Conservation of mechanical energy proof

Question

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?

Answer 1

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.

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$^*$ 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, $$W_c=-\int \mathbf\nabla U\cdot\text d\mathbf x=-\Delta U$$