A hoop of radius 1.5 m has mass of 50 kg. It rolls along a horizontal surface with speed of center of mass 25 cms^-1. Work done to stop it is

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3.125 J

6.25 J

zero

12.5 J

answer is A.

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Detailed Solution

$W = Δ K E$

$\begin{array}{l} = \frac{1}{2} m v^{2} (1 + \frac{k^{2}}{r^{2}}) \\ = \frac{1}{2} \times 50 \times {(\frac{1}{4})}^{2} (1 + 1) \\ = 3.125 J \end{array}$

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