Re: derivation of kinetic energy
- From: jem <xxx@xxxxxxx>
- Date: Sun, 16 Jul 2006 08:27:03 -0400
castertroy14@xxxxxxxxxxx wrote:
can anyone show me the process of getting equation (2) from equation
(1).
(1) D(mv / (1-v^2/c^2)^0.5) (2) m(1 - v^2/c^2)^-1.5 Dv
Let g = 1/(1-v^2/c^2)^0.5
then Dg = vg^3 Dv, so
D(mvg) = m(g + v^2 g^3)Dv
= mg^3 Dv
.
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- derivation of kinetic energy
- From: castertroy14
- derivation of kinetic energy
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