A mass of 1 kg is kept on an inclined plane with 30° inclination...

Since the assembly moves with a constant velocity, the acceleration of the mass is zero. The mass is in equilibrium relative to the incline.

The frictional force $f$ acting on the mass balances the component of gravity along the inclined plane. Therefore, the magnitude of the frictional force is:

$f = mg \sin 30^\circ$

Substituting the given values ($m = 1\text{ kg}$, $g = 10\text{ m/s}^2$):

$f = 1 \times 10 \times \dfrac{1}{2} = 5\text{ N}$

The direction of the frictional force is upwards along the incline. Since the incline makes an angle of $30^\circ$ with the horizontal, the frictional force makes an angle of $90^\circ - 30^\circ = 60^\circ$ with the vertical upward direction.

The displacement $s$ of the assembly in time $t = 2\text{ s}$ is vertically upwards:

$s = vt = 4 \times 2 = 8\text{ m}$

The work done by the frictional force is:

$W = f s \cos 60^\circ$

$W = 5 \times 8 \times \dfrac{1}{2} = 20\text{ J}$

Answer: $20$