# ch08 - Chapter 8 1 1. The potential energy stored by the...

This preview shows pages 1–4. Sign up to view the full content.

303 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. We use Eq. 7-12 for W g and Eq. 8-9 for U . (a) The displacement between the initial point and A is horizontal, so = 90.0° and 0 g W (since cos 90.0° = 0). (b) The displacement between the initial point and B has a vertical component of h /2 downward (same direction as F g ), so we obtain 2 5 1 1 (825 kg)(9.80 m/s )(50.0 m) 2.02 10 J 2 2 g g W F d mgh . (c) The displacement between the initial point and C has a vertical component of h downward (same direction as F g ), so we obtain 2 5 (825 kg)(9.80 m/s )(50.0 m) 4.04 10 J g g W F d mgh . (d) With the reference position at C , we obtain 2 5 1 1 2 2 B U . (e) Similarly, we find 2 5 A U . (f) All the answers are proportional to the mass of the object. If the mass is doubled, all answers are doubled. 3. (a) Noting that the vertical displacement is 10.0 m – 1.50 m = 8.50 m downward (same direction as F g ), Eq. 7-12 yields

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
CHAPTER 8 304 2 cos (2.00 kg)(9.80 m/s )(8.50 m)cos0 167 J. g W mgd   (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 upward 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. 8-1, we must again find U = – W g = –167 J. (g) With this new information (that U 0 = 100 J where y = 0) we have U i = mgy i + U 0 = 296 J. (h) With this new information (that U 0 = 100 J where y = 0) we have U f = mgy f + U 0 = 129 J. We can check part (f) by subtracting the new U i from this result. 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 (0.382 kg)(9.80 m/s )(0.498 m) 1.86 J W mgL . (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 (0.382 kg)(9.80 m/s )(0.498 m) 1.86 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.
305 (d) The force of gravity is conservative. The change in the gravitational potential energy

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This document was uploaded on 04/22/2011.

### Page1 / 66

ch08 - Chapter 8 1 1. The potential energy stored by the...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online