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JEE Mains · Physics · STD 11 - System of particles and rotational motion
  • JEE Mains 2024
  • Hard
  • Numerical

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 _______ \(\mathrm{J}\).

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Answer & Solution

Correct Answer

(D) \(6\)

Step-by-step Solution

Detailed explanation

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 \(=+6 \mathrm{~J}\)

\(W=-6 J \quad|W|=6 J\)

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