Question
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.
Answer: Option B
:
B
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 sinθacting 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))=mgsin2θvt
Substituting the Values we get Work = 9.8 J
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:
B
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 sinθacting 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))=mgsin2θvt
Substituting the Values we get Work = 9.8 J
Was this answer helpful ?
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