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JenniferSmart1Best ResponseYou've already chosen the best response.0
Mechanical energy vertical displacent energy
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
I've been up since 5 am :'( can't think anymore. what happens to energy in this situation?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
now we have potential energy correct?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
\[\frac 12 mv^2\]
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
what happens to the energy in terms of mechanics?
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
well lets see it starts with a potential energy and no kinetic
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
we're only looking at it before and after it finished moving, so: no K I think
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
so what happened to the energy you are trying to ask?
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
since you end up with no kinetic or potential?
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
oh it's lifting the object
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
the capacitor i'm assuming is using the energy to do work , and potential
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
so C=Q/V voltage from where? battery? outlet?
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
where rae you getting it? lol outlet and battery are different lol
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
I guess I have to relate \[C=QV\] to \[U=\frac 12 CV^2\] to \[mgh\] what are we solving for? the displacement?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
we know V \[mgh=\frac 12 CV^2\] we know mg and V
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
how can we get C? where is work? LOL
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
\[W=\int F\cdot dr\] or E dr
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
http://en.wikipedia.org/wiki/Work_%28electrical%29
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
http://en.wikipedia.org/wiki/Work_%28electrical%29 work and potential energy!!!!
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
would work be electrical in this case though since it's mechanical work?
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
is that yes or no lol
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
potential energy
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
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
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
we dont have to integrate i'm assuming
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
no it's straight up
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
now we need the work done
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
yes isn't that what it's asking lol
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
so you're lifting the mass meaning there is work being done = energy being used to lift it
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
I'm tired dude.... ok so we used up E
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
ok so E is actually your capacitor right?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
I hope so....let's say it is
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
Alright so E is equation 248 in page you showed me
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
since \[E_1=E_2\] \[E_1=mgh+\frac{1}{2}mv^2+U_c\]
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
\[mg(0)+\frac{1}{2}m(0^2)+U_c\]
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
so all energy is coming from the capacitor
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
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
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
what's W app? and W d?
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
but since we're looking at it at rest in two places you have \[mgh+0+Fh\] \[U_c=mgh+Fh=(mg+F)h\]
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
this is what i'm thinking but i'm only like 50% certain lol
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
if it isn't that much don't worry about it
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
\[F=ma\] \[(mg+ma)h=(m(g+a))h\]
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
\[\frac{U_c}{m(g+a)}=h\] maybe!?!?!?! lol
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
W_d is work done by the drag force... which is going to be nothing probably since the velocity will be minimal
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
that's sort of a toughie question... we haven't gotten to capacitors and such yet
 one year ago
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