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QJEEM24Ph32d7: JEE Main 2024 Physics - Class 11 JEEM24Ph32d7 Solution: A solid circular disc of mass rolls along a horizontal floor so …

A solid circular disc of mass 50 kg50 \mathrm{~kg} rolls along a horizontal floor so that its center of mass has a speed of 0.4 m/s0.4 \mathrm{~m} / \mathrm{s} . The absolute value of work done on the disc to stop it is ________ J.

Step-by-Step Verified Solution

Using work energy theorem

amp;W=ΔKE=0−(12mv2+12Iω2)amp;W=0−12mv2(1+K2R2)amp;=−12×50×0.42(1+12)=−6 J\begin{aligned}& \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}\end{aligned}

Absolute work =+6 J=+6 \mathrm{~J}

W=−6J∣W∣=6JW=-6 J \quad|W|=6 J

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