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


I wonder if there is an intuitive explanation for this

edit: It’s E= 1/2 m v^2 actually

In: 0

Well, energy is the force over distance. A faster moving object will cover more distance in the same time.

So let’s imagine an impact. An object has some energy that must be absorbed over a distance. This distance is fixed. If it is moving twice as fast, it has twice as much momentum and so you’d expect twice us much force be needed to stop it. However, it also has to dissipate this force in half as much time, so we double this yet again. Four times the force over the same distance is four times the energy for double the speed.

Energy is a force applied over a certain distance, or F*d

So if E = F*d and F = m*a, then E = m*a*d.

But a*d is just v^(2) (acceleration is distance over time^(2) so multiplying that by distance gives you distance^(2)/time^(2) or (distance/time)^(2) and distance/time is velocity)

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.