JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Work, Power & Energy

While the block rests on the incline, static friction balances the component of gravity along the surface, $f=mg\sin 30^{\circ }=1\times 10\times 0.5=5\text{N}$, directed up the incline. When the whole assembly is carried vertically upward at constant speed the block stays in equilibrium, so this friction force is unchanged in magnitude and direction. In $2\text{s}$ the vertical displacement is $d=4\times 2=8\text{m}$. The friction force points up the incline, which makes an angle of $60^{\circ }$ with the vertical displacement, so the work done is $W=fd\cos 60^{\circ }=5\times 8\times 0.5=20\text{J}$. The work done by the frictional force is $20\text{J}$.