Work done by frictional force on a sliding block 
A block slides across a table horizontally with an initial velocity $V$. The frictional force $F$ brings it to rest after its Centre of Mass covers distance $S$.  
What is the work done by the frictional force?  

According to me, it should be zero because the point of contact of the block and table on which friction acts is always at rest.  
But common sense tells, the loss in KE is the work done by the frictional force. 
Where am I going wrong?
 A: According to Newtonian mechanics, it is true that the table exerts an equal but opposite force against gravity that results in $\Delta y = 0$, where $y$ is the up/down dimension. However, sliding block is clearly moving in the $x$ dimension (i.e. horizontally across the table). And it is also acted on by a force, namely friction. The block does not generate a force of its own to oppose friction nor is some other force (besides gravity and the normal force) acting on it. As a result, friction causes the block to decelerate until it comes to a halt. At this point, all the kinetic energy the block had has dissipated. Hence, $$KE = \frac{1}{2}m(0)^2 = 0$$. 
I'll leave you to solve the rest. Hopefully that takes care of the confusion as it sounds like you were mixing up dimensions and confusing $\Delta y= 0$ and $\Delta x \neq 0$. 
A: No the point of contact is not at rest. It moves with the block. You are probably confusing with rolling motion in which the point of contact is always at rest. There the point of contact is rest because the lowest point on the disk has two contributions, one due to forward motion of disk as a whole (v) and one in the backward direction due to rotation ($\omega r$). If $v=\omega r$ the point of contact is at rest and the motion is pure rolling. However in this case, there is no rotation. The point (or rather points in most cases) of contact is moving only with the rest of the block with velocity v. Then power dissipated is:
$P=\mathbf{F}_{fric} \cdot \mathbf{v}$ and the work done is $W = \int Pdt$.
