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) $20$
Concept. The block does not slide on the incline, so static friction exactly balances the component of gravity along the slope. The whole assembly moves straight up, so the displacement is vertical; only the vertical component of the friction force does work, and because it points up (with the motion) the work is positive.
Step 1. Friction force. Along the slope, friction balances $mg\sin\theta$ and acts up the slope:
$$f=mg\sin30^\circ=1\times10\times\frac12=5\ \text{N}$$
Step 2. Vertical component of friction. Friction acts along the incline, which makes $30^\circ$ with the horizontal, so its vertical part is
$$f\sin30^\circ=5\times\frac12=2.5\ \text{N (upwards)}$$
Step 3. Displacement. At constant velocity $4$ m/s for $2$ s, the block rises
$$s=4\times2=8\ \text{m (vertically up)}$$
Step 4. Work by friction.
$$W_f=(f\sin30^\circ)\,s=2.5\times8=20\ \text{J}$$
Answer. Option A, $20$ J. Check: friction and normal force together balance $mg$, so their total work is $mgs=80$ J; the normal force supplies $N\cos30^\circ\cdot s=mg\cos^230^\circ\cdot s=60$ J and friction the remaining $20$ J.
Solution by Sreeraj P, M.Sc Physics
