JEE MAIN - Physics (2024 - 31st January Morning Shift - No. 21)
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 ________ J.
Answer
6
Explanation
Using work energy theorem
$$\begin{aligned} & \mathrm{W}=\Delta \mathrm{KE}=0-\left(\frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{I} \omega^2\right) \\ & \mathrm{W}=0-\frac{1}{2} \mathrm{mv}^2\left(1+\frac{\mathrm{K}^2}{\mathrm{R}^2}\right) \\ & =-\frac{1}{2} \times 50 \times 0.4^2\left(1+\frac{1}{2}\right)=-6 \mathrm{~J} \end{aligned}$$
Absolute work $$=+6 \mathrm{~J}$$
$$W=-6 J \quad|W|=6 J$$
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