On the other hand the force of gravity has magnitude

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in the same direction as the displacement. On the other hand, the force of gravity has magnitude g F mg and is opposite in direction to the displacement. (a) Since the force of the cable F and the displacement d are in the same direction, the work done by F is 2 4 4 11 11 (72 kg)(9.8 m/s )(15 m) 1.164 10 J 1.2 10 J 10 10 F mgd W Fd . (b) Using Eq. 7-7, the work done by gravity is 2 4 4 (72 kg)(9.8 m/s )(15 m) 1.058 10 J 1.1 10 J g g W F d mgd           (c) The total work done is the sum of the two works: 4 4 3 3 net 1.164 10 J 1.058 10 J 1.06 10 J 1.1 10 J F g W W W . Since the astronaut started from rest, the work-kinetic energy theorem tells us that this is her final kinetic energy. (d) Since K mv 1 2 2 , her final speed is 3 2 2(1.06 10 J) 5.4 m/s 72 kg K v m . Note: For a general upward acceleration a , the net work done is net ( ) F g g W W W Fd F d m g a d mgd mad . Since 2 net / 2, W K mv   by the work-kinetic energy theorem, the speed of the astronaut would be 2 v ad , which is independent of the mass of the astronaut. 5. P. 7-23. The fact that the applied force a F causes the box to move up a frictionless ramp at a constant speed implies that there is no net change in the kinetic energy: 0 K . Thus, the work
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done by a F must be equal to the negative work done by gravity: a g W W   . Since the box is displaced vertically upward by 0.150 m h , we have 2 (3.00 kg)(9.80 m/s )(0.150 m) 4.41 J a W mgh   6. P. 7-27. From Eq. 7-25, we see that the work done by the spring force is given by 2 2 1 ( ) 2 s 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 3 360 N 90 N/cm 9.0 10 N/m 4.0 cm k .
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(a) When the block moves from 5.0 cm i x   to 3.0 cm x   , we have 3 2 2 1 (9.0 10 N/m)[(0.050 m) (0.030 m) ] 7.2 J. 2 s W (b) Moving from 5.0 cm i x   to 3.0 cm x   , we have 3 2 2 1 (9.0 10 N/m)[(0.050 m) ( 0.030 m) ] 7.2 J. 2 s W  
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(c) Moving from 5.0 cm i x   to 5.0 cm x   , we have 3 2 2 1 (9.0 10 N/m)[(0.050 m) ( 0.050 m) ] 0 J. 2 s W   (d) Moving from 5.0 cm i x   to 9.0 cm x   , we have 3 2 2 1 (9.0 10 N/m)[(0.050 m) ( 0.090 m) ] 25 J. 2 s W     7. P. 7-30. Hooke’s law and the work done by a spring is discussed in the chapter. We apply the work-kinetic energy theorem, in the form of K W W a s , to the points in Figure 7-35 at x = 1.0 m and x = 2.0 m, respectively. The ―applied‖ work W a is that due to the constant force P . 2 2 1 4 J (1.0 m) (1.0 m) 2 1 0 (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. 8. P. 7-33. There are two forces with nonzero work during motion of the block, the constant force F and the spring (elastic) force, F s . The normal force, F N , and weight, mg , are perpendicular to the displacement of the block so their works are both zero.
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