1. Given data

   • Mass of disc: $m = 50\,\text{kg}$
   • Speed of center of mass: $v = 0.4\,\text{m/s}$
   • The disc is rolling without slipping.

2. Total kinetic energy of a rolling body For pure rolling, $K = K_{\text{trans}} + K_{\text{rot}}$ $K = \frac12 mv^2 + \frac12 I\omega^2$

3. Moment of inertia of a solid disc about its center $I = \frac12 mR^2$

4. Rolling condition $v = \omega R \quad \Rightarrow \quad \omega = \frac{v}{R}$

5. Rotational kinetic energy $K_{\text{rot}} = \frac12 \left(\frac12 mR^2\right)\left(\frac{v}{R}\right)^2$ $K_{\text{rot}} = \frac14 mv^2$

6. Translational kinetic energy $K_{\text{trans}} = \frac12 mv^2$

7. Total kinetic energy $K = \frac12 mv^2 + \frac14 mv^2 = \frac34 mv^2$

8. Substitute values $K = \frac34 \times 50 \times (0.4)^2$ $K = \frac34 \times 50 \times 0.16$ $K = \frac34 \times 8$ $K = 6\,\text{J}$

9. Work required to stop the disc To stop the disc, the magnitude of work done on it must equal its initial total kinetic energy: $|W| = 6\,\text{J}$

Therefore, the required integer answer is 6.