Physics (10th Edition)

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

Chapter 6 - Work and Energy - Problems - Page 171: 84


a) $W_{f_k}=-268.4J$ b) $f_k=141.3N$

Work Step by Step

a) As the hill makes an angle of $25^o$ with the horizontal, a distance of $1.9m$ up the hill corresponds to a $1.9\times\sin25=0.8m$ rise in height. We assume kinetic friction is the only non-conservative force here. According to the work-energy theorem, $$W_{nc}=E_f-E_0$$ $$W_{f_k}=\frac{1}{2}m(v_f^2-v_0^2)+mg(h_f-h_0)$$ We have $m=63kg, v_0=6.6m/s, v_f=4.4m/s, h_f-h_0=0.8m, g= 9.8m/s^2$ The work done by $f_k$ on the skis is $$W_{f_k}=-268.4J$$ b) The magnitude of the kinetic frictional force can be found by $$f_k=\frac{W_{f_k}}{s\cos\theta}$$ The distance of the skis is $s=1.9m$. $f_k$ opposes the motion along its way, so $\cos\theta=\cos180=-1$ $$f_k=\frac{-268.4J}{-1.9m}=141.3N$$
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