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

Published by Pearson

# Chapter 9 - Work and Kinetic Energy - Exercises and Problems - Page 229: 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.

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