# On work done by internal forces which is coming out to be not equal to zero

1) Let us consider a block which explodes due to some internal mechanism into two smaller fragments of equal masses.The system was initially at rest and now is having some finite kinetic energy(due to momentum conservation).We can hence comment that the work has been done by the internal force by the Work-energy Theorem since there are no other forces acting on the system.But this seems to contradict the fact that work done by internal forces is always 0.Where am I going wrong? I have researched similar questions on stack and other site but to no avail. Also,textbooks for some reason do not consider a lot of theory on this matter for some reason which adds to my woes.

2) I have another question that in a two mass spring block system does the spring do any work?It should be 0 according to me as it is an internal force when solving from COM frame but is this also true from a ground frame?While writing the work energy theorem on this system, would the spring work show up even in the form of potntial energy?

• Find where you heard that internal forces do no work and look for the (stated or unstated) restrictions to the claim. Because it simply isn't true in general (as you have observed) but it is true for a useful special class of situations. Commented Oct 8, 2019 at 19:03
• Can you sketch the 2nd question? Commented Oct 8, 2019 at 19:06
• What @dmckee?My teachers have been religiously saying that it's true forever Commented Oct 8, 2019 at 19:23
• @Luka Mandic it's just two blocks connected by a spring performing shm +translation Commented Oct 8, 2019 at 19:25
• "My teachers have been religiously saying that" That probably shouldn't happen, but it does because it is easier (and perhaps less confusing) than making a distinction between rigid and non-rigid systems at the start. However, this is a time to trust yourself: you have shown that internal forces can do non-zero work if the system changing shape. Keep the special case in mind, however, because it simplifies a lot of things. Commented Oct 8, 2019 at 19:26

The answer to your first question is that work done by internal forces only sums to zero in the case of rigid bodies, so the principle does not apply to an exploding body.

The same is true for the two blocks linked by a spring- it is not a rigid body.

The two cases are analogous. In the case of the exploding block, potential energy stored in the explosive was converted into the KE of the two moving parts. Likewise the spring can store PE which is converted into the KE of the blocks.

• Hold on there...while writing the term of kinetic energy in the work energy theorem do we consider kinetic energy of the two broken off chunks or kinetic energy of the COM? Because if we consider that of the COM the internal work does beocme 0!!!(as COM is at rest)? Commented Oct 8, 2019 at 19:37
• No, in the Com reference frame the KE is the sum of the KEs of the individual parts. Commented Oct 8, 2019 at 19:53
• There are no additional forces acting in the COM frame in this case,sometimes we would have like pseudo forces in certain cases in COM frame Commented Oct 8, 2019 at 19:58
• Work done by the spring relative to the COM and ground would still be zero on the system right?Because it does equal and opposite work on both the locks which effectively cancel out for system? Commented Oct 9, 2019 at 18:00

Something to keep in mind is that "internal force" is a subjective term. It completely depends on what we say the system is, and therefore what we say is "internal" and "external". However, the work done by a force is not dependent on this distinction. Therefore, we should not expect that the label of "internal" or "external" should influence how much work the force actually does.

As you have shown, internal forces can certainly do work. As long as a force is applied to an object over some distance (i.e. $$\int\mathbf F\cdot\text d\mathbf x\neq0$$), work is being done.

• Hold on there...while writing the term of kinetic energy in the work energy theorem do we consider kinetic energy of the two broken off chunks or kinetic energy of the COM? Because if we consider that of the COM the internal work does beocme 0!!!(as COM is at rest)? Commented Oct 8, 2019 at 19:42
• @SchwarzKugelblitz Then it sounds like you already have your answer. There are instances where it is helpful to think of the COM, and other times where it is not. As you have seen, the kinetic energy of a system of particles is not determined by the motion of the COM. Even for rigid bodies this is true (think of a rigid spinning wheel whose COM is stationary). You should always start with thinking of each body in the system separately. If further simplification arises, then that is great. But I would not try to start with wishful simplifications, should they turn out to be invalid. Commented Oct 8, 2019 at 19:43
• Ok..so it is not correct to do so always..but like doesn't COM model the system? Commented Oct 8, 2019 at 19:48
• @SchwarzKugelblitz What do you mean by "model the system"? Model what about the system? The COM is certainly an important coordinate to know in certain analysis. In determining the total kinetic energy of the system it is useless though. Commented Oct 8, 2019 at 19:50
• Hmm..ya I saw a post rn on stack which say KEsys=KEcom+KEparticles Commented Oct 8, 2019 at 19:51