0
$\begingroup$

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 $TS$?

$\endgroup$
  • $\begingroup$ 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 $Ts -Fs$ ... $\endgroup$ – user182687 Mar 19 '18 at 14:58
  • $\begingroup$ Please sir give the name of any good book or web site to clear this concept $\endgroup$ – Gilll Mar 20 '18 at 0:49
  • 1
    $\begingroup$ Look into University Physics by Young , Mechanics by Kleppner , lectures of Walter lewin .. $\endgroup$ – user182687 Mar 23 '18 at 15:59
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.