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 ms−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 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 ms−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.
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 sinθ acting along the plane.
Distance moved by the particle (or lift) in time t=vt
Work done in time t=(mg sinθ)vt(cos(90∘- θ))=mg sin2θ vt
Substituting the Values we get Work = 9.8 J
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 sinθ acting along the plane.
Distance moved by the particle (or lift) in time t=vt
Work done in time t=(mg sinθ)vt(cos(90∘- θ))=mg sin2θ vt
Substituting the Values we get Work = 9.8 J
