A150 kg hoop rolls along a horizontal floor so that the hoop's center of mass has a speed of 0.130 m/s. how much work must be done on the hoop to stop it?

The work needed is equal to the KE of the hoop which is = 150 * 0.130^2 / 2 ...(1) 
Let the following be:r be the radiuscenter of mass in the moment of inertia is denoted by:
I = 150 r^2 kg m^2. 
The angular velocity relative to the center of mass is: 
w = 0.130 / r rad / s. 
Rotational KE = Iw^2 / 2 
= (150 * r^2) (0.130 / r)^2 / 2 
= 150 * 0.130^2 / 2 J ...(2) 
Adding (1) and (2): 
Total KE = 130 * 0.130^2 
= 2.2 J.

Hello how was your day? ? i hope it was good