Since block doesn't move with respect towedge, so friction force should balance the component of weight along the inclined plane.
$f=\mathrm{m}\mathrm{g}\mathrm{ }\sin \mathrm{\theta }$

Component of displacement along the inclined plane in time $t$ is given by $vt\cos \left(\theta \right)$
Work done by $\mathrm{f}=–\mathrm{ }\mathrm{m}\mathrm{g}\mathrm{ }\sin \mathrm{\theta }\mathrm{ }\cos \mathrm{\theta }\mathrm{ }\mathrm{v}\mathrm{t}$

$=\mathrm{ }–\mathrm{ }\frac{\mathrm{m}\mathrm{g}\mathrm{ }\sin 2\mathrm{\theta }}{2}\left(\mathrm{v}\mathrm{ }\times \mathrm{t}\right)$