# 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?

• @StanShunpike The comment was a sort of additional but related question. The confusion got cleared. – Adrian Apr 26 '15 at 7:50

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 $W = \int Pdt$.

• Thanks. So is F times S the magnitude of the frictional work? My book says it isn't. – Adrian Apr 26 '15 at 6:55
• No no, as I have written $W = \int \mathbf{F} \cdot \mathbf{v} dt$ where time varies from start to rest of the motion. This would mean $W = \int \mathbf{F} \cdot d\mathbf{s}$ from start to endpoint. But that is not equal to F times S. This is because F may be varying as a function of s. – gautam1168 Apr 26 '15 at 7:40
• But isn't that the same thing? F remains constant. – Adrian Apr 26 '15 at 7:51
• No F will change as energy is lost in many cases. For example take a drag force, which will keep reducing as the velocity of the particle decreases. – gautam1168 Apr 26 '15 at 12:25
• I am ignoring all those forces. High school level :) – Adrian Apr 26 '15 at 12:28

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 $\Delta x \neq 0$.

• Actually I was aware of the fact that there is no motion in the y-direction. But thanks for answering :) – Adrian Apr 26 '15 at 7:02
• Yeah, so work is just $W = F\Delta x$. It is just a number we just to quantify how much effort or energy something takes to move. – Stan Shunpike Apr 26 '15 at 7:09
• Another common problem I encountered when I was starting was some books assume you know what a vector is. So they will write $W = F \Delta x$ but you may also see $W = \vec{F} \cdot \Delta \vec{x}$\$. The latter version involves the dot product. Its really all just book keeping. You can find great info on this if you watch videos on YouTube. Example: m.youtube.com/watch?v=Lm7UL0XU74U – Stan Shunpike Apr 26 '15 at 7:13