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(C) The block is at rest with respect to the lift. The lift moves upward with a constant velocity $v$. In time $t$,the displacement of the lift (and t

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$mgtv \sin^2 	heta$

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(C) The block is at rest with respect to the lift. The lift moves upward with a constant velocity $v$. In time $t$,the displacement of the lift (and the block) is $d = vt$ in the upward direction.The forces acting on the block are gravity ($mg$ downwards),normal force $(N)$,and static friction $(f)$.Since the block is at rest relative to the inclined plane,the friction force $f$ must balance the component of gravity along the incline: $f = mg \sin 	heta$.The direction of the friction force $f$ is along the inclined plane,pointing upwards.The angle between the friction force $f$ and the displacement vector $d$ (which is vertical) is $(90^\circ - 	heta)$.The work done by the friction force is given by $W_f = f \cdot d \cdot \cos(90^\circ - 	heta)$.Substituting the values: $W_f = (mg \sin 	heta) \cdot (vt) \cdot \sin 	heta$.Therefore,$W_f = mgvt \sin^2 	heta$.

$A$ block of mass $m$ is at rest with respect to a lift when placed on an inclined plane of inclination $\theta$ inside the lift. If the lift moves upward at a constant velocity $v$,then the work done by the friction force on the block in time $t$ is:

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