Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 9 - Work and Kinetic Energy - Exercises and Problems: 46

Answer

(a) The work done on the particle from x = 0 to x = L is $\frac{1}{2}mv_0^2$ (b) The work done on the particle from x = 0 to x = 2L is zero.

Work Step by Step

(a) The work done on the particle is equal to the change in kinetic energy. We can find the speed at x = 0 $v_x = v_0~sin(\frac{\pi~x}{2L})$ $v_x = v_0~sin(\frac{\pi~(0)}{2L})$ $v_x = 0$ We can find the speed at x = L $v_x = v_0~sin(\frac{\pi~x}{2L})$ $v_x = v_0~sin(\frac{\pi~(L)}{2L})$ $v_x = v_0~sin(\frac{\pi~}{2})$ $v_x = v_0$ We can find the change in kinetic energy. $\Delta KE = KE_2-KE_1$ $\Delta KE = \frac{1}{2}mv_0^2-0$ $\Delta KE = \frac{1}{2}mv_0^2$ The work done on the particle from x = 0 to x = L is $\frac{1}{2}mv_0^2$ (b) The work done on the particle is equal to the change in kinetic energy. We can find the speed at x = 0 $v_x = v_0~sin(\frac{\pi~x}{2L})$ $v_x = v_0~sin(\frac{\pi~(0)}{2L})$ $v_x = 0$ We can find the speed at x = 2L $v_x = v_0~sin(\frac{\pi~x}{2L})$ $v_x = v_0~sin(\frac{\pi~(2L)}{2L})$ $v_x = v_0~sin(\pi)$ $v_x = 0$ We can find the change in kinetic energy. $\Delta KE = KE_2-KE_1$ $\Delta KE = 0-0$ $\Delta KE = 0$ The work done on the particle from x = 0 to x = 2L is zero.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.