Answer
$K.E_t=170J$
$K.E_r=0.072J$
Work Step by Step
We can find the translational energy as
$K.E_t=\frac{1}{2}Mv^2$
We plug in the known values to obtain:
$K.E_t=\frac{1}{2}(0.15)(48)$
$K.E_t=170J$
Now we can find the rotational kinetic energy as below
$K.E_r=\frac{1}{2}I\omega^2$
$\implies K.E_r=\frac{1}{2}(\frac{2}{5}MR^2)\omega^2=\frac{1}{5}MR^2\omega^2$
We plug in the known values to obtain:
$K.E_r=\frac{1}{5}(0.15)(0.037)(42)=0.072J$
