A small block of mass `2kg` is kept on a rough inclined surface of inclination `theta=30^@` fixed in a lift. The lift goes up with a uniform speed of `1ms^-1` and the block does not slide relative to the inclined surface. The work done by the force of friction on the block in a time interval of `2s` is

Options

A. Zero
B. `9.8J`
C. `29.4J`
D. `16.9J`

Answer

B

Explanation

Since the lift's speed is constant, in the absence of acceleration, there will be no pseudo-force (referring to the lift as the frame of reference) or additional reaction/thrust due to inclined plane (referring to ground as the reference frame) on the particle.

Therefore, friction=`mgsintheta` acting along the plane.
Distance moved by the particle (or lift) in time `t=vt`
Work done in time `t=(mg sin theta)vt (cos 90-theta)=mgsin^2thetavt`