A solid circular disc of mass 50 kg 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. Numerical answer type. Work it out, then open the answer.

Answer: 6. KE of rolling disc = 1/2mv²(1+1/2) = 3/4mv². = 3/4× 50 × (0.4)² = 6 J. Work-energy theorem: work needed to stop it = 6 J.

A solid circular disc of mass 50 kg rolls along a horizontal floor so… | JEE Main 2024 (31 Jan, Shift 1)

Q34JEE Main 2024 · 31 JAN · SHIFT 1Medium

Rolling Motion

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

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 6

KE of rolling disc $= \frac{1}{2}mv^2\left(1+\frac{1}{2}\right) = \frac{3}{4}mv^2$ .
$= \frac{3}{4}\times 50 \times (0.4)^2 = 6$ J.
Work-energy theorem: work needed to stop it $= 6$ J.

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