Because he was a coward and deleted his own thread, I'm continuing it here:
E[sub]k[/sub] = (1/2)mv^2.
Originally posted by CaptBevvil:
Kinetic energy (Ek = (1/2)mv^2) is an expression of Work, not an expression of Force. Work is also force applied over a distance. Instantaneous Force is useful to determine the rate that energy is applied to an object, but it is most certainly not energy.A while back I made a statement about Force sub instantaneous equals mass times velocity where 'Force sub Instantaneous' was a 'place-holder' varible for that which Force measure for any unit of t such that Ft=F sub i. At the time, I was merely focused on the time value as it relates to acceleration. However, since my discussions with Mr. Novak, it is clear that my 'place-holder' (F sub i) is actually KE (Kinetic Energy). We agreed that the following definitions were true:
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Kinetic Energy is the amount of Work that was applied to a specific mass to cause it to move at a specific velocity.Kinetic Energy - Is the specific Mass and Velocity of an object at any given moment in time.
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No, that is momentum. p=mv.KE = mv
E[sub]k[/sub] = (1/2)mv^2.
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Force is the rate that kinetic energy is applied to a mass.Force - Force is the measure of the change in mass and the change in velocity over a period of time.
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F = ma. Mass does not change significantly enough for the equation to matter. Additionally, as mass increases momentum increases. I believe these terms cancel cout but I'm not going to waste my time writing a proof for it.F = (m1 - m2)(v2-v1)/(t2-t1) or, more simply:
F = delta ma OR F = delta KE/t
F = delta ma OR F = delta KE/t
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Are we talking about a rocket or are we talking about your perpetual motion machine?Note: delta m is used because in the case of a rocket, the mass (hull + fuel) is reduced over the period of time in which it is measured.