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: 17


(a) $W = 0.20~J$ (b) $v = 2.98~m/s$

Work Step by Step

(a) We can find the work done by the wind as; $W = F\cdot \Delta r$ $W = F_x~\Delta r_x+F_y~\Delta r_y$ $W = (4.0\times 10^{-2}~N)(2.0~m)+(-6.0\times 10^{-2}~N)(-2.0~m)$ $W = 0.20~J$ (b) The kinetic energy will be equal to the work done by the wind. Therefore; $KE = W$ $\frac{1}{2}mv^2 = W$ $v^2 = \frac{2W}{m}$ $v = \sqrt{\frac{2W}{m}}$ $v = \sqrt{\frac{(2)(0.20~J)}{0.045~kg}}$ $v = 2.98~m/s$
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