# 1/2m*v^2 and E=mc^2 ?

#### brian.green

The kinetic energy equ. is 1/2m*v^2 but why just 1/2m and why v^2? I understand why m*v but the rest of it not make sense for me.
There is the well known E=mc^2 where c is v.light but the mass is not half here. Why?

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#### jbriggs444

Homework Helper
One way of seeing it is that the energy that goes into accelerating an object is given by the work done on that object. Work = force times distance:

E = Fd
If you accelerate an object up to a velocity of v with a constant force then it will have a constant acceleration. Its average velocity during the acceleration will be half of its maximum velocity [right there is your factor of two]. The distance it will cover during the process of accelerating to a velocity of v over a time t will be equal to elapsed time times average velocity:

d = vt/2.​

The acceleration required to reach velocity v in time t is:

a = v/t​

The force required to achieve that (F=ma) is:

F = mv/t​

Put it together and you have

E = Fd = vt/2 * mv/t = mv2/2​

#### DrStupid

[]
The kinetic energy equ. is 1/2m*v^2 but why just 1/2m and why v^2? I understand why m*v but the rest of it not make sense for me.
If you understand m*v then let's start with it. In classical mechanics (and that's what we are talking about here) momentum is defined as

$p: = m \cdot v$

force is defined as the change of momentum with time:

$F: = \frac{{dp}}{{dt}} = m \cdot \frac{{dv}}{{dt}}$

and mechanical work is defined as the product of force and displacement:

$dW: = F \cdot ds = m \cdot v \cdot dv$

Integration of the work gives the change of kinetic energy:

$E_{kin} = \int {m \cdot v \cdot dv} = {\textstyle{1 \over 2}}m \cdot v^2$

That's where 1/2 and v^2 come from.

There is the well known E=mc^2 where c is v.light but the mass is not half here. Why?
That's something completely different because
1. It's not classical mechanics but relativity.
2. It's not kinetic energy but rest energy.

Gold Member

#### brian.green

Thanks Dave, I understand now! The well known equ. is not correct, not the mass is half actually.

#### jbriggs444

Homework Helper
Thanks, I understand now! The well known equ. is not correct, not the mass is half actually.
$E_0=mc^2$ is correct. But it is the formula for rest energy, not for kinetic energy. It is the energy equivalent of an object's mass when the object is just sitting there.

$E^2 = m^2c^4 + p^2c^2$ is a generalization that gives total energy E in terms of mass m and momentum p.

$E = \frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$ is a generalization that gives total energy E in terms of mass m and velocity v.

If you extract kinetic energy KE = Total energy - Rest energy = $\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}} - mc^2$ then you get something for which $KE=\frac{1}{2}mv^2$ is a very good approximation.

So the two formulas are not in conflict. They are, in fact, compatible.

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