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Suppose two charged particles $a$(+) and $b$(-) exert force on each other.

The work by $F_{ab}$ plus the work by $F_{ba}$ is equal to the increase in total kinetic energy of the two particles. But we know $F_{ab}= -F_{ba}$ according to Newton's third law. So that $F_{ab} + F_{ba} = 0$. But the work done by $F_{ab} + F_{ba}$ can't be $0$.

Why? The forces are in opposite directions. Plus, does this theorem apply to internal forces?

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The work energy theorem is indeed a tricky beast and it is easy to misuse it. Unfortunately, most derivations are rather lax about identifying the assumptions in the derivation. Most derivations assume that the system being analyzed is a point particle with no internal degrees of freedom. Your system here has internal degrees of freedom and an internal potential energy. So it doesn’t apply.

When used on systems with internal degrees of freedom it is important to understand that the “net work” of the work energy theorem is not the same as the sum of the work done by all forces acting on the system, which is the total thermodynamic work.

In this case there is no external force, so the “net work” is zero. But because the system has internal degrees of freedom the work energy theorem does not apply and we cannot infer that the KE is unchanged.

What we can do is examine the total thermodynamic work. Since there are no external forces the total thermodynamic work is also zero. That means that there is no overall change in energy and any increase in KE must be associated with a corresponding decrease in internal PE. This is exactly what is observed.

So, the bottom line is that the work energy theorem is very limited in its application. For systems with internal degrees of freedom the concept of total thermodynamic work is more useful.

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$F_{ab}+F_{ba}=0$ tells us that there is no net force on the system consisting of the two particles. That is, there is no external force on the system, and no work done on the system. The total mechanical energy of the system is constant.

But there are internal forces in play in this system, and work is done internally on both particles. But the total energy of the system remains constant. The internal forces increase the kinetic energy of the particles (work-kinetic energy theorem).

What gives?

Don't forget that for systems of particles interacting by conservative forces we can define potential energy $$ \Delta PE = -W_\mathrm{internal}$$

With the introduction of potential energy we see that as the kinetic energy of the particles increase, the potential energy of the system decreases in exactly the right amount to keep the total system energy constant. This is perfectly consistent with the conclusion based on zero net force on the system.

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  • $\begingroup$ I'm sorry if it is not clear to you, but these forces are internal only i.e., Fab + Fba = 0 $\endgroup$ – Madiha Nazir Mar 13 at 13:41
  • $\begingroup$ Yes, exactly. Internal forces do internal work which changes the internal kinetic energy. At the same time, potential energy decreases. The work-energy theorem does apply to internal forces. There is no external work ($F_{ab}+F_{ba}=0$ $\endgroup$ – garyp Mar 13 at 13:54
  • $\begingroup$ Oh. Thanks for explaining $\endgroup$ – Madiha Nazir Mar 13 at 14:31
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It's because the work depends both on force and displacement. Let's say that the first body was displaced by $\vec d_1$ and the second body by $\vec d_2$, then the work: $$ A=\vec F_1\cdot \vec d_1+\vec F_2\cdot \vec d_2=\vec F_1\cdot \vec d_1-\vec F_1\cdot \vec d_2=\vec F_1\cdot(\vec d_1-\vec d_2). $$ Thus, unless the displacement of one body relative to another $\vec d_1-\vec d_2$ is perpendicular to the line of forces $\vec F_{1,2}$, the work done is non-zero.

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