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Question
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 cos2θ
mgvt sin2θ
mgvt sin 2θ
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Solution
mgvt sin2θ
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
\[W = Fd\cos(90 - \theta)\]
\[ \Rightarrow W = \text{ mg } \sin\theta \times d \times \cos(90 - \theta)\]
\[ \Rightarrow W = \text{ mgd } \sin^2 \theta\]
\[ \therefore W = \text{ mgvt } \sin^2 \theta\]
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