Re: derivation of kinetic energy

Dear castertroy14:

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castertro...@xxxxxxxxxxx wrote:
N:dlzc D:aol T:com (dlzc) wrote:
Dear castertroy14:

<castertroy14@xxxxxxxxxxx> wrote in message
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N:dlzc D:aol T:com (dlzc) wrote:
Dear castertroy14:

<castertroy14@xxxxxxxxxxx> wrote in message
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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

I don't see how that relates to kinetic energy.

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/relmom.html#c4

David A. Smith

K = integral(vDp) = integral(vD(mv / (1-v^2/c^2)^0.5))
D(mv / (1-v^2/c^2)^0.5) = m(1 - v^2/c^2)^-1.5 Dv
therefore
K = integral(m(1-v^2/c^2)^-1.5 vDv) =
mc^2(1/(1-(v^2/c^2)^0.5 -
1)

so my question is how do you get from
D(mv / (1-v^2/c^2)^0.5) to m(1 - v^2/c^2)^-1.5 Dv

What text do you find this in? I have issues
with the very first:
K = integral(vDp)
I think this should be:
K = integral(pDv)

K = integral(fDs) = integral([Dp/Dt]Dx) = integral(vDp)

correction:
K = integral(fDx) = integral([Dp/Dt]Dx) = integral(vDp)

I ask again, what text did you get this from?
http://scienceworld.wolfram.com/physics/KineticEnergy.html

Note that f, Dx, Dp, v, and p are vector quantities, and the
product represented is a dot product. If you are going to spend
time learning relativity, you will find that force is not well
defined. It is much easier to stick with momentum, where
possible.
K = integral(pDv)
.... has none of the issues you are trying to have with
derivatives, when you are intending to *integrate* anyway.

David A. Smith


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