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. $-\frac{mgvt}{2}\sin 2\theta$
• C. $-\frac{mgvt}{2}$
• D. $-\frac{mgvt}{2}\sin \theta$

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 $f=-mg\sin \theta \cos \theta vt$
$=-\frac{mg\sin 2\theta }{2}\left(v\times t\right)$