@Outkast3r09

- anonymous

@Outkast3r09

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- anonymous

i see boxes =P

- anonymous

Mechanical energy
vertical displacent
energy

- anonymous

umm? ok question?

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## More answers

- anonymous

I've been up since 5 am :'( can't think anymore.
what happens to energy in this situation?

- anonymous

now we have potential energy correct?

- anonymous

V?

- anonymous

mgh?

- anonymous

E= U + K

- anonymous

\[\frac 12 mv^2\]

- anonymous

C=Q/V?

- anonymous

what do you mean?

- anonymous

what happens to the energy in terms of mechanics?

- anonymous

yes

- anonymous

well lets see it starts with a potential energy and no kinetic

- anonymous

we're only looking at it before and after it finished moving, so:
no K I think

- anonymous

so what happened to the energy you are trying to ask?

- anonymous

since you end up with no kinetic or potential?

- anonymous

oh it's lifting the object

- anonymous

yes

- anonymous

sorry

- anonymous

the capacitor i'm assuming is using the energy to do work , and potential

- anonymous

sounds right

- anonymous

so
C=Q/V
voltage from where? battery? outlet?

- anonymous

yes battery

- anonymous

where rae you getting it? lol outlet and battery are different lol

- anonymous

alright

- anonymous

I guess I have to relate
\[C=QV\]
to
\[U=\frac 12 CV^2\]
to \[mgh\]
what are we solving for?
the displacement?

- anonymous

we know V
\[mgh=\frac 12 CV^2\]
we know mg and V

- anonymous

how can we get C?
where is work? LOL

- anonymous

what about work?

- anonymous

\[W=\int F\cdot dr\]
or E dr

- anonymous

http://en.wikipedia.org/wiki/Work_%28electrical%29

- anonymous

http://en.wikipedia.org/wiki/Work_%28electrical%29
work and potential energy!!!!

- anonymous

would work be electrical in this case though since it's mechanical work?

- anonymous

sure

- anonymous

is that yes or no lol

- anonymous

yes?

- anonymous

potential energy

- anonymous

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

- anonymous

also dr = h

- anonymous

;)

- anonymous

W=Fd=Fh
\[mgh+Fh=

- anonymous

\[h(mg+F)\]?

- anonymous

we dont have to integrate i'm assuming

- anonymous

no it's straight up

- anonymous

alright

- anonymous

now we need the work done

- anonymous

?

- anonymous

yes isn't that what it's asking lol

- anonymous

so you're lifting the mass meaning there is work being done = energy being used to lift it

- anonymous

I'm tired dude....
ok so we used up E

- anonymous

E= U+0

- anonymous

ok so E is actually your capacitor right?

- anonymous

I hope so....let's say it is

- anonymous

Alright so E is equation 24-8 in page you showed me

- anonymous

since
\[E_1=E_2\]
\[E_1=mgh+\frac{1}{2}mv^2+U_c\]

- anonymous

\[mg(0)+\frac{1}{2}m(0^2)+U_c\]

- anonymous

so all energy is coming from the capacitor

- anonymous

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

- anonymous

what's W app? and W d?

- anonymous

but since we're looking at it at rest in two places you have
\[mgh+0+Fh\]
\[U_c=mgh+Fh=(mg+F)h\]

- anonymous

this is what i'm thinking but i'm only like 50% certain lol

- anonymous

if it isn't that much don't worry about it

- anonymous

\[F=ma\]
\[(mg+ma)h=(m(g+a))h\]

- anonymous

\[\frac{U_c}{m(g+a)}=h\]
maybe!?!?!?! lol

- anonymous

W_d is work done by the drag force... which is going to be nothing probably since the velocity will be minimal

- anonymous

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

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