Question

@Outkast3r09

Answer 1

i see boxes =P

Answer 2

Mechanical energy
vertical displacent
energy

Answer 3

umm? ok question?

Answer 4

I've been up since 5 am :'( can't think anymore.
what happens to energy in this situation?

Answer 5

now we have potential energy correct?

Answer 6

V?

Answer 7

mgh?

Answer 8

E= U + K

Answer 9

\[\frac 12 mv^2\]

Answer 10

C=Q/V?

Answer 11

what do you mean?

Answer 12

what happens to the energy in terms of mechanics?

Answer 13

yes

Answer 14

well lets see it starts with a potential energy and no kinetic

Answer 15

we're only looking at it before and after it finished moving, so:
no K I think

Answer 16

so what happened to the energy you are trying to ask?

Answer 17

since you end up with no kinetic or potential?

Answer 18

oh it's lifting the object

Answer 19

yes

Answer 20

sorry

Answer 21

the capacitor i'm assuming is using the energy to do work , and potential

Answer 22

sounds right

Answer 23

so
C=Q/V
voltage from where? battery? outlet?

Answer 24

yes battery

Answer 25

where rae you getting it? lol outlet and battery are different lol

Answer 26

alright

Answer 27

I guess I have to relate
\[C=QV\]
to
\[U=\frac 12 CV^2\]
to \[mgh\]
what are we solving for?
the displacement?

Answer 28

we know V
\[mgh=\frac 12 CV^2\]
we know mg and V

Answer 29

how can we get C?
where is work? LOL

Answer 30

what about work?

Answer 31

\[W=\int F\cdot dr\]
or E dr

Answer 32

http://en.wikipedia.org/wiki/Work_%28electrical%29

Answer 33

http://en.wikipedia.org/wiki/Work_%28electrical%29
work and potential energy!!!!

Answer 34

would work be electrical in this case though since it's mechanical work?

Answer 35

sure

Answer 36

is that yes or no lol

Answer 37

yes?

Answer 38

potential energy

Answer 39

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

Answer 40

also dr = h

Answer 41

;)

Answer 42

W=Fd=Fh

\[mgh+Fh=

Answer 43

\[h(mg+F)\]?

Answer 44

we dont have to integrate i'm assuming

Answer 45

no it's straight up

Answer 46

alright

Answer 47

now we need the work done

Answer 48

?

Answer 49

yes isn't that what it's asking lol

Answer 50

so you're lifting the mass meaning there is work being done = energy being used to lift it

Answer 51

I'm tired dude....
ok so we used up E

Answer 52

E= U+0

Answer 53

ok so E is actually your capacitor right?

Answer 54

I hope so....let's say it is

Answer 55

Alright so E is equation 24-8 in page you showed me

Answer 56

since

\[E_1=E_2\]

\[E_1=mgh+\frac{1}{2}mv^2+U_c\]

Answer 57

\[mg(0)+\frac{1}{2}m(0^2)+U_c\]

Answer 58

so all energy is coming from the capacitor

Answer 59

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

Answer 60

what's W app? and W d?

Answer 61

but since we're looking at it at rest in two places you have

\[mgh+0+Fh\]
\[U_c=mgh+Fh=(mg+F)h\]

Answer 62

this is what i'm thinking but i'm only like 50% certain lol

Answer 63

if it isn't that much don't worry about it

Answer 64

\[F=ma\]
\[(mg+ma)h=(m(g+a))h\]

Answer 65

\[\frac{U_c}{m(g+a)}=h\]

maybe!?!?!?! lol

Answer 66

W_d is work done by the drag force... which is going to be nothing probably since the velocity will be minimal

Answer 67

that's sort of a toughie question... we haven't gotten to capacitors and such yet