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Derivation of Common Formula

  • Thread starter Haftred
  • Start date
Can someone please explain the derivation of the common formula for average kinetic energy: K = 0.5mv^2?
 
Derivation of KE?

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

ΣWork = ΣF * d = mad = 0.5m(v^2 - v0^2) = ke
 

HallsofIvy

Science Advisor
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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!
 
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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!
 
Originally posted by PrudensOptimus


ΣWork = ΣF * d = mad = 0.5m(v^2 - v0^2) = ke
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.
 
656
0
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:

[tex]
W=\int_A^B \mathbf{F}\cdot d\mathbf{s}
[/tex]

but by Newton
[tex]
\mathbf{F}=m\mathbf{a}
[/tex]
so
[tex]
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
[/tex]

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:

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