Kinetic Energy & E=mc^2: A Physics Query

[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?

 

[jbriggs444]

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.

 

[davenn]

and it if still isn't clear from either of those 2 posts
try this ...

http://www.citycollegiate.com/workpowerenergy_Xc.htmDave

 

[brian.green]

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

 

[jbriggs444]

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.