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The question arises from my confusion over the two definitions of work (relevant to classical mechanics) I've encountered:

  1. $W= \int \vec F\cdot\mathrm{d}\vec x$
  2. $W=$ net change in energy of a system

We define the mechanical energy for a state to be the sum of the kinetic and potential energies, and then define a conservative force as one that doesn't change/conserves a system's mechanical energy. By the second definition, this implies that conservative forces like gravitational force do no work on a mechanical system, but this conflicts with the statements in my reference book and the ones I have read on the internet. They reason, with the (aforementioned) first definition: if a non-zero conservative force $\vec F$ acts along and throughout a body's non-zero displacement, the work ($W=\int\vec F\cdot\mathrm{d}\vec x$) should be non-zero, and therefore a conservative force does work.

Which of the aforesaid is actually true, and why?

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3 Answers 3

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Your second definition is incorrect. The total work done on a system by all forces is equal to the change in kinetic energy, not total energy. So, gravity pulling a mass down does work that increases its kinetic energy. Non-conservative forces change total mechanical energy, conservative forces do not. They can both do work by changing the kinetic energy of a mass or system of masses.

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  • $\begingroup$ I may be misunderstanding the definition in your second sentence but it seem you're suggesting that if I have two bodies attracted to each other (whether by gravity, electrostatic force or just by virtue of being connected by a spring) and I pull them apart in such a way that their initial and final velocity is zero, I haven't done any work because the kinetic energy hasn't changed. That sounds wrong to me. Can you please clarify? $\endgroup$
    – aekmr
    Nov 5, 2021 at 13:49
  • $\begingroup$ @aekmr the force you have exerted is just one of those acting on the body you have pulled. Note that Mark said total work and all forces. $\endgroup$
    – Alchimista
    Nov 5, 2021 at 13:58
  • $\begingroup$ @Alchimista That's true - when defining work, it's important to specify which forces are doing the work. I agree that total work done by all forces, including internal ones, is equal to the change in kinetic energy. But when people say "work", they quite often mean "work done on a closed system by external forces". I do not see anywhere in OP's question that they're asking specifically about the total work. If the $W$ in their second definition is to be understood as work done on a system by external forces then that definition is perfectly fine - wouldn't you agree? $\endgroup$
    – aekmr
    Nov 5, 2021 at 14:07
  • $\begingroup$ @aekmr because the point is what is conservative and what's not. Not the definition of work. $\endgroup$
    – Alchimista
    Nov 5, 2021 at 14:13
  • $\begingroup$ @Alchimista Definitions are important - I'd argue we should understand what is meant by $W$ in OP's second definition before we can dismiss it as incorrect. I now actually agree with everything in Mark's answer except for the first sentence. That one is only true if $W$ is the total work, including internal forces, which the OP did not specify. Anyway, I feel we're straying from the topic, so we should perhaps stop the discussion here. Thanks for your input! :) $\endgroup$
    – aekmr
    Nov 5, 2021 at 14:29
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A conservative force can indeed do work on a system, just with the stipulation that this work is path-independent. This is equivalent to the statement that: $$W=\oint\mathbf{F}\cdot\mathrm{d}\mathbf{\ell}=0$$ However, this integral doesn't necessarily vanish along a non-closed line: $$W=\int_L\mathbf{F}\cdot\mathrm{d}\mathbf{\ell}$$ Since the integral is path-independent, potential energy can be well-defined. For instance, it can be defined as the work necessary to bring a system into some point, starting from infinity. We can say $W=\Delta K=-\Delta V$ such that $\Delta U=\Delta K+\Delta V=0$.

You may have confused work with change in mechanical energy; generally, the work is equal to the change in kinetic energy.

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Often if one does not define the system being considered apparent inconsistencies can arise.

Consider a system consisting of a point mass moving in a gravitational field produced by another mass.
There is one external force acting on the system - gravitational attraction which is a conservative force.
The work done by the gravitational field is equal to the change in the kinetic energy of the system.

Now consider the same situation but now the system consists of both masses with no external forces acting.
In such a system the sum of the kinetic energy and the gravitational potential energy of the system (mechanical energy) stays constant.
However, there are internal forces (Newton third law pairs) at play and those internal forces do work on the constituent parts of the system.
Both masses have work done on them by the gravitational field produced by the other mass.
With this statement you will perhaps note that you can treat the two-mass system as two one-mass systems with the internal forces for the two-mass system being external forces for the two one-mass systems.

In the examples that I have used, conservative (gravitational) forces are doing work.

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