Answer
The wheel's rotational kinetic energy is $~~10~J$
Work Step by Step
We can find the speed of the box:
$K = \frac{1}{2}mv^2 = 6.0~J$
$v^2 = \frac{(2)(6.0~J)}{m}$
$v = \sqrt{\frac{(2)(6.0~J)}{m}}$
$v = \sqrt{\frac{(2)(6.0~J)}{6.0~kg}}$
$v = 1.414~m/s$
We can find the angular speed of the wheel:
$\omega = \frac{v}{r}$
$\omega = \frac{1.414~m/s}{0.20~m}$
$\omega = 7.07~rad/s$
We can find the wheel's rotational kinetic energy:
$K = \frac{1}{2}~I~\omega^2$
$K = \frac{1}{2}~(0.40~kg~m^2)~(7.07~rad/s)^2$
$K = 10~J$
The wheel's rotational kinetic energy is $~~10~J$
