College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 10 - Problems - Page 402: 80


In ten cycles, the energy decreases by 9.75%

Work Step by Step

We can write an expression for the initial value of energy: $E_1 = \frac{1}{2}m~A_1^2~\omega^2$ Note that $A_2 = 0.950~A_1$. We can find the new energy of the pendulum: $E_2 = \frac{1}{2}m~A_2^2~\omega^2$ $E_2 = \frac{1}{2}m~(0.950~A_1)^2~\omega^2$ $E_2 = 0.9025\times \frac{1}{2}m~A_1^2~\omega^2$ $E_2 = 0.9025\times E_1$ We can find the percentage decrease in the energy: 100% - 90.25% = 9.75% In ten cycles, the energy decreases by 9.75%.
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