Question

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.

Answer 1

Using the work-energy theorem:

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

Since the disc rolls without slipping, we use:

$I = \frac{K^2}{R^2}$

Thus:

$W = 0 - \frac{1}{2}mv^2 \left( 1 + \frac{K^2}{R^2} \right)$

Substituting the values:

$W = -\frac{1}{2} \times 50 \times 0.4^2 \left( 1 + \frac{1}{2} \right) = -6J$

Therefore, the absolute value of work is:

$|W| = 6J$