## calculusxy one year ago MEDAL!!!! A 1000 kg rollercoaster is released from rest from point D on the rollercoaster track. Assume no friction and no air resistance. How fast is the rollercoaster moving at point F? (Point D is 75 m high ; point F is 35 m high) Diagram attached below.

1. calculusxy

2. calculusxy

@jim_thompson5910

3. calculusxy

So would this be related to the Law of Conservation of Energy: (PE + KE) = (PE + KE)

4. jim_thompson5910

I think once you get 392,000 J for the KE at point F, you use that to find the velocity v

5. calculusxy

so like: 392,000 = 1/2mv^2

6. jim_thompson5910

yes

7. calculusxy

and then: 392,000 = 1/2(1000)v^2

8. calculusxy

392,000 = 500v^2 784 = v^2 v= 28 m/s

9. jim_thompson5910

very good

10. calculusxy

so acceleration is $\frac{ \Delta V }{ time }$

11. jim_thompson5910

over change in time

12. calculusxy

i have to find the two velocities?

13. calculusxy

how would i find the time?

14. jim_thompson5910

the first velocity is 0 since it starts at rest

15. jim_thompson5910

I'm not sure about the time part

16. calculusxy

can i use: distance = 1/2gt^2

17. jim_thompson5910

maybe there's another formula tying together distance, velocity and acceleration

18. calculusxy

distance for point d = 1/2gt^2 75 m = 1/2(9.8m/s^2)t^2

19. jim_thompson5910

20. jim_thompson5910

idk if that applies along the curve

21. calculusxy

do u know a formula that i can use to find the time?

22. jim_thompson5910

23. jim_thompson5910

do they give any other info?

24. calculusxy

no...

25. calculusxy

but can we do the second question?

26. jim_thompson5910

I'm not sure. They want the acceleration at point F right?

27. calculusxy

yes

28. jim_thompson5910

I'd get a second opinion from a physics expert because I'm not sure about the second part.

29. calculusxy

here is the second question: 2. a 400 kg rollercoaster is released from rest at point B on the rollercoaster track above. assume no friction and no air resistance. if it is moving at 20 m/s at point c ( 60 m high), how high is point B?

30. jim_thompson5910

PE at point C PE = m*g*h PE = 400*9.8*60 PE = 235,200 KE at point C KE = (1/2)*m*v^2 KE = (1/2)*400*20^2 KE = 80,000 Total Energy (TE) at point C TE = PE + KE TE = 235,200 + 80,000 TE = 315,200 ------------------------------------------------------- TE at point B TE = PE + KE TE = m*g*h + (1/2)*m*v^2 315,200 = 400*9.8*h + (1/2)*400*0^2 solve for h

31. phi

For Q1, where does it ask for the acceleration? the only acceleration is due to gravity, and you need more info to calculate it (though at the top of a "hill" gravity is perpendicular to direction, and acceleration is 0)

32. calculusxy

i think i got. since $(PE + KE)_D = (PE + KE)_F$ and we already know the potential energy, which is 735,000 joules and the kinetic energy would be 0 joules. $(735,000 + 0)_D = (343,000 + KE)_F$ wouldn't we have to subtract 735,000 - 343,000 to get the kinetic energy?

33. jim_thompson5910

are you referring to problem #1 or #2 ?

34. calculusxy

#1

35. jim_thompson5910

yeah I agree with the work you wrote on the paper (in the attachment)

36. calculusxy

movement = kinetic energy (which doesn't necessarily have to mean acceleration)

37. jim_thompson5910

then you did a bit more work to find that v = 28 m/s

38. calculusxy

yeah i just realized that :)

39. calculusxy

we can work on problem #2 now

40. jim_thompson5910

I wrote out #2 above

41. calculusxy

oh okay i will look on that right after i write down my work on the paper. here is another question: a 600 kg rollercoaster is released from rest at point A (150m high). what is its potential energy at A? how fast is it going at point C (40 m high)? what is its kinetic energy at point C? what is its potential energy at point D?

42. jim_thompson5910

a 600 kg rollercoaster is released from rest at point A (150m high). what is its potential energy at A? use PE = m*g*h m = 600 is the mass g = 9.8 is the approximate acceleration of gravity h = height

43. calculusxy

44. jim_thompson5910

I'm getting h = 80.40816327

45. jim_thompson5910

46. calculusxy

PE for c = m x g x h PE for c = 400 kg x 9.8 m/s^2 x 60m PE for c = 235,200 J KE for c = 1/2mv^2 KE for c = 1/2(400 kg)(20 m / s)^2 KE for c = 80,000J KE for b = 0 J 80,000J + 235,200J = 315,000J PE for b = 315,000J

47. calculusxy

am i correct so far?

48. jim_thompson5910

the error made is here 80,000J + 235,200J = 315,000J

49. jim_thompson5910

you should get 80,000J + 235,200J = 315,200J

50. calculusxy

oh yes

51. calculusxy

got that :)

52. jim_thompson5910

then solve 315,200 = 400*9.8*h for h to get h = 80.40816327

53. calculusxy

for problem 3, how would u find how fast it goes? is it like the velocity?

54. jim_thompson5910

yeah and you use the KE for that

55. calculusxy

okay

56. calculusxy

@jim_thompson5910 how would i find the potential energy for point D?

57. jim_thompson5910

you did that back in #1

58. jim_thompson5910

PE = m*g*h point D is 75 m off the ground

59. calculusxy

oh no this is another graph. sorry i will attach the picture to u .

60. calculusxy

61. jim_thompson5910

So point D is on the ground?

62. calculusxy

yeah

63. calculusxy

oh so there isn't any potential energy because distance = 0 meters, right?

64. calculusxy

but that doesn't make much sense as well

65. jim_thompson5910

why not? if it's on the ground, then it has no potential to fall further, so it doesn't have any potential energy

66. calculusxy

so potential energy is equal to 0J?

67. jim_thompson5910

when it's on the ground, it's all kinetic energy

68. jim_thompson5910

yeah

69. calculusxy

oh yes potential energy decreases and kinetic energy increases.

70. jim_thompson5910

correct

71. calculusxy

okay thank you so much :)

72. jim_thompson5910

you're welcome