A solid circular disc of mass 50 kg rolls along a horizontal floor so that its center of mass has a speed of $0.4 \mathrm{~m} / \mathrm{s}$. The absolute value of work done on the disc to stop it is ______ J.

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Work, Power & Energy

A solid circular disc of mass 50 kg rolls along a horizontal floor so that its center of mass has a speed of

0.4 m / s

. The absolute value of work done on the disc to stop it is ______ J.

Answer: 6

Solution

By the work–energy theorem, the work done to stop the disc equals (minus) its total kinetic energy:

W = 0 - (\frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}) = - \frac{1}{2} m v^{2} (1 + \frac{k ^{2}}{R ^{2}})

For a solid disc,

k^{2} / R^{2} = \frac{1}{2}

, so

W = - \frac{1}{2} (50) (0.4)^{2} (1 + \frac{1}{2}) = - \frac{1}{2} (50) (0.16) (1.5) = - 6 J

The magnitude of the work done on the disc is

6 J

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Integer answer

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