# Derivation of Kinetic energy formula and energy principle

1. Jun 9, 2014

### albertrichardf

Hi all,

Here is the derivation of kinetic energy from Work:

W = ∫Fds
From the second law of motion F = dp/dt, which is equal to mdv/dt, so:

W = m∫dvdx/dt which = m∫dv x v because dx/dt = v
Therefore W = 1/2mv2, when integrated.
However from simple algebra derivation, W = Δ1/2mv2.

Did I skip something in the derivation? I know that the actual integration would be 1/2mv2
+ C, where C is a constant. However I omitted the C. Is it because of that that I did not end up with the proper equation? If so, why would C = -1/2vi2. (vi because Δv = v - vi and the rest comes from kinetic energy equation.) Or is it because I wrongly integrated?

Any answers would be appreciated. Thanks

2. Jun 9, 2014

### Staff: Mentor

Do it as a definite integral between vi and vf, which corresponds to doing the original integral of Fds as a definite integral between an initial position si and final position sf.

3. Jun 9, 2014

### stevendaryl

Staff Emeritus
The formula for work is not an indefinite integral, it's a definite integral. That means that there is no arbitrary constant $C$. The definite integral gives:

$W = 1/2 m (v^2 - v_i^2)$

where $v_i$ is the initial velocity. The work is only equal to the kinetic energy in the case of a force accelerating a particle from rest ($v_i = 0)$.

4. Jun 9, 2014

### dauto

You integrated correctly and C = -1/2vi2. To see that just keep in mind that the final time tf can be any time at all and in particular we can chose tf = ti in which case no time has elapsed no displacement has occurred and no work was done, so you have

0 = W = 1/2 m vf2 + C = 1/2 m vi2 + C,

hence C = -1/2 m vi2

5. Jun 9, 2014

### albertrichardf

Thanks for the answers. However, there is something else I wish to clear up now. Why would work be a definite integral?
Thanks

6. Jun 9, 2014

### UltrafastPED

Because it is the work done in moving from point A to point B.