A solid circular disc of mass 50kg rolls along a horizontal floor...

Using work-energy theorem, we get

$W = ∆ KE$

$⇒ W = 0 - \left(\dfrac{1}{2}m{v}^{2} + \dfrac{1}{2}I{\omega}^{2}\right)$

$⇒ W = 0 - \dfrac{1}{2}m{v}^{2}\left(1 + \dfrac{{K}^{2}}{{R}^{2}}\right)$

$= - \dfrac{1}{2} \times 50 \times 0 . {4}^{2}\left(1 + \dfrac{1}{2}\right) = - 6J$

Absolute work $= + 6J$

Therefore, required value of $W = 6J$