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. $-\frac{mgvt}{2}sin2\theta$
• 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.
$f=mgsin\theta$
Component of displacement along the inclined plane in time $t$ is given by $vtcos\left(\theta \right)$
Work done by $f=-mgsin\theta cos\theta vt$
$=-\frac{mgsin2\theta }{2}\left(v\times t\right)$