A solid circular disc of mass 50 mathrm{~kg} rolls along a horizontal floor so that its center of ma

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6.System of Particles and Rotational Motion

hard

A solid circular disc of mass $50 \mathrm{~kg}$ 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____ $\mathrm{J}$.

A

$3$

B

$4$

C

$5$

D

$6$

(JEE MAIN-2024)

Solution

Using work energy theorem

$\mathrm{W}=\Delta \mathrm{KE}=0-\left(\frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{I} \omega^2\right)$

$\mathrm{W}=0-\frac{1}{2} \mathrm{mv}^2\left(1+\frac{\mathrm{K}^2}{\mathrm{R}^2}\right) $

$=-\frac{1}{2} \times 50 \times 0.4^2\left(1+\frac{1}{2}\right)=-6 \mathrm{~J}$

Absolute work $=+6 \mathrm{~J}$

$W=-6 J \quad|W|=6 J$

Standard 11

Physics

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