
#1
Sep2906, 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.




#2
Sep2906, 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.




#3
Sep2906, 08:33 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

And energy is conserved as long as the are no "nonconservative" forces i.e. as long as there is no friction.




#4
Sep3006, 02:51 AM

P: 8

mgh = 1/2mv^2
O.o
So when there is no friction P.E = K.E? 



#5
Sep3006, 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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