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I recently came across an MCAT question that confused me, and I haven't been able to understand it. The answer explanation claims that the work energy theorem is not valid when there are nonconservative forces or a change in potential energy. Is that true? I have looked at other posts, but as far as I can tell, non-conservative forces and potential energy shouldn't invalidate the work energy theorem. The question is included below enter image description here

"Knowing the speed of an RBC means that the initial and final kinetic energies of the RBC can be found, and therefore that the change in kinetic energy can be calculated. However, realize that the work energy theorem is valid only when there is no energy lost to non-conservative forces. Since in moving from the aorta to the capillaries, energy of the RBCs will be dissipated due to frication and the potential energy will change, the work energy theorem is invalid." According to Kaplan

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  • $\begingroup$ The work-energy theorem doesn't work with non-conservative forces, like friction. $\endgroup$
    – Jon Custer
    Jun 3, 2017 at 16:56
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    $\begingroup$ @JonCuster Work energy theorem does work when non-conservative forces are involved. $W_{conservative} + W_{non-conservative} + W_{other} = \Delta K$. Although, the change in potential energy is equal to the work done by conservative forces $W_{conservative}=-\Delta U$ $\endgroup$
    – Mitchell
    Jun 3, 2017 at 17:24
  • $\begingroup$ Aside: I'm all for gender inclusive pronoun style , but the switch from "he" to "she" between the prompt and the answers is really distracting. Are there two people involved? $\endgroup$ Jun 3, 2017 at 19:21

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The difficulty with the problem shown is the ambiguity in the phrase "the work done on [...]" because it does not the specify by what force or forces the work is done.

If that is to be understood as "the net work" then the answer is a resounding "yes".

However, it is clear that the author of the question is thinking of some particular subset of forces. Though I am unsure what subset it might be aside from excluding frictive forces.


If I found that I had written a question that ambiguously stated on an exam for my own students I would credit all defensible answers or simply throw the question out wholesale.

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  • $\begingroup$ The thing is, I don't think the answer is a matter of ambiguity. The answer explanation says outright that the work energy theorem is invalid. Are they wrong? $\endgroup$ Jun 4, 2017 at 17:27
  • $\begingroup$ "Knowing the speed of an RBC means that the initial and final kinetic energies of the RBC can be found, and therefore that the change in kinetic energy can be calculated. However, realize that the work energy theorem is valid only when there is no energy lost to non-conservative forces. Since in moving from the aorta to the capillaries, energy of the RBCs will be dissipated due to frication and the potential energy will change, the work energy theorem is invalid." According to Kaplan $\endgroup$ Jun 4, 2017 at 17:29
  • $\begingroup$ The statement "However, realize that the work energy theorem is valid only when there is no energy lost to non-conservative forces." is utterly incorrect. The work-energy theorem depends on only two concepts—work and energy—and is often used to find the work done by non-conservative forces (Example: A 100 kg crate moving at 5 m/s slides to a halt on the a concrete floor; how much work is done by friction?"). The writer seems to be confusing the work-energy theorem with a restricted form of conservation of energy. $\endgroup$ Jun 4, 2017 at 20:11
  • $\begingroup$ Thank you very much! I wish I could have upvoted your comment! $\endgroup$ Jun 5, 2017 at 3:49
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There's a lot of confusion about the Work-Kinetic Energy theorem. It applies only to single particle systems, and therefore applies only to ideal systems. That is, in real life it is an approximation.

The restriction to single particle systems implies that there can be no potential energy associated with the system because potential energy is defined only for system comprising more than one particle. It is thus also technically not applicable to any object made of atoms, but it can be used in such cases if the idealization is taken that the object is rigid (no internal energy). Non-conservative forces are allowable in principle, but in practice non-conservative forces come with complications that might preclude its validity. For example, some of the work done by friction inevitably raises the temperature of the object. A portion of the work goes into thermal rather than kinetic energy. Since the temperature rise is unavoidable, the theorem is not valid for the friction force

Update after comments by dmckee

I'm still trying to wrap my head around Goldstein, but I think the issue is this: as you say, his statement that you quote talks about all the KE, and includes all internal forces. However, this is rarely a useful thing to discuss, and note that the equation you point to is really the "point particle" W - KE theorem applied to a collection of point particles taken individually. If one tries to apply it to a collection of interacting points (a real deformable object) taken as a whole it does not work. The internal PE is not taken into account. Goldstein alludes to this at the end of that section

Simple 1d example: two masses connected by a spring. I can do work on one of the masses, but the resulting total KE of the system is not even constant. The total work done is equal to neither the COM KE nor the total (not even constant) KE. Goldstein also points out (by inference), as I did above, that if the object is rigid, the theorem works for an extended body. (I admit that I'm still working through Goldstein's detail.) I understand your point. My primary concern is the temptation by beginning students to use the theorem where it does not apply.

I'm also trying to grok Drost's answer that you cite, but I suspect that it is consistent with my interpretation of things. I don't quite understand the implications of his final equation and comments.

Possibly related: I see references to a Work-Energy Theorem and a Work-Kinetic Energy Theorem and it's possible that one is meant when the other is written.

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  • $\begingroup$ The work-energy theorem works just fine when applied to compound system. The only caveat is that we're talking about all the kinetic energy (including rotational KE and that associated with changing morphology as well.). I really don't get this strange insistence that many people have that it only applies to point particles. Goldstein disagrees, and stated in words ("the net work done equals the change in kinetic energy") it is perfectly happy to apply to systems. $\endgroup$ Jun 3, 2017 at 19:16
  • $\begingroup$ If I do work on an elastic system some of the work becomes potential energy, so $\Delta K \neq W$, no? What am I misunderstanding? $F\Delta x_\mathrm{com} = 1/2 m v^2_\mathrm{com}$, but $F\Delta x_\mathrm{com}$ is not the work done on the system. The work not accounted for by this goes into KE of constituents and PE of the system $\endgroup$
    – garyp
    Jun 3, 2017 at 19:31
  • $\begingroup$ The "net work" of the theorem is the work due to all the forces (including internal forces), and it affects the total kinetic energy (accounting for all forms). Seriously, look in Goldstein or look at CR Drost's answer and mine in the question I linked. $\endgroup$ Jun 3, 2017 at 21:19
  • $\begingroup$ @dmckee I've responded in an edit to my answer. $\endgroup$
    – garyp
    Jun 11, 2017 at 21:33
  • $\begingroup$ grok? Have I learnt a new word today? Or is is a case of "Thar's nowt wrang with owt what mitherin clutterbucks don't barely grummit"? $\endgroup$ Jul 31, 2017 at 5:29

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