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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.

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  1. calculusxy
    • one year ago
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  2. calculusxy
    • one year ago
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    @jim_thompson5910

  3. calculusxy
    • one year ago
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    So would this be related to the Law of Conservation of Energy: (PE + KE) = (PE + KE)

  4. jim_thompson5910
    • one year ago
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    I think once you get 392,000 J for the KE at point F, you use that to find the velocity v

  5. calculusxy
    • one year ago
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    so like: 392,000 = 1/2mv^2

  6. jim_thompson5910
    • one year ago
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    yes

  7. calculusxy
    • one year ago
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    and then: 392,000 = 1/2(1000)v^2

  8. calculusxy
    • one year ago
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    392,000 = 500v^2 784 = v^2 v= 28 m/s

  9. jim_thompson5910
    • one year ago
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    very good

  10. calculusxy
    • one year ago
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    so acceleration is \[\frac{ \Delta V }{ time }\]

  11. jim_thompson5910
    • one year ago
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    over change in time

  12. calculusxy
    • one year ago
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    i have to find the two velocities?

  13. calculusxy
    • one year ago
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    how would i find the time?

  14. jim_thompson5910
    • one year ago
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    the first velocity is 0 since it starts at rest

  15. jim_thompson5910
    • one year ago
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    I'm not sure about the time part

  16. calculusxy
    • one year ago
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    can i use: distance = 1/2gt^2

  17. jim_thompson5910
    • one year ago
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    maybe there's another formula tying together distance, velocity and acceleration

  18. calculusxy
    • one year ago
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    distance for point d = 1/2gt^2 75 m = 1/2(9.8m/s^2)t^2

  19. jim_thompson5910
    • one year ago
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    I think that's for free fall only?

  20. jim_thompson5910
    • one year ago
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    idk if that applies along the curve

  21. calculusxy
    • one year ago
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    do u know a formula that i can use to find the time?

  22. jim_thompson5910
    • one year ago
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    no I don't sadly

  23. jim_thompson5910
    • one year ago
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    do they give any other info?

  24. calculusxy
    • one year ago
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    no...

  25. calculusxy
    • one year ago
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    but can we do the second question?

  26. jim_thompson5910
    • one year ago
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    I'm not sure. They want the acceleration at point F right?

  27. calculusxy
    • one year ago
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    yes

  28. jim_thompson5910
    • one year ago
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    I'd get a second opinion from a physics expert because I'm not sure about the second part.

  29. calculusxy
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    are you referring to problem #1 or #2 ?

  34. calculusxy
    • one year ago
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    #1

  35. jim_thompson5910
    • one year ago
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    yeah I agree with the work you wrote on the paper (in the attachment)

  36. calculusxy
    • one year ago
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    movement = kinetic energy (which doesn't necessarily have to mean acceleration)

  37. jim_thompson5910
    • one year ago
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    then you did a bit more work to find that v = 28 m/s

  38. calculusxy
    • one year ago
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    yeah i just realized that :)

  39. calculusxy
    • one year ago
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    we can work on problem #2 now

  40. jim_thompson5910
    • one year ago
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    I wrote out #2 above

  41. calculusxy
    • one year ago
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    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
    • one year ago
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    `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
    • one year ago
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    @jim_thompson5910 Would the answer for #2 be about 80.36 meters?

  44. jim_thompson5910
    • one year ago
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    I'm getting h = 80.40816327

  45. jim_thompson5910
    • one year ago
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    double check your work

  46. calculusxy
    • one year ago
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    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
    • one year ago
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    am i correct so far?

  48. jim_thompson5910
    • one year ago
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    the error made is here `80,000J + 235,200J = 315,000J`

  49. jim_thompson5910
    • one year ago
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    you should get 80,000J + 235,200J = 315,200J

  50. calculusxy
    • one year ago
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    oh yes

  51. calculusxy
    • one year ago
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    got that :)

  52. jim_thompson5910
    • one year ago
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    then solve 315,200 = 400*9.8*h for h to get h = 80.40816327

  53. calculusxy
    • one year ago
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    for problem 3, how would u find how fast it goes? is it like the velocity?

  54. jim_thompson5910
    • one year ago
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    yeah and you use the KE for that

  55. calculusxy
    • one year ago
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    okay

  56. calculusxy
    • one year ago
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    @jim_thompson5910 how would i find the potential energy for point D?

  57. jim_thompson5910
    • one year ago
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    you did that back in #1

  58. jim_thompson5910
    • one year ago
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    PE = m*g*h point D is 75 m off the ground

  59. calculusxy
    • one year ago
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    oh no this is another graph. sorry i will attach the picture to u .

  60. calculusxy
    • one year ago
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  61. jim_thompson5910
    • one year ago
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    So point D is on the ground?

  62. calculusxy
    • one year ago
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    yeah

  63. calculusxy
    • one year ago
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    oh so there isn't any potential energy because distance = 0 meters, right?

  64. calculusxy
    • one year ago
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    but that doesn't make much sense as well

  65. jim_thompson5910
    • one year ago
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    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
    • one year ago
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    so potential energy is equal to 0J?

  67. jim_thompson5910
    • one year ago
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    when it's on the ground, it's all kinetic energy

  68. jim_thompson5910
    • one year ago
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    yeah

  69. calculusxy
    • one year ago
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    oh yes potential energy decreases and kinetic energy increases.

  70. jim_thompson5910
    • one year ago
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    correct

  71. calculusxy
    • one year ago
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    okay thank you so much :)

  72. jim_thompson5910
    • one year ago
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    you're welcome

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