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anonymous
 3 years ago
@Outkast3r09
anonymous
 3 years ago
@Outkast3r09

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Mechanical energy vertical displacent energy

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now we have potential energy correct?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what happens to the energy in terms of mechanics?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well lets see it starts with a potential energy and no kinetic

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so what happened to the energy you are trying to ask?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0since you end up with no kinetic or potential?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh it's lifting the object

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the capacitor i'm assuming is using the energy to do work , and potential

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so C=Q/V voltage from where? battery? outlet?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0where rae you getting it? lol outlet and battery are different lol

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0we know V \[mgh=\frac 12 CV^2\] we know mg and V

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how can we get C? where is work? LOL

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[W=\int F\cdot dr\] or E dr

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0would work be electrical in this case though since it's mechanical work?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0is that yes or no lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So 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
 3 years ago
Best ResponseYou've already chosen the best response.0we dont have to integrate i'm assuming

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now we need the work done

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes isn't that what it's asking lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so you're lifting the mass meaning there is work being done = energy being used to lift it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'm tired dude.... ok so we used up E

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok so E is actually your capacitor right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I hope so....let's say it is

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Alright so E is equation 248 in page you showed me

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0since \[E_1=E_2\] \[E_1=mgh+\frac{1}{2}mv^2+U_c\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[mg(0)+\frac{1}{2}m(0^2)+U_c\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so all energy is coming from the capacitor

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now 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
 3 years ago
Best ResponseYou've already chosen the best response.0what's W app? and W d?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0this is what i'm thinking but i'm only like 50% certain lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if it isn't that much don't worry about it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[F=ma\] \[(mg+ma)h=(m(g+a))h\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{U_c}{m(g+a)}=h\] maybe!?!?!?! lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0W_d is work done by the drag force... which is going to be nothing probably since the velocity will be minimal

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that's sort of a toughie question... we haven't gotten to capacitors and such yet
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