A disc of radius \( 2\text{ m} \) and mass \( 100\text{ kg} \) rolls on a horizontal floor. Its centre of mass has speed of \( 20\text{ cm/s} \). How much work is needed to stop it?

Total kinetic energy of rolling disc is \( K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{3}{4}mv^2 \). Given \( m = 100\text{ kg} \), \( v = 0.2\text{ m/s} \). \( K = \frac{3}{4} \times 100 \times (0.2)^2 = 75 \times 0.04 = 3\text{ J} \). Work needed equals total kinetic energy, i.e., \( 3\text{ J} \).