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While explaining Centre of Mass, my teacher told this exact statement that "If net $F=0$ on a system then KE of the system may change due to internal forces or due to internal work done."

Let's take 2 positive charges "q" and place them at distance "d". My query was it's right that KE of the system can change due to internal electrostatic forces which contribute to the increase in KE when the particles are released. But is it right to say that because of work done by internal forces KE of the system increases? Because I studied that work done by internal forces is equal to zero if we concider a system.

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  • $\begingroup$ Consider an explosive device. No external force is applied to it, but when it goes off the kinetic energy of its parts increases dramatically. $\endgroup$
    – The Photon
    Commented Nov 5, 2020 at 18:55
  • $\begingroup$ The work is not zero for your system. For each force the displacement is in the direction of the force so you add two positive contributions. $\endgroup$
    – nasu
    Commented Nov 5, 2020 at 19:08
  • $\begingroup$ @nasu I related this "making a system" Concept to the pully block system... Where we used to concider some blocks as a system and the work done by tension becomes 0 as it became internal force... I was confused by that, that if there tension became internal force if we considered some blocks as system then why not here we can assign the two charges as a system and finalize the electrostatic work done as 0 as it's internal.. $\endgroup$
    – Faham J
    Commented Nov 5, 2020 at 19:10
  • $\begingroup$ The work done by a force does not depend on what you consider to be your system. And anyway the idea that the work of internal forces is zero is wrong. Forget about it. $\endgroup$
    – nasu
    Commented Nov 5, 2020 at 23:21

2 Answers 2

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The internal forces of a system cannot change the momentum of that system because of Newton's third law. They can, however, change the kinetic energy of a system. Consider two masses connected by a compressed spring. When the system is released, the blocks gain kinetic energy, but maintain zero total momentum.

In the case where the internal forces rigidly constrain the motion of the parts with respect to each other (rigid bars or ropes that are never slack), then you can say that the internal forces do no work. That is because every change is motion caused by an internal force causes an equal and opposite change in motion somewhere else.

In your problems with ropes and pulleys, the tension of the rope is always decelerating one block and accelerating another. The result is zero total work because the blocks travel the same distance and the tension is the same at both ends of the rope.

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Your two charges lose electrostatic potential energy and each of them gains kinetic energy. Total energy is conserved.

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  • $\begingroup$ I related this "making a system" Concept to the pully block system... Where we used to concider some blocks as a system and the work done by tension becomes 0 as it became internal force... I was confused by that, that if there tension became internal force if we considered some blocks as system then why not here we can assign the two charges as a system and finalize the electrostatic work done as 0 as it's internal.. $\endgroup$
    – Faham J
    Commented Nov 5, 2020 at 19:09
  • $\begingroup$ Please correct me where I am going wrong.. $\endgroup$
    – Faham J
    Commented Nov 5, 2020 at 19:09
  • $\begingroup$ Usually, if the masses in a pulley system are moving, the tension is caused by an external force. With your two charges, the electrostatic forces are doing work as the charges move. $\endgroup$
    – R.W. Bird
    Commented Nov 5, 2020 at 21:43
  • $\begingroup$ You are wrong considering that the work of internal forces is always zero. $\endgroup$
    – nasu
    Commented Nov 5, 2020 at 23:23
  • $\begingroup$ nasu: Who told you that? If the two charges are allowed to accelerate toward each other, they are clearly gaining kinetic energy. $\endgroup$
    – R.W. Bird
    Commented Nov 6, 2020 at 14:47

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