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%
