Q 11-06-099JEE MainJEE Main 2024 (31 Jan, Shift 1)Medium
A solid circular disc of mass $50\ \text{kg}$ rolls along a horizontal floor so that its centre of mass has a speed of $0.4\ \text{m s}^{-1}$. The absolute value of work done on the disc to stop it is ______ J.
Numerical value type. Enter your answer.
Answer: 6
For a rolling disc $K = \tfrac12mv^2\left(1 + \dfrac{k^2}{R^2}\right) = \tfrac34mv^2$.
$$K = \tfrac34\times50\times0.16 = 6\ \text{J}$$
By the work–energy theorem $|W| = 6\ \text{J}$.
Solution by Sreeraj P, M.Sc Physics