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) mgut cos²θ 

(c) mgvt sin²θ 

(d) mgvt sin2θ.

Answer 1

(c) mgvt sin²θ 

Explanation:- 

In time t, vertical distance covered =vt The vertical component of frictional force = F.sinθ 

Work done by this component 

= F.sinθ .vt 

= µmg.cosθ .sinθ .vt 

= (sinθ /cosθ) mgvt.cosθ.sinθ    (Since µ=tanθ) 

= mgvt.sin ²θ 

See Figure below,