Work exchanged by a system, and internal forces' work

[MathError]

Homework Statement

Hi to everyone, I'm studying thermodynamics, especially I'm focused on the work exchanged by a system. Book's explanation says that when the system received a force from the environment [F(e-s)] (on its surface) it reacts with a force from the system to the environment [F(s-e)]. Because of Third Newton's law F(e-s)=-F(s-e).
These forces act for the same displacement so
W(e-s)=-W(s-e).

I think I understood these points but I'm not so clear about the following part of explanation.

The equation W(e-s)=-W(s-e) is true if there are no forces of friction.
In general you have to consider the work of these forces.
In presence of forces of friction the equation W(e-s)=-W(s-e) becomes:
W(e-s)=-W(s-e)-W(T(s-e)) [W(T(s-e)) represents the work of force of friction from the system to the environment].
The book doesn't explain how it comes to this equation so I tried to write an example, but I dubt it's correct.

In all these cases the system is rigid so the internal forces of the system doesn't do work. However if the system is not a solid body, internal forces do work.

This is the point where I find more difficulties.
From dinamics I know the resultant of the internal forces [R] is zero but they can do work beacuse internal points can cover different displacements.

Now I'm a bit confused.

The book distinguish the internal forces from forces from the system to the environment
[F(e-s)]. My question is: are F(e-s) produced by internal forces?
I don't think so, because R=0, but at the same time I imagine that internal stress of the system causes these forces on the surface.
If F(e-s) are produced by internal forces, does the work of F(e-s) coincide with the work of internal forces?
Sorry for the bad english but I tried to as clear as possible.
I hope you can help me.
Thanks

Homework Equations

F(e-s)=forces from the environment to the system
F(s-e)=forces from the system to the environment
T(s-e)=forces of friction, from the system to the environment
T(e-s)=forces of friction, from the environment to the system
W(e-s)=-W(s-e)
W(e-s)=-W(s-e)-W(T(s-e))

The Attempt at a Solution

These are the examples I did to try to understand the different cases
https://ibb.co/n3sy9Pr

 

[jbsteele]

https://ibb.co/njK8R2TIn case1 the forces of friction don't do work, so we have W(e-s)=-W(s-e).In case2 the point 1 is at rest and the point 2 is displaced with the same velocity of point A, so the internal forces don't do work (because points are displaced with the same velocity), so we have again W(e-s)=-W(s-e).In case3 the internal forces do work because point 1 and point 2 covers different displacements. This is the part where I'm more confused.If F(e-s) are produced by internal forces, does the work of F(e-s) coincide with the work of internal forces?I think that in this case we have:W(e-s)=-W(s-e)-W(internal forces).