Homework Help: Derivation of Common Formula

1. Nov 11, 2003

Haftred

Can someone please explain the derivation of the common formula for average kinetic energy: K = 0.5mv^2?

2. Nov 11, 2003

PrudensOptimus

Derivation of KE?

a = v^2 - v0^2 / 2d

&Sigma;Work = &Sigma;F * d = mad = 0.5m(v^2 - v0^2) = ke

3. Nov 11, 2003

HallsofIvy

First, K = 0.5mv^2 is not "average" kinetic energy- it is kinetic energy- in fact it is the definition of kinetic energy- there is no "derivation". At some time, back at the beginning of "physics" someone notice that, in many dynamics problems that quantity was conserved- so they gave it a name!

4. Nov 16, 2003

mani

First, K = 0.5mv^2 is not "average" kinetic energy- it is kinetic energy- in fact it is the definition of kinetic energy- there is no "derivation". At some time, back at the beginning of "physics" someone notice that, in many dynamics problems that quantity was conserved- so they gave it a name!

Quantities, mv^2, 35mv^2, are also conserved!

5. Nov 17, 2003

physics247

So ummmm ... This is relative kinetic energy not average kinetic energy?

That is, relative to the kinetic energy of the object when v0 = 0

Interesting derivation. Thanks.

6. Nov 17, 2003

lethe

energy is the ability to do work.

work is force times distance.

kinetic energy is that energy a body has due to its motion.

so lets see:

$$W=\int_A^B \mathbf{F}\cdot d\mathbf{s}$$

but by Newton
$$\mathbf{F}=m\mathbf{a}$$
so
$$W=\int_A^Bm\mathbf{a}\cdot d\mathbf{s} = m\int_{t_{A}}^{t_{B}}\mathbf{a}\cdot\mathbf{v}dt\\ = \frac{m}{2}\int_{t_{A}}^{t_{B}}\frac{d(v^2)}{dt}\! dt = \frac{1}{2}mv_B^2 - \frac{1}{2}mv_A^2$$

so you see that the work done is equal to the change in kinetic energy between the two points, and this is a derivation of the formula for that kinetic energy.

Last edited: Nov 17, 2003