According to the work-energy theorem, the work required to stop the rolling disc is equal to its total initial kinetic energy. For a disc rolling without slipping, its total kinetic energy is the sum of its translational and rotational kinetic energies: $K_{total} = K_{trans} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$ The moment of inertia of a disc about its central axis is $I = \frac{1}{2}mr^2$. In pure rolling, the relation between linear speed and angular speed is $v = r\omega$, so $\omega = \frac{v}{r}$. Substituting these values: $K_{total} = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{1}{2}mr^2\right)\left(\frac{v}{r}\right)^2 = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$ Given: Mass, $m = 100 \text{ kg}$ Velocity, $v = 20 \text{ cm/s} = 0.2 \text{ m/s}$ $K_{total} = \frac{3}{4} \times 100 \times (0.2)^2 = 75 \times 0.04 = 3 \text{ J}$ Therefore, the work needed to stop the disc is $3 \text{ J}$.