Question

A block of mass $m$ is stationary with respect to the wedge of mass $M$ moving with uniform speed $v$ on horizontal surface. Work done by friction force on the block in $t$ seconds is

• A. zero
• B. $-\frac{m g v t}{2} \, sin 2 \theta $
• C. $- \, \frac{m g v t}{2}$
• D. $-\frac{m g v t}{2} \, sin \theta $

Answer 1

Answer: (B)

Since block doesn't move with respect towedge, so friction force should balance the component of weight along the inclined plane.
$f=mg \, sin \theta $
Component of displacement along the inclined plane in time $t$ is given by $vtcos\left(\theta \right)$
Work done by $f=- \, mg \, sin \theta \, cos⁡\theta \, vt$
$= \, - \, \frac{m g \, sin 2 \theta }{2}\left(v \, \times t\right)$