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
(a) zero
(b) mgvt cos2θ
(c) mgvt sin2θ
(d) mgvt sin 2θ

Answer 1

Answer:

c zero

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Answer 2

Option (c) ${mgvt} \sin ^{2} \theta$ .

The work done by the force of friction on the block in time t will be ${mgvt} \sin ^{2} \theta$.      

Explanation:          

Work done by friction force, W = f d cosθ          

Friction force :                                                

Friction is a force that retards a slipping object's travel.

That is it, friction is straight forward.

Distance travelled by the elevator in time t = vt

Block is not sliding on the wedge.    

Friction force is mg $\sin \theta$.    

Work done by friction force on the block in time t is given by,

                 $W=F d \cos (90-\theta)$        

                 $W={mg} \sin \theta \times d \times \cos (90-\theta)$

                 $W={mgd} \sin ^{2} \theta$

                 $W={mgvt} \sin ^{2} \theta$

Therefore, option (c) is the correct answer.