STANDARD Physics Kinetic Energy MCQ Question
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⁻¹. How much work has to be done to stop it?
2 J
4 J
6 J
8 J
Correct Answer
Detailed Explanation
The work done to stop the hoop is equal to its kinetic energy. The kinetic energy of the hoop is given by KE = 1/2 mv² + 1/2 Iω². Since I = mr² for a hoop and v = rω, the total kinetic energy is KE = mv². Substituting the given values, KE = 100 kg × (0.2 m/s)² = 4 J.
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