Question

A small block of mass 2 kg is kept on a rough inclined surface of inclination $\theta ={30}^{\circ }$ fixed in a lift. The lift goes up with a uniform speed of 1 m$s^{-1}$ 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 2 s is.

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

Answer 1

The correct option is B

9.8 J

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 =mg sin$\theta$ 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$({90}^{\circ }$- $\theta$))=mg $si{n}^2\theta$ vt

Substituting the Values we get Work = 9.8 J