#9, your assertion is not proven. PROVE that the mass doesn't change when you're performing the calculus. I know I lose a noticeable amount of weight, therefore the mass factor changes, which can be integrated.
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Anonymous2007-06-26 20:57 ID:hyCvRRE1
W = ∫ F dr
= ∫ ma dr
= ∫ m dv/dt dr
= ∫ m dr/dt dv
= ∫ m v dv
= 1/2 m v^2
P = ∫ F dt
= ∫ ma dt
= ∫ m dv/dt dt
= ∫ m dv
= mv
Name:
Anonymous2007-06-27 3:20 ID:IC423zu6
RedCream thinks he's a genius. >>11 just proved him wrong.
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Anonymous2007-06-27 3:46 ID:UUH2v25D
>>11
momentum is lowercase p. P is power which is dW/dt.
In the Newtonian limit, it's momentum^2/(2m)
In general, it's sqrt[(pc)^2 + (mc^2)^2]
The most important difference is that momentum has direction, while kinetic energy does not. In addition, for a given system, momentum is conserved in the absence of external forces. The same is not necessarily true for kinetic energy.
Name:
Anonymous2009-03-18 3:09
I'm feeling really keen, for some of that good ol' green