anonymous
  • anonymous
@Outkast3r09
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
i see boxes =P
anonymous
  • anonymous
Mechanical energy vertical displacent energy
anonymous
  • anonymous
umm? ok question?

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anonymous
  • anonymous
I've been up since 5 am :'( can't think anymore. what happens to energy in this situation?
anonymous
  • anonymous
now we have potential energy correct?
anonymous
  • anonymous
V?
anonymous
  • anonymous
mgh?
anonymous
  • anonymous
E= U + K
anonymous
  • anonymous
\[\frac 12 mv^2\]
anonymous
  • anonymous
C=Q/V?
anonymous
  • anonymous
what do you mean?
anonymous
  • anonymous
what happens to the energy in terms of mechanics?
anonymous
  • anonymous
yes
anonymous
  • anonymous
well lets see it starts with a potential energy and no kinetic
anonymous
  • anonymous
we're only looking at it before and after it finished moving, so: no K I think
anonymous
  • anonymous
so what happened to the energy you are trying to ask?
anonymous
  • anonymous
since you end up with no kinetic or potential?
anonymous
  • anonymous
oh it's lifting the object
anonymous
  • anonymous
yes
anonymous
  • anonymous
sorry
anonymous
  • anonymous
the capacitor i'm assuming is using the energy to do work , and potential
anonymous
  • anonymous
sounds right
anonymous
  • anonymous
so C=Q/V voltage from where? battery? outlet?
anonymous
  • anonymous
yes battery
anonymous
  • anonymous
where rae you getting it? lol outlet and battery are different lol
anonymous
  • anonymous
alright
anonymous
  • anonymous
I guess I have to relate \[C=QV\] to \[U=\frac 12 CV^2\] to \[mgh\] what are we solving for? the displacement?
anonymous
  • anonymous
we know V \[mgh=\frac 12 CV^2\] we know mg and V
anonymous
  • anonymous
how can we get C? where is work? LOL
anonymous
  • anonymous
what about work?
anonymous
  • anonymous
\[W=\int F\cdot dr\] or E dr
anonymous
  • anonymous
http://en.wikipedia.org/wiki/Work_%28electrical%29
anonymous
  • anonymous
http://en.wikipedia.org/wiki/Work_%28electrical%29 work and potential energy!!!!
anonymous
  • anonymous
would work be electrical in this case though since it's mechanical work?
anonymous
  • anonymous
sure
anonymous
  • anonymous
is that yes or no lol
anonymous
  • anonymous
yes?
anonymous
  • anonymous
potential energy
anonymous
  • anonymous
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
anonymous
  • anonymous
also dr = h
anonymous
  • anonymous
;)
anonymous
  • anonymous
W=Fd=Fh \[mgh+Fh=
anonymous
  • anonymous
\[h(mg+F)\]?
anonymous
  • anonymous
we dont have to integrate i'm assuming
anonymous
  • anonymous
no it's straight up
anonymous
  • anonymous
alright
anonymous
  • anonymous
now we need the work done
anonymous
  • anonymous
?
anonymous
  • anonymous
yes isn't that what it's asking lol
anonymous
  • anonymous
so you're lifting the mass meaning there is work being done = energy being used to lift it
anonymous
  • anonymous
I'm tired dude.... ok so we used up E
anonymous
  • anonymous
E= U+0
anonymous
  • anonymous
ok so E is actually your capacitor right?
anonymous
  • anonymous
I hope so....let's say it is
anonymous
  • anonymous
Alright so E is equation 24-8 in page you showed me
anonymous
  • anonymous
since \[E_1=E_2\] \[E_1=mgh+\frac{1}{2}mv^2+U_c\]
anonymous
  • anonymous
\[mg(0)+\frac{1}{2}m(0^2)+U_c\]
anonymous
  • anonymous
so all energy is coming from the capacitor
anonymous
  • anonymous
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
anonymous
  • anonymous
what's W app? and W d?
anonymous
  • anonymous
but since we're looking at it at rest in two places you have \[mgh+0+Fh\] \[U_c=mgh+Fh=(mg+F)h\]
anonymous
  • anonymous
this is what i'm thinking but i'm only like 50% certain lol
anonymous
  • anonymous
if it isn't that much don't worry about it
anonymous
  • anonymous
\[F=ma\] \[(mg+ma)h=(m(g+a))h\]
anonymous
  • anonymous
\[\frac{U_c}{m(g+a)}=h\] maybe!?!?!?! lol
anonymous
  • anonymous
W_d is work done by the drag force... which is going to be nothing probably since the velocity will be minimal
anonymous
  • anonymous
that's sort of a toughie question... we haven't gotten to capacitors and such yet

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