Prepare-se para as provas

Estude fácil! Tem muito documento disponível na Docsity

Ganhe pontos para baixar

Ganhe pontos ajudando outros esrudantes ou compre um plano Premium

Guias e Dicas

Prepare-se para as provas

Estude fácil! Tem muito documento disponível na Docsity

Encontrar documentos

Prepare-se para as provas com trabalhos de outros alunos como você, aqui na Docsity

Pesquisar documentos Store

Os melhores documentos à venda: Trabalhos de alunos formados

Videoaulas

Prepare-se com as videoaulas e exercícios resolvidos criados a partir da grade da sua Universidade

QuizNEW

Responda perguntas de provas passadas e avalie sua preparação.

Ganhe pontos para baixar

Ganhe pontos ajudando outros esrudantes ou compre um plano Premium

Comunidade

Pergunte à comunidade

Peça ajuda à comunidade e tire suas dúvidas relacionadas ao estudo

Ranking universidades

Descubra as melhores universidades em seu país de acordo com os usuários da Docsity

Guias grátis

Os eBooks que salvam estudantes!

Baixe gratuitamente nossos guias de estudo, métodos para diminuir a ansiedade, dicas de TCC preparadas pelos professores da Docsity

RESOLUÇÃO HALLIDAY, 9ªEDIÇÃO, VOLUME 1, CAPITULO 07

Tipologia: Exercícios

2019

1 / 80

Baixe RESOLUÇÃO HALLIDAY, 9ªEDIÇÃO, VOLUME 1, CAPITULO 07 e outras Exercícios em PDF para Física, somente na Docsity! 1. (a) The change in kinetic energy for the meteorite would be ( )( )22 6 3 141 1 4 10 kg 15 10 m/s 5 10 J2 2f i i i iK K K K m vΔ = − = − = − = − × × = − × , or 14| | 5 10 JKΔ = × . The negative sign indicates that kinetic energy is lost. (b) The energy loss in units of megatons of TNT would be ( )14 151 megaton TNT5 10 J 0.1megaton TNT.4.2 10 JK−Δ = × =× (c) The number of bombs N that the meteorite impact would correspond to is found by noting that megaton = 1000 kilotons and setting up the ratio: 0.1 1000kiloton TNT 8. 13kiloton TNT N ×= = 2. With speed v = 11200 m/s, we find 2 5 2 131 1 (2.9 10 kg) (11200 m/s) 1.8 10 J. 2 2 K mv= = × = × 5. We denote the mass of the father as m and his initial speed vi. The initial kinetic energy of the father is K Ki = 1 2 son and his final kinetic energy (when his speed is vf = vi + 1.0 m/s) is K Kf = son . We use these relations along with Eq. 7-1 in our solution. (a) We see from the above that K Ki f= 12 which (with SI units understood) leads to ( )221 1 1 1.0 m/s 2 2 2i i mv m v= + . The mass cancels and we find a second-degree equation for vi : 1 2 1 2 02v vi i− − = . The positive root (from the quadratic formula) yields vi = 2.4 m/s. (b) From the first relation above K Ki = 12 sonb g , we have 2 2 son 1 1 1 ( /2) 2 2 2i mv m v= and (after canceling m and one factor of 1/2) are led to v vison = 2 = 4.8 m s. 6. We apply the equation 210 0 2( )x t x v t at= + + , found in Table 2-1. Since at t = 0 s, x0 = 0 and 0 12 m/sv = , the equation becomes (in unit of meters) 21 2( ) 12x t t at= + . With 10 mx = when 1.0 st = , the acceleration is found to be 24.0 m/sa = − . The fact that 0a < implies that the bead is decelerating. Thus, the position is described by 2( ) 12 2.0x t t t= − . Differentiating x with respect to t then yields ( ) 12 4.0dxv t t dt = = − . Indeed at t =3.0 s, ( 3.0) 0v t = = and the bead stops momentarily. The speed at 10 st = is ( 10) 28 m/sv t = = − , and the corresponding kinetic energy is 2 2 21 1 (1.8 10 kg)( 28 m/s) 7.1 J. 2 2 K mv −= = × − = 7. By the work-kinetic energy theorem, ( )2 2 2 21 1 1 (2.0kg) (6.0m/s) (4.0m/s) 20 J.2 2 2f iW K mv mv= Δ = − = − = We note that the directions of v f and vi play no role in the calculation. 10. Using Eq. 7-8 (and Eq. 3-23), we find the work done by the water on the ice block: 3 ˆ ˆ ˆ ˆ(210 N) i (150 N) j (15 m) i (12 m) j (210 N)(15 m) ( 150 N)( 12 m) 5.0 10 J. W F d= ⋅ = − ⋅ − = + − − = × 11. We choose +x as the direction of motion (so a and F are negative-valued). (a) Newton’s second law readily yields 2(85kg) ( 2.0m/s )F = − so that 2| | 1.7 10 NF F= = × . (b) From Eq. 2-16 (with v = 0) we have ( ) ( ) 2 2 2 0 2 37 m/s 0 2 3.4 10 m 2 2.0 m/s v a x x= + Δ Δ = − = × − . Alternatively, this can be worked using the work-energy theorem. (c) Since F is opposite to the direction of motion (so the angle φ between F and d x= Δ is 180°) then Eq. 7-7 gives the work done as 45.8 10 JW F x= − Δ = − × . (d) In this case, Newton’s second law yields ( ) ( )285kg 4.0m/sF = − so that 2| | 3.4 10 NF F= = × . (e) From Eq. 2-16, we now have ( ) ( ) 2 2 2 37 m/s 1.7 10 m. 2 4.0m/s xΔ = − = × − (f) The force F is again opposite to the direction of motion (so the angle φ is again 180°) so that Eq. 7-7 leads to 45.8 10 J.W F x= − Δ = − × The fact that this agrees with the result of part (c) provides insight into the concept of work. 2 21 1( ) (2 ) 2 2f i K m v v m a x ma xΔ = − = Δ = Δ where we have used 2 2 2f iv v a x= + Δ from Table 2-1. From Fig. 7-27, we see that (0 30) J 30 JKΔ = − = − when 5 mxΔ = + . The acceleration can then be obtained as 2( 30 J) 0.75 m/s . (8.0 kg)(5.0 m) Ka m x Δ −= = = − Δ The negative sign indicates that the mass is decelerating. From the figure, we also see that when 5 mx = the kinetic energy becomes zero, implying that the mass comes to rest momentarily. Thus, 2 2 2 2 2 0 2 0 2( 0.75 m/s )(5.0 m) 7.5 m /sv v a x= − Δ = − − = , or 0 2.7 m/sv = . The speed of the object when x = −3.0 m is 2 2 2 2 0 2 7.5 m /s 2( 0.75 m/s )( 3.0 m) 12 m/s 3.5 m/sv v a x= + Δ = + − − = = . 12. The change in kinetic energy can be written as 15. Using the work-kinetic energy theorem, we have cosK W F d Fd φΔ = = ⋅ = In addition, 12 NF = and 2 2 2(2.00 m) ( 4.00 m) (3.00 m) 5.39 md = + − + = . (a) If 30.0 JKΔ = + , then 1 1 30.0 Jcos cos 62.3 (12.0 N)(5.39 m) K Fd φ − −Δ= = = ° . (b) 30.0 JKΔ = − , then 1 1 30.0 Jcos cos 118 (12.0 N)(5.39 m) K Fd φ − −Δ −= = = ° 16. The forces are all constant, so the total work done by them is given by W F x= netΔ , where Fnet is the magnitude of the net force and Δx is the magnitude of the displacement. We add the three vectors, finding the x and y components of the net force: net 1 2 3 net 2 3 sin 50.0 cos35.0 3.00 N (4.00 N)sin 35.0 (10.0 N)cos35.0 2.13N cos50.0 sin 35.0 (4.00 N) cos50.0 (10.0 N)sin 35.0 3.17 N. x y F F F F F F F = − − ° + ° = − − ° + ° = = − ° + ° = − ° + ° = The magnitude of the net force is 2 2 2 2 net net net (2.13 N) (3.17 N) 3.82 N.x yF F F= + = + = The work done by the net force is net (3.82 N) (4.00m) 15.3 JW F d= = = where we have used the fact that d Fnet|| (which follows from the fact that the canister started from rest and moved horizontally under the action of horizontal forces — the resultant effect of which is expressed by Fnet ). = mg/10, so F = 11 mg/10. Since the force F and the displacement d are in the same direction, the work done by F is 2 411 11 (72 kg)(9.8 m/s )(15 m) 1.164 10 J 10 10F mgdW Fd= = = = × which (with respect to significant figures) should be quoted as 1.2 × 104 J. (b) The force of gravity has magnitude mg and is opposite in direction to the displacement. Thus, using Eq. 7-7, the work done by gravity is 2 4 (72 kg)(9.8 m/s )(15 m) 1.058 10 JgW mgd= − = − = − × which should be quoted as – 1.1 × 104 J. (c) The total work done is W = × − × = ×1164. 10 J 1.058 10 J 1.06 10 J4 4 3 . Since the astronaut started from rest, the work-kinetic energy theorem tells us that this (which we round to 1 1. ×10 J3 ) is her final kinetic energy. (d) Since K mv= 12 2 , her final speed is v K m = = × =2 2 106 10 5 4 3( . .J) 72 kg m / s. 17. (a) We use F to denote the upward force exerted by the cable on the astronaut. The force of the cable is upward and the force of gravity is mg downward. Furthermore, the acceleration of the astronaut is g/10 upward. According to Newton’s second law, F – mg 20. (a) Using notation common to many vector capable calculators, we have (from Eq. 7- 8) W = dot([20.0,0] + [0, −(3.00)(9.8)], [0.500 ∠ 30.0º]) = +1.31 J. (b) Eq. 7-10 (along with Eq. 7-1) then leads to v = 2(1.31 J)/(3.00 kg) = 0.935 m/s. 21. The fact that the applied force aF causes the box to move up a frictionless ramp at a constant speed implies that there is no net change in the kinetic energy: 0KΔ = . Thus, the work done by aF must be equal to the negative work done by gravity: a gW W= − . Since the box is displaced vertically upward by 0.150 mh = , we have 2(3.00 kg)(9.80 m/s )(0.150 m) 4.41 JaW mgh= + = = 2 2 0 2 2(40.0 J) 5.00 kg. (4.00 m/s) sKm v = = = Thus, the normal force is 2 2 2 2 2 2( ) (5.0 kg) (9.8 m/s ) (20 N) 44.7 N 45 N.y xF mg F= − = − = ≈ 22. From the figure, one may write the kinetic energy (in units of J) as a function of x as 20 40 20sK K x x= − = − Since xW K F x= Δ = ⋅ Δ , the component of the force along the force along +x is / 20 N.xF dK dx= = − The normal force on the block is N yF F= , which is related to the gravitational force by 2 2( )x ymg F F= + − . (Note that NF points in the opposite direction of the component of the gravitational force.) With an initial kinetic energy 40.0 JsK = and 0 4.00 m/sv = , the mass of the block is 2 23.00 N (0.250 kg)(9.80 m/s ) 2.20 m/s 0.250 kgN F mg ma a −− = = = . Thus the force from the cable is 4( )( ) 1.08 10 NNF m M a g F= + + − = × , and the work done by the cable on the cab is 4 4 1 (1.80 10 N)(2.40 m) 2.59 10 J.W Fd= = × = × (b) If 92.61 kJW = and 2 10.5 md = , the magnitude of the normal force is 4 2 2 9.261 10 J( ) (0.250 kg 900 kg)(9.80 m/s ) 2.45 N. 10.5 mN WF m M g d ×= + − = + − = 25. (a) The net upward force is given by ( ) ( )NF F m M g m M a+ − + = + where m = 0.250 kg is the mass of the cheese, M = 900 kg is the mass of the elevator cab, F is the force from the cable, and 3.00 NNF = is the normal force on the cheese. On the cheese alone, we have 26. The spring constant is k = 100 N/m and the maximum elongation is xi = 5.00 m. Using Eq. 7-25 with xf = 0, the work is found to be 2 2 31 1 (100 N/m)(5.00 m) 1.25 10 J. 2 2i W kx= = = × 3 2 21 (9.0 10 N/m)[(0.050 m) ( 0.050 m) ] 0 J. 2s W = × − − = (d) Moving from 5.0 cmix = + to 9.0 cmx = − , we have 3 2 21 (9.0 10 N/m)[(0.050 m) ( 0.090 m) ] 25 J. 2s W = × − − = − 27. From Eq. 7-25, we see that the work done by the spring force is given by 2 21 ( ) 2s i f W k x x= − . The fact that 360 N of force must be applied to pull the block to x = + 4.0 cm implies that the spring constant is 3360 N 90 N/cm 9.0 10 N/m 4.0 cm k = = = × . (a) When the block moves from 5.0 cmix = + to 3.0 cmx = + , we have 3 2 21 (9.0 10 N/m)[(0.050 m) (0.030 m) ] 7.2 J. 2s W = × − = (b) Moving from 5.0 cmix = + to 3.0 cmx = − , we have 3 2 21 (9.0 10 N/m)[(0.050 m) ( 0.030 m) ] 7.2 J. 2s W = × − − = (c) Moving from 5.0 cmix = + to 5.0 cmx = − , we have 30. The work done by the spring force is given by Eq. 7-25: 2 21 ( ) 2s i f W k x x= − . Since xF kx= − , the slope in Fig. 7-36 corresponds to the spring constant k. Its value is given by 380 N/cm=8.0 10 N/mk = × . (a) When the block moves from 8.0 cmix = + to 5.0 cmx = + , we have 3 2 21 (8.0 10 N/m)[(0.080 m) (0.050 m) ] 15.6 J 16 J. 2s W = × − = ≈ (b) Moving from 8.0 cmix = + to 5.0 cmx = − , we have 3 2 21 (8.0 10 N/m)[(0.080 m) ( 0.050 m) ] 15.6 J 16 J. 2s W = × − − = ≈ (c) Moving from 8.0 cmix = + to 8.0 cmx = − , we have 3 2 21 (8.0 10 N/m)[(0.080 m) ( 0.080 m) ] 0 J. 2s W = × − − = (d) Moving from 8.0 cmix = + to 10.0 cmx = − , we have 3 2 21 (8.0 10 N/m)[(0.080 m) ( 0.10 m) ] 14.4 J 14 J. 2s W = × − − = − ≈ − 3 2 21 (2.0 10 N/m)[(0.050 m) (0.040 m) ] 0.90 J. 2s W = × − = (b) Moving from 5.0 cmix = + to 2.0 cmx = − , we have 3 2 21 (2.0 10 N/m)[(0.050 m) ( 0.020 m) ] 2.1 J. 2s W = × − − = (c) Moving from 5.0 cmix = + to 5.0 cmx = − , we have 3 2 21 (2.0 10 N/m)[(0.050 m) ( 0.050 m) ] 0 J. 2s W = × − − = 31. The work done by the spring force is given by Eq. 7-25: 2 21 ( ) 2s i f W k x x= − . The spring constant k can be deduced from Fig. 7-37 which shows the amount of work done to pull the block from 0 to x = 3.0 cm. The parabola 2 / 2aW kx= contains (0,0), (2.0 cm, 0.40 J) and (3.0 cm, 0.90 J). Thus, we may infer from the data that 32.0 10 N/mk = × . (a) When the block moves from 5.0 cmix = + to 4.0 cmx = + , we have 32. Hooke’s law and the work done by a spring is discussed in the chapter. We apply work-kinetic energy theorem, in the form of ΔK W Wa s= + , to the points in Figure 7-38 at x = 1.0 m and x = 2.0 m, respectively. The “applied” work Wa is that due to the constant force P . 2 2 14 J (1.0 m) (1.0 m) 2 10 (2.0 m) (2.0 m) 2 P k P k = − = − (a) Simultaneous solution leads to P = 8.0 N. (b) Similarly, we find k = 8.0 N/m. 35. (a) The graph shows F as a function of x assuming x0 is positive. The work is negative as the object moves from x x x= =0 0 to and positive as it moves from x x x x= =0 02 to . Since the area of a triangle is (base)(altitude)/2, the work done from x x x= =0 0 to is 0 0( )( ) / 2x F− and the work done from x x x x= =0 02 to is 0 0 0 0 0(2 )( ) / 2 ( )( ) / 2x x F x F− = The total work is the sum, which is zero. (b) The integral for the work is 0 0 2 22 0 00 0 0 0 1 0. 2 x x x xW F dx F x x x = − = − = 36. According to the graph the acceleration a varies linearly with the coordinate x. We may write a = αx, where α is the slope of the graph. Numerically, α = = −20 8 0 2 5 2 m / s m s 2 . . . The force on the brick is in the positive x direction and, according to Newton’s second law, its magnitude is given by .F ma m xα= = If xf is the final coordinate, the work done by the force is 2 2 2 2 0 0 (10 kg)(2.5 s ) (8.0 m) 8.0 10 J. 2 2 f fx x f mW F dx m x dx xαα − = = = = = × v dx dt t t= = − +30 8 0 30 2. . . in SI units. Thus, the initial speed is vi = 3.0 m/s and the speed at t = 4 s is vf = 19 m/s. The change in kinetic energy for the object of mass m = 3.0 kg is therefore ( )2 21 528 J2 f iK m v vΔ = − = which we round off to two figures and (using the work-kinetic energy theorem) conclude that the work done is 25.3 10 J.W = × 37. We choose to work this using Eq. 7-10 (the work-kinetic energy theorem). To find the initial and final kinetic energies, we need the speeds, so 2.0 2 3 0 1(2.5 ) 0 (2.5)(2.0) (2.0) 2.3 J. 3f i K K x dx= + − = + − = (b) For a variable end-point, we have Kf as a function of x, which could be differentiated to find the extremum value, but we recognize that this is equivalent to solving F = 0 for x: 20 2.5 0F x= − = . Thus, K is extremized at 2.5 1.6 mx = ≈ and we obtain 2.5 2 3 0 1(2.5 ) 0 (2.5)( 2.5) ( 2.5) 2.6 J. 3f i K K x dx= + − = + − = Recalling our answer for part (a), it is clear that this extreme value is a maximum. 40. (a) Using the work-kinetic energy theorem 41. As the body moves along the x axis from xi = 0 m to xf = 3.00 m the work done by the force is 3 2 2 3 2 3 0 ( 3.00 ) (3.00) (3.00) 2 2 4.50 27.0. f f i i x x xx x c cW F dx cx x dx x x c = = − = − = − = − However, (11.0 20.0) 9.00 JW K= Δ = − = − from the work-kinetic energy theorem. Thus, 4.50 27.0 9.00c − = − or 4.00 N/mc = . 42. We solve the problem using the work-kinetic energy theorem which states that the change in kinetic energy is equal to the work done by the applied force, K WΔ = . In our problem, the work done is W Fd= , where F is the tension in the cord and d is the length of the cord pulled as the cart slides from x1 to x2. From Fig. 7-42, we have 2 2 2 2 2 2 2 2 1 2 (3.00 m) (1.20 m) (1.00 m) (1.20 m) 3.23 m 1.56 m 1.67 m d x h x h= + − + = + − + = − = which yields (25.0 N)(1.67 m) 41.7 J.K FdΔ = = = 45. (a) The power is given by P = Fv and the work done by F from time t1 to time t2 is given by W P t Fv t t t t t d d= = zz . 1 2 1 2 Since F is the net force, the magnitude of the acceleration is a = F/m, and, since the initial velocity is v0 0= , the velocity as a function of time is given by v v at F m t= + =0 ( ) . Thus 2 1 2 2 2 2 2 1 1( / ) d ( / )( ). 2 t t W F m t t F m t t= = − For t1 0= and 2 1.0s,t = 2 21 (5.0 N) (1.0 s) = 0.83 J. 2 15 kg W = (b) For 1 1.0s,t = and 2 2.0s,t = 2 2 21 (5.0 N) [(2.0 s) (1.0 s) ] 2.5 J. 2 15 kg W = − = (c) For 1 2.0st = and 2 3.0s,t = 2 2 21 (5.0 N) [(3.0 s) (2.0 s) ] 4.2 J. 2 15 kg W = − = (d) Substituting v = (F/m)t into P = Fv we obtain P F t m= 2 for the power at any time t. At the end of the third second P (5.0 N) (3.0 s) 15 kg 5.0 W. 2 = F HG I KJ = 46. (a) Since constant speed implies ΔK 0,= we require W Wa g= − , by Eq. 7-15. Since Wg is the same in both cases (same weight and same path), then 29.0 10aW = × J just as it was in the first case. (b) Since the speed of 1.0 m/s is constant, then 8.0 meters is traveled in 8.0 seconds. Using Eq. 7-42, and noting that average power is the power when the work is being done at a steady rate, we have 2900 J 1.1 10 W. 8.0 s WP t = = = × Δ (c) Since the speed of 2.0 m/s is constant, 8.0 meters is traveled in 4.0 seconds. Using Eq. 7-42, with average power replaced by power, we have 900 J 4.0 s WP t = = Δ = 225 W 22.3 10 W≈ × . 47. The total work is the sum of the work done by gravity on the elevator, the work done by gravity on the counterweight, and the work done by the motor on the system: WT = We + Wc + Ws. Since the elevator moves at constant velocity, its kinetic energy does not change and according to the work-kinetic energy theorem the total work done is zero. This means We + Wc + Ws = 0. The elevator moves upward through 54 m, so the work done by gravity on it is 2 5(1200 kg)(9.80 m/s )(54 m) 6.35 10 J.e eW m gd= − = − = − × The counterweight moves downward the same distance, so the work done by gravity on it is 2 5(950 kg)(9.80 m/s )(54 m) 5.03 10 J.c cW m gd= = = × Since WT = 0, the work done by the motor on the system is 5 5 56.35 10 J 5.03 10 J 1.32 10 J.s e cW W W= − − = × − × = × This work is done in a time interval of Δt 3.0 min 180 s,= = so the power supplied by the motor to lift the elevator is 5 21.32 10 J 7.4 10 W. 180 s sWP t ×= = = × Δ 50. (a) Since the force exerted by the spring on the mass is zero when the mass passes through the equilibrium position of the spring, the rate at which the spring is doing work on the mass at this instant is also zero. (b) The rate is given by P F v Fv = ⋅ = − , where the minus sign corresponds to the fact that F and v are anti-parallel to each other. The magnitude of the force is given by F = kx = (500 N/m)(0.10 m) = 50 N, while v is obtained from conservation of energy for the spring-mass system: 2 2 2 21 1 1 110 J (0.30 kg) (500 N/m)(0.10 m) 2 2 2 2 E K U mv kx v= + = = + = + which gives v = 7.1 m/s. Thus, 2(50 N)(7.1 m/s) 3.5 10 W.P Fv= − = − = − × (3.00 N)( 8.00 m) (7.00 N)(6.00 m) (7.00 N)(2.00 m) 32.0 J.W F d= ⋅ = − + + = (b) The average power is given by Eq. 7-42: avg 32.0 8.00 W. 4.00 WP t = = = (c) The distance from the coordinate origin to the initial position is 2 2 2(3.00 m) ( 2.00 m) (5.00 m) 6.16 m,id = + − + = and the magnitude of the distance from the coordinate origin to the final position is 2 2 2( 5.00 m) (4.00 m) (7.00 m) 9.49 mfd = − + + = . Their scalar (dot) product is 2(3.00 m)( 5.00 m) ( 2.00 m)(4.00 m) (5.00 m)(7.00 m) 12.0 m .i fd d⋅ = − + − + = Thus, the angle between the two vectors is 1 1 12.0cos cos 78.2 . (6.16)(9.49) i f i f d d d d φ − − ⋅ = = = ° 51. (a) The object’s displacement is ˆ ˆ ˆ( 8.00 m) i (6.00 m) j (2.00 m) k .f id d d= − = − + + Thus, Eq. 7-8 gives 2 33 2 3 2 8 8 3 9 2 9 T Tmv mvP PTL P m P m = = = which implies that 3 29 constant. 8 PT mL= = Differentiating the above equation gives 3 23 0,dPT PT dT+ = or . 3 TdT dP P = − 52. According to the problem statement, the power of the car is 21 constant. 2 dW d dvP mv mv dt dt dt = = = = The condition implies /dt mvdv P= , which can be integrated to give 2 0 0 2 TT v Tmvmvdvdt T P P = = where Tv is the speed of the car at .t T= On the other hand, the total distance traveled can be written as 3 2 0 0 0 . 3 T TT v v Tmvmvdv mL vdt v v dv P P P = = = = By squaring the expression for L and substituting the expression for T, we obtain 55. One approach is to assume a “path” from ri to rf and do the line-integral accordingly. Another approach is to simply use Eq. 7-36, which we demonstrate: 4 3 2 3 (2 ) (3) f f i i x y x yx y W F dx F dy x dx dy − − = + = + with SI units understood. Thus, we obtain W = 12 J – 18 J = – 6 J. 56. (a) The force of the worker on the crate is constant, so the work it does is given by W F d FdF = ⋅ = cosφ , where F is the force, d is the displacement of the crate, and φ is the angle between the force and the displacement. Here F = 210 N, d = 3.0 m, and φ = 20°. Thus, WF = (210 N) (3.0 m) cos 20° = 590 J. (b) The force of gravity is downward, perpendicular to the displacement of the crate. The angle between this force and the displacement is 90° and cos 90° = 0, so the work done by the force of gravity is zero. (c) The normal force of the floor on the crate is also perpendicular to the displacement, so the work done by this force is also zero. (d) These are the only forces acting on the crate, so the total work done on it is 590 J. (98 N)(0.040 m) 3.9 J.W F d= ⋅ = = (d) Since the force of gravity Fg (with magnitude mg) is opposite to the displacement dc = 0 020. m (up) of the canister, Eq. 7-7 leads to (196 N) (0.020 m) 3.9 J.g cW F d= ⋅ = − = − This is consistent with Eq. 7-15 since there is no change in kinetic energy. 57. There is no acceleration, so the lifting force is equal to the weight of the object. We note that the person’s pull F is equal (in magnitude) to the tension in the cord. (a) As indicated in the hint, tension contributes twice to the lifting of the canister: 2T = mg. Since F T= , we find 98 N.F = (b) To rise 0.020 m, two segments of the cord (see Fig. 7-44) must shorten by that amount. Thus, the amount of string pulled down at the left end (this is the magnitude of d , the downward displacement of the hand) is d = 0.040 m. (c) Since (at the left end) both F and d are downward, then Eq. 7-7 leads to 60. The acceleration is constant, so we may use the equations in Table 2-1. We choose the direction of motion as +x and note that the displacement is the same as the distance traveled, in this problem. We designate the force (assumed singular) along the x direction acting on the m = 2.0 kg object as F. (a) With v0 = 0, Eq. 2-11 leads to a = v/t. And Eq. 2-17 gives Δx vt12= . Newton’s second law yields the force F = ma. Eq. 7-8, then, gives the work: 21 1 2 2 vW F x m vt mv t = Δ = = as we expect from the work-kinetic energy theorem. With v = 10 m/s, this yields 21.0 10 JW = × . (b) Instantaneous power is defined in Eq. 7-48. With t = 3.0 s, we find 67 W.vP Fv m v t = = = (c) The velocity at 1.5st′ = is v at' ' .= = 50 m s . Thus, ′ = ′ =P Fv 33 W. 61. The total weight is (100)(660 N) = 6.60 × 104 N, and the words “raises … at constant speed” imply zero acceleration, so the lift-force is equal to the total weight. Thus P = Fv = (6.60 × 104)(150 m/60.0 s) = 1.65 × 105 W. 62. (a) The force F of the incline is a combination of normal and friction force which is serving to “cancel” the tendency of the box to fall downward (due to its 19.6 N weight). Thus, F mg= upward. In this part of the problem, the angle φ between the belt and F is 80°. From Eq. 7-47, we have cos (19.6 N)(0.50 m/s) cos 80P Fv φ= = ° = 1.7 W. (b) Now the angle between the belt and F is 90°, so that P = 0. (c) In this part, the angle between the belt and F is 100°, so that P = (19.6 N)(0.50 m/s) cos 100° = –1.7 W. 65. (a) To hold the crate at equilibrium in the final situation, F must have the same magnitude as the horizontal component of the rope’s tension T sin θ , where θ is the angle between the rope (in the final position) and vertical: θ = FHG I KJ = ° −sin . . . .1 4 00 12 0 19 5 But the vertical component of the tension supports against the weight: T cos θ = mg . Thus, the tension is T = (230 kg)(9.80 m/s2)/cos 19.5° = 2391 N and F = (2391 N) sin 19.5° = 797 N. An alternative approach based on drawing a vector triangle (of forces) in the final situation provides a quick solution. (b) Since there is no change in kinetic energy, the net work on it is zero. (c) The work done by gravity is W F d mghg g= ⋅ = − , where h = L(1 – cos θ ) is the vertical component of the displacement. With L = 12.0 m, we obtain Wg = –1547 J which should be rounded to three figures: –1.55 kJ. (d) The tension vector is everywhere perpendicular to the direction of motion, so its work is zero (since cos 90° = 0). (e) The implication of the previous three parts is that the work due to F is –Wg (so the net work turns out to be zero). Thus, WF = –Wg = 1.55 kJ. (f) Since F does not have constant magnitude, we cannot expect Eq. 7-8 to apply. K K W K x x x x x f f f f − = − = − − < <3 3 12 4 3 0( )( . ) so that the requirement 8.0 JxfK = leads to x f = 4 0. m. (c) As long as the work is positive, the kinetic energy grows. The graph shows this situation to hold until x = 1.0 m. At that location, the kinetic energy is 1 0 0 1 16 J 2.0 J 18 J.xK K W < <= + = + = 66. From Eq. 7-32, we see that the “area” in the graph is equivalent to the work done. We find the area in terms of rectangular [length × width] and triangular [ 12 base × height] areas and use the work-kinetic energy theorem appropriately. The initial point is taken to be x = 0, where v0 = 4.0 m/s. (a) With K mvi = =12 0 2 16 J, we have 3 0 0 1 1 2 2 3 4.0 Jx x xK K W W W< < < < < <− = + + = − so that K3 (the kinetic energy when x = 3.0 m) is found to equal 12 J. (b) With SI units understood, we write 3 as ( 4.0 N)( 3.0 m)fx x x fW F x x< < Δ = − − and apply the work-kinetic energy theorem: 67. (a) Noting that the x component of the third force is F3x = (4.00 N)cos(60º), we apply Eq. 7-8 to the problem: W = [5.00 N – 1.00 N + (4.00 N)cos 60º](0.20 m) = 1.20 J. (b) Eq. 7-10 (along with Eq. 7-1) then yields v = 2W/m = 1.10 m/s. 70. After converting the speed to meters-per-second, we find K = 12 mv 2 = 667 kJ. 71. (a) Hooke’s law and the work done by a spring is discussed in the chapter. Taking absolute values, and writing that law in terms of differences Δ ΔF xand , we analyze the first two pictures as follows: N N mm 40 mm) | | | | ( Δ ΔF k x k = − = −240 110 60 which yields k = 6.5 N/mm. Designating the relaxed position (as read by that scale) as xo we look again at the first picture: 110 40 N mm o= −k x( ) which (upon using the above result for k) yields xo = 23 mm. (b) Using the results from part (a) to analyze that last picture, we find W k x= − =( )30 45mm N .o 72. (a) Using Eq. 7-8 and SI units, we find ˆ ˆ ˆ ˆ(2 i 4 j) (8 i j) 16 4W F d c c= ⋅ = − ⋅ + = − which, if equal zero, implies c = 16/4 = 4 m. (b) If W > 0 then 16 > 4c, which implies c < 4 m. (c) If W < 0 then 16 < 4c, which implies c > 4 m. 75. (a) The plot of the function (with SI units understood) is shown below. Estimating the area under the curve allows for a range of answers. Estimates from 11 J to 14 J are typical. (b) Evaluating the work analytically (using Eq. 7-32), we have 22 / 2 / 2 0 0 10 20 12.6 J 13 J.x xW e dx e− −= = − = ≈ (c) Replacing θ with 180º – θ, and still using vi = 1, we find vf = [1 + cos(180º – θ )]1/2 = (1 – cosθ )1/2. (d) The graphs are shown on the right. Note that as θ is increased in parts (a) and (b) the force provides less and less of a positive acceleration, whereas in part (c) the force provides less and less of a deceleration (as its θ value increases). The highest curve (which slowly decreases from 1.4 to 1) is the curve for part (b); the other decreasing curve (starting at 1 and ending at 0) is for part (a). The rising curve is for part (c); it is equal to 1 where θ = 90º. 76. (a) Eq. 7-10 (along with Eq. 7-1 and Eq. 7-7) leads to vf = (2 d m F cosθ ) 1/2= (cosθ )1/2, where we have substituted F = 2.0 N, m = 4.0 kg and d = 1.0 m. (b) With vi = 1, those same steps lead to vf = (1 + cosθ )1/2. 77. (a) We can easily fit the curve to a concave-downward parabola: x = 110 t(10 – t), from which (by taking two derivatives) we find the acceleration to be a = –0.20 m/s2. The (constant) force is therefore F = ma = –0.40 N, with a corresponding work given by W = Fx = 250 t(t – 10). It also follows from the x expression that vo = 1.0 m/s. This means that Ki = 12 mv 2 = 1.0 J. Therefore, when t = 1.0 s, Eq. 7-10 gives K = Ki + W = 0.64 J 0.6 J≈ , where the second significant figure is not to be taken too seriously. (b) At t = 5.0 s, the above method gives K = 0. (c) Evaluating the W = 250 t(t – 10) expression at t = 5.0 s and t = 1.0 s, and subtracting, yields –0.6 J. This can also be inferred from the answers for parts (a) and (b).