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While studying energy conservation on Morin I found this explanation about the work-energy theorem for a system.

The work–energy theorem stated before is relevant to one particle. What if we are dealing with the work done on a system that is composed of various parts? The general work–energy theorem states that the work done on a system by external forces equals the change in energy of the system. This energy may come in the form of (1) overall kinetic energy, (2) internal potential energy, or (3) internal kinetic energy (heat falls into this category, because it’s simply the random motion of molecules). So we can write the general work–energy theorem as $$W_\textrm{external} = \Delta K +\Delta V +\Delta K_\textrm{internal}.$$ For a point particle, there is no internal structure, so we have only the first of the three terms on the right-hand side.

Using Koenig theorem $$\Delta K_\textrm{system}=\Delta K +\Delta K_\textrm{internal}$$ so we have

$$W_\textrm{external} = \Delta K_\textrm{system} +\Delta V$$

Nevertheless, considering a system of $n$ material points the following holds.

$$\sum W=\Delta K_\textrm{system}$$

But here $$\sum W=\sum W_{i}=\sum \left(W_{i}^{(\textrm{ext})}+W_{i}^{(\textrm{int})}\right)$$ : the amount of work considered is the sum of the work done on each point (both from external and internal forces).

And in general we do not have that $\sum W_{i}^{(\textrm{int})}=0$.

Counterexample: two masses attracting each other gravitationally.

I'm confused about this, is one of these two in contrast with the other?

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$W=\Delta K_\text{system}$ and $W_\text{external}=\Delta K_\text{system}+\Delta V$ are consistent with each other iff $\Delta V=-W_\text{internal}$. The latter is the definition of the potential energy for conservative forces.

The cited equation is thus valid iff all internal forces are conservative.

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  • $\begingroup$ Why does necessarily holds that $W_{int}=-\Delta V$? Internal force can be non-conservative $\endgroup$
    – Sørën
    Commented Apr 4, 2016 at 14:24
  • $\begingroup$ Of course it can. The decomposition of energy into overall, potential, and internal contributions, plus the reference to heat, made me think your textbook was addressing "energy-conserving" systems only, e.g. a gas in thermodynamics. If your textbook does not mention this hypothesis, change textbook! I'll rephrase my answer to make this clear. $\endgroup$
    – L. Levrel
    Commented Apr 4, 2016 at 17:11
  • $\begingroup$ we have learnt that internal forces, which includes constraining forces between parts of a system does not contribute to work as the displacements are perpendicular to the constraining forces; if we further consider a particle acting on another one ,then the force of second one on the first will be equal and opposite to the first and thereby the net force can be zero-and such pairs of internal forces can always be found. $\endgroup$
    – drvrm
    Commented Apr 4, 2016 at 17:39
  • $\begingroup$ @drvrm: the constraining forces are just one example of internal forces, you cannot draw a general conclusion from them. Second, cancellation of internal forces does not imply cancellation of internal works, because the velocities of your two particles are not equal. $\endgroup$
    – L. Levrel
    Commented Apr 4, 2016 at 19:42

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