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 229: 37


(a) $W = 206~J$ (b) $P = 68.7~watts$

Work Step by Step

(a) Since the block moves at a steady speed, the applied force $F$ must be equal in magnitude to the force of kinetic friction. We can find the magnitude of the force $F$ which is applied by the person. We can use $\mu_k = 0.70$. Therefore; $F = F_f$ $F = mg~\mu_k$ $F = (10~kg)(9.80~m/s^2)(0.70)$ $F = 68.6~N$ We can find the distance $d$ the block moves in 3.0 seconds. $d = v~t$ $d = (1.0~m/s)(3.0~s)$ $d = 3.0~m$ We can find the work done by the person. $W = F~d$ $W = (68.6~N)(3.0~m)$ $W = 206~J$ (b) We can find the power output. $P = \frac{W}{t}$ $P = \frac{206~J}{3.0~s}$ $P = 68.7~watts$
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