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When exactly is the Net work done by internal forces on a system 0, and what is the intuition behind it?

I have heard it is valid for rigid bodies but I am unaware of what they are as of this point, since i am only in high school.

Secondly, what is the reason behind it? I know that by newton’s 3rd law, internal forces occur in equal and opposite reaction pairs, but that doesnt say anything about the distance moved in the direction of force.

Lastly, in what situations can I correct apply this principle? The normal contact force between a block sliding over a slope and the tension in a pulley system?

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2 Answers 2

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I remember when my teacher explained this with a funny example.

Imagine you are in a car and you have superhuman strength. You jump in the air and kick the car from inside with such a force that both you and the car move in opposite directions. If you see the initial and final kinetic energy, the system was at rest initially. But now it has kinetic energy. As there are no external forces here (Assuming no external friction) this is entirely done by internal forces that you exerted on the car and by newton's third law the force the car exerted on you.

Anyway you don't need any superstrength for such examples.

Getting if from a boat, both initally at rest, and finally both you and the boat attain velocity is a similar example for the same.

You can find such examples in so many circumstances.

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  • $\begingroup$ so there no specific cases where the net work by external forces is bound to be 0? I thought that it was always zero for a massless string pulley system, for reasons i admit i do not know. $\endgroup$ Commented Jun 29, 2020 at 10:43
  • $\begingroup$ @OVERWOOTCH Why would the work done by the force itself by 0 here? If you mean net work done, that's 0 as the electrostatic and external forces do equal and opposite works. $\endgroup$ Commented Jun 29, 2020 at 10:53
  • $\begingroup$ @OVERWOOTCH Well, in the string pulley system, work done by which force are you talking about? $\endgroup$ Commented Jun 29, 2020 at 10:55
  • $\begingroup$ very sorry for mistyping. i meant internal forces, not external $\endgroup$ Commented Jun 29, 2020 at 10:55
  • $\begingroup$ internl forces; tension to be specific $\endgroup$ Commented Jun 29, 2020 at 10:56
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The key phrase in most work/energy conundrums is "on the system".

If I am riding in a car I can have a spring that I stretch. I am doing work on the spring, but none of that work goes to changing the kinetic energy of the system of me and the car and the spring moving down the road.

As you said all of the internal forces have equal and opposite pairs. So if you try to speed the car up by pushing on the dashboard your force on the dashboard and the dashboards force back on you are equal and opposite and so forces dotted with your displacement will be equal but of opposite sign, hence a net of zero.

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