Physics (10th Edition)

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

Chapter 6 - Work and Energy - Problems - Page 167: 37


The speed of the gymnast at the bottom of the swing is $6.57m/s$

Work Step by Step

From the bottom of the swing to the top, the gymnast's increases by $\Delta h=1.1\times2=2.2m$ This change in height leads to an increase in PE: $\Delta PE=mg\Delta h=2.2mg (J)$ According to the principle of energy conservation, an increase in PE means a decrease in KE by the same amount: $\Delta KE=-\Delta PE=-2.2mg$ We know that $$\Delta KE=\frac{1}{2}m\Delta v^2=-2.2mg$$ $$\frac{1}{2}(v_f^2-v_0^2)=-2.2g$$ $$v_f^2-v_0^2=-4.4g=-43.12$$ When the gymnast is at the top, his speed is zero, so $v_f=0$ Therefore, $$v_0=\sqrt{v_f^2+43.12}=6.57m/s$$ which is the speed of the gymnast at the bottom of the swing.
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