## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 10 - Problems - Page 401: 78

#### Answer

(a) $E = 6.125\times 10^{-3}~J$ (b) The average percentage of energy lost during one cycle is 1.1%

#### Work Step by Step

(a) We can find the energy in the pendulum: $E = \frac{1}{2}~m~A^2~\omega^2$ $E = \frac{m~A^2~g}{2~L}$ $E = \frac{(0.50~kg)~(0.050~m)^2~(9.80~m/s^2)}{(2)~(1.0~m)}$ $E = 6.125\times 10^{-3}~J$ (b) We can find the period: $T = 2\pi~\sqrt{\frac{L}{g}}$ $T = 2\pi~\sqrt{\frac{1.0~m}{9.80~m/s^2}}$ $T = 2.0~s$ We can find the number of cycles in one week: $\frac{(7)(24)(3600~s)}{2.0~s} = 302,400~cycles$ We can find the magnitude of the change in potential energy when a 2.0-kg mass drops $1.0~m$: $\Delta U_g = mgh$ $\Delta U_g = (2.0~kg)(9.80~m/s^2)(1.0~m)$ $\Delta U_g = 19.6~J$ We can find the amount of energy lost each cycle: $\frac{19.6~J}{302,400~cycles} = 6.48\times 10^{-5}~J$ We can find the fraction of the energy that is lost each cycle: $\frac{6.48\times 10^{-5}~J}{6.125\times 10^{-3}~J} = 0.011$ The average percentage of energy lost during one cycle is 1.1%

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