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 Question mgh 1 2mv^2 ( Physics Help and Math Help - Physics Forums Introductory Physics )Updated: 2008-11-09 05:10:13 (4)
 mgh 1 2mv^2 I just wanna know in what conditions does mgh = 1/2mv^2 when a guy goes down a slope and in what conditions does mgh =/= 1/2mv^2. Thanks in advance.
 Answers: mgh 1 2mv^2 ( Physics Help and Math Help - Physics Forums Introductory Physics )
 mgh 1 2mv^2 Yes, decrease in PE= increase in KE and vice versa. If there's friction, decrease in PE= increase in KE + energy 'lost' to surroundings, so increase in KE
 mgh 1 2mv^2 O.o So when there is no friction P.E = K.E? ZGMF - X20A
 mgh 1 2mv^2 And energy is conserved as long as the are no "non-conservative" forces- i.e. as long as there is no friction. HallsofIvy
 mgh 1 2mv^2 according to COE, (initial)PE + KE = PE + KE(final) so i think that its better to use this instead of your equation.For example,if the object were to be released with some initial velocity,the KE(initial) would not be 0 so total initial energy is mgh+1/2mv^2 instead of mgh.Btw,it is more precise to write change of PE = -change of KE as this shows that energy is conserved. semc
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