JEE mains 2024 PYQ Of Rotational motion

Q. A solid circular disc of mass 50kg rolls along a horizontal floor so that its center of mass has a speed of 0.4 m/s. The absolute value of work done on the disc to stop it is _______ J.

(jee mains 2024)

Ans:- (6)

Here, we will use the Work-Energy Theorem:

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

Simplifying,

W = - \left( \frac{1}{2} \times 50 \times 0.4^2 \times \left( 1 + \frac{1}{2} \right) \right)

Absolute work = +6 J

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