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Consider a point-mass $m$ having constant velocity but undergoing influence from two forces, $F_1$, $F_2$, having equal magnitude but opposite directions. Because the forces' magnitudes are equal, I would expect no net acceleration of $m$, but if $m$ is moving, the forces are doing work on $m$ ($F_1$'s work being the inverse of $F_2$'s work), whereas if $m$ is stationary, no work is done by either $F_1$ or $F_2$; why is that? In neither case is there net acceleration, and in neither case is there net work, but why is work calculated differently in one case than in the other?

Note: This is a conceptual question, so I'm afraid that simply appealing to definitions, i.e., "because work is defined that way", will be unsatisfactory.

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  • $\begingroup$ The net power to the system is always zero. I don't understand your question. What you consider zero energy is entirely arbitrary and dependent of the frame of reference. $\endgroup$
    – John Alexiou
    Commented Jun 23, 2018 at 1:54

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Because the forces' magnitudes are equal, I would expect no net acceleration of $m$

Correct. Even if $m$ is moving, there will be no acceleration, since there is no net force.

if $m$ is moving, the forces are doing work on $m$ ($F_1$'s work being the inverse of $F_2$'s work), whereas if $m$ is stationary, no work is done by either $F_1$ or $F_2$; why is that?

Minor but important point: $F_1$'s work is the negative of $F_2$'s work, rather than the inverse of $F_2$'s work. So the net work done by the forces in both cases is zero, as you said.

You seem to be confused by the fact that two different situations (one with moving $m$, one with stationary $m$) both have the same answer (no net work is done in either case). But this follows directly from the fact that the net force is zero; if you have one person pulling your left arm and another pulling equally strong on your right arm, it doesn't matter whether you're moving or not, you're not going to be sped up, and so the kinetic energy transfer to your body is zero in both cases.

why is work calculated differently in one case than in the other?

Because they are different situations. In one case, $m$ is moving. In the other, $m$ is not. It shouldn't be surprising that the math differs between the two cases; they just coincidentally happen to have the same answer.

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  • $\begingroup$ Put differently, $F_1$ is doing work on $F_2$ (Or if you like energy is being transferred from one to the other) and $m$ Is just sitting there... $\endgroup$
    – Floris
    Commented Apr 22, 2014 at 0:10
  • $\begingroup$ What is the inverse of a scalar if not the scalar's negative? Would you please elaborate on why you think I'm describing two situations? I believe I'm describing two cases of one situation: $m$ has constant velocity. In many/most-if-not-all other areas of kinematics, what matters most is whether or not the velocity is constant, not the value of the constant if it is constant. $\endgroup$
    – BlueBomber
    Commented Apr 22, 2014 at 3:31
  • $\begingroup$ Also, I think your second paragraph is misleading. Let's say the person pulling your left arm is applying a force $F_1$, and the person pulling your right arm is pulling with a force $F_2$, and $F_1$ is slightly greater than $F_2$. You could calculate the work done by the (leftward) net force, but you could also calculate the work done individually by both forces, and those calculations would depend on the entire magnitude of each force individually, right? $\endgroup$
    – BlueBomber
    Commented Apr 22, 2014 at 3:36
  • $\begingroup$ @BlueBomber: Normally when people say the "inverse of a number $a$", they mean $a^{-1}$. I say two situations because they're two situations: one is moving, the other is not. I'm not sure what you mean by "what matters most is whether or not the velocity is constant, not the value of the constant if it is constant"; I've never heard of that principle, and irregardless it's not always true (depending on the specifics of the problem), and thus it's not necessarily safe to assume. $\endgroup$
    – DumpsterDoofus
    Commented Apr 22, 2014 at 4:26
  • $\begingroup$ @BlueBomber: For your second comment, I'm also not quite sure what you're getting at. Are you confused by the fact that there can be two equally valid ways to get the right answer? You can calculate work based on net force, or you can calculate the individual works and add them; you will get the same result either way. For example, the quadratic reciprocity theorem in number theory has been proved in over 120 different ways; it's still the same theorem, but there's lots of ways of going about the proof. $\endgroup$
    – DumpsterDoofus
    Commented Apr 22, 2014 at 4:29
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What may confuses you is the fact that there are two opposing forces doing work against each other if the body is moving and no forces at all if the body is at rest.

This is a actually a nice example of the principle of relativity: Whether the body is at rest or whether it is moving with constant velocity shouldn't change the physics. :-)

'Two equal forces opposing each other' or 'no forces at all' shouldn't change the outcome. That means that these 'forces' are allowed to do work if the work done cancels each other.

What that tells us is, that some physical terms like 'force' or 'work' do depend on the observer's point of view (or more physically 'reference frame') but that does not change the physics of the system. There are always the conserved quantities that are independent of the reference frame and, once you agree on a reference frame, you can safely use all other quantities as well.

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