JenniferSmart1
@Outkast3r09



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Outkast3r09
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i see boxes =P

JenniferSmart1
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Mechanical energy
vertical displacent
energy

Outkast3r09
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umm? ok question?

JenniferSmart1
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I've been up since 5 am :'( can't think anymore.
what happens to energy in this situation?

JenniferSmart1
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now we have potential energy correct?

JenniferSmart1
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V?

JenniferSmart1
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mgh?

JenniferSmart1
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E= U + K

JenniferSmart1
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\[\frac 12 mv^2\]

JenniferSmart1
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C=Q/V?

Outkast3r09
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what do you mean?

Outkast3r09
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what happens to the energy in terms of mechanics?

JenniferSmart1
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yes

Outkast3r09
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well lets see it starts with a potential energy and no kinetic

JenniferSmart1
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we're only looking at it before and after it finished moving, so:
no K I think

Outkast3r09
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so what happened to the energy you are trying to ask?

Outkast3r09
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since you end up with no kinetic or potential?

Outkast3r09
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oh it's lifting the object

JenniferSmart1
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yes

JenniferSmart1
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sorry

Outkast3r09
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the capacitor i'm assuming is using the energy to do work , and potential

JenniferSmart1
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sounds right

JenniferSmart1
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so
C=Q/V
voltage from where? battery? outlet?

JenniferSmart1
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yes battery

Outkast3r09
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where rae you getting it? lol outlet and battery are different lol

Outkast3r09
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alright

JenniferSmart1
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I guess I have to relate
\[C=QV\]
to
\[U=\frac 12 CV^2\]
to \[mgh\]
what are we solving for?
the displacement?

JenniferSmart1
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we know V
\[mgh=\frac 12 CV^2\]
we know mg and V

JenniferSmart1
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how can we get C?
where is work? LOL

Outkast3r09
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what about work?

JenniferSmart1
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\[W=\int F\cdot dr\]
or E dr



Outkast3r09
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would work be electrical in this case though since it's mechanical work?

JenniferSmart1
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sure

Outkast3r09
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is that yes or no lol

JenniferSmart1
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yes?

JenniferSmart1
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potential energy

Outkast3r09
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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

Outkast3r09
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also dr = h

Outkast3r09
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;)

Outkast3r09
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W=Fd=Fh
\[mgh+Fh=

Outkast3r09
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\[h(mg+F)\]?

JenniferSmart1
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we dont have to integrate i'm assuming

Outkast3r09
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no it's straight up

Outkast3r09
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alright

Outkast3r09
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now we need the work done

JenniferSmart1
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?

Outkast3r09
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yes isn't that what it's asking lol

Outkast3r09
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so you're lifting the mass meaning there is work being done = energy being used to lift it

JenniferSmart1
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I'm tired dude....
ok so we used up E

JenniferSmart1
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E= U+0

Outkast3r09
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ok so E is actually your capacitor right?

JenniferSmart1
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I hope so....let's say it is

Outkast3r09
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Alright so E is equation 248 in page you showed me

Outkast3r09
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since
\[E_1=E_2\]
\[E_1=mgh+\frac{1}{2}mv^2+U_c\]

Outkast3r09
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\[mg(0)+\frac{1}{2}m(0^2)+U_c\]

Outkast3r09
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so all energy is coming from the capacitor

Outkast3r09
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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

JenniferSmart1
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what's W app? and W d?

Outkast3r09
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but since we're looking at it at rest in two places you have
\[mgh+0+Fh\]
\[U_c=mgh+Fh=(mg+F)h\]

Outkast3r09
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this is what i'm thinking but i'm only like 50% certain lol

Outkast3r09
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if it isn't that much don't worry about it

Outkast3r09
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\[F=ma\]
\[(mg+ma)h=(m(g+a))h\]

Outkast3r09
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\[\frac{U_c}{m(g+a)}=h\]
maybe!?!?!?! lol

Outkast3r09
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W_d is work done by the drag force... which is going to be nothing probably since the velocity will be minimal

Outkast3r09
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that's sort of a toughie question... we haven't gotten to capacitors and such yet