Question Detail
Question
A 140 kg hoop rolls along a horizontal floor so that the hoop's centre of mass has a speed of 0.150 m/s. How much work must be done on the hoop to stop it?
Correct Answer
Option A is the correct answer.
Detailed Explanation
A hoop rolling has kinetic energy from translation and rotation. For a hoop, I = mR². Total KE = ½mv² + ½Iω² = ½mv² + ½(mR²)(v²/R²) = ½mv² + ½mv² = mv². So KE = 140 kg * (0.15 m/s)² = 140 * 0.0225 = 3.15 J. The work needed to stop it equals its initial kinetic energy, 3.15 J.
Hint
The work-energy theorem states that the work done to stop the hoop equals its initial kinetic energy. Remember to include rotational kinetic energy for a rolling object.