Practicing Success
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 : |
19.6 J 4.9 J 9.8 J \(zero\) |
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. \(\Rightarrow friction = mg \sin {\theta}\) 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 |