$\newcommand{\force}{\mathbf{F}}$ $\newcommand{\velocity}{\mathbf{v}}$ $\newcommand{\position}{\mathbf{r}}$

Here's an incomplete proof of the work-energy theorem:

$$W = \int_1^2\force \cdot d\position = m\int_1^2\frac{d\velocity}{dt} \cdot d\position = m \int_1^2 \velocity \cdot \frac{d\velocity}{dt} dt.$$

Allow $m$ to be time-dependent. Isn't it the case that the equation is not correct for $m$ and $\position$ both depend on time and thus $m$ is not allowed to be pulled out of the integral?

If we're unable to pull $m$ out of the integral then the work-energy theorem (presumably) doesn't hold when $m$ is time-dependent.

Edit: Please note the "Newtonian-mechanics" tag.

  • $\begingroup$ It depends on what you consider the mass of your system to be and how you take into account the change in mass. Do you have a specific example you are looking at? $\endgroup$ – Aaron Stevens Apr 6 at 1:09
  • $\begingroup$ @AaronStevens The mass of the system is implicitly the mass of a single particle (by what the work-energy theorem looks like). The change in mass isn't accounted for. $\endgroup$ – PiKindOfGuy Apr 6 at 1:25
  • $\begingroup$ @AaronStevens And if you're going to point out that that's non-physical, I will refer you to the edit to this question of mine that you've seen before: physics.stackexchange.com/q/465176 $\endgroup$ – PiKindOfGuy Apr 6 at 1:27
  • $\begingroup$ You are considering a system where the particle emits mass equally in opposite directions so as to not change its velocity relative to some frame? $\endgroup$ – Aaron Stevens Apr 6 at 1:30
  • $\begingroup$ @AaronStevens No, it's just an example to illustrate that one needn't always care what happens to the mass. I'm considering a particle for which the reduction or gain in mass doesn't impart any momentum. To me that's just that "the particle's mass is time-dependent", but others don't seem to like throwing the conservation of mass out the window. $\endgroup$ – PiKindOfGuy Apr 6 at 1:40

Based on our discussion in the comments and chat, it looks like you are considering a situation where the mass is lost by the system in such a way that the forces on the system caused by the loss of mass cancels out, so all that matters is the net force acting on the object. Then Newton's second law has the form $$\mathbf F(t)=m(t)a(t)$$

Now, the work-energy theorem for constant mass systems is $W=\Delta K$ i.e. the net work is equal to the change in kinetic energy. But is this valid in our case? Let's say I see an object moving by me at a constant velocity. There are no external forces acting on the object, and it is emitting/absorbing mass in such a way that the change in mass doesn't change its velocity (as discussed above).

Now, this means that I see the object's mass changing with a constant velocity, which means it's kinetic energy must be changing. Yet there is no work being done on the object since there is no external force and the forces supplied by the change in mass cancels out.

Therefore it must be concluded that the work energy theorem does not work for a variable mass system. The change in mass can add or remove energy without any work being done on the "main system". Of course this doesn't mean that energy conservation is invalid here. This just means our system isn't large enough. If you consider the mass that is entering or leaving system as well you will find that the total energy will remain constant.

  • $\begingroup$ In the case of time dependent mass should you start from ${\bf F} = \frac{d {\bf p}}{d t} = m(t) {\bf a}(t) + \dot{m(t)} {\bf v}$? $\endgroup$ – jim Apr 6 at 10:19
  • 1
    $\begingroup$ @jim No, you should not. This is explained here under the variable mass section for Newton's second law $\endgroup$ – Aaron Stevens Apr 6 at 11:32

You are right. The correct version of work energy theorem states that the work done equals change in total relativistic energy and not just kinetic energy:

$W = \Delta(mc^2)$

Here m is not rest mass $m_0$ but the total relativistic mass $m$:

$m^2 = m_0^2 + P^2/c^2$

Where $P$ is relativistic momentum. If velocity is small compared to light the above theorem reduces to simple work-energy theorem. See Robert Resnick’s “Special relativity”.

  • $\begingroup$ I'm talking about Newtonian mechanics. $\endgroup$ – PiKindOfGuy Apr 6 at 2:47

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