Answer
(a) $a = 1.12~m/s^2$
(b) $T = 14.0~N$
Work Step by Step
(a) We can set up a torque equation for the flywheel.
$\tau = I\alpha$
$T~R = I(\frac{a}{R})$
$T = \frac{Ia}{R^2}$
We can use this expression in the force equation for the block. Let $m$ be the mass of the block.
$\sum F = ma$
$mg~sin(\theta)-F_f-T = ma$
$mg~sin(\theta)-mg~\mu_k~cos(\theta) = ma+T$
$mg~sin(\theta)-mg~\mu_k~cos(\theta) = ma+ \frac{Ia}{R^2}$
$a = \frac{mg~sin(\theta)-mg~\mu_k~cos(\theta)}{m+ \frac{I}{R^2}}$
$a = \frac{(5.00~kg)(9.80~m/s^2)~sin(36.9^{\circ})-(5.00~kg)(9.80~m/s^2)(0.25)~cos(36.9^{\circ})}{5.00~kg+ \frac{0.500~kg~m^2}{(0.200~m)^2}}$
$a = 1.12~m/s^2$
(b) $T = \frac{Ia}{R^2}$
$T = \frac{(0.500~kg~m^2)(1.12~m/s^2)}{(0.200~m)^2}$
$T = 14.0~N$
