Physics (10th Edition)

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

Chapter 6 - Work and Energy - Problems - Page 170: 75


The speed of the object at $s=20m$ is $7.07m/s$

Work Step by Step

The net work done by the force is the area of the red shape as portrayed in the force-displacement graph. $$\sum W=\frac{1}{2}(10m+20m)\times10N=150J$$ The work-energy theorem states that $$\sum W=\frac{1}{2}m(v_f^2-v_0^2)$$ We know $v_0=0$ and $m=6kg$. Therefore, $$3v_f^2=150$$ $$v_f=7.07m/s$$ which is the final speed of the object at $s=20m$
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