- #71

etotheipi

atyy said:Yes, I would prefer to stress the generality of the work-energy theorem in Newtonian mechanics, rather than inferring its failure from macroscopic examples

Yes I think the main take-away from this discussion is that, whilst the work-energy theorem holds conceptually for any arbitrary body (i.e. assuming that we take

*all*works [including internal] into account), it is only useful as a problem-solving tool so long as we have some means of computing these internal works.

For a simple system, like two bodies interacting gravitationally, it's very easy to apply the work energy theorem to the entire system because the total internal work is simply ##W = \int \vec{F}_{12} \cdot d\vec{x}_1 + \int \vec{F}_{21} \cdot d\vec{x}_2 = -\Delta U_{\text{system}}##. The theorem yields ##-\Delta U_{\text{system}} =\Delta T_{\text{system}} \implies \Delta U_{\text{system}} + \Delta T_{\text{system}} = 0## immediately.

For a more complicated system, like a ball of putty, or a spring, the work energy theorem is not so easily applied. With the spring, for instance, if we compress it slowly from its unstretched length through a distance ##\delta##, we do external work ##\frac{1}{2}k \delta^2## on the spring. However it's change in kinetic energy is zero! If we step back and think about general energy considerations, it's clear that this is because the work we did is now locked up in elastic potential energy, and not kinetic.

What we infer is that exactly ##-\frac{1}{2}k\delta^2## of

*internal*work was done by the spring on itself (by the many internal forces between the many particles in the spring, as they were moved closer together), so that the spring has no change in kinetic energy. We could go one step further to note that this internal work is done by conservative forces, which means the change in potential energy of the spring is ##\Delta U_{\text{spring}} = -\left( - \frac{1}{2} k \delta^2 \right) = \frac{1}{2}k\delta^2##, as we expect.

So everything works out if we are careful