Question

A small block of mass $m$ is kept on a rough inclined surface of inclination $\theta$ 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 a time $t$ will be

• A. zero
• B. $mgvt{\cos }^2\theta$
• C. $mgvt{\sin }^2\theta$
• D. $\frac{1}{2}mgvt\sin 2\theta$

Answer 1

The correct option is C $mgvt{\sin }^2\theta$


Block does not slide, so estabilishing conditions of equilibrium along the incline we have,

$f=mg\sin \theta$

In time $t$, displacement $S=vt$

The horizontal component of friction does zero work, but vertical component $f\sin \theta$ does work on the block.

$W_f=\left(f\sin \theta \right)vt$

$W_f=mgvt{\sin }^2\theta (\because f=mg\sin \theta )$

Why this question? Tip: If the block is observed in the reference frame of elevator then there is no work done on it by the friction as its displacement is zero in this referance frame.