## anonymous 3 years ago @Outkast3r09

1. anonymous

i see boxes =P

2. anonymous

Mechanical energy vertical displacent energy

3. anonymous

umm? ok question?

4. anonymous

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

5. anonymous

now we have potential energy correct?

6. anonymous

V?

7. anonymous

mgh?

8. anonymous

E= U + K

9. anonymous

$\frac 12 mv^2$

10. anonymous

C=Q/V?

11. anonymous

what do you mean?

12. anonymous

what happens to the energy in terms of mechanics?

13. anonymous

yes

14. anonymous

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

15. anonymous

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

16. anonymous

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

17. anonymous

since you end up with no kinetic or potential?

18. anonymous

oh it's lifting the object

19. anonymous

yes

20. anonymous

sorry

21. anonymous

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

22. anonymous

sounds right

23. anonymous

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

24. anonymous

yes battery

25. anonymous

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

26. anonymous

alright

27. anonymous

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

28. anonymous

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

29. anonymous

how can we get C? where is work? LOL

30. anonymous

31. anonymous

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

32. anonymous
33. anonymous

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

34. anonymous

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

35. anonymous

sure

36. anonymous

is that yes or no lol

37. anonymous

yes?

38. anonymous

potential energy

39. 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

40. anonymous

also dr = h

41. anonymous

;)

42. anonymous

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

44. anonymous

we dont have to integrate i'm assuming

45. anonymous

no it's straight up

46. anonymous

alright

47. anonymous

now we need the work done

48. anonymous

?

49. anonymous

yes isn't that what it's asking lol

50. anonymous

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

51. anonymous

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

52. anonymous

E= U+0

53. anonymous

ok so E is actually your capacitor right?

54. anonymous

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

55. anonymous

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

56. anonymous

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

57. anonymous

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

58. anonymous

so all energy is coming from the capacitor

59. 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

60. anonymous

what's W app? and W d?

61. anonymous

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

62. anonymous

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

63. anonymous

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

64. anonymous

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

65. anonymous

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

66. anonymous

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

67. anonymous

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