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i see boxes =P

Mechanical energy
vertical displacent
energy

umm? ok question?

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

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

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

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

that's sort of a toughie question... we haven't gotten to capacitors and such yet