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  4. 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 -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

Q. 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

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Solution:

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 $=m g \sin \theta$ acting along the plane.
Distance moved by the particle (or lift) in time $t=v t$
Work done in time $t=(m g \sin \theta) v t\left(\cos 90^{\circ}-\theta\right)=m g \sin ^{2} \theta v t$

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