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Conservation of Mechanical Energy on a Roller Coaster

  1. Feb 18, 2015 #1
    1. The problem statement, all variables and given/known data
    A roller coaster at an amusement park is at rest on top of a 30 m hill (point A). The car starts to roll down the hill and reaches point B which is 10 m above the ground, and then rolls up the track to point C, which is 20 m above the ground.
    (A) A student assumes no energy is lost, and solves for how fast the car is moving at point C using energy arguments. What answer does he get?
    (B) If the final speed at C is actually measured to be 2 m/s, what percentage of energy was "lost" and where did it go?
    2u8xq2v.jpg
    2. Relevant equations
    Kf + Uf = Ki + Uf
    K = 1/2mv2
    Ug = mgh

    3. The attempt at a solution
    I believe there is a typo in the book for Part A, I think they got the height wrong, their answer is 20 m/s. For Part B, I think I did something wrong because the height error has been corrected.

    My Work for Part A:
    I set potential energy at A equal to kinetic energy at C. My reference point is considering the height of C as ground level.

    mgh = 1/2mv2
    gh = 1/2v2
    10*10 = 1/2v2
    100 = 1/2v2
    200 = v2
    14 = v

    Their Work for Part A
    2ldkawy.jpg

    My Work for Part B

    Find ratio of Final Energy to Initial Energy and subtract from 100%
    [itex]\frac{\frac{1}{2}mv^2}{mgh}[/itex]

    [itex]\frac{\frac{1}{2}v^2}{gh}[/itex]

    [itex]\frac{\frac{1}{2}*2^2}{10*10}[/itex]

    [itex]\frac{2}{100}[/itex]

    2%

    So 98% is lost to heat. This is radically different from their answer:

    2qsxlpl.jpg
    For this part I again used a reference height of the height at C, but am I not allowed to do that?
     
    Last edited: Feb 18, 2015
  2. jcsd
  3. Feb 18, 2015 #2

    Nathanael

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    They seem to have messed up; they solved for the speed at point B. I agree with your answer for the speed at point C

    You still messed up for part B, though. You seem to be saying the total energy of the system is m*g*10
     
  4. Feb 18, 2015 #3
    If I set my reference height of C as ground level, is that not the total energy of the system? It has no Kinetic at point A, all Potential, right?
     
  5. Feb 18, 2015 #4

    Nathanael

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    Ah, good point; the question is ambiguous because the answer depends on where you choose the zero for your potential. The reason the answer differs so much is because they chose their PE=0 at the actual ground.
     
  6. Feb 18, 2015 #5
    view the attached file please for your answer
     

    Attached Files:

  7. Feb 19, 2015 #6
    I hate to bother you Arjun but I'm having trouble reading it because of the quality, do you have another picture you could upload?
     
  8. Feb 19, 2015 #7
    Please do check this. I have attached a clearer image. Please do reply if the information was good. Thank you. Always happy to help. :) the variable of gravity g changes with height so the h should not be changed
    DSCN0180.JPG
     
    Last edited: Feb 19, 2015
  9. Feb 19, 2015 #8

    BvU

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    My calculator has something else for 202/300
     
  10. Feb 19, 2015 #9

    BvU

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    The answer does not depend on the choice for zero potential. That's the fun of potentials in general: the only thing that matters is the difference in potential. It's about ##\Delta h##, not about ##h## itself.
     
  11. Feb 19, 2015 #10
    The textbook and I both agree that there was a change in height of 10 m, but where we differ is that they are calculating the potential energy from the height above the ground(30m) while I am calculating the potential energy from the height above point C(10m). That's why their ratio is different from mine, at least I think.
     
  12. Feb 19, 2015 #11

    Nathanael

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    Yes, but in this case it's a bit different, isn't it? Because we're calculating the percentage change in energy... which is ("final energy") / ("initial energy"), but this is ambiguous because as you said, only changes in potential energy make sense, so neither final energy nor initial energy are clearly defined (and the ambiguity does not cancel out).

    The answer would be, (KE+C) / (10*mg+C) Where "C" represents the arbitrary potential energy due to the choice of where the zero potential is. This depends on C; the bigger C is, the closer the expression will be to 1. Their solution chose C=20*mg and the OP chose C=0.
     
    Last edited: Feb 19, 2015
  13. Feb 20, 2015 #12

    BvU

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    I humbly agree. If there was any percentage worth mentioning here it is indeed 98%.
    Kinetic energy ideally 10 m x mg turns out to be only 0.5 x (2m/s)2.

    It's not the book answer that is wrong, it's the book question that is disastrously bad. Kudos for Foopy for not going with that nonsense -- and idem for Nathaniel
     
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