A small block of mass m is placed on a rough inclined surface of...

Since the elevator ascends with a uniform velocity $v$, the acceleration of the block is zero. The block is at rest relative to the incline, so it is in equilibrium.

The force of friction $f$ acts upwards along the inclined plane to balance the component of gravity. Its magnitude is given by:

$f = mg \sin \theta$

The displacement of the block in time $t$ is $s = vt$ in the vertically upward direction.

The angle between the upward inclined plane (direction of friction) and the vertical displacement is $90^{{}^{{}^{\circ}}} - \theta$.

The work done by the force of friction is:

$W = \vec{f} \cdot \vec{s} = f s \cos(90^{{}^{{}^{\circ}}} - \theta)$

Substituting the values of $f$ and $s$:

$W = (mg \sin \theta)(vt)(\sin \theta)$

$W = mgvt \sin^2 \theta$