Answer
(a) h = 0.673 m
(b) The pulley has 45.5% of the total kinetic energy.
Work Step by Step
Let $v$ be the speed of the stone after it drops a height $h$. We can calculate the required speed $v$.
$KE = \frac{1}{2}I\omega^2 = 4.50~J$
$\frac{1}{2}(\frac{1}{2}m_p~r^2)(\frac{v}{r})^2 = 4.50~J$
$v^2 = \frac{(4)(4.50~J)}{m_p}$
$v^2 = \frac{(4)(4.50~J)}{2.50~kg}$
$v^2 = 7.20~m^2/s^2$
We can find the required height $h$ for the stone to fall.
$m_sgh = \frac{1}{2}m_sv^2+\frac{1}{2}I\omega^2$
$m_sgh = \frac{1}{2}m_sv^2+\frac{1}{2}(\frac{1}{2}m_p~r^2)(\frac{v}{r})^2$
$m_sgh = \frac{1}{2}m_sv^2+\frac{1}{4}m_p~v^2$
$h = \frac{v^2~(\frac{1}{2}m_s+\frac{1}{4}m_p)}{m_s~g}$
$h = \frac{(7.20~m^2/s^2)~[\frac{1}{2}(1.50~kg)+\frac{1}{4}(2.50~kg)]}{(1.50~kg)(9.80~m/s^2)}$
$h = 0.673~m$
(b) We can find the kinetic energy of the stone.
$KE = \frac{1}{2}m_sv^2$
$KE = \frac{1}{2}(1.50~kg)(7.20~m^2/s^2)$
$KE = 5.40~J$
We can find the percent of kinetic energy in the pulley.
$\frac{4.50~J}{4.50~J+5.40~J}\times 100\% = 45.5\%$
The pulley has 45.5% of the total kinetic energy.
