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

JenniferSmart1
 2 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?

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

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

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

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

JenniferSmart1
 2 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

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

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

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

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

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

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

JenniferSmart1
 2 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?

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

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

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

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

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

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

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0potential energy

Outkast3r09
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0we dont have to integrate i'm assuming

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

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

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

Outkast3r09
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0I'm tired dude.... ok so we used up E

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

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

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

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

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

Outkast3r09
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0what's W app? and W d?

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

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

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

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

Outkast3r09
 2 years 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
 2 years 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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