College Physics (4th Edition)

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

Chapter 4 - Problems - Page 148: 17


(a) The skier's speed at the bottom is $22.8~m/s$ (b) $\mu_k = 0.19$

Work Step by Step

(a) We can find the skier's acceleration down the slope: $\sum F = ma$ $mg~sin~\theta = ma$ $a = g~sin~\theta$ $a = (9.80~m/s^2)~sin~32^{\circ}$ $a = 5.19~m/s^2$ We can find the speed at the bottom: $v_f^2 = v_0^2+2ad$ $v_f = \sqrt{v_0^2+2ad}$ $v_f = \sqrt{0+(2)(5.19~m/s^2)(50~m)}$ $v_f = 22.8~m/s$ The skier's speed at the bottom is $22.8~m/s$ (b) We can find the deceleration on the horizontal surface: $v_f^2 = v_0^2+2ad$ $a = \frac{v_f^2-v_0^2}{2d}$ $a = \frac{0-(22.8~m/s)^2}{(2)(140~m)}$ $a = -1.86~m/s^2$ We can use the magnitude of acceleration to find the coefficient of kinetic friction: $F_f = ma$ $mg~\mu_k = ma$ $\mu_k = \frac{a}{g}$ $\mu_k = \frac{1.86~m/s^2}{9.80~m/s^2}$ $\mu_k = 0.19$
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