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Unformatted text preview: Chapter 8 1. The potential energy stored by the spring is given by U kx = 1 2 2 , where k is the spring constant and x is the displacement of the end of the spring from its position when the spring is in equilibrium. Thus k U x = = = 2 2 25 0 075 8 9 10 2 2 3 J m N m b g b g . . . 2. (a) Noting that the vertical displacement is 10.0 1.5 = 8.5 m downward (same direction as F g ), Eq. 7-12 yields W mgd g = = = cos ( . ) ( . ) ( . ) cos 2 00 9 8 8 5 0 167 J . (b) One approach (which is fairly trivial) is to use Eq. 8-1, but we feel it is instructive to instead calculate this as U where U = mgy (with upwards understood to be the + y direction). U mgy mgy f i =- =- = - ( . )( . )( . ) ( . )( . )( . ) 2 00 9 8 15 2 00 9 8 10 0 167J. (c) In part (b) we used the fact that U i = mgy i =196 J. (d) In part (b), we also used the fact U f = mgy f = 29 J. (e) The computation of W g does not use the new information (that U = 100 J at the ground), so we again obtain W g = 167 J. (f) As a result of Eq. 8-1, we must again find U = W g = 167 J. (g) With this new information (that U = 100 J where y = 0) we have U i = mgy i + U = 296 J. (h) With this new information (that U = 100 J where y = 0) we have U f = mgy f + U = 129 J. We can check part (f) by subtracting the new U i from this result. 3. (a) The force of gravity is constant, so the work it does is given by W F d = , where 309 CHAPTER 8 F is the force and d is the displacement. The force is vertically downward and has magnitude mg , where m is the mass of the flake, so this reduces to W = mgh , where h is the height from which the flake falls. This is equal to the radius r of the bowl. Thus W mgr = = = --- ( . )( . 2 00 10 22 0 10 3 2 kg)(9.8 m s m) 4.31 10 J. 2 3 (b) The force of gravity is conservative, so the change in gravitational potential energy of the flake-Earth system is the negative of the work done: U = W = 4.31 10 3 J. (c) The potential energy when the flake is at the top is greater than when it is at the bottom by | U |. If U = 0 at the bottom, then U = +4.31 10 3 J at the top. (d) If U = 0 at the top, then U = 4.31 10 3 J at the bottom. (e) All the answers are proportional to the mass of the flake. If the mass is doubled, all answers are doubled. 4. (a) The only force that does work on the ball is the force of gravity; the force of the rod is perpendicular to the path of the ball and so does no work. In going from its initial position to the lowest point on its path, the ball moves vertically through a distance equal to the length L of the rod, so the work done by the force of gravity is 2 (0.341 kg)(9.80 m/s )(0.452 m) 1.51 J W mgL = = = ....
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