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Mgh = 1/2mv^2

by ZGMF - X20A
Tags: 1 or 2mv2
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ZGMF - X20A
#1
Sep29-06, 07:42 AM
P: 8
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.
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semc
#2
Sep29-06, 08:08 AM
P: 329
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.
HallsofIvy
#3
Sep29-06, 08:33 AM
Math
Emeritus
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Thanks
PF Gold
P: 39,348
And energy is conserved as long as the are no "non-conservative" forces- i.e. as long as there is no friction.

ZGMF - X20A
#4
Sep30-06, 02:51 AM
P: 8
Mgh = 1/2mv^2

O.o
So when there is no friction P.E = K.E?
Tomsk
#5
Sep30-06, 04:17 AM
P: 227
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<decrease in PE.


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