Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 10 - Energy - Exercises and Problems: 27

Answer

(a) At point A, the minimum speed is 7.7 m/s (b) At point B, the minimum speed is 10 m/s

Work Step by Step

(a) At the midpoint between point A and point B, the potential energy is at a maximum of 5.0 J. For the particle at point A to reach point B, the total energy must be at least 5.0 J. We can find the minimum kinetic energy required at point A. $K+U = E$ $K = E-U$ $K = 5.0~J - 2.0~J$ $K = 3.0~J$ We can find the minimum speed of the particle at point A. $K = 3.0~J$ $\frac{1}{2}mv^2 = 3.0~J$ $v^2 = \frac{(2)(3.0~J)}{m}$ $v = \sqrt{\frac{(2)(3.0~J)}{m}}$ $v = \sqrt{\frac{(2)(3.0~J)}{0.10~kg}}$ $v = 7.7~m/s$ At point A, the minimum speed is 7.7 m/s. (b) At the midpoint between point A and point B, the potential energy is at a maximum of 5.0 J. For the particle at point B to reach point A, the total energy must be at least 5.0 J. We can find the minimum kinetic energy required at point B. $K+U = E$ $K = E-U$ $K = 5.0~J - 0$ $K = 5.0~J$ We can find the minimum speed of the particle at point B. $K = 5.0~J$ $\frac{1}{2}mv^2 = 5.0~J$ $v^2 = \frac{(2)(5.0~J)}{m}$ $v = \sqrt{\frac{(2)(5.0~J)}{m}}$ $v = \sqrt{\frac{(2)(5.0~J)}{0.10~kg}}$ $v = 10~m/s$ At point B, the minimum speed is 10 m/s.
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