ZGMF - X20A
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I just want to 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.
Understanding Mgh = 1/2mv^2: Conditions for Equal and Unequal Energy on Slopes
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semc
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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.
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And energy is conserved as long as the are no "non-conservative" forces- i.e. as long as there is no friction.
ZGMF - X20A
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So when there is no friction P.E = K.E?
Tomsk
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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.