Why is the change in kinetic energy equal to force times distance moved, rather than force times time?

Question

Yes, this question has been asked a few times, and will be downvoted to hell, but I still haven't seen a good answer.

Answer 1

We define concepts which are useful.

If you take force times time, you get the concept of momentum, which is conserved, and useful, as made evident by all the physics problems you can solve by considering conservation of momentum.

Imagine no one forced you to define kinetic energy as force times distance.

But you notice that many of the forces in nature depend on distance. Gravity, electric fields, springs. You notice that many forces are conservative. Moreover, you find that you have the option of doing path integrals of the form $F \cdot dr$, and these path integrals end up not depending on the path, but only on the endpoints. This (I hope) is starting to seem very useful to you (both in this reality, and in the imagined reality wherein we do not yet have such a thing as energy), as it gives you knowledge about a situation which only depends on the start and endpoints, and not on what's happened in between.

One day, sitting in a bath, eureka. What does $F \cdot dr$ make you think of? Let's consider time, now. $dr = v dt$. $F = \frac{dp}{dt} = m\frac{dv}{dt}$. These integrals we were playing with, the ones that seemed to have so much potential to be useful, can be written in another form. $\int F \cdot dr = \int m\frac{dv}{dt}vdt = \int \frac{1}{2}m\frac{d}{dt}v^{2}dt = \frac{1}{2}mv^{2} |_{start}^{end}$. This is essentially the derivation here.

After noticing how useful a construct like this might be, you start applying some names to them, since $F \cdot dr$ and $\frac{1}{2}mv^{2}$ is a mouthful. So you call one the potential energy, the other the kinetic energy, and you notice that the two combined are conserved, which has proven to be a very useful definition.