57. We choose the potential energy to be zero at the ground (y = 0). We convert the speeds: (500 km/h)/(3.6 ks/h) = 139 m/s; (00 km/h)/(3.6 ks/h) = 55.6 m/s. (a) I there were no air resistance, energy would be conserved: 0 = ÆKE + ÆPE = ( i ) + (h h i ); 0 =!(1000 kg)[(v (139 m/s) ] + (1000 kg)(9.80 m/s )( m), which gives v = 97 m/s = km/h. (b) With air resistance we have W NC = ÆKE + ÆPE = ( i ) + (h h i ); F(h i / sin θ) =!m(v v i ) + (0 h i ); F(3500 m)/sin 10 =!(1000 kg)[(55.6 m/s) (139 m/s) ] + (1000 kg)(9.80 m/s )( 3500 m) which gives F = N. 58. The amount o work required is the increase in potential energy: W = y. We ind the time rom P = W/t = y/t; 1750 W = (85 kg)(9.80 m/s )(16.0 m)/t, which gives t = 5.5 s. 59. We ind the equivalent orce exerted by the engine rom P = Fv; (18 hp)(746 W/hp) = F(90 km/h)/(3.6 ks/h), which gives F = N. At constant speed, this orce is balanced by the average retarding orce, which must be N. 60. (a) 1 hp = (550 t lb/s)(4.45 N/lb)/(3.81 t/m) = 746 W. (b) P = (100 W)/(746 W/hp) = hp. Page 6 17

18 61. (a) 1 kwh = (1000 Wh)(3600 s/h) = J. (b) W = Pt = (500 W)(1 kw/1000w)(1 mo)(30 day/mo)(4 h/day) = 360 kwh. (c) W = (360 kwh)( J/kWh) = J. (d) Cost = W rate = (360 kwh)($0.1/kwh) = $43.0. The charge is or the amount o energy used, and thus is independent o rate. 6. We ind the average resistance orce rom the acceleration: R = ma = m Æv/Æt = (1000 kg)(70 km/h 90 km/h)/(3.6 ks/h)(6.0 s) = 933 N. I we assume that this is the resistance orce at 80 km/h, the engine must provide an equal and opposite orce to maintain a constant speed. We ind the power required rom P = Fv = (933 N)(80 km/h)/(3.6 ks/h) = W = ( W)/(746 W/hp) = 8 hp. 63. We ind the work rom W = Pt = (3.0 hp)(746 W/hp)(1 h)(3600 s/h) = J. 64. The work done by the shot-putter increases the kinetic energy o the shot. We ind the power rom P = W/t = ÆKE/t = ( i )/t =!(7.3 kg)[(14 m/s) 0]/(.0 s) = W (about 0.5 hp). 65. The work done by the pump increases the potential energy o the water. We ind the power rom P = W/t = ÆPE/t = (h h i )/t = (m/t)g(h h i ) = [(8.00 kg/min)/(60 s/min)](9.80 m/s )(3.50 m 0) = 4.57 W. 66. The work done increases the potential energy o the player. We ind the power rom P = W/t = ÆPE/t = (h h i )/t = (105 kg)(9.80 m/s )[(140 m) sin 30 0]/(61 s) = W (about 1.6 hp). 67. The work done increases the potential energy o the player. We ind the speed rom P = W/t = ÆPE/t = (h h i )/t = (L sin θ 0)/t = v sin θ (0.5 hp)(746 W/hp) = (70 kg)(9.80 m/s )v sin 6.0, which gives v =.6 m/s. 68. From the orce diagram or the car, we have: x-component: F F r = sin θ. Because the power output is P = Fv, we have (P/v) F r = sin θ. The maximum power determines the maximum angle: (P max /v) F r = sin θ max ; (10 hp)(746 W/hp)/[(70 km/h)/(3/6 ks/h)] 600 N = (1000 kg)(9.80 m/s ) sin θ max, which gives sin θ max = 0.409, or θ max = 4. r F N θ F d y x Page 6 18

19 69. The work done by the lits increases the potential energy o the people. We assume an average mass o 70 kg and ind the power rom P = W/t = ÆPE/t = (h h i )/t = (m/t)g(h h i ) = [(47,000 people/h)(70 kg/person)/(3600 s/h)](9.80 m/s )(00 m 0) = W (about hp). 70. For the work-energy principle applied to coasting down the hill a distance L, we have W NC = ÆKE + ÆPE = ( i ) + (h h i ); F r L = ( ) + (0 L sin θ), which gives F r = sin θ. Because the climb is at the same speed, we assume the resisting orce is the same. For the work-energy principle applied to climbing the hill a distance L, we have W NC = ÆKE + ÆPE = ( i ) + (h h i ); FL F r L = ( ) + (0 L sin θ); (P/v) sin θ = sin θ, which gives P = v sin θ = (75 kg)(9.80 m/s )(5.0 m/s) sin 7.0 = W (about 1. hp). 71. (a) I we ignore the small change in potential energy when the snow brings the paratrooper to rest, the work done decreases the kinetic energy: W = ÆKE = i =!(80 kg)[0 (30 m/s) ] = J. (b) We ind the average orce rom F = W/d = ( J)/(1.1 m) = N. (c) With air resistance during the all we have W NC = ÆKE + ÆPE = ( i ) + (h h i ) =!(80 kg)[(30 m/s) 0] + (80 kg)(9.80 m/s )(0 370 m) = J. 7. For the motion during the impact until the car comes momentarily to rest, we use energy conservation: KE i + PE springi = KE + PE spring ; i + 0 = 0 +!kx ; (1400 kg kg)[(8 km/h)/(3/6 ks/h)] = k(0.015 m), which gives k = N/m. 73. We let N represent the number o books o mass m that can be placed on a shel. For each book the work increases the potential energy and thus is times the distance the center is raised. From the diagram we see that the work required to ill the nth shel is W n = N[D +!h + (n 1)H]. Thus or the ive shelves, we have W = W 1 + W + W 3 + W 4 + W 5 = N[5(D +!h) + H + H + 3H + 4H] = N[5(D +!h) + 10H] = (5)(1.5 kg)(9.80 m/s ){5[0.100 m +!(0.0 m)] + 10(0.300 m)} = J. H H D +y h y = 0 Page 6 19

20 74. We choose the potential energy to be zero at the ground (y = 0). We ind the minimum speed by ignoring any rictional orces. Energy is conserved, so we have E = KE i + PE i = KE + PE ; i + y i = + y ; i + m(9.80 m/s )(0) =!m(6.5 m/s) + m(9.80 m/s )(1.1 m), which gives v i = 8.0 m/s. Note that the initial velocity will not be horizontal, but will have a horizontal component o 6.5 m/s. 75. We choose the reerence level or the gravitational potential energy at the ground. (a) With no air resistance during the all we have 0 = ÆKE + ÆPE = ( i ) + (h h i ), or!(v 0) = (9.80 m/s )(0 18 m), which gives v = 19 m/s. (b) With air resistance during the all we have W NC = ÆKE + ÆPE = ( i ) + (h h i ); F air (18 m) =!(0.0 kg)[(10.0 m/s) 0] + (0.0 kg)(9.80 m/s )(0 18 m), which gives F air = 1.4 N. 76. We choose the reerence level or the gravitational potential energy at the lowest point. The tension in the cord is always perpendicular to the displacement and thus does no work. (a) With no air resistance during the all, we have 0 = ÆKE + ÆPE = ( 1 0 ) + (h 1 h 0 ), or!(v 1 0) = g(0 L), which gives v 1 = (gl) 1/. (b) For the motion rom release to the rise around the peg, we have 0 = ÆKE + ÆPE = ( 0 ) + (h h 0 ), or!(v 0) = g[(l h) L] = g(h L) = 0.60gL, which gives v = (1.gL) 1/. y = 0 L h F T y peg 77. (a) The work done against gravity is the increase in the potential energy: W = ÆPE = (h h i ) = (65 kg)(9.80 m/s )(3900 m 00 m) = J. (b) We ind the power rom P = W/t = ( J)/(5.0 h)(3600 s/h) = 60 W = hp. (c) We ind the power input rom P input = P/eiciency = (60 W)/(0.15) = W = 0.54 hp. 78. The potential energy is zero at x = 0. (a) Because energy is conserved, the maximum speed occurs at the minimum potential energy: KE i + PE i = KE + PE ; 0 +!kx 0 = max + 0, which gives v max = [v 0 + (kx 0 /m)] 1/. (b) The maximum stretch occurs at the minimum kinetic energy: KE i + PE i = KE + PE ; 0 +!kx 0 = 0 +!kx max, which gives x max = [x 0 + (mv 0 /k)] 1/. Page 6 0

21 79. (a) With y = 0 at the bottom o the circle, we call the start point A, the bottom o the circle B, and the top o the circle C. A At the top o the circle we have the orces and F Ntop, both F Ntop downward, that provide the centripetal acceleration: + F Ntop = mv C /r. C The minimum value o F Ntop is zero, so the minimum speed H at C is ound rom v C min = gr. From energy conservation or the motion rom A to C we have r KE A + PE A = KE C + PE C ; 0 + h = C + (r), B thus the minimum height is ound rom gh =!v C min + gr =!gr + gr, which gives h =.5r. (b) From energy conservation or the motion rom A to B we have KE A + PE A = KE B + PE B ; 0 + h = 5r = B + 0, which gives v B = 10gr. At the bottom o the circle we have the orces down and F Nbottom up that provide the centripetal acceleration: + F Nbottom = mv B /r. I we use the previous result, we get F Nbottom = mv B /r + = 11. (c) From energy conservation or the motion rom A to C we have KE A + PE A = KE C + PE C ; 0 + h = 5r = C + (r), which gives v C = 6gr. At the top o the circle we have the orces and F Ntop, both down, that provide the centripetal acceleration: + F Ntop = mv C /r. I we use the previous result, we get F Ntop = mv C /r = 5. (d) On the horizontal section we have F N =. F Nbottom 80. (a) The work done by gravity is the decrease in the potential energy: W grav = ÆPE = (h h i ) = (900 kg)(9.80 m/s )(0 30 m) = J. (b) The work done by gravity increases the kinetic energy: W grav = ÆKE; J =!(900 kg)v 0, which gives v = 4 m/s. (c) For the motion rom the break point to the maximum compression o the spring, we use energy conservation: KE i + PE gravi + PE springi = KE + PE grav + PE spring ; 0 + h + 0 = 0 + ( x max ) +!kx max ; (900 kg)(9.80 m/s )(30 m) = (900 kg)(9.80 m/s )x max +!( N/m)x max. This is a quadratic equation or x max, which has the solutions x max = 1.13 m, 1.17 m. Because x max must be positive, the spring compresses 1. m. h v = 0 y = 0 +y x Page 6 1

22 81. We choose the reerence level or the gravitational potential energy at the lowest point. (a) With no air resistance during the all, we have 0 = ÆKE + ÆPE = ( 0 ) + (h h 0 ), or!(v 0) = g(0 H), which gives v 1 = (gh) 1/ = [(9.80 m/s )(80 m)] = 40 m/s. (b) I 60% o the kinetic energy o the water is transerred, we have P = (0.60) /t = (0.60)!(m/t)v = (0.60)!(550 kg/s)(40 m/s) ) = W (about hp). 8. We convert the speeds: (10 km/h)/(3.6 ks/h) =.78 m/s; (30 km/h)/(3.6 ks/h) = 8.33 m/s. We use the work-energy principle applied to coasting down the hill a distance L to ind b: W NC = ÆKE + ÆPE = ( i ) + (h h i ); bv 1 L = ( 1 1 ) + (0 L sin θ), which gives b = ( /v ) sin θ = [(75 kg)(9.80 m/s )/(.78 m/s) ] sin 4.0 = 6.63 kg/m. For the work-energy principle applied to speeding down the hill a distance L, the cyclist must provide a orce so we have W NC = ÆKE + ÆPE = ( i ) + (h h i ); F L bv L = ( ) + (0 L sin θ), which gives F = bv sin θ. The power supplied by the cyclist is P = F v = [(6.63 kg/m)(8.33 m/s) (75 kg)(9.80 m/s ) sin 4.0 ](8.33 m/s) = W. For the work-energy principle applied to climbing the hill a distance L, the cyclist will provide a orce F 3 = P/v 3, so we have W NC = ÆKE + ÆPE = ( i ) + (h h i ); (P/v 3 )L bv 3 L = ( 3 3 ) + (L sin θ 0), which gives [( W)/v 3 ] (6.63 kg/m)v 3 = (75 kg)(9.80 m/s ) sin 4.0. This is a cubic equation or v 3, which has one real solution: v 3 = 5.54 m/s. The speed is (5.54 m/s)(3.6 ks/h) = 0 km/h. 83. We choose the reerence level or the gravitational potential energy at the bottom. From energy conservation or the motion v top rom top to bottom, we have KE top + PE top = KE bottom + PE bottom ; top + r = bottom + 0, which gives F Ntop v bottom = v top + 4gr. At the bottom o the circle we have the orces down and F Nbottom up that provide the centripetal acceleration: R + F Nbottom = mv bottom /r, which gives F Nbottom = (mv bottom /r) +. At the top o the circle we have the orces and F Ntop, both down, that provide the centripetal acceleration: + F Ntop = mv top /r, which gives F Nbottom F Ntop = (mv top /r). I we subtract the two equations, we get F Nbottom F Ntop = (mv bottom /r) + [(mv top /r) ] = (m/r)(v bottom v top ) + = 4 + = 6. v bottom The speed must be above the minimum at the top so the roller coaster does not leave the track. From Problem 44, we know that we must have h >.5r. Page 6

23 The result we ound does not depend on the radius or speed. Page 6 3

24 84. We choose y = 0 at the scale. We ind the spring constant rom the orce (your weight) required to compress the spring: k = F 1 /x 1 = ( 700 N)/( m) = N/m. We apply conservation o energy or the jump to the scale. I we ignore the small change in gravitational potential energy when the scale compresses, we have KE i + PE i = KE + PE ; 0 + H = 0 +!kx ; (700 N)(1.0 m) =!( N/m)x, which gives x = 0.03 m. The reading o the scale is F = kx = ( N/m)(0.03 m) = N. 85. We choose the potential energy to be zero at the lowest point (y = 0). (a) Because the tension in the vine does no work, energy is conserved, so we have KE i + PE i = KE + PE ; θ = 0 + h = (L L cos θ) = L(1 cos θ); L F T!m(5.0 m/s) = m(9.80 m/s )(10.0 m)(1 cos θ) which gives cos θ = 0.87, or θ = 9. (b) The velocity is zero just beore he releases, so there is no centripetal h y = 0 acceleration. There is a tangential acceleration which has been decreasing his tangential velocity. For the radial direction we have F T cos θ = 0; or v 0 F T = cos θ = (75 kg)(9.80 m/s )(0.87) = N. (c) The velocity and thus the centripetal acceleration is maximum at the bottom, so the tension will be maximum there. For the radial direction we have F T = mv 0 /L, or F T = + mv 0 /L = (75 kg)[(9.80 m/s ) + (5.0 m/s) /(10.0 m)] = N. 86. We choose the potential energy to be zero at the loor. The work done increases the potential energy o the athlete. We ind the power rom P = W/t = ÆPE/t = (h h i )/t = (70 kg)(9.80 m/s )(5.0 m 0)/(9.0 s) = W (about 0.5 hp). Page 6 4