A small block of mass 2 kg is kept on a rough inclined surface of inclination θ = 30∘ fixed in a lift. The lift goes up with a uniform speed of 1 m/s 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 :
Answer & explanation
Correct answer: option 3
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.
\(\Rightarrow friction = mg \sin {\theta}\) acting along the plane.
Distance moved by the particle (or lift) in time t = vt
Work done in time \(W = mg \sin {\theta} vt \cos ({90^o - \theta})\)
= \(vt mg \sin^2 {\theta}\)
= 1 m/s x 2 s x 2 kg x 9.8 m/s2 x sin2 30o
= 9.8 J