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 - Page 228: 21


The particle's velocity at x = 3 m is 11.1 m/s

Work Step by Step

The work done on the particle is equal to the area under the force versus position graph. We can find the work done during the interval 0 - 3 m: $W = (15~N)(1~m)+\frac{1}{2}(15~N)(2~m)$ $W = 30~J$ We can use the work-energy theorem to find the particle's velocity at x = 3 m. $KE_2 = KE_1+W$ $\frac{1}{2}mv_2^2 = \frac{1}{2}mv_1^2+W$ $v_2^2 = \frac{mv_1^2+2W}{m}$ $v_2 = \sqrt{\frac{mv_1^2+2W}{m}}$ $v_2 = \sqrt{\frac{(0.50~kg)(2.0~m/s)^2+(2)(30~J)}{0.50~kg}}$ $v_2 = 11.1~m/s$ The particle's velocity at x = 3 m is 11.1 m/s
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