Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 4 - Forces and Newton's Laws of Motion - Problems - Page 119: 108


(a) The deceleration of the skater is $-0.98m/s^2$ (b) The skater will travel $29.5m$ before coming to rest.

Work Step by Step

(a) Since the skater stops propelling himself, the only force influencing the motion is kinetic friction $f_k$, which slows the motion. We call the deceleration caused by kinetic friction $a$, and we have $$f_k=m_{skater}a$$ $$\mu_k F_N=m_{skater}a$$ Because there is no vertical movement, $F_N$ equals the skater's weight, so $$\mu_k m_{skater}g=m_{skater}a$$ $$a=\mu_kg=0.1\times9.8=0.98m/s^2$$ Since $a$ is deceleration, we take $a\lt 0$, therefore, $a=-0.98m/s^2$ (b) We have $v_0=7.6m/s$, $v=0$ and $a=-0.98m/s^2$ The distance the skater will travel before coming to rest is $$x=\frac{v^2-v_0^2}{2a}=29.5m$$
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