Q 11-05-052JEE MainJEE Main 2026 (5 Apr, Shift 2)Medium
A mass of $1$ kg is kept on a inclined plane with $30^\circ$ inclination with respect to horizontal plane and it is at rest initially. Then the whole assembly is moved up with constant velocity of $4$ m/s. The work done by the frictional force in time $2$ s is ______ J. (Take $g = 10\ \text{m/s}^2$)
Answer: (A) $20$
The block stays at rest on the incline, so static friction balances the component of gravity along the slope: $f = mg\sin 30^\circ = 5$ N, directed up the slope.
The assembly rises vertically by $4 \times 2 = 8$ m. The vertical component of friction is $f\sin 30^\circ = 2.5$ N, so
$$W_f = 2.5 \times 8 = 20\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics