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 11 - Work - Exercises and Problems: 46

Answer

$v = 2.4~m/s$

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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