A  very  large  plank  P  of  some  unknown  mass  is  being  moved with a constant velocity  $v_0\widehat{i}$ under application of an external force (not shown in figure). Simultaneously a block B of mass m placed on the plank is also moving with velocity $v_0\widehat{k}$ . All  these  velocities  are  with  respect  to  ground frame  and  at  t  =  0.  Coefficient  of  friction  between  the plank and the block is $\mu$ .  Choose the CORRECT option(s).

From frame of reference of P , velocity of block is ${\overrightarrow{v}}_{net}=\left(-v_0\hat{i}+v_0\hat{k}\right)$

So, kinetic friction acts exactly in opposite direction of ${\overrightarrow{v}}_{net}$.

Magnitude of $f_k$ is $\mu mg$ but to express in vector form, we have to take its components.

$\begin{array}{l}{\overrightarrow{f}}_{k }on the block B=\mu mg(\frac{1}{\sqrt{2}}\widehat{i}-\frac{1}{\sqrt{2}}\widehat{k})\\ \text{Option (A) is incorrect.}\\ \text{Now, from ground frame, power is}\\ P_{f_k}=\mu mg[\frac{\widehat{i}}{\sqrt{2}}-\frac{\widehat{k}}{\sqrt{2}}]\cdot \left(v_0\widehat{k}\right)\text{}\\ \Rightarrow P_{f_k}=-\frac{\mu mg{v}_0}{\sqrt{2}}\text{}\\ \text{To find heat dissipation we have to observe from plank frame}\\ \Rightarrow \frac{dQ}{dt}=\mu mg[\frac{\widehat{i}}{\sqrt{2}}-\frac{\widehat{k}}{\sqrt{2}}]\cdot (-v_0\widehat{i}+v_0\widehat{k})=-\mu mg{v}_0\frac{2}{\sqrt{2}}=-\sqrt{2}\mu mg{v}_0\\ \text{Negative sign indicates dissipation.}\end{array}$