College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 2 - Problems - Page 69: 85


(a) $F_f = 158.9~N$ (b) $\mu_k = 0.19$

Work Step by Step

(a) Since the skier moves at a constant velocity, the force of kinetic friction is equal in magnitude to the component of the skier's weight that is directed down the slope. We can find the force of kinetic friction: $F_f = mg~sin~\theta$ $F_f = (85~kg)(9.80~m/s^2)~sin~11^{\circ}$ $F_f = 158.9~N$ (b) We can find the coefficient of kinetic friction: $F_N~\mu_k = F_f$ $\mu_k = \frac{F_f}{F_N}$ $\mu_k = \frac{mg~sin~\theta}{mg~cos~\theta}$ $\mu_k = tan~\theta$ $\mu_k = tan~11^{\circ}$ $\mu_k = 0.19$
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