Answer
$W=-3.15J$
Work Step by Step
Work is equal to the change in kinetic energy: $W=∆KE$
The initial kinetic energy has translational and rotational components:
$KE_i=\frac{1}{2}mv^2+\frac{1}{2}I_{hoop}\omega^2$
We know that:
$I_{hoop}=mr^2$
and
$\omega=\frac{v}{r}$
Therefore, the formula becomes:
$KE_i=\frac{1}{2}mv^2+\frac{1}{2}(mr^2)(\frac{v}{r})^2=\frac{1}{2}mv^2+\frac{1}{2}mv^2=mv^2=(140kg)(0.150m/s)^2=3.15J$
The final kinetic energy is $0J$ because the hoop stops: $KE_f=0J$
$W=KE_f-KE_i=0J-3.15J=-3.15J$
