JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
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 \(6\)
Answer & Solution
Correct Answer
(D) \(6\)
Step-by-step Solution
Detailed explanation
Using work energy theorem \(\mathrm{W}=\Delta \mathrm{KE}=0-\left(\frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{I} \omega^2\right)\) \(\mathrm{W}=0-\frac{1}{2} \mathrm{mv}^2\left(1+\frac{\mathrm{K}^2}{\mathrm{R}^2}\right) \)…
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