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For example, If I rub my finger on a table (which doesn't move due to friction from ground), does friction between my finger and table does work (positive or negative) on it?

Or, a even simpler example - A block is moving, surely the friction is doing (negative) work on the block. But what about the ground? Although it doesn't move but the friction force itself is moving, does it count as work?

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The work of the friction force on the ground is zero because the application point has zero velocity even if it continuously changes. So, it is not the work of the ground on the block up to its sign!

(This is the reason why a perfect non-sliding wheel due to the presence of frictional force caused by a perfectly flat ground would run forever: The contact point (just one point because the wheel and the ground are perfect) of the wheel on the ground has zero speed and continuously changes. This way there is no work of the ground on the wheel and it preserves its kinetic energy in spite of the presence of a friction force.)

To handle these generalized cases one may use the general form of work-kinetic energy theorem of continuous mechanics $$\frac{d}{dt} \int_{V_t} \frac{1}{2} \mu(t,{\bf x}) {\bf V}(t,{\bf x})^2 d^3x = \int_{V_t} \mu(t,{\bf x}) {\bf f}(t,{\bf x})\cdot {\bf V}(t,{\bf x}) \: d^3x+ \int_{\partial V_t} {\bf s}(t,{\bf x})\cdot {\bf V}(t,{\bf x})\: dS\:.\tag{1}$$

The left-hand side is the time derivative of the kinetic energy of a portion of continuous body $V_t$ evolving in time, preserving the number of particles along its story, with mass density $\mu(t,{\bf x})$ and velocity field ${\bf V}(t,{\bf x})$.

The two integrals in the right-hand side are the instantaneous power of the volume density of force ${\bf f}(t,{\bf x})$ at time $t$ and applied on ${\bf x}\in V_t$ and of the surface density forces ${\bf s}(t,{\bf x})$ at time $t$ and applied on ${\bf x}\in \partial V_t$ respectively.

You see that (1) is valid at time $t$ and it does not matter if application points of forces "change".

In the case of the block moving along the rigid ground at rest in a reference frame, $V_t$ is that piece of ground with ${\bf V}(t,{\bf x})=0$ and the friction force of the block on the ground can be represented as a surface density of force ${\bf s}(t,{\bf x})$ which, at fixed $t$, is different from $0$ only in a surface region given by the surface of contact of the block and the ground. This region changes in time and this change is accommodated by the time dependence of ${\bf s}(t,{\bf x})$. However, the last integral in (1) vanishes because ${\bf V}(t,{\bf x})=0$ by hypothesis.

The total surface works due to the friction force of the block on the ground is $$W_s = \int_{t_1}^{t_2} dt\int_{\partial V_t} {\bf s}(t,{\bf x})\cdot {\bf V}(t,{\bf x})\: dS =0\:.$$

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  • $\begingroup$ This is assuming a rigid ground, correct? It should probably also apply with as long as the ground deforms elastically; but if the ground is able to deform (or completely dislodge itself in the case of gravel surfaces/something that can wear out), then it can also have a velocity after the friction interaction. It just seems worth mentioning since this answer does seem very thorough. $\endgroup$ – JMac May 7 '17 at 14:19
  • $\begingroup$ Yes I am assuming a rigid ground. If the ground deforms the picture is much more complicated and one has to deal with the work of internal stresses... $\endgroup$ – Valter Moretti May 7 '17 at 14:33
  • $\begingroup$ I appreciate your answer, although it seems beyond my knowledge at present. I am pretty surprised actually to see such advanced mathematical form of WE theorem as I thought most of the mechanics (even advanced) was already covered in our curriculum. Guess I was brutally wrong. Anyways, thanks for the insight, one day.. maybe.. I would come back here to understand what I can not at the present moment. $\endgroup$ – Sarthak123 May 7 '17 at 16:28
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Corrected answer as a result of the comments made by @ValterMoretti

I have found it easier to understand what is happening by applying the constant kinetic frictional forces on the “sides” of the surface and the block rather than on top or underneath.
I think that this way of presentation helps to understand the displacements which are involved.

When discussing or answering such question the first thing to do is to define the system under consideration.

enter image description here

Consider a block as the system and it is sliding along a rough horizontal surface with an external kinetic frictional force $-f \hat i$ acting on the block due to the surface.
If the displacement of the block is $+y\hat i$ then the work done on the block by the frictional force is $-fy$ and if you were using the work-energy theorem this would produce a change (reduction) in the kinetic energy of the block.

Consider a surface (and the Earth) as the system.
There is an external kinetic frictional force $+f \hat i$ acting on the surface due to the block.
However the displacement of that frictional force is now $x \hat i$ and so the work done on the surface by that frictional force is $+fx$.
This produces an increase in the kinetic energy of the surface but because the displacement of the surface is less than that of the block, the gain in kinetic energy of the surface is less than the decrease in kinetic energy of the block.
The difference is the heat produced which enters both bodies.

The difference in the displacements can be shown as follows.
Assume that the mass of the surface (and the Earth) is greater than the mass of the block.
The magnitude of the impulse $f \Delta t$ on the surface and the block must be the same so the magnitude of the change in momentum of the surface and the block must be the same.
Since the mass of the surface is more massive than the block the magnitude of the acceleration of the surface must be less than that of the block.
As the interaction times $\Delta t$ are the same for both the surface and the block the velocity vs time graphs for the two objects show the displacement of the surface (area under graph) to be less than that of the block.

enter image description here

In practice because the Earth is so massive the displacement $x$ is taken to be zero.


A good example of the point of contact not moving and work done by a static frictional force is you sitting on a seat with no back rest in a car which is accelerating.
Now in this case the displacement $x$ of the two static frictional forces on you and on the car seat is the same.
You and the car seat accelerate but whilst doing so you do not move relative to the car seat because there is a static frictional force between you and the car seat.

If you are the system then you have an external static frictional force acting on you $+f\hat i$ due to the car seat and after a displacement of $x \hat i$ that static frictional force does $fx$ amount of work on you.
Your kinetic energy increases.

At the same time there is $-fx$ amount of work done on the car seat by the static frictional force that you exert on the car seat.

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  • $\begingroup$ When the block is decelerated by the friction force with the ground, the work done by the frictional force on the ground (remaining at rest) is zero, while the work done by the frictional force on the block is -fx<0. It is false that the sum of works is zero. $\endgroup$ – Valter Moretti May 7 '17 at 10:39
  • $\begingroup$ The sum of the works of a pair action-reaction forces is zero when the two application points have the same velocity. There are many situations, many of them involving friction forces, where it is not the case. $\endgroup$ – Valter Moretti May 7 '17 at 10:42
  • $\begingroup$ @ValterMoretti Not my finest hour and I thank you for putting me straight. I have changed my answer which is not as general as yours but, I hope, conveys the right message? $\endgroup$ – Farcher May 7 '17 at 13:43
  • $\begingroup$ Yes I agree with your improved answer. The only point I cannot understand completely is "the gain in kinetic energy of the surface is less than the decrease in kinetic energy of the block, the difference manifesting itself as heat". $\endgroup$ – Valter Moretti May 7 '17 at 14:41
  • $\begingroup$ I think that the only thing we can say about the total heat produced (entering both bodies) is $Q = -(\Delta K_1 + \Delta K_2)$ because the first principle says $W_i + Q_i = \Delta U_i + \Delta K_i$ (where $W_i$ is the work done on the system $i$), while from mechanics we know that $W_i = \Delta K_i$ so that $Q_1+Q_2 = \Delta U_1 + \Delta U_2$ and finally energy is conserved $\Delta U_1+ \Delta K_1 + \Delta U_2+ \Delta K_2=0$...Putting all together $Q_1+Q_2 = - (\Delta K_1+ \Delta K_2)$. $\endgroup$ – Valter Moretti May 7 '17 at 14:44

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