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JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.0Mechanical energy vertical displacent energy

JenniferSmart1
 one year 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?

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.0now we have potential energy correct?

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac 12 mv^2\]

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1what happens to the energy in terms of mechanics?

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1well lets see it starts with a potential energy and no kinetic

JenniferSmart1
 one year 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

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1so what happened to the energy you are trying to ask?

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1since you end up with no kinetic or potential?

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1oh it's lifting the object

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1the capacitor i'm assuming is using the energy to do work , and potential

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.0so C=Q/V voltage from where? battery? outlet?

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1where rae you getting it? lol outlet and battery are different lol

JenniferSmart1
 one year 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?

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.0we know V \[mgh=\frac 12 CV^2\] we know mg and V

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.0how can we get C? where is work? LOL

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.0\[W=\int F\cdot dr\] or E dr

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.0http://en.wikipedia.org/wiki/Work_%28electrical%29 work and potential energy!!!!

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1would work be electrical in this case though since it's mechanical work?

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1is that yes or no lol

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.0potential energy

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1So what is the actual question... i haven't actually gotten a question.. I've gotten 2 formulas and a procedure it sounds like lol

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.0we dont have to integrate i'm assuming

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1no it's straight up

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1now we need the work done

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1yes isn't that what it's asking lol

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1so you're lifting the mass meaning there is work being done = energy being used to lift it

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.0I'm tired dude.... ok so we used up E

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1ok so E is actually your capacitor right?

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.0I hope so....let's say it is

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1Alright so E is equation 248 in page you showed me

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1since \[E_1=E_2\] \[E_1=mgh+\frac{1}{2}mv^2+U_c\]

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1\[mg(0)+\frac{1}{2}m(0^2)+U_c\]

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1so all energy is coming from the capacitor

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1now 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
 one year ago
Best ResponseYou've already chosen the best response.0what's W app? and W d?

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1but since we're looking at it at rest in two places you have \[mgh+0+Fh\] \[U_c=mgh+Fh=(mg+F)h\]

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1this is what i'm thinking but i'm only like 50% certain lol

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1if it isn't that much don't worry about it

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1\[F=ma\] \[(mg+ma)h=(m(g+a))h\]

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{U_c}{m(g+a)}=h\] maybe!?!?!?! lol

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1W_d is work done by the drag force... which is going to be nothing probably since the velocity will be minimal

Outkast3r09
 one year ago
Best ResponseYou've already chosen the best response.1that's sort of a toughie question... we haven't gotten to capacitors and such yet
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