MCQ

A small block of mass m is kept on a rough inclined surface of inclination θ fixed in an elevator. the elevator goes up with a uniform velocity v and the block does not slide on the wedge. The work done by the force of friction on the block in time t will be

#### Options

zero

mgvt cos

^{2}θmgvt sin

^{2}θmgvt sin 2θ

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#### Solution

mgvt sin^{2}θ

Distance (d) travelled by the elevator in time t = vt

The block is not sliding on the wedge.

Then friction force (f) = mg sin \[\theta\] Work done by the friction force on the block in time t is given by

\[W = Fd\cos(90 - \theta)\]

\[ \Rightarrow W = \text{ mg } \sin\theta \times d \times \cos(90 - \theta)\]

\[ \Rightarrow W = \text{ mgd } \sin^2 \theta\]

\[ \therefore W = \text{ mgvt } \sin^2 \theta\]

Is there an error in this question or solution?

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