JEE Main 2026 — Laws of Motion Question with Solution

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

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

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:

$W=fs\cos 60^{\circ }$

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

Answer: $20$