Why in E=mv^2 the kinetic energy increases to the square of the velocity???

Question

I wonder if there is an intuitive explanation for this

edit: It’s E= 1/2 m v^2 actually

In: 0

Answer 1

Yes, there’s an intuitive explanation.

The amount of work you do on something when you push it is the force times the distance…it doesn’t matter how fast you go, it’s always the force times the distance. And the kinetic energy gain is how much work I did on the thing to get it up to speed.

If we push a thing from 0 up to speed v, we’re going to cover a particular distance. If you calculate the work done over that distance, that’s going to be equal to the kinetic energy. So it’s really just E = work = F*d. If you assume constant acceleration (because it’s convenient and doesn’t change the end result), F=ma, so E = (ma)*d. The relationship between a and d is just integration. When you do the algebra for the relationship between a, v, and c, out will fall 1/2 m v^2.

So it’s just the work done to push something from 0 up to speed v.