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:
\[ |W| = 6J \]
A force \( \vec{f} = x^2 \hat{i} + y \hat{j} + y^2 \hat{k} \) acts on a particle in a plane \( x + y = 10 \). The work done by this force during a displacement from \( (0,0) \) to \( (4m, 2m) \) is Joules (round off to the nearest integer).