Answer
$h = 14.0 \, \mathrm{m}$
Work Step by Step
Total energy at the bottom of the hill:
$$K = \frac{Mv_{cm}^2}{2} + \frac{I\omega^2}{2} = \frac{M\omega^2R^2}{2} + \frac{0.800MR^2\omega^2}{2} = 0.900M\omega^2R^2$$
Total energy at height $h$:
$$ E = U - W_{f} = Mgh + W_f $$
Conservation of energy yields:
$$0.900M\omega^2R^2 = Mgh + W_f \Rightarrow$$
$$h = \frac{0.900\omega^2R^2}{g} - \frac{W_f}{Mg} = \frac{0.900\omega^2R^2}{g} - \frac{W_f}{w} = $$
$$ \frac{0.900(25.0 \, \mathrm{rad/s})^2(0.600 \, \mathrm{m})^2}{9.80 \, \mathrm{m/s^2}} - \frac{2600 \, \mathrm{J}}{392 \, \mathrm{N}} = 14.0 \, \mathrm{m} $$
