$A$ hoop of radius $2 \; m$ weighs $100 \; kg$. It rolls along a horizontal floor so that its centre of mass has a speed of $20 \; cm/s$. How much work has to be done to stop it?

(4 J) Radius of the hoop,$r = 2 \; m$.
Mass of the hoop,$m = 100 \; kg$.
Velocity of the centre of mass,$v = 20 \; cm/s = 0.2 \; m/s$.
The total kinetic energy $(K)$ of a rolling hoop is the sum of its translational kinetic energy and rotational kinetic energy.
$K = K_{\text{trans}} + K_{\text{rot}} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.
For a hoop,the moment of inertia about its centre is $I = mr^2$.
Since the hoop is rolling without slipping,$v = r\omega$,which implies $\omega = v/r$.
Substituting $I$ and $\omega$ into the energy equation:
$K = \frac{1}{2}mv^2 + \frac{1}{2}(mr^2)(v/r)^2 = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$.
The work required to stop the hoop is equal to its total kinetic energy.
$W = K = mv^2 = 100 \; kg \times (0.2 \; m/s)^2 = 100 \times 0.04 = 4 \; J$.