Answer
a) $K_f=15J$
b) $K_f=\frac{3}{4}mv^2$
c) $K_r=4.9J$
$K_t=9.8J$
Work Step by Step
(a) We know that
$U_i+K_i=U_f+K_f$
$\implies mgh+0=0+K_f$
$K_f=mgh$
We plug in the known values to obtain:
$K_f=2.0(9.81)(0.75)$
$K_f=15J$
(b) As $K_f=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2$
$\implies K_f=\frac{1}{2}mv^2+\frac{1}{2}(\frac{1}{2}mr^2)(\frac{v}{r})^2=\frac{3}{4}mv^2$
(c) We can find the translational and rotational kinetic energy as follows:
$K_r=\frac{1}{3}K_f$
$\implies K_r=\frac{1}{3}(14.7)=4.9J$
$K_t=\frac{2}{3}(14.7)=9.8J$
