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Unformatted text preview: 309 Chapter 8 1. (a) Noting that the vertical displacement is 10.0 m – 1.50 m = 8.50 m downward (same direction as r F g ), Eq. 712 yields 2 cos (2.00 kg)(9.80 m/s )(8.50 m)cos 0 167 J. g W mgd φ = = ° = (b) One approach (which is fairly trivial) is to use Eq. 81, but we feel it is instructive to instead calculate this as ∆ U where U = mgy (with upwards understood to be the + y direction). The result is 2 ( ) (2.00 kg)(9.80 m/s )(1.50 m 10.0 m) 167 J. f i U mg y y ∆ = = =  (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. 81, 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. 2. (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 = = = . CHAPTER 8 310 (b) In going from its initial position to the highest point on its path, the ball moves vertically through a distance equal to L , but this time the displacement is upward, opposite the direction of the force of gravity. 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 =  =  =  (c) The final position of the ball is at the same height as its initial position. The displacement is horizontal, perpendicular to the force of gravity. The force of gravity does no work during this displacement. (d) The force of gravity is conservative. The change in the gravitational potential energy of the ballEarth system is the negative of the work done by gravity: 2 (0.341 kg)(9.80 m/s )(0.452 m) 1.51 J U mgL ∆ =  =  =  as the ball goes to the lowest point. (e) Continuing this line of reasoning, we find 2 (0.341 kg)(9.80 m/s )(0.452 m) 1.51 J U mgL ∆ = + = = as it goes to the highest point. (f) Continuing this line of reasoning, we have ∆ U = 0 as it goes to the point at the same height. (g) The change in the gravitational potential energy depends only on the initial and final positions of the ball, not on its speed anywhere. The change in the potential energy is the same since the initial and final positions are the same....
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 Summer '10
 Prof.Yang
 Physics

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