11.1. Visualize: r r Please refer to Figure Ex11.1.
(a) A ⋅ B = AB cos α = ( 4)(5)cos 40° = 15.3. r r C ⋅ D = CD cos α = (2)( 4)cos120° = −4.0. r r E ⋅ F = EF cos α = (3)( 4)cos 90° = 0.
Solve:
(b) (c)
11.2. Visualize: r r Please refer to Figure Ex11.2.
(a) A ⋅ B = AB cos α = (3)( 4)cos110° = −4.1. r r C ⋅ D = CD cos α = ( 4)(5)cos180° = −20. r r E ⋅ F = EF cos α = ( 4)(3)cos 30° = 10.4.
Solve:
(b) (c)
11.3. Solve: (a) (b)
r r A ⋅ B = Ax Bx + Ay By = (3)( −2) + ( −4)(6) = −30.
r r A ⋅ B = Ax Bx + Ay By = (2)(6) + (3)( −4) = 0 .
11.4. Solve: (a) (b)
r r A ⋅ B = Ax Bx + Ay By = (5)(2) + ( −3)( −2) = 16.
r r A ⋅ B = Ax Bx + Ay By = ( 4)( −3) + ( −2)(6) = −24.
r r W = F ⋅ ∆r = (6.0iˆ − 3.0 jˆ ) ⋅ (2.0iˆ ) N ⋅ m = (12.0iˆ ⋅ iˆ − 3.0 jˆ ⋅ iˆ ) J = 12.0 J. r r W = F ⋅ ∆r = (6.0iˆ − 3.0 jˆ ) ⋅ (2.0 jˆ ) N ⋅ m = (12.0iˆ ⋅ jˆ − 6.0 jˆ ⋅ jˆ ) J = −6.0 J.
11.5. Solve: (a) (b)
r r r W = F ⋅ ∆r = ( −5.0iˆ + 4.0 jˆ ) N ⋅ (3.0iˆ ) m = ( −15.0iˆ ⋅ iˆ + 12.0 jˆ ⋅ iˆ ) J = −15.0 J. r r r W = F ⋅ ∆r = ( −5.0iˆ + 4.0 jˆ ) N ⋅ ( −3.0 jˆ ) m = (15.0iˆ ⋅ jˆ − 12.0 jˆ ⋅ jˆ ) J = −12.0 J.
11.6. Solve: (a) (b)
11.7. Model: Use the work-kinetic energy theorem to find the work done on the particle. Visualize:
Solve:
From the work-kinetic energy theorem,
W = ∆K =
1 2 1 2 1 1 mv1 − mv0 = m(v12 − v02 ) = (0.020 kg)[(30 m /s) 2 − ( −30 m /s) 2 ]2 = 0 J 2 2 2 2
Assess: Negative work is done in slowing down the particle to rest, and an equal amount of positive work is done in bringing the particle to the original speed but in the opposite direction.
r
r
r
r
r
11.8. Model: Work done by a force F on a particle is defined as W = F ⋅ ∆r , where ∆r is the particle’s displacement. Visualize:
Solve:
(a) The work done by gravity is r r Wg = w ⋅ ∆r = ( − mgjˆ ) N ⋅ (2.25 − 0.75) jˆ m = −(2.0 kg)(9.8 m/s 2 )(1.50 m ) J = −29.4 J r r (b) The work done by hand is WH = Fhand on book ⋅ ∆r . As long as the book does not accelerate, r r Fhand on book = − Fearth on book = −( − mgjˆ ) = mgjˆ
⇒ WH = ( mgjˆ ) ⋅ (2.25 − 0.75) jˆ m = (2.0 kg)(9.8 m/s 2 )(1.50 m ) = 29.4 J
r
r
r
11.9. Model: Model the piano as a particle and use W = F ⋅ ∆r , where W is the work done by the force F r through the displacement ∆r . Visualize:
Solve:
r For the force w:
r For the tension T1 : r For the tension T2 :
r r r r W = F ⋅ ∆r = w ⋅ ∆r = ( w )( ∆r )cos 0° = (2500 N )(5.0 m )(1) = 12, 500 J
r r W = T1 ⋅ ∆r = (T1 )( ∆r )cos(150°) = (1830 N )(5.0 N )( −0.8660) = −7920 J
r r W = T2 ⋅ ∆r = (T2 )( ∆r )cos(135°) = (1295 N )(5.0 m )( −0.7071) = −4580 J r Assess: Note that the displacement ∆r in all the above cases is directed downwards along − jˆ.
r
r
r
11.10. Model: Model the crate as a particle and use W = F ⋅ ∆r , where W is the work done by a force F on a
r particle and ∆r is the particle’s displacement. Visualize:
Solve:
r For the force fk :
r For the tension T1 : r For the tension T 2 :
r r W = fk ⋅ ∆r = fk ( ∆r )cos(180°) = (500 N)(3.0 m)( −1) = −1500 J r r W = T1 ⋅ ∆r = (T1 )( ∆r )cos 20° = (326 N)(3.0 m)(0.9397) = 919 J
r r W = T2 ⋅ ∆r = (T2 )( ∆r )cos 30° = (223 N )(3.0 m )(0.866) = 579 J r Assess: Negative work done by the force of kinetic friction ( fk ) means that 1500 J of energy has been transferred out of the crate.
11.11. Model: Model the 2.0 kg object as a particle, and use the work-kinetic energy theorem. Visualize: Please refer to Figure Ex11.11. For each of the five intervals the velocity-versus-time graph gives the initial and final velocities. The mass of the object is 2.0 kg. Solve: According to the work-kinetic energy theorem:
1 2 1 2 1 mvf = mvi = m(vf2 − vi2 ) 2 2 2 1 vf = 2 m /s ⇒ W = (2.0 kg)[(2 m /s) 2 − (0 m /s) 2 ] = +4.0 J 2 1 vf = 0 m /s ⇒ W = (2.0 kg)[(0 m /s) 2 − (2 m /s) 2 ] = −4.0 J 2 1 vf = −2 m /s ⇒ W = (2.0 kg)[( −2 m /s) 2 − (0 m /s) 2 ] = +4.0 J 2 1 vf = −2 m /s ⇒ W = (2.0 kg)[( −2 m /s) 2 − ( −2 m /s) 2 ] = 0 J 2 1 vf = −1 m /s ⇒ W = (2.0 kg)[( −1 m /s) 2 − ( −2 m /s) 2 ] = −3.0 J 2
W = ∆K = Interval AB: vi = 0 m /s, Interval BC: vi = 2 m /s, Interval CD: vi = 0 m /s, Interval DE: vi = −2 m /s, Interval EF: vi = −2 m /s,
Assess: The work done is the same in intervals AB and CD. It is not whether v is positive or negative that counts because K ∝ v 2 . What is important is the magnitude of v and how v changes.
11.12. Model: Use the definition of work. Visualize: Please refer to Figure Ex11.12. Solve: Work is defined as the area under the force-versus-position graph: sf
W = ∫ Fs ds = area under the force curve si
Interval 0 – 1 m: W = ( 4.0 N )(1.0 m − 0.0 m ) = 4.0 J Interval 1 – 2 m: W = ( 4.0 N )(0.5 m ) + ( − 4.0 N )(0.5 m ) = 0 J Interval 2 – 3 m: W =
1 ( −4.0 N )(1 m ) = −2.0 J 2
11.13. Model: Use the work-kinetic energy theorem to find velocities. Visualize: Please refer to Figure Ex11.13. Solve: The work-kinetic energy theorem is
∆K =
1 2 1 2 mvf − mvi = W = 2 2
xf
∫ F dx = x
area under the force curve from x i to x f
xi
x
⇒
f 1 2 1 1 mvf − (0.500 kg)(2.0 m /s) 2 = mvf2 − 1.0 J = ∫ Fx dx = area from 0 to x 2 2 2 0m
1 (0.500 kg)vf2 − 1.0 J = 12.5 J ⇒ vf = 7.35 m /s 2 1 At x = 2 m: (0.500 kg)vf2 − 1.0 J = 20 J ⇒ vf = 9.17 m /s 2 1 At x = 3 m: (0.500 kg)vf2 − 1.0 J = 22.5 J ⇒ vf = 9.70 m /s 2 At x = 1 m:
11.14. Model: Use the work-kinetic energy theorem. Visualize: Please refer to Figure Ex11.14. Solve: The work-kinetc energy theorem is
∆K =
1 2 1 2 mvf − mvi = W = 2 2
xf
∫ F dx = area under the force curve from x x
i
to x f
xi
x
⇒
f 1 2 1 1 mvf − (2 kg)( 4.0 m /s) 2 = mvf2 − 16 J = ∫ Fx dx = area from 0 to x 2 2 2 0m
1 1 (2.0 kg)vf2 − 16 J = ( −10 N )(2 m ) = −10 J ⇒ vf = 2.45 m /s 2 2 1 At x = 4 m: (2.0 kg)vf2 − 16 J = 0 J ⇒ vf = 4.0 m /s 2 At x = 2 m:
11.15. Model: Use the work-kinetic energy theorem. Visualize: Please refer to Figure Ex11.15. Solve: The work-kinetic energy theorem is xf
∆K = W = ∫ Fx dx = area of the Fx -versus- x graph between x i and x f xi
1 2 1 2 1 mvf − mvi = ( Fmax )(2 m ) 2 2 2 Using m = 0.500 kg, vf = 6.0 m/s, and vi = 2.0 m/s, the above equation yields Fmax = 8.0 N. Assess: Problems in which the force is not a constant can not be solved using constant-acceleration kinematic equations.
11.16. Model: Use the definition Fs = − dU / ds. Visualize: Please refer to Figure Ex11.16. Solve: Fx is the negative of the slope of the potential energy graph at position x. Between x = 0 cm and x = 10 cm the slope is slope = (U f − U i ) / ( x f − x i ) = (0 J − 10 J ) / (0.1 m − 0.0 m ) = −100 N Thus, Fx = 100 N at x = 5 cm. The slope between x = 10 cm and x = 20 cm is zero, so Fx = 0 N at x = 15 cm. Between 20 cm and 40 cm, slope = (10 J − 0 J ) / (0.40 m − 0.20 m ) = 50 N At x = 25 cm and x = 35 cm, therefore, Fx = −50 N.
11.17. Model: Use the definition Fs = − dU ds . Visualize: Please refer to Figure Ex11.17. Solve: Fx is the negative of the slope of the potential energy graph at position x. dU Fx = − dx Between x = 0 m and x = 2 m, the slope is slope = (U f − U i ) / ( x f − x i ) = (60 J − 0 J ) / (2 m − 0 m ) = 30 N and between x = 2 m and x = 5 m, the slope is slope = (U f − U i ) / ( x f − x i ) = (0 J − 60 J ) / (5 m − 2 m ) = −20 J Thus, Fx = –30 N at x = 1 m and Fx = 20 N at x = 3 m.
11.18. Model: Use the negative derivative of the potential energy to determine the force acting on a particle. Solve:
(a)
The graph of the potential energy U = 4 y 3 is shown. (b) The y-component of the force is dU d Fy = − = − ( 4 y 3 ) = −12 y 2 dy dy At y = 0 m, Fy = 0 N; at y = 1 m, Fy = – 12 N; and at y = 2 m, Fy = − 48 N.
11.19. Model: Use the negative derivative of the potential energy to determine the force acting on a particle. Solve:
(a)
The graph of the potential energy U = 10 / x is shown. (b) The x-component of the force is
Fx = − Fx =2 m =
dU d 10 10 =− = dx dx x x 2
10 x2
= 2.5 N x =2 m
Fx =5 m =
10 x2
= 0.40 N x =5 m
Fx =8 m =
10 x2
= 0.156 N x =8 m
11.20. Model: Assume the carbon-carbon bond acts like an ideal spring that obeys Hooke’s law. Visualize:
The quantity ( x − x e ) is the stretching relative to the spring’s equilibrium length. In the present case, bond stretching is analogous to spring stretching. Solve: (a) The kinetic energy of the carbon atom is
K=
1 2 1 mv = (2.00 × 10 −26 kg)(500 m /s) 2 = 2.5 × 10 −21 J 2 2
(b) The energy of the spring is given by
1 k( x − xe )2 = K 2 N 2K 2(2.50 × 10 −21 J ) ⇒k= = = 2.0 2 ( x − xe ) (0.050 × 10 −9 m ) 2 m
Us =
11.21. Visualize: One mole of helium atoms in the gas phase contains N A = 6.02 × 10 23 atoms. Solve:
If each atom moves with the same speed v, the microscopic total kinetic energy will be
1 K micro = N A mv 2 = 3700 J ⇒ v = 2
2 K micro = mN A
2(3700 J ) = 1360 m /s (6.68 × 10 −27 kg)(6.02 × 10 23 )
11.22. Solve: (a) The car has an initial kinetic energy Ki. That energy is transformed into thermal energy of the car and the road surface. The gravitational potential energy does not change and no work is done by external forces, so during the skid Ki transforms entirely into thermal energy Eth. This energy transfer and transformation is shown on the energy bar chart. Note that Ki =
1 2 1 mv = (1500 kg)(20 m /s) 2 = 3.0 Ă— 10 5 J 2 2
(b) The change in the thermal energy of the car and the road surface is 3.0 Ă— 10 5 J .
11.23. Visualize:
1 2 mv0 = 0 J U i = U g 0 = mgy0 = (20 kg)(9.8 m /s 2 )(3 m ) = 588 J 2 1 1 Wext = 0 J K f = K1 = mv12 = (20 kg)(2.0 m /s) 2 = 40 J U f = U g1 = mgy1 = 0 J 2 2 At the top of the slide, the child has gravitational potential energy of 588 J. This energy is transformed into thermal energy of the child’s pants and the slide and the kinetic energy of the child. This energy transfer and transformation is shown on the energy bar chart. (b) Solve:
(a)
Ki = K0 =
The change in the thermal energy of the slide and of the child’s pants is 588 J – 40 J = 548 J.
11.24. Visualize: The system loses 1000 J of potential energy. In the process of losing this energy, it does 500 J of work on the environment, which means Wext = –500 J. Since the thermal energy increases 100 J, we have ∆Eth = 100 J. This is shown in the energy bar chart.
11.25. Visualize:
Note that the conservation of energy equation K i + U i + Wext = K f + U f + ∆Eth requires that Wext be equal to +400 J.
11.26. Solve: Please refer to Figure Ex11.26. The energy conservation equation yields K i + U i + Wext = K f + U f + ∆Eth ⇒ 4 J + 1 J + Wext = 1 J + 2 J + 1 J ⇒ Wext = −1 J Thus, the work done to the environment is –1 J. In other words, 1 J of energy is transferred from the system into the environment. This is shown in the energy bar chart.
11.27. Visualize: The tension of 20 N in the cable is an external force that does work on the block Wext =
(20 N)(2.0 m) = 40 J, increasing the gravitational potential energy of the block. We placed the origin of our coordinate system on the initial resting position of the block, so we have Ui = 0 J and U f = mgyf = (1.02 kg)(9.8 m /s 2 )(2.0 m ) = 20 J. Also, Ki = 0 J, and ∆Eth = 0 J. The energy bar chart shows the energy transfers and transformations.
Solve:
The conservation of energy equation is
K i + U i + Wext = K f + U f + ∆Eth ⇒ 0 J + 0 J + 40 J =
1 2 mvf + 20 J + 0 J 2
⇒ vf = (20 J )(2) / 1.02 kg = 6.26 m /s
11.28. Model: Model the elevator as a particle, and apply the conservation of energy equation. Solve: The tension in the cable does work on the elevator to lift it. Because the cable is pulled by the motor, we say that the motor does the work of lifting the elevator. (a) The energy conservation equation is K i + U i + Wext = K f + U f + ∆Eth . Using Ki = 0 J, Kf = 0 J, and ∆Eth = 0 J gives
Wext = (U f − U i ) = mg( yf − yi ) = (1000 kg)(9.8 m /s 2 )(100 m ) = 9.8 × 10 5 J (b) The power required to give the elevator this much energy in a time of 50 s is
P=
Wext 9.8 × 10 5 J = = 1.96 × 10 4 W ∆t 50 s
Assess: Since 1 horsepower (hp) is 746 W, the power of the motor is 26 hp. This is a reasonable amount of power to lift a mass of 1000 kg to a height of 100 m in 50 s.
11.29. Model: Model the steel block as a particle subject to the force of kinetic friction and use the energy conservation equation. Visualize:
r r r Solve: (a) The work done on the block is Wnet = Fnet ⋅ ∆r where ∆r is the displacement. We will find the displacement using kinematic equations and the force using Newton’s second law of motion. The displacement in the x-direction is 1 ∆x = x1 = x 0 + v0 x (t1 − t 0 ) + a x (t1 − t 0 ) 2 = 0 m + (1.0 m /s)(3.0 s − 0 s) + 0 m = 3.0 m 2 r Thus ∆r = 3.0iˆ m. The equations for Newton’s second law along the x and y components are
( Fnet ) y = n − w = 0 N ⇒ n = w = mg = (10 kg)(9.8 m /s 2 ) = 98.0 N r r ( Fnet ) x = F − fk = 0 N ⇒ F = fk = µ k n = (0.6)(98.0 N) = 58.8 N
r r ⇒ Wnet = Fnet ⋅ ∆r = F∆ x cos 0° = (58.8 N )(3.0 m )(1) = 176.4 J
(b) The power required to do this much work in 3.0 s is
P=
W 176.4 J = = 58.8 W t 3.0 s
11.30. Solve: The power of the solar collector is the solar energy collected divided by time. The intensity of the solar energy striking the earth is the power divided by area. We have W ∆E 150 × 10 6 J P= = = 41, 667 W and intensity = 1000 2 3600 s m ∆t 41, 667 W ⇒ Area of solar collector = = 41.7 m 2 1000W / m 2
11.31. Solve: The night light consumes more energy than the hair dryer. The calculations are 1.2 kW × 10 min = 1.2 × 10 3 × 10 × 60 J = 7.2 × 10 5 J 10 W × 24 hours = 10 × 24 × 60 × 60 J = 8.64 × 10 5 J
11.32. Solve: (a) A kilowatt hour is a kilowatt multiplied by 3600 seconds. It has the dimensions of energy. (b)
One kilowatt hour is energy
1 kwh = (1000 J /s)(3600 s) = 3.6 × 10 6 J Thus
3.6 × 10 6 J 9 500 kwh = (500 kwh) = 1.8 × 10 J 1 kwh
11.33. Model: Model the sprinter as a particle, and use the constant-acceleration kinematic equations and the definition of power in terms of velocity. Visualize:
Solve: (a) We can find the acceleration from the kinematic equations and the horizontal force from Newton’s second law. We have 1 1 x = x 0 + v0 x (t1 − t 0 ) + a x (t1 − t 0 ) 2 ⇒ 50 m = 0 m + 0 m + a x (7.0 s − 0 s) 2 ⇒ a x = 2.04 m/s 2 2 2 2 ⇒ Fx = ma x = (50 kg)(2.04 m/s ) = 102 N r r r (b) We obtain the sprinter’s power output by using P = F ⋅ v , where v is the sprinter’s velocity. At t = 2 s the power is
P = ( Fx )[v0 x + a x (t − t 0 )] = (102 N )[0 m /s + (2.04 m /s 2 )(2.0 s − 0 s)] = 416 W The power at t = 4 s is 832 W, and at t = 6 s the power is 1248 W.
r r r r r Visualize: Please refer to Figure P11.34. The force F = (6iˆ + 8 jˆ ) N on the particle is a constant. r r r r Solve: (a) WABD = WAB + WBD = F ⋅ ( ∆s ) AB + F ⋅ ( ∆s ) BD
11.34. Model: Use the definition of work for a constant force F, W = F ⋅ ∆s , where ∆s is the displacement.
(b)
WACD = WAC
= (6iˆ + 8ˆ j ) N ⋅ (3iˆ ) m + (6iˆ + 8 jˆ ) N ⋅ ( 4 jˆ ) m = 18 J + 32 J = 50 J r r r r + WCD = F ⋅ ( ∆s ) AC + F ⋅ ( ∆s ) CD
= (6iˆ + 8ˆ j ) N ⋅ ( 4 jˆ ) m + (6iˆ + 8 jˆ ) N ⋅ (3 jˆ ) m = 32 J + 18 J = 50 J r r (c) WAD = F ⋅ ( ∆s ) AD = (6iˆ + 8ˆ j ) N ⋅ (3iˆ + 4 jˆ ) m = 18 J + 32 J = 50 J The force is conservative because the work done is independent of the path.
11.35. Model:
The force is conservative, so it has a potential energy. Visualize: Please refer to Figure P11.35 for the graph of the force. Solve: The definition of potential energy is ∆U = –W(i → f). In addition, work is the area under the force-versusdisplacement graph. Thus ∆U = Uf – Ui = –(area under the force curve). Since Ui = 0 at x = 0 m, the potential energy at position x is U(x) = –(area under the force curve from 0 to x). From 0 m to 3 m, the area increases linearly from 0 N m to –60 N m, so U increases from 0 J to 60 J. At x = 4 m, the area is –70 J. Thus U = 70 J at x = 4 m, and U doesn’t change after that since the force is then zero. Between 3 m and 4 m, where F changes linearly, U must have a quadratic dependence on x (i.e., the potential energy curve is a parabola). This information is shown on the potential energy graph below.
(b) Mechanical energy is E = K + U. From the graph, U = 20 J at x = 1 m. The kinetic energy is K = 12 mv2 = 12 (0.10 kg) (25 m/s) 2 = 31.25 J. Thus E = 51.25 J. (c) The total energy line at 51.25 J is shown on the graph above. (d) The turning point occurs where the total energy line crosses the potential energy curve. We can see from the graph that this is at approximately 2.5 m. For a more accurate value, the potential energy function is U = 20x J. The TE line crosses at the point where 20x = 51.25, which is x = 2.56 m.
11.36. Model: Use the relationship between force and potential energy and the work-kinetic energy theorem. Visualize:
Please refer to Figure P11.36. We will find the slope in the following x regions: 0 cm < x < 1 cm, 1 < x < 3 cm, 3 < x < 5 cm, 5 < x < 7 cm, and 7 < x < 8 cm. Solve: (a) Fx is the negative slope of the U-versus-x graph, for example, for 0 m < x < 2 m dU −4 J = = −400 N ⇒ Fx = +400 N dx 0.01 m Calculating the values of Fx in this way, we can draw the force-versus-position graph as shown. (b) Since W = ∫ xxfi Fx dx = area of the Fx-versus-x graph between xi and xf, the work done by the force as the particle moves from xi = 2 cm to xf = 6 cm is –2 J. (c) The conservation of energy equation is Kf + Uf = Ki + Ui. We can see from the graph that Ui = 0 J and Uf = 2 J in moving from x = 2 cm to x = 6 cm. The final speed is vf = 10 m/s, so
2 J + 12 (0.010 kg)(10.0 m /s) 2 = 0 J + 12 (0.010 kg)vi 2 ⇒ vi = 22.4 m /s
11.37. Model: Use the relationship between a conservative force and potential energy. Visualize: Please refer to Figure P11.37. We will obtain U as a function of x and Fx as a function of x by using the calculus techniques of integration and differentiation. Solve: (a) For the interval 0 m < x < 0.5 m, Fx = (4x) N, where x is in meters. This means
dU = − Fx = −4 x ⇒ U = −2 x 2 + C1 = −2 x 2 dx where we have used U = 0 J at x = 0 m to obtain C 1 = 0. For the interval 0.5 m < x < 1 m, Fx = ( −4 x + 4) N. Likewise,
dU = 4 x − 4 ⇒ U = 2 x 2 − 4 x + C2 dx Since U should be continuous at the junction, we have the continuity condition
( −2 x 2 ) x = 0.5 m = (2 x 2 − 4 x + C2 ) x = 0.5 m ⇒ −0.5 = 0.5 − 2 + C2 ⇒ C2 = 1 U remains constant for x ≥ 1 m. (b) For the interval 0 m < x < 0.5 m, U = +4x, and for the interval 0.5 m < x < 1.0 m, U = −4x + 4, where x is in meters. The derivatives give Fx = −4 N and Fx = +4 N , respectively. The slope is zero for x ≥ 1 m.
dv x , x = ∫ v x dt, K = 12 mv x2 , and F = ma x . dt Visualize: Please refer to Figure P11.38. We know a x = slope of the vx -versus-t graph and x = area under the vx-versus-x graph between 0 and x. Solve: Using the above definitions and methodology, we can generate the following table: x(m) K(J) F(N) t(s) ax (m/s2) 0 10 0 0 5 0.5 10 1.25 6.25 5 1.0 10 5 25 5 1.5 10 11.25 56.25 5 2.0 +10 or −10 20 100 −5 or +5 2.5 28.75 56.25 −10 −5 3.0 −10 or 0 35 25 −5 or 0 3.5 40 25 0 0 4.0 45 25 0 0
11.38. Model: Use a x =
(e) Let J1 be the impulse from t = 0 s to t = 2 s and J2 be the impulse from t = 2 s to t = 4s. We have
J1 = (f)
2s
4s
0s
0s
∫ Fx dt = (5 N)(2 s) = 10 N ⋅ s and J2 =
∫F
x
dt = ( −5 N )(1 s) = −5 N ⋅ s
J = ∆p = mvf − mvi ⇒ vf = vi + J / m At t = 2 s, v x = 0 m /s + (10 N ⋅ s) / 0.5 kg = 0 m /s + 20 m /s = 20 m /s
At t = 4 s, v x = 20 m /s + ( −5 N ⋅ s) / 0.5 kg = 20 m /s − 10 m /s = 10 m /s The vx-versus-t graph also gives vx = 20 m/s at t = 2 s and vx = 10 m/s at t = 4 s. (g)
(h)
From t = 0 s to t = 2 s, W = ∫ Fx dx = (5 N )(20 m ) = 100 J From t = 2 s to t = 4 s, W = ∫ Fx dx = ( −5 N )(15 m ) = −75 J
(i) At t = 0 s, vx = 0 m/s so the work-kinetic energy theorem for calculating vx at t = 2 s is
W = ∆K =
1 2 1 2 1 1 mvf − mvi ⇒ 100 J = (0.5 kg)v x2 − (0.5 kg)(0 m /s) 2 ⇒ v x = 20 m /s 2 2 2 2
To calculate vx at t = 4 s, we use vx at t = 2 s as the initial velocity:
−75 J =
1 1 (0.5 kg)v x2 − (0.5 kg)(20 m /s) 2 ⇒ v x = 10 m /s 2 2
Both of these values agree with the values on the velocity graph.
11.39. Model: Model the elevator as a particle. Visualize:
Solve:
(a) The work done by gravity on the elevator is
Wg = − ∆U g = mgy0 − mgy1 = − mg( y1 − y0 ) = −(1000 kg)(9.8 m /s 2 )(10 m ) = −9.8 × 10 4 J (b) The work done by the tension in the cable on the elevator is
WT = T ( ∆y)cos 0° = T ( y1 − y0 ) = T (10 m ) To find T we write Newton’s second law for the elevator:
∑F
y
= T − w = ma y ⇒ T = w + ma y = m( g + a y ) = (1000 kg)(9.8 m /s 2 + 1.0 m /s 2 ) = 1.08 × 10 4 N ⇒ WT = (1.08 × 10 5 N )(10 m ) = 1.08 × 10 5 J
(c) The work-kinetic energy theorem is
1 2 1 mv0 ⇒ K f = Wg + WT + mv02 2 2 1 4 5 2 ⇒ K f = ( −9.8 × 10 J ) + (1.08 × 10 J ) + (1000 kg)(0 m /s) = 1.0 × 10 4 J 2 Wnet = Wg + WT = ∆K = K f − K i = K f −
(d)
Kf =
1 2 1 mvf ⇒ 1.0 × 10 4 J = (1000 kg)v 2f ⇒ vf = 4.47 m /s 2 2
11.40. Model: Model the rock as a particle, and apply the work-kinetic energy theorem. Visualize:
Solve:
(a) The work done by Bob on the rock is
WBob = ∆K =
1 2 1 2 1 2 1 mv1 − mv0 = mv1 = (0.50 kg)(30 m / s) 2 = 225 J 2 2 2 2
(b) For a constant force, WBob = FBob ∆x ⇒ FBob = WBob / ∆x = 225 N . (c) Bob’s power output is PBob = FBob vrock and will be a maximum when the rock has maximum speed. This is just as he releases the rock with vrock = v1 = 30 m/s. Thus, Pmax = FBob v1 = 6750 W = 6.75 kW.
11.41. Model: Model the crate as a particle, and use the work-kinetic energy theorem. Visualize:
Solve: (a) The work-kinetic energy theorem is ∆K = 12 mv12 − 12 mv02 = 12 mv12 = Wtotal . Three forces act on the box, so Wtotal = Wgrav + Wn + Wpush. The normal force is perpendicular to the motion, so Wn = 0 J. The other two forces do the following amount of work: r r r r Wpush = Fpush ⋅ ∆r = Fpush ∆x cos 20° = 137.4 J Wgrav = W ⋅ ∆r = Wx ∆ x = ( − mg sin 20°)∆ x = −98.0 J Thus, Wtotal = 39.4 J, leading to a speed at the top of the ramp equal to
v1 =
2Wtotal = m
2(39.4 J ) = 3.97 m /s 5 kg
(b) The x-component of Newton’s second law is
ax = a =
( Fnet ) x Fpush cos 20° − w sin 20° Fpush cos 20° − mg sin 20° = = = 1.347 m /s 2 m m m
Constant-acceleration kinematics with x1 = h / sin 20° = 5.85 m gives the final speed
v12 = v02 + 2 a( x1 − x 0 ) = 2 ax1 ⇒ v1 = 2 ax1 = 2(1.347 m / s 2 )(5.85 m ) = 3.97 m /s
11.42. Model: Model Sam strapped with skis as a particle, and apply the law of conservation of energy. Visualize:
Solve:
(a) The conservation of energy equation is
K1 + U g1 + ∆ Eth = K 0 + U g 0 + Wext The snow is frictionless, so ∆E th = 0 J. However, the wind is an external force doing work on Sam as he moves down the hill. Thus,
Wext = Wwind = ( K1 + U g1 ) − ( K 0 + U g 0 ) 1 1 1 1 = mv12 + mgy1 − mv02 + mgy0 = mv12 + 0 J − (0 J + mgy0 ) = mv12 − mgy0 2 2 2 2 ⇒ v1 = 2 gy0 +
2Wwind m
We compute the work done by the wind as follows: r r Wwind = Fwind ⋅ ∆r = Fwind ∆r cos160° = (200 N )(146 m )cos160° = −27, 400 J where we have used ∆r = h / sin 20° = 146 m. Now we can compute
v1 = 2(9.8 m / s 2 )(50 m ) +
2( −27,400 J ) = 15.7 m / s 75 kg
(b) We will use a tilted coordinate system, with the x-axis parallel to the slope. Newton’s second law for Sam is w sin 20° − Fwind cos 20° mg sin 20° − Fwind cos 20° (F ) a x = a = net x = = m m m (75 kg)(9.8 m /s 2 )sin 20° − (200 N )cos 20° = = 0.846 m /s 2 75 kg Now we can use constant-acceleration kinematics as follows:
v12 = v02 + 2 a( x1 − x 0 ) = 2 ax1 ⇒ v1 = 2 ax1 = 2(0.846 m /s 2 )(146 m ) = 15.7 m /s Assess: We used a vertical y-axis for energy analysis, rather than a tilted coordinate system, because Ug is determined by its vertical position.
11.43. Model: Model Paul and the mat as a particle, assume the mat to be massless, use the model of kinetic friction, and apply the work-kinetic energy theorem. Visualize:
We define the x-axis along the floor and the y-axis perpendicular to the floor. Solve: We need to first determine fk. Newton’s second law in the y-direction is n + T sin30° = w = mg ⇒ n = mg − T sin 30° = (10 kg)(9.8 m /s 2 ) − (30 N )(sin 30°) = 83.0 N. Using n and the model of kinetic friction, fk = µ k n = (0.2)(83.0 N ) = 16.60 N. Thus, r r r r Wnet = fk ⋅ ∆r + T ⋅ ∆r = ( fk )( ∆x )(cos180°) + (T )( ∆x )(cos 30°)
= (16.60 N )(3.0 m )( −1) + (30.0 N )(3.0 m )(0.8660) = 28.14 J r r r The other forces n and w make an angle of 90° with ∆r and do zero work. We can now use the work-kinetic energy theorem to find the final velocity as follows:
Wnet = K f − K i = K f − 0 J =
1 2 mvf ⇒ vf = 2Wnet / m = 2(28.14 J )/10 kg = 2.37 m /s 2
Assess: A speed of 2.37 m/s or 5.3 mph is reasonable for the present problem.
11.44. Model: Assume an ideal spring that obeys Hooke’s law. Model the box as a particle and use the model of kinetic friction. Visualize:
Solve:
When the horizontal surface is frictionless, conservation of energy means 1 1 1 k ( x 0 − x e ) 2 = mv12x = K1 ⇒ K1 = (100 N / m )(0.20 m − 0 m ) 2 = 2.0 J 2 2 2 That is, the box is launched with 2.0 J of kinetic energy. It will lose 2.0 J of kinetic energy on the rough surface. The work-kinetic energy theorem is r r Wnet = fk ⋅ ∆r = K 2 − K1 = 0 J − 2.0 J = −2.0 J
⇒ fk ( x 2 − x1 )(cos180°) = − µ k mg( x 2 − x1 ) = −2.0 J ⇒ ( x 2 − x1 ) =
2.0 J 2.0 J = = 54.4 cm ( µ k )( mg) (0.15)(2.5 kg)(9.8 m /s 2 )
Assess: Because the force of friction does negative work on the box, energy is transferred out of the box into the environment. In response, the box slows down and comes to rest.
11.45. Model: Model the suitcase as a particle, use the model of kinetic friction, and use the work-kinetic energy theorem. Visualize:
r r r The only force that does work on the suitcase is the force of kinetic friction, fk . The forces w and n are perpendicular to the displacement, and therefore do not do work on the suitcase. Solve: The work-kinetic energy theorem is r r 1 1 1 1 Wnet = ∆K = mv12 − mv02 ⇒ fk ⋅ ∆r = 0 J − mv02 ⇒ ( fk )( x1 − x 0 )cos180° = − mv02 2 2 2 2 v02 1 2 (1.2 m /s) 2 ⇒ − µ k mg( x1 − x 0 ) = − mv0 ⇒ µ k = = = 0.037 2 2 g( x1 − x 0 ) 2(9.8 m /s 2 )(2.0 m − 0 m ) Assess: Friction does negative work on the suitcase, and thus transfers energy out of the suitcase. In response, the suitcase slows down and comes to rest.
11.46. Model: Identify the truck and the loose gravel as the system. We need the gravel inside the system because friction increases the temperature of the truck and the gravel. We will also use the model of kinetic friction and the conservation of energy equation. Visualize:
We place the origin of our coordinate system at the base of the ramp in such a way that the x-axis is along the ramp and the y-axis is vertical so that we can calculate potential energy. The free-body diagram of forces on the truck is shown. Solve: The conservation of energy equation is K1 + U g1 + ∆ Eth = K 0 + U g 0 + Wext . In the present case, Wext = 0 J, v1x = 0 m/s, Ug0 = 0 J, v0x = 35 m/s. The thermal energy created by friction is
∆Eth = ( fk )( x1 − x 0 ) = ( µ k n)( x1 − x 0 ) = µ k ( mg cos 6.0°)( x1 − x 0 ) = (0.40)(15,000 kg)(9.8 m /s 2 )(cos 6.0°)( x1 − x 0 ) = (58,478 J / m )( x1 − x 0 ) Thus, the energy conservation equation simplifies to
1 2 mv0 x + 0 J + 0 J 2 1 (15,000 kg)(9.8 m /s 2 )[( x1 − x 0 )sin 6.0°] + (58,478 J / m )( x1 − x 0 ) = (15000 kg)(35 m /s) 2 2 ⇒ ( x1 − x 0 ) = 124.4 m 0 J + mgy1 + (58,478 J / m )( x1 − x 0 ) =
Assess: A length of 124.4 m at a slope of 6º seems reasonable.
11.47. Model: We will use the spring, the package, and the ramp as the system. We will model the package as a particle. Visualize:
We place the origin of our coordinate system on the end of the spring when it is compressed and is in contact with the package to be shot. Model: (a) The energy conservation equation is
K1 + U g1 + Us1 + ∆ Eth = K 0 + U g 0 + Us 0 + Wext 1 2 1 1 1 mv1 + mgy1 + k ( x e − x e ) 2 + ∆ Eth = mv02 + mgy0 + k ( ∆x ) 2 + Wext 2 2 2 2 Using y1 = 1 m, ∆Eth = 0 J (note the frictionless ramp), v0 = 0 m/s, y0 = 0 m, ∆x = 30 cm, and Wext = 0 J, we get
1 2 1 mv1 + mg(1 m ) + 0 J + 0 J = 0 J + 0 J + k (0.30 m ) 2 + 0 J 2 2 1 1 2 2 (2.0 kg)v1 + (2.0 kg)(9.8 m /s )(1 m ) = (500 N / m )(0.30 m ) 2 2 2 ⇒ v1 = 1.70 m /s (b) How high can the package go after crossing the sticky spot? If the package can reach y1 ≥ 1.0 m before stopping (v1 = 0), then it makes it. But if y1 < 1.0 m when v1 = 0, it does not. The friction of the sticky spot generates thermal energy r r ∆Eth = −Wdiss = − fk ⋅ ∆r = −( − µ k mg)∆x = (0.30)(2.0 kg)(9.8 m / s 2 )(0.50 m) = 2.94 J The energy conservation equation is now 1 2
mv12 + mgy1 + ∆Eth = 12 k ( ∆x ) 2
If we set v1 = 0 m/s to find the highest point the package can reach, we get
y1 = ( 12 k ( ∆x ) 2 − ∆Eth ) / mg = ( 12 (500 N / m) (0.30 m) 2 − 2.94 J ) / (2.0 kg)(9.8 m / s 2 ) = 0.998 m The package does not make it. It just barely misses.
11.48. Model: Model the two blocks as particles. The two blocks make our system. Visualize:
We place the origin of our coordinate system at the location of the 3.0 kg block. Solve: (a) The conservation of energy equation is K f + U gf + ∆Eth = K i + U gi + Wext . Using ∆Eth = 0 J and Wext = 0 J we get
1 1 1 1 m2 (vf ) 22 + m3 (vf ) 32 + m2 g( yf ) = m2 (vi ) 22 + m3 (vi ) 32 + m2 g( yi ) 2 2 2 2 Noting that (vf ) 2 = (vf ) 3 = vf and (vi ) 2 = (vi ) 3 = 0 m / s, this becomes
1 ( m2 + m3 )vf2 = − m2 g( yf − yi ) 2 vf =
2 m 2 g( y i − y f ) = m 2 + m3
2(2.0 kg)(9.8 m /s 2 )(1.50 m ) = 3.43 m /s (2.0 kg + 3.0 kg)
(b) We will use the same energy conservation equation. However, this time
∆Eth = ( fk )( ∆x ) = ( µ k n)( ∆ x ) = µ k ( m3 g)( ∆ x) = (0.15)(3.0 kg)(9.8 m /s 2 )(1.50 m ) = 6.615 J The energy conservation equation is now
1 1 1 1 m2 vf2 + m3 vf2 + m2 gyf + 6.615 J = m2 (vi ) 22 + m3 (v1 ) 32 + m2 gyi + 0 J 2 2 2 2 1 2 ( m2 + m3 )vf2 + 6.615 J = m2 g( yi − yf ) ⇒ vf = [ m2 g( yi − yf ) − 6.615 J ] 2 m 2 + m3 2 2 = [(2.0 kg)(9.8 m /s )(1.50 m ) − 6.615 J ] = 3.02 m /s 5.0 kg Assess: A reduced speed when friction is present compared to when there is no friction is reasonable.
r
r
11.49. Model: Use the particle model, the definition of work W = F ⋅ ∆s , and the model of kinetic friction. Visualize: We place the coordinate frame on the incline so that its x-axis is along the incline.
r r WT = T ⋅ ∆r = (T )( ∆x )cos18° = (120 N )(5.0 m )cos18° = 571 J r r Wg = ( w ⋅ ∆r ) = mg( ∆x )cos(120°) = (8 kg)(9.8 m /s 2 )(5.0 m )cos120° = −196 J r r Wfriction = fk ⋅ ∆r = ( fk )( ∆x )cos180° = −( µ k n)( ∆x )
Solve: (b) (c)
(a)
To find n, we write Newton’s second law as follows:
∑F
y
= n − w cos 30° + T sin 18° = 0 N ⇒ n = w cos 30° − T sin 18°
⇒ n = mg cos 30° − T sin 18° = (8.0 kg)(9.8 m /s 2 )cos 30° − (120 N )sin 18° = 30.814 N Thus, Wfriction = −(0.25)(30.814 N )(5.0 m ) = −38.5 J . r r (d) Wn = (n ⋅ ∆r ) = (n)( ∆ x )cos 90° = 0 J Assess: Any force that acts perpendicular to the displacement does no work.
11.50. Model: Assume the spring is ideal so that Hooke’s law is obeyed, and model the weather rocket as a particle. Visualize:
The origin of the coordinate system is placed on the free end of the spring. Note that the bottom of the spring is anchored to the ground. Solve: (a) The rocket is initially at rest. The free-body diagram on the rocket helps us write Newton’s second law as
(∑ F ) = 0 N ⇒ F y
sp
= w = mg ⇒ k∆y = mg
mg (10.2 kg)(9.8 m /s 2 ) = = 20 cm k (500 N / m ) (b) The thrust does work. Using the energy conservation equation when y2 − y0 = 40 cm = 0.40 m: K 2 + U g 2 + Usp 2 = K1 + U g1 + Usp1 + Wext ⇒ ∆y =
1 1 1 1 Wext = F ( y2 − y1 ) ⋅ mv22 + mgy2 + k ( y2 − ye ) 2 = mv12 + mgy1 + k ( y1 − ye ) 2 + (200 N )(0.60 m ) 2 2 2 2 (5.10 kg)v22 + 40.0 J + 40.0 J = 0 − 20.0 J + 10.0 J + 120 J ⇒ v2 = 2.43 m /s If the rocket were not attached to the spring, the energy conservation equation would not involve the spring energy term Usp2. That is, K 2 + U g 2 = K1 + U g1 + Usp1 + Wext
1 (10.2 kg)v22 + (10.2 kg)(9.8 m /s 2 )(0.40 m ) = 0 J − (10.2 kg)(9.8 m /s 2 )(0.20 m ) 2 1 + (500 N / m )(0.20 m ) 2 + (200 N )(0.60 m ) 2 ⇒ (5.1 kg)v22 = 70.0 J ⇒ v2 = 3.70 m /s
11.51. Model: Use the particle model for the ice skater, the model of kinetic/static friction, and the work-kinetic energy theorem. Visualize:
Solve:
(a) The work-kinetic energy theorem is 1 1 ∆K = mv12 − mv02 = Wwind + Wfric 2 2 There is no kinetic friction along her direction of motion. Static friction acts to prevent her skates from slipping sideways on the ice, but this force is perpendicular to the motion and does no work: Wfric = 0 J. The angle between r r Fwind and ∆r is θ = 135°, so r r Wwind = Fwind ⋅ ∆r = Fwind ∆y cos135° = ( 4 N )(100 m )cos135° = −282.8 J Thus, her final speed is
2Wwind = 2.16 m/s m (b) If the skates don’t slip, she has no acceleration in the x-direction and so (Fnet)x = 0 N. That is: v1 = v02 +
fs − Fwind cos 45° = 0 N ⇒ fs = Fwind cos 45° = 2.83 N Now there is an upper limit to the static friction: fs ≤ ( fs ) max = µ s mg. To not slip requires
µs ≥
fs 2.83 N = = 0.0058 mg (50 kg)(9.8 m /s 2 )
Thus, the minimum value of µs is 0.0058. Assess: The work done by the wind on the ice skater is negative, because the wind slows the skater down.
11.52. Model: Model the ice cube as a particle, the spring as an ideal that obeys Hooke’s law, and the law of conservation of energy. Visualize:
Solve: (a) The normal force does no work and the slope is frictionless, so mechanical energy is conserved. We’ve drawn two separate axes: a vertical y-axis to measure potential energy and a tilted s-axis to measure distance along the slope. Both have the same origin which is at the point where the spring is not compressed. Thus, the two axes are related by y = s sin θ . Also, this choice of origin makes the elastic potential energy simply Us = 12 k ( s − s0 ) 2 = 2 1 2 ks . Because energy is conserved, we can relate the initial point—with the spring compressed—to the final point where the ice cube is at maximum height. We do not need to find the speed with which it leaves the spring. We have
K 2 + U g 2 + Us 2 = K1 + U g1 + Us1 1 2 1 1 1 mv2 + mgy2 + ks02 = mv12 + mgy1 + ks12 2 2 2 2 It is important to note that at the final point, when the ice cube is at y2, the end of the spring is only at s0. The spring does not stretch to s2, so Us2 is not 12 ks22 . Three of the terms are zero, leaving
mgy2 = + mgy1 +
ks 2 1 2 ks1 ⇒ y2 − y1 = ∆y = height gained = 1 = 0.255 m = 25.5 cm 2 2 mg
The distance traveled is ∆s = ∆y/sin 30° = 51.0 cm. (b) Using the energy equation and the model for kinetic friction:
K 2 + U g 2 + Us 2 + ∆Eth = K1 + U g1 + Us1 + Wext From the free-body diagram,
∆Eth = fk ∆s = µ k n∆s
r ( Fnet ) y = 0 N = n − mg cos 30° ⇒ n = mg cos 30°
Now, having found ∆Eth = µ k ( mg cos 30°)∆s, the energy equation can be written
0 J + mgy2 + 0 J + µ k ( mg cos 30°)∆s = 0 J + mgy1 + ⇒ mg( y2 − y1 ) −
1 2 ks1 + 0 J 2
1 2 ks1 + µ k mg cos 30°∆s = 0 2
Using ∆y = ( ∆s)sin 30°, the above equation simplifies to
mg∆s sin 30° + µ k mg cos 30°∆s =
ks12 1 2 ks1 ⇒ ∆s = = 0.379 m = 37.9 cm 2 2 mg(sin 30° + µ k cos 30°)
11.53. Model: Assume an ideal spring, so Hooke’s law is obeyed. Treat the box as a particle and apply the energy conservation law. Box, spring, and the ground make our system, and we also use the model of kinetic friction. Visualize: We place the origin of the coordinate system on the ground directly below the box’s starting position.
Solve:
(a) The energy conservation equation is
K1 + U g1 + Us1 + ∆Eth = K 0 + U g 0 + Us 0 + Wext 1 2 1 1 mv1 + mgy1 + 0 J + 0 J = mv02 + mgy0 + 0 J + 0 J ⇒ mv12 + 0 J = 0 J + mgy0 2 2 2 ⇒ v1 = 2 gy0 = 2(9.8m /s 2 )(5 m ) = 9.90 m /s (b) The work of friction creates thermal energy. The energy conservation equation for this part of the problem is
1 2 1 mv2 + 0 J + 0 J + ( fk )( x 2 − x1 ) = mv12 + 0 J + 0 J + 0 J 2 2 1 2 1 2 1 2 1 2 mv2 + µ k n( x 2 − x1 ) = mv1 ⇒ mv2 + µ k ( mg)( x 2 − x1 ) = mv1 2 2 2 2
K 2 + U g 2 + Us 2 + ∆Eth = K1 + U g1 + Us1 + Wext
⇒ v2 = v12 − 2 µ k g( x 2 − x1 ) = (9.90 m /s) 2 − 2(0.25)(9.8 m /s 2 )(2.0 m ) = 9.39 m /s (c) To find how much the spring is compressed, we apply the energy conservation once again: 1 1 K 3 + U g 3 + Us 3 + ∆Eth = K 2 + U g 2 + Us 2 + Wext 0 J + 0 J + k ( x3 − x 2 ) 2 + 0 J = mv22 + 0 J + 0 J + 0 J 2 2 Using v2 = 9.39 m/s, k = 500 N/m and m = 5.0 kg, the above equation yields ( x 3 − x 2 ) = ∆x = 93.9 cm. (d) The initial energy = mgy0 = (5.0 kg)(9.8 m / s 2 )(5.0 m ) = 245 J. The energy transformed to thermal energy during each passage is fk ( x 2 − x1 ) = µ k mg( x 2 − x1 ) = (0.25)(5.0 kg)(9.8 m / s 2 )(2.0 m ) = 24.5 J The number of passages is equal to 245 J / 24.5 J or 10.
11.54. Model: Assume an ideal spring, so Hooke’s law is obeyed. Treat the physics student as a particle and apply the law of conservation of energy. Our system is comprised of the spring, the student, and the ground. We also use the model of kinetic friction. Visualize: We place the origin of the coordinate system on the ground directly below the end of the compressed spring that is in contact with the student.
Solve:
(a) The energy conservation equation is K1 + U g1 + Us1 + ∆Eth = K 0 + U g 0 + Us 0 + Wext
1 2 1 1 1 mv1 + mgy1 + k ( x1 − x e ) 2 + 0 J = mv02 + mgy0 + k ( x1 − x 0 ) 2 + 0 J 2 2 2 2 Since y1 = y0 = 10 m, x1 = xe, v0 = 0 m/s, k = 80,000 N/m, m = 100 kg, and (x1 − x0) = 0.5 m,
1 2 1 mv1 = k ( x1 − x 0 ) 2 ⇒ v1 = 2 2
k ( x1 − x 0 ) = 14.14 m/s m
(b) The work of friction creates thermal energy. Applying the conservation of energy equation once again:
K 2 + Ug 2 + Us 2 + ∆Eth = K 0 + U g 0 + Us 0 + Wext 1 2 1 mv2 + mgy2 + 0 J + fk ∆s = 0 J + mgy0 + k ( x1 − x 0 ) 2 + 0 J 2 2 With v2 = 0 m/s and y2 = (∆s)sin30º, the above equation is simplified to 1 mg( ∆s)sin 30° + µ k n∆s = mgy0 + k ( x1 − x 0 ) 2 2 From the free-body diagram for the physics student, we see that n = w cos 30°. Thus, the conservation of energy equation gives 1 ∆s( mg sin 30° + µ k mg cos 30°) = mgy0 + k ( x1 − x 0 ) 2 2 Using m = 100 kg, k = 80,000 N/m, (x1 − x0) = 0.50 m, y0 = 10 m, and µk = 0.15, we get
1 k ( x1 − x 0 ) 2 2 ∆s = = 32.1 m mg(sin 30° + µ k cos 30°) mgy0 +
Assess: y2 = ( ∆s)sin 30° = 16.05 m, which is greater than y0 = 10 m. The higher value is due to the transformation of the spring energy into gravitational potential energy.
11.55. Model: Treat the block as a particle, use the model of kinetic friction, and apply the energy conservation law. The block and the incline comprise our system. Visualize: We place the origin of the coordinate system directly below the block’s starting position at the same level as the horizontal surface. On the horizontal surface the model of kinetic friction applies.
Solve:
(a) For the first incline, the conservation of energy equation gives 1 2 K1 + U g1 + ∆Eth = K 0 + U g 0 + Wext mv1 + 0 J + 0 J = 0 J + mgy0 + 0 J ⇒ v1 = 2 gy0 = 2 gh 2 (b) The work of friction creates thermal energy. Applying once again the conservation of energy equation, we have 1 2 1 K 3 + U g 3 + ∆Eth = K1 + U g1 + Wext mv3 + mgy3 + fk ( x 2 − x1 ) = mv12 + mgy1 + Wext 2 2 Using v3 = 0 m/s, y1 = 0 m, Wext = 0 J, fk = µkmg, v1 = 2 gh , and (x2 – x1) = L, we get 1 mgy3 + µ k mgL = m(2 gh) ⇒ y3 = h − µ k L 2 Assess: For µ k = 0, y3 = h which is predicted by the law of the conservation of energy.
11.56. Model: Assume an ideal spring, so Hooke’s law is obeyed. Visualize:
Solve: For a conservative force the work done on a particle as it moves from an initial to a final position is independent of the path. We will show that WA→C→B = WA→B for the spring force. Work done by a spring force F = −kx is given by xf
W = ∫ Fdx = − ∫ kx dx xi
This means xB
WA→B = − ∫ kx dx = − xA
x
x
C B k 2 k k x B − x A2 ), WA→C = − ∫ kx dx = − ( x C2 − x A2 ), and WC→B = − ∫ kx dx = − ( x B2 − x C2 ) ( 2 2 2 xA xC
Adding the last two:
WA→C→B = WA→C + WC→B = −
k 2 ( xC − xA2 + xB2 − xC2 ) = WA→B 2
11.57. Model: A “sprong” that obeys the force law Fx = − q( x − x e )3 , where q is the sprong constant and xe is the equilibrium position. Visualize: We place the origin of the coordinate system on the free end of the sprong, that is, xe = xf = 0 m.
Solve: (a) The units of q are N/m3. (b) A cubic curve rises more steeply than a parabola. The force increases by a factor of 8 every time x increases by a factor of 2.
x dU qx 4 , we have U ( x ) = − ∫ Fx dx = − ∫ ( − qx 3 )dx = . 0 dx 4 (d) Applying the energy conservation equation to the ball and sprong system: Kf + Uf = Ki + Ui
(c) Since Fx = −
qx 4 1 2 mvf + 0 J = 0 J + i 2 4 ⇒ vf =
q x4 = m 2
( 40,000 N / m 3 ) ( −0.10 m ) 4 ⋅ = 10 m /s (0.020 kg) 2
11.58. Solve: (a) Because sin (cx) is dimensionless, F0 must have units of force in newtons. (b) The product cx is an angle because we are taking the sine of it. An angle has no real physical units. If x has units of m and the product cx is unitless, then c has to have units of m−1. (c) At x0 = 0 m, the force is F0 sin(0) = 0 N. The particle is able to leave x0 = 0 m only because it has an initial velocity. (d) The force is a maximum when sin(cx) = 1. This occurs when cx = π/2, or for xmax = π/2c. (e) The graph is the first quarter of a sine curve.
(f) We can find the velocity vf at xf = xmax from the work-kinetic energy theorem:
1 2 1 2 1 2 1 2 2W mvf − mvi = mvf − mv0 = W ⇒ vf = v02 + 2 2 2 2 m This is a variable force. As the particle moves from xi = 0 m to xf = xmax = π/2c, the work done on it is ∆K =
W=
∫
xf
xi
F ( x ) dx = F0 ∫
π / 2c
0
sin(cx ) dx = −
F0 F F π cos(cx )|π0 / 2 c = − 0 cos − cos 0 = 0 2 c c c
Thus, the particle’s speed at xf = xmax = π/2c is vf = v02 + 2 F0 / mc .
11.59. Visualize: We place the origin of the coordinate system at the base of the stairs on the first floor.
Solve: (a) We might estimate y2 − y1 ≈ 4.0 m ≈ 12 ft ≈ y3 − y2 , thus, y3 – y1 ≈ 8.0 m. (b) We might estimate the time to run up these two flights of stairs to be 20 s. (c) Estimate your mass as m ≈ 70 kg ≈ 150 lb. Your power output while running up the stairs is work done by you change in potential energy mg( y3 − y1 ) = = time time time 2 (70 kg)(9.8 m /s )(8.0 m ) 1 hp = ≈ 270 W = (270 W ) ≈ 0.35 hp 746 W 20 s Assess: Your estimate may vary, depending on your mass and how fast you run.
11.60. Model: Model the lawnmower as a particle and use the model of kinetic friction. Visualize:
We placed the origin of our coordinate system on the lawnmower and drew the free-body r diagram of forces. r Solve: The normal force n, which is related to the frictional force, is not equal to w. This is due to the presence r r r of F. The rolling friction is fr = µ r n, or n = fr µ r . The lawnmower moves at constant velocity, so Fnet = 0. The two components of Newton’s second law are
(∑ F ) = n − w − F sin 37° = ma = 0 N ⇒ f / µ − mg − F sin 37° = 0 N ⇒ f (∑ F ) = F cos 37° − f = 0 N ⇒ F cos 37° − µ mg − µ F sin 37° = 0 N y
y
x
r
⇒F=
r
r
r
r
= µ r mg + µ r F sin 37°
r
µ r mg (0.15)(12 kg)(9.8 m /s 2 ) = = 29.4 N cos 37° − µ sin 37° (0.7986) − (0.15)(0.6018)
Thus, the power supplied by the gardener in pushing the lawnmower at a constant speed of 1.2 m/s is P = Fv = (24.9 N )(1.2 m / s) = 29.9 W.
11.61. Solve: Power output during the push-off period is equal to the work done by the cat divided by the time the cat applied the force. Since the force on the floor by the cat is equal in magnitude to the force on the cat by the floor, work done by the cat can be found using the work-kinetic energy theorem during the push-off period, Wnet = Wfloor = ∆K . We do not need to explicitly calculate Wcat, since we know that the cat’s kinetic energy is transformed into its potential energy during the leap. That is,
∆U g = mg( y2 − y1 ) = (5.0 kg)(9.8 m / s 2 )(0.95 m ) = 46.55 J Thus, the average power output during the push-off period is W 46.55 J P = net = = 233 W 0.20 s t
11.62. Solve: Using the conversion 746 W = 1 hp, we have a power of 1492 J/s. This means W = Pt = (1492 J/s)(1 hr) = 5.3712 × 106 J is the total work done by the electric motor in one hour. Furthermore,
Wmotor = − Wg = U gf − U gi = mg( yf − yi ) = mg(10 m ) m=
Wmotor 5.3712 × 10 6 J 1 liter = = 5.481 × 10 4 kg = 5.481 × 10 4 kg × = 54, 800 liters g(10 m ) (9.8 m /s 2 )(10 m ) 1 kg
11.63. Solve: (a) The change in the potential energy of 1.0 kg of water in falling 25 m is ∆Ug = − mgh = −(1.0 kg)(9.8 m /s 2 )(25 m ) = −245 J. (b) The power required of the dam is
W W = = 50 × 10 6 Watts ⇒ W = 50 × 10 6 J t 1s That is, 50 × 10 6 J of energy is required per second for the dam. Out of the 245 J of lost potential energy, (245 J)(0.80) = 196 J is converted to electrical energy. Thus, the amount of water needed per second is 50 × 10 6 J(1 kg /196 J) = 255, 000 kg . P=
11.64. Solve: The force required to tow a water skier at a speed v is Ftow = Av. Since power P = Fv, the power required to tow the water skier is Ptow = Ftow v = Av 2 . We can find the constant A by noting that a speed of v = 2.5 mph requires a power of 2 hp. Thus, hp (2 hp) = A(2.5 mph ) 2 ⇒ A = 0.32 ( mph ) 2 Now, the power required to tow a water skier at 7.5 mph is hp Ptow = Av 2 = 0.32 ⋅ (7.5 mph ) 2 = 18 hp ( mph ) 2 Assess: Since P ∝ v 2 , a three-fold increase in velocity leads to a nine-fold increase in power.
11.65. Solve: By definition, the maximum powerroutput of a horse is P ≈ 1 hp = 746 W. At maximum speed,
r when the horse is running at constant speed, Fnet = 0 . The propulsion force Fhorse, provided by the horse pushing against the ground, is balanced by the drag force of air resistance: Fhorse = D. We learned in Chapter 5 that a reasonable model for drag is D ≈ 14 Av2, where A is the cross section area. Since the power needed for force F to push an object at velocity v is P = Fv, we have P ≈ 1 hp = 746 W = Fhorse v = ( 14 Av 2 )v = 14 Av 3 4P v≈ A
1/ 3
4(746 W) = (0.5 m)(1.8 m)
1/ 3
= 15 m / s
Assess: 15 m/s ≈ 30 mph is a reasonable top speed for a well-trained horse.
11.66. Solve: The net force on a car moving at a steady speed is zero. The motion is opposed both by rolling friction and by air resistance. Thus the propulsion force provided by the drive wheels must be Fcar = µ r mg + 14 Av 2 , where µr is the rolling friction, m is the mass, A is the cross-section area, and v is the car’s velocity. The power required to move the car at speed v is 1 P = Fcar v = µ r mgv + Av 3 4 Since the maximum power output is 200 hp and 75% of the power reaches the drive wheels, P = (200 hp)(0.75) = 150 hp. Thus,
746 W 1 2 3 (150 hp) = (0.02)(1500 kg)(9.8 m /s )v + (1.6 m )(1.4 m )v 4 1 hp ⇒ 0.56 v 3 + 294 v − 111900 = 0 ⇒ v = 55.5 m /s The easiest way to solve this equation is through iterations by trial and error. Assess: A speed of 55.5 m/s ≈ 110 mph is very reasonable.
11.67. Model: Use the model of static friction, kinematic equations, and the definition of power.
Solve: (a) The rated power of the Porsche is 217 hp = 161,882 W and the weight of the car is (1480 kg)(9.8 m/s2) = 14504 N. The weight of the car on the drive wheels is (14504)(2/3) = 9670 N. Because the static friction of the tires on road pushes the car forward,
Fmax = fs max = µ s n = µ s mg = (1.0)(9670 N ) = mamax ⇒ amax =
9670 N = 6.53 m/s 2 1480 kg
P 161882 W = = 16.7 m /s F 9670 N (c) Using the kinematic equation, vmax = v0 + amax (t min − t 0 ) with v0 = 0 m/s and t0 = 0 s, we obtain (b)
P = Fvmax ⇒ vmax =
t min =
vmax 16.7 m /s = = 2.56 s amax 6.53 m /s 2
Assess: An acceleration time of 2.56 s for the Porsche to reach a speed of ≈35 mph from rest is reasonable.
11.68. (a) A student uses a string to pull her 2.0 kg physics book, starting from rest, across a 2.0-m-long lab bench. The coefficient of kinetic friction between the book and the lab bench is 0.15. If the book’s final speed is 4.0 m/s, what is the tension in the string? (b)
(c) The tension does external work Wext. This work increases the book’s kinetic energy and also causes an increase ∆Eth in the thermal energy of the book and the lab bench. Solving the equation gives T = 10.9 N.
11.69. (a) A 20 kg chicken crate slides down a 2.5-m-high, 40째 ramp from the back of a truck to the ground. The coefficient of kinetic friction between the crate and the ramp bench is 0.15. How fast are the chickens going at the bottom of the ramp? (b)
(c) v1 = 6.34 m/s.
11.70. (a) If you expend 75 W of power to push a 30 kg sled on a surface where the coefficient of kinetic friction between the sled and the surface is µk = 0.20, what speed will you be able to maintain? (b)
(c)
Fpush = (0.20)(30 kg)(9.8 m /s 2 ) = 58.8 N ⇒ 75 W = (58.9 N)v ⇒ v =
75 W = 1.28 m /s 58.8 N
11.71. (a) A 1500 kg object is being accelerated upward at 1.0 m/s2 by a rope. How much power must the motor supply at the instant when the velocity is 2.0 m/s? (b)
(c)
T = (1500 kg)(9.8 m/s 2 ) + 1500 kg(1.0 m/s 2 ) = 16, 200 N P = T (2 m/s) = (16200 N )(2.0 m/s) = 32, 400 W = 32.4 kW
11.72. Model: Model the water skier as a particle, apply the law of conservation of mechanical energy, and use the constant-acceleration kinematic equations. Visualize:
We placed the origin of the coordinate system at the base of the frictionless ramp. Solve: We’ll start by finding the smallest speed v1 at the top of the ramp that allows her to clear the shark tank. From the vertical motion for jumping the shark tank,
1 a y (t2 − t1 ) 2 2 1 ⇒ 0 m = (2.0 m ) + 0 m + ( −9.8 m /s 2 )(t2 − t1 ) 2 ⇒ (t2 − t1 ) = 0.639 s 2
y2 = y1 + v1y (t2 − t1 ) +
From the horizontal motion,
x 2 = x1 + v1 x (t2 − t1 ) +
1 a x (t2 − t1 ) 2 2
⇒ ( x1 + 5.0 m ) = x1 + v1 (0.639 s) + 0 m ⇒ v1 =
5.0 m = 7.825 m /s 0.639 s
Having found the v1 that will take the skier to the other side of the tank, we now use the energy equation to find the minimum speed v0. We have
K1 + U g1 = K 0 + U g 0 ⇒
1 2 1 mv1 + mgy1 = mv02 + mgy0 2 2
v0 = v12 + 2 g( y1 − y0 ) = (7.825 m /s) 2 + 2(9.8 m /s 2 )(2.0 m ) = 10.0 m /s
11.73. Model: Assume the spring to be ideal that obeys Hooke’s law, model the block as a particle, and apply the model of kinetic friction. Visualize: We placed the origin of the coordinate system on the free end of the compressed spring which is in contact with the block. Because the horizontal surface at the bottom of the ramp is frictionless, the spring energy appears as kinetic energy of the block until the block begins to climb up the incline.
Solve: Although we could find the speed v1 of the block as it leaves the spring, we don’t need to. We can use energy conservation to relate the initial potential energy of the spring to the energy of the block as it begins projectile motion at point 2. However, friction requires us to calculate the increase in thermal energy. The energy equation is
K 2 + U g2 + ∆Eth = K 0 + U g0 + Wext ⇒ 12 mv22 + mgy2 + fk ∆s = 12 k ( x 0 − x e ) 2 The distance along the slope is ∆s = y2/sin45°. The friction force is fk = µkn, and we can see from the free-body diagram that n = mgcos45°. Thus
k v2 = ( x 0 − x e ) 2 − 2 gy2 − 2 µ k gy2 cos 45° / sin45° m
1/ 2
1000 N / m = (0.15 m) 2 − 2(9.8 m / s 2 )(2.0 m) − 2(0.20)(9.8 m / s 2 )(2.0 m) cos 45° / sin45° 0.20 kg
1/ 2
= 8.091 m / s
Having found the velocity v2, we can now find (x3 − x2) = d using the kinematic equations of projectile motion:
1 a2 y ( t 3 − t 2 ) 2 2 1 2.0 m = 2.0 m + (v2 sin 45°)(t3 − t2 ) + ( −9.8 m / s 2 )(t3 − t2 ) 2 2 y3 = y 2 + v 2 y ( t 3 − t 2 ) +
⇒ (t3 − t2 ) = 0 s and 1.168 s Finally,
1 a2 x ( t 3 − t 2 ) 2 2 d = ( x3 − x 2 ) = (v2 cos 45°)(1.168 s) + 0 m = 6.68 m x 3 = x 2 + v2 x (t 3 − t 2 ) +
11.74. Model: Assume an ideal spring that obeys Hooke’s law, the particle model for the ball, and the model of kinetic friction. The ball, the spring, and the barrel comprise our system. Visualize:
We placed the origin of the coordinate system on the free end of the spring which is in contact with the ball. r Solve: Force F does external work Wext. This work compresses the spring, gives the ball kinetic energy, and is partially dissipated to thermal energy by friction. (a) The energy equation for our system is
K1 + Us1 + ∆Eth = K 0 + Us 0 + Wext 1 2 1 1 1 mv1 + k ( x1 − x 0 ) 2 + ( fk )( x1 − x 0 ) = mv02 + k ( x e − x e ) 2 + F ( x1 − x 0 ) 2 2 2 2 1 1 (1.0 kg)(2.0 m /s) 2 + (3000 N / m )(0.3 m ) 2 + µ k ( mg)(0.30 m ) = 0 J + 0 J + F (0.30 m ) 2 2 The friction force is fk = µkmg. With µk = 0.30 and m = 1.0 kg, we can solve this equation to obtain F = 460 N. (b) Using the energy equation for our system once again, we have 1 2 1 1 1 K 2 + Us 2 + ∆Eth = K1 + Us1 + Wext mv2 + k ( x e − x e ) 2 + ( fk )( x 2 − x1 ) = mv12 + k ( x1 − x 0 ) 2 + 0 J 2 2 2 2 1 2 1 2 mv2 + µ k mg( x 2 − x1 ) = 0 J + k ( x1 − x 0 ) 2 2 1 1 (1.0 kg)v22 + (0.30)(1.0 kg)(9.8 m /s 2 )(1.5 m ) = (3000 N / m )(0.30 m ) 2 2 2 (0.5 kg)v22 + 4.41 J = 135 J ⇒ v2 = 16.2 m /s
11.75. Solve: (a)
The graph is a hyperbola. (b) The separation for zero potential energy is r = ∞, since
U=−
Gm1 m2 → 0 J as r → ∞ r
This makes sense because two masses don’t interact at all if they are infinitely far apart. (c) Due to the absence of nonconservative forces in our system of two particles, the mechanical energy is conserved.
The equations of energy and momentum conservation are
K f + U gf = K i + U gi
Gm1 m2 Gm1 m2 1 1 1 1 2 2 m1v12f + m2 v22f + − = m1v1i + m2 v2 i + − 2 2 2 rf 2 ri 1 1 1 1 m1v12f + m2 v22f = Gm1 m2 − 2 2 rf ri
pf = pi ⇒ m1v1f + m2 v2 f = 0 kg m / s ⇒ v2 f = −
m1 v1f m2
Substituting this expression for v2f into the energy equation, we get 2
1 1 1 1 2Gm2 1 1 m m1v12f + m2 1 v1f = Gm1 m2 − ⇒ v12f = − (1 + m1 / m2 ) rf ri 2 2 m2 rf ri With G = 6.67 × 10 −11 N ⋅ m( kg) −2 , rf = R1 + R2 = 18 × 10 8 m, ri = 1.0 × 1014 m, m1 = 8.0 × 10 30 kg, and m2 = 2.0 × 1030 kg, the above equation can be simplified to yield
v1f = 1.72 × 10 5 m /s, and v2 f = −
8.0 × 10 30 kg m1 5 5 v1f = × 1.72 × 10 m /s = 6.89 × 10 m /s m2 2.0 × 10 30 kg
(
The speed of the heavier star is 172,000 m/s. That of the lighter star is 689,000 m/s.
)