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Using Newtonian mechanics, how does one prove that the change in mechanical energy of a classical system of particles is equal to the work done by non-conservative forces? What assumptions are made? For example, is it assumed that the particles have a constant mass?

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It just comes from the definitions of work, kinetic energy, potential energy, and mechanical energy.

The work done by a force is defined by $$W=\int \mathbf F\cdot\text d \mathbf x$$

The work done by all forces is given by$^*$ $$W_{net}=\int \mathbf F_{net}\cdot\text d \mathbf x=\Delta K$$

where $\Delta K$ is the change in kinetic energy. But we can also break our net force up into a sum of conservative and non-conservative forces so that $$W_{net}=\int (\mathbf F_{c}+\mathbf F_{nc})\cdot\text d \mathbf x=W_c+W_{nc}=\Delta K$$

Furthermore, by definition of potential energy, $W_c=-\Delta U$ ,therefore$^{**}$ $$W_c+W_{nc}=-\Delta U+W_{nc}=\Delta K$$ or $$W_{nc}=\Delta K+\Delta U$$

Mechanical energy is defined as the sum of kinetic and potential energies: $$E=K+U$$ so that $$\Delta E=\Delta K+\Delta U$$

Therefore we get to what we want $$W_{nc}=\Delta E$$

This doesn't depend on assumptions of mass. If the particles' masses are changing, that mass must be coming from or going to somewhere, and so there must be interactions taking place through either conservative or non-conservative forces. Therefore, this derivation covers that scenario.

Note that this also applies to systems of particles as well, since you can apply this analysis to each particle individually, then add up everything for the entire system.


$^*$ To show why $W_{net}=\Delta K$: By Newton's second law, $\mathbf F_{net}=m\mathbf a$, and we can also treat the acceleration as constant during the interval $\text d \mathbf x$ that we integrate over. Therefore the relation $|\mathbf v+\text d\mathbf v|^2=|\mathbf v|^2+2\mathbf a\cdot\text d\mathbf x$ holds.

Now, $$|\mathbf v+\text d\mathbf v|^2=(\mathbf v+\text d\mathbf v)\cdot(\mathbf v+\text d\mathbf v)=|\mathbf v|^2+2\mathbf v\cdot\text d\mathbf v+|\text d\mathbf v|^2$$

Since $|\text d\mathbf v|^2\approx0$, then we end up with $$\mathbf v\cdot\text d\mathbf v=\mathbf a\cdot\text d\mathbf x$$

Therefore, our work integral becomes $$W_{net}=\int m\mathbf a\cdot\text d \mathbf x=\int \mathbf v\cdot\text d\mathbf v=\frac12m\Delta|\mathbf v|^2=\Delta K$$

$^{**}$ Potential energy is defined through its relation to a conservative force: $$\mathbf F_c=-\mathbf\nabla U$$ therefore, $$W_c=-\int \mathbf\nabla U\cdot\text d\mathbf x=-\Delta U$$

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  • $\begingroup$ Suppose that some of the particles spontaneously change their mass without any interactions going on. Does the conservation of mechanical energy hold then? I know that this scenario is perhaps artificial, but it may be instructive to consider such a situation to solve problems concerning physically realizable configurations. $\endgroup$
    – user113773
    Commented Nov 25, 2018 at 19:59
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    $\begingroup$ @PiKindOfGuy If you have a particle whose mass is changing without interactions then yes, mechanical energy is not conserved. This is because $K=\frac12mv^2$, so the kinetic energy would be changing without the energy going to or coming from anywhere. $\endgroup$ Commented Nov 25, 2018 at 21:04
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    $\begingroup$ @PiKindOfGuy right... But it's true for each particle in the system. It's all just energy accounting. If the energy for each particle is correctly accounted for, then the energy of the entire system is correctly accounted for. I'm not sure what more to say about it. $\endgroup$ Commented Nov 28, 2018 at 4:39
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    $\begingroup$ @PiKindOfGuy Yes the potential energy of the system can depend on things external to the system. But that's fine. Your question wasn't asking about tracking particles or actually calculating these energies. At the end of the day each particle will experience conservative and non-conservative forces, each of which can be taken into account in terms of their effect on the total energy of the system. $\endgroup$ Commented Nov 28, 2018 at 12:17
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    $\begingroup$ @PiKindOfGuy Yes you do. $\endgroup$ Commented Nov 28, 2018 at 14:44

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