Does work energy theorem fail while dealing with friction?

[NoahCygnus]

Let's consider a setup consisting of a table with friction, and a block on top of it. Suppose we drag the block across the table with a constant speed. The applied force $f_{app}$ acting through a distance $d$ does a work $f_{app}d$. The frictional force $\mu N$ is equal to $f_{app}$ since there is no acceleration. So the total work done on the block by these external forces will be $\Sigma W = W_{app} + W_{friction} = F_{app}d - f_{Kinetic Friction}d = 0$

According to law of conservation of energy, if there is an energy change in the system , it is because the energy is being transferred across the system boundary by a transfer mechanism (work, heat, mechanical waves, matter transfer, electromagnetic radiation etc.)
So, $\Delta E_{system} = \Sigma T$, where $\Delta E_{system} = \Delta K + \Delta U + \Delta E_{internal}$ and $\Sigma T$ is an energy transfer mechanism.

In our case, we have $W_{app} + W_{friction} = \Delta K + \Delta U+ \Delta E_{internal}$, as work is the only mechanism of transfer of energy. Also $\Delta K = 0$ as there is no change in speed and $\Delta U = 0$. Then $W_{app} + W_{friction} = \Delta E_{internal}$ , but as $W_{app} + W_{friction} = 0 \Longrightarrow \Delta E_{internal} = 0$ So that means there is no increase in the internal energy, but clearly the block heats up. Can anyone explain to me what's happening here?

 

[Dale]

The work energy theorem assumes a rigid body. A rigid body cannot heat up since it has no internal degrees of freedom. In your example, it is only the table that heats up.

 

[NoahCygnus]

The work energy theorem assumes a rigid body. A rigid body cannot heat up since it has no internal degrees of freedom. In your example, it is only the table that heats up.

If work energy theorem assumes a rigid body, and the points of contact between the block and table deform rendering work energy theorem useless, then why do we say work done by kinetic friction on a body for a displacement $d$ is $W = -f_{kinetic}d$? Shouldn't the work done by kinetic friction be incalculable as the points of contact get deformed locally?

 

[Dale]

I wouldn't say it is useless. It is useful for describing the mechanical energy. It just doesn't describe thermal energy. If you are interested only in the mechanical energy then it makes calculations very easy.

Lots of physics is like that.

 

[Doc Al]

If work energy theorem assumes a rigid body, and the points of contact between the block and table deform rendering work energy theorem useless,

You can extend the "Work"-energy theorem to non-rigid bodies by introducing 'center of mass' work (or pseudo-work). It makes some calculations of mechanical energy very easy (in line with what Dale said above).

You can read more about it in this thread: https://www.physicsforums.com/threads/work-done-by-friction.732127/

 

[NoahCygnus]

You can extend the "Work"-energy theorem to non-rigid bodies by introducing 'center of mass' work (or pseudo-work). It makes some calculations of mechanical energy very easy (in line with what Dale said above).

You can read more about it in this thread: https://www.physicsforums.com/threads/work-done-by-friction.732127/

What I don't understand is why do we say work done by frictional force on something is $W_{friction} = -fd$ for a displacement $d$ if frictional force is not concentrated on a single point , and is distributed over the entire contact area. The individual contact points should deform so how did we come to the result $W_{friction} = -fd$ ?

 

[Doc Al]

What I don't understand is why do we say work done by frictional force on something is $W_{friction} = -fd$ for a displacement $d$ if frictional force is not concentrated on a single point , and is distributed over the entire contact area. The individual contact points should deform so how did we come to the result $W_{friction} = -fd$ ?

The real work done by friction is difficult to calculate. But if you're only interested in mechanical energy and rigid bodies that can be treated as particles (as is usually the case in intro physics courses), then friction X displacement does correctly describe the change in kinetic energy of the body. But you are correct, that it is not really a true "work" (as in conservation of energy).

Bruce Sherwood wrote a classic paper that I think directly addresses your concerns. Here it is: Work and heat transfer in the presence of sliding friction (Linked from Prof Sherwood's website.)