A solid circular disc of mass $$50$$ kg rolls along a horizontal floor so that its center of mass has a speed of $$0.4 \text{ m s}^{-1}$$. The absolute value of work done on the disc to stop it is ______ J.

A solid circular disc of mass $$m = 50$$ kg rolls along a horizontal floor with center-of-mass speed $$v = 0.4$$ m/s. We need to find the work done to stop it.

We start by recalling that for a body rolling without slipping, the total kinetic energy is the sum of translational and rotational kinetic energies:

$$ KE_{total} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 $$

Here, $$I$$ is the moment of inertia about the center of mass and $$\omega$$ is the angular velocity.

Next, the rolling condition gives $$v = R\omega$$, so $$\omega = \frac{v}{R}$$, and for a solid disc, $$I = \frac{1}{2}mR^2$$.

Substituting into the rotational kinetic energy term yields:

$$ \frac{1}{2}I\omega^2 = \frac{1}{2} \cdot \frac{1}{2}mR^2 \cdot \frac{v^2}{R^2} = \frac{1}{4}mv^2 $$

Therefore, the total kinetic energy becomes:

$$ KE_{total} = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2 $$

Substituting the numerical values gives:

$$ KE_{total} = \frac{3}{4} \times 50 \times (0.4)^2 = \frac{3}{4} \times 50 \times 0.16 = \frac{3}{4} \times 8 = 6 \text{ J} $$

By the work-energy theorem, the work done to stop the disc equals the total kinetic energy (in magnitude).

The answer is $$\boxed{6}$$ J.