A small block of mass 2 kg is kept on a rough inclined surface of inclination $\mathrm{\theta }={30}^{\circ }$ 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

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 $=\mathrm{mgsin}\mathrm{\theta }$ acting along the plane. 

Distance moved by the particle (or lift) in time t = vt 

Work done in time $\mathrm{t} \mathrm{is} \mathrm{W}=(\mathrm{mgsin}\mathrm{\theta })\mathrm{vt}\left(\cos {90}^{\circ }-\mathrm{\theta }\right)={\mathrm{mgsin}}^2\mathrm{\theta vt}=2\times 9.8\times \frac{1}{4}\times 1\times 2=9.8 \mathrm{J}$