Physics (10th Edition)

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

Chapter 6 - Work and Energy - Problems - Page 167: 41


a) $KE=52.2J$ b) $v=48.77m/s$

Work Step by Step

a) Take the final point to be the highest point the ball reaches. As the ball starts to rise, its KE decreases while its PE increases. From the principle of energy conservation, we have $$KE_f-KE_0 = -(PE_f-PE_0)$$ $$KE_f-\frac{1}{2}mv_0^2=-mg\Delta h$$ $$KE_f=\frac{1}{2}m v_0^2-mg\Delta h$$ We have $g=9.8m/s^2$, $\Delta h=24.6m$, $m=47g=0.047kg$ and $v_0=52m/s$. Therefore, $$KE_f=52.2J$$ b) Similarly, from the principle of energy conservation, we have $$\Delta KE = -\Delta PE$$ $$\frac{1}{2}m\Delta v^2=-mg\Delta h$$ $$\frac{1}{2}\Delta v^2=-g\Delta h$$ We have $g=9.8m/s^2$, $\Delta h=24.6-8=16.6m$ and $\Delta v=v_{f}^2-v_{0}^2=v_{f}^2-52^2=v_{f}^2-2704$. Thus: $$\frac{1}{2}(v_{f}^2-2704)=-162.68$$ $$v_f=\sqrt{-162.68\times2+2704}=48.77m/s$$
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