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i see boxes =P
Mechanical energy vertical displacent energy
umm? ok question?

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I've been up since 5 am :'( can't think anymore. what happens to energy in this situation?
now we have potential energy correct?
V?
mgh?
E= U + K
\[\frac 12 mv^2\]
C=Q/V?
what do you mean?
what happens to the energy in terms of mechanics?
yes
well lets see it starts with a potential energy and no kinetic
we're only looking at it before and after it finished moving, so: no K I think
so what happened to the energy you are trying to ask?
since you end up with no kinetic or potential?
oh it's lifting the object
yes
sorry
the capacitor i'm assuming is using the energy to do work , and potential
sounds right
so C=Q/V voltage from where? battery? outlet?
yes battery
where rae you getting it? lol outlet and battery are different lol
alright
I guess I have to relate \[C=QV\] to \[U=\frac 12 CV^2\] to \[mgh\] what are we solving for? the displacement?
we know V \[mgh=\frac 12 CV^2\] we know mg and V
how can we get C? where is work? LOL
what about work?
\[W=\int F\cdot dr\] or E dr
http://en.wikipedia.org/wiki/Work_%28electrical%29
http://en.wikipedia.org/wiki/Work_%28electrical%29 work and potential energy!!!!
would work be electrical in this case though since it's mechanical work?
sure
is that yes or no lol
yes?
potential energy
So what is the actual question... i haven't actually gotten a question.. I've gotten 2 formulas and a procedure it sounds like lol
also dr = h
;)
W=Fd=Fh \[mgh+Fh=
\[h(mg+F)\]?
we dont have to integrate i'm assuming
no it's straight up
alright
now we need the work done
?
yes isn't that what it's asking lol
so you're lifting the mass meaning there is work being done = energy being used to lift it
I'm tired dude.... ok so we used up E
E= U+0
ok so E is actually your capacitor right?
I hope so....let's say it is
Alright so E is equation 24-8 in page you showed me
since \[E_1=E_2\] \[E_1=mgh+\frac{1}{2}mv^2+U_c\]
\[mg(0)+\frac{1}{2}m(0^2)+U_c\]
so all energy is coming from the capacitor
now the other side you have \[mgh+\frac{1}{2}mv^2+W_{app}+W_d\]i'm assuming drag will be really low so it's not taken into account
what's W app? and W d?
but since we're looking at it at rest in two places you have \[mgh+0+Fh\] \[U_c=mgh+Fh=(mg+F)h\]
this is what i'm thinking but i'm only like 50% certain lol
if it isn't that much don't worry about it
\[F=ma\] \[(mg+ma)h=(m(g+a))h\]
\[\frac{U_c}{m(g+a)}=h\] maybe!?!?!?! lol
W_d is work done by the drag force... which is going to be nothing probably since the velocity will be minimal
that's sort of a toughie question... we haven't gotten to capacitors and such yet

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