Physics (10th Edition)

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

Chapter 6 - Work and Energy - Problems - Page 171: 89


$\sum W=1474J$

Work Step by Step

The work done by force $F$ on an object has the following formula $$W=(F\cos\theta)s$$ The sign of $W$ depends on the angle $\theta$. - Work done by weight $W$: since $\theta=65^o$, $\cos\theta\gt0$, so $W_W\gt0$ - Work done by friction $f$: since $\theta=180^o$, $\cos\theta\lt0$, so $W_f\lt0$ - Work done by normal force $F_N$: since $\theta=90^o$, $\cos\theta=0$, so $W_N=0$ The net work done by 3 forces over a distance of $s=9.2m$ is $$\sum W=(W\cos65+f\cos180+0)s$$ We have $W=675N$ and $f=125N$, so $$\sum W=1474J$$
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