## JenniferSmart1 2 years ago @Outkast3r09

1. Outkast3r09

i see boxes =P

2. JenniferSmart1

Mechanical energy vertical displacent energy

3. Outkast3r09

umm? ok question?

4. JenniferSmart1

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

5. JenniferSmart1

now we have potential energy correct?

6. JenniferSmart1

V?

7. JenniferSmart1

mgh?

8. JenniferSmart1

E= U + K

9. JenniferSmart1

$\frac 12 mv^2$

10. JenniferSmart1

C=Q/V?

11. Outkast3r09

what do you mean?

12. Outkast3r09

what happens to the energy in terms of mechanics?

13. JenniferSmart1

yes

14. Outkast3r09

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

15. JenniferSmart1

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

16. Outkast3r09

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

17. Outkast3r09

since you end up with no kinetic or potential?

18. Outkast3r09

oh it's lifting the object

19. JenniferSmart1

yes

20. JenniferSmart1

sorry

21. Outkast3r09

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

22. JenniferSmart1

sounds right

23. JenniferSmart1

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

24. JenniferSmart1

yes battery

25. Outkast3r09

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

26. Outkast3r09

alright

27. JenniferSmart1

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

28. JenniferSmart1

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

29. JenniferSmart1

how can we get C? where is work? LOL

30. Outkast3r09

31. JenniferSmart1

$W=\int F\cdot dr$ or E dr

32. JenniferSmart1
33. JenniferSmart1

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

34. Outkast3r09

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

35. JenniferSmart1

sure

36. Outkast3r09

is that yes or no lol

37. JenniferSmart1

yes?

38. JenniferSmart1

potential energy

39. Outkast3r09

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

40. Outkast3r09

also dr = h

41. Outkast3r09

;)

42. Outkast3r09

W=Fd=Fh $mgh+Fh= 43. Outkast3r09 \[h(mg+F)$?

44. JenniferSmart1

we dont have to integrate i'm assuming

45. Outkast3r09

no it's straight up

46. Outkast3r09

alright

47. Outkast3r09

now we need the work done

48. JenniferSmart1

?

49. Outkast3r09

yes isn't that what it's asking lol

50. Outkast3r09

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

51. JenniferSmart1

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

52. JenniferSmart1

E= U+0

53. Outkast3r09

ok so E is actually your capacitor right?

54. JenniferSmart1

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

55. Outkast3r09

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

56. Outkast3r09

since $E_1=E_2$ $E_1=mgh+\frac{1}{2}mv^2+U_c$

57. Outkast3r09

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

58. Outkast3r09

so all energy is coming from the capacitor

59. Outkast3r09

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

60. JenniferSmart1

what's W app? and W d?

61. Outkast3r09

but since we're looking at it at rest in two places you have $mgh+0+Fh$ $U_c=mgh+Fh=(mg+F)h$

62. Outkast3r09

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

63. Outkast3r09

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

64. Outkast3r09

$F=ma$ $(mg+ma)h=(m(g+a))h$

65. Outkast3r09

$\frac{U_c}{m(g+a)}=h$ maybe!?!?!?! lol

66. Outkast3r09

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

67. Outkast3r09

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