# Work done in equilibrium positions

I am confused in work done against friction. Let $F$ be the friction when a big stone is pushed by person on ground and $T$ be the component of force applied by a person in the direction of displacement $S$.

I think $T$ should be greater than $F$, otherwise stone will not move. Then $T$ is not equal to $F$. But why do we check work done against friction as $FS$? I mean why isn't it $TS$?

• Friction applies force $F$ and you have to overcome it ... All the way during your travel at a distance $s$ , friction is active ... And work done by friction is $Fs$ ... But total work is $Ts -Fs$ ... – user182687 Mar 19 '18 at 14:58
• Please sir give the name of any good book or web site to clear this concept – Gilll Mar 20 '18 at 0:49
• Look into University Physics by Young , Mechanics by Kleppner , lectures of Walter lewin .. – user182687 Mar 23 '18 at 15:59

If $T$ > $F$ , the stone will have an acceleration ,say $a$ .
$$M_{stone}a=T-F$$
So , $a=\frac{T-F}{M_{stone}} >0$ .
The stone will have the speed $v(t)=at$ , so a kinetic energy $E_k=\frac{M_{stone}v^2}{2}=\frac{M_{stone}(at)^2}{2}=\frac{t^2(T-F)^2}{2M_{stone}}$ .
The idea is that you have done the work $TS$ , which goes in overcoming friction ( $FS$ ) and accelerating the stone ( $E_k$) .
As an exercise , try to write the equation $TS=FS+E_k$ and see if it's checked.