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: 45


$v = 2.4~m/s$

Work Step by Step

The sum of the work done by the tension and the friction will be equal to the kinetic energy. We can find the work done by the tension. $W_T = T~d~cos(\theta)$ $W_T = (30~N)(3.0~m)~cos(30^{\circ})$ $W_T = 77.94~J$ We can find the work done by friction. $W_f = F_f~d~cos(180^{\circ})$ $W_f = F_N~\mu_k~d~cos(180^{\circ})$ $W_f = [mg-T~sin(30^{\circ})]~\mu_k~d~cos(180^{\circ})$ $W_f = [(10~kg)(9.80~m/s^2)-(30~N)~sin(30^{\circ})](0.20)(3.0~m)~cos(180^{\circ})$ $W_f = -49.8~J$ We can find Paul's speed. $KE = W_T+W_f$ $\frac{1}{2}mv^2 = W_T+W_f$ $v^2 = \frac{2(W_T+W_f)}{m}$ $v = \sqrt{\frac{2(W_T+W_f)}{m}}$ $v = \sqrt{\frac{(2)(77.94~J-49.8~J)}{10~kg}}$ $v = 2.4~m/s$
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