Answer
$K.E_i=133J$
$K.E_f=34.8J$
Work Step by Step
We can find the required initial and final kinetic energies as follows:
$K.E_i=\frac{1}{2}mv_f^2$
We plug in the known values to obtain:
$K.E_i=\frac{1}{2}(34.0)(2.80)$
$K.E_i=133J$
Now $K.E_f=\frac{1}{2}I\omega^2=\frac{1}{2}(I+mr^2)\omega^2$
We plug in the known values to obtain:
$K.E_f=\frac{1}{2}[512+(34.0)(2.31)^2](0.317)$
$K.E_f=34.8J$