1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Roller Coaster Physics and Energy

  1. May 25, 2010 #1
    1. The problem statement, all variables and given/known data
    1. Assume the mass of the roller coaster is 750 kg. Answer the following.

    a) If the height of the first hill is 35 meters (measured from the ground), calculate the potential and kinetic energies at this highest position on the track.

    b) When the riders are at the first "dip" in the track, the height is 10 meters (measured from the ground), calculate the potential and kinetic energies at position on the track (assume no loss of energy due to friction).

    c) At a hill right before a loop, the riders are 20 meters above the ground. Calculate the potential and kinetic energies at position on the track (assume no loss of energy due to friction).




    2. Relevant equations

    KE= (1/2)*m*(v^2)
    PE= m*g*h
    PEi+KEi=PEf+KEf



    3. The attempt at a solution

    For A, I used PE=mgh which is (750kg)*(9.8)*(35) = 257250 joules. Also for A, KE is 0 because it is at the top of the hill. Is this correct? And how do I do b and c? I have no clue!

    Thanks!
     
  2. jcsd
  3. May 25, 2010 #2

    rock.freak667

    User Avatar
    Homework Helper

    Good, you've found this correctly. Now you know the total energy at A = Total energy at B= Total energy at C right? (conservation of mechanical energy)

    Since at point B, the roller coaster is at a height of 10m, what is the potential energy? Now your third equation posted

    (KE+PE)1=(KE+PE)2
    (1 = point 1, 2 = point 2)

    can you find the KE at B using it?
     
  4. May 25, 2010 #3
    No friction=> PE+KE is constant.
    For b, you can calculate PE the same way you did for a. Since you know, that in total PE+KE are constant, and you know this value (for you've already calculate it in a). c - the very same way.
     
  5. May 25, 2010 #4

    Point B PE: m*g*h
    :750*9.8*10=73500 joules
     
  6. May 25, 2010 #5

    rock.freak667

    User Avatar
    Homework Helper

    right at point B, PE= 73500 J.

    KE+PE (at B) = KE+PE (At A)

    So what is the KE at B?
     
  7. May 25, 2010 #6
    257250 joules + 0 joules = 73500 joules + KE (b)
    so..
    257250 joules = KE (b) + 73500 joules
    so..
    KE (b) = 183750 joules correct?
     
  8. May 25, 2010 #7

    rock.freak667

    User Avatar
    Homework Helper

    that should be correct, it is a similar exercise for part c)
     
  9. May 25, 2010 #8
    So:

    PE for c = m*g*h
    =750*9.8*20= 14700 joules

    I use PEi+KEi=PEf+KEf right?

    PEi would be 14700 right?
    KEi would be what I calculated in B right?
    Whats PEf and Kef?
     
  10. May 25, 2010 #9

    rock.freak667

    User Avatar
    Homework Helper

    No. Remember KE+PE = constant. This means that (KE+PE)A=(KE+PE)B=(KE+PE)C

    So if you wanted to find at C, using what was found at B, you add the PE and KE at B and put that equal to the KE and PE at C
     
  11. May 25, 2010 #10
    So
    73500+183750=Pe+Ke

    Pe at C= m*g*h = 750*9.8*20=147000

    so
    73500+183750=14700(pe) + ke
    solve for ke = 242550 joules
    right?
     
  12. May 25, 2010 #11

    rock.freak667

    User Avatar
    Homework Helper

    Yes that should be correct.
     
  13. May 25, 2010 #12
    Thank you rock freak! My savior!
     
  14. May 25, 2010 #13
    One more question rockfreak, sorry about all this!

    What would be the speed of the car at the bottom of the first hill?
     
  15. May 25, 2010 #14

    rock.freak667

    User Avatar
    Homework Helper

    Same process as before. Except that at the bottom of the hill, what is the height relative to the ground?
     
  16. May 25, 2010 #15
    I would solve for velocity here right? The height is 10 m.

    so v= sq root(2ke[at b]/m)
    so v= sq root(2(183750)/(750))

    so velocity = 22.13594362 m/s?
     
  17. May 25, 2010 #16

    rock.freak667

    User Avatar
    Homework Helper

    Yes and no, remember the bottom of the hill is where you are measuring your heights from, if you are at the bottom of the hill, you have no height!.

    So your PE =?

    Remember the total energy is what you found at A. So you don't always need to pick the previous point, you can just pick a convenient point. (Else you will always have to add numbers together and might make an error)
     
  18. May 25, 2010 #17
    I do not understand, so what do I have to solve for or what should I do?
     
  19. May 25, 2010 #18

    rock.freak667

    User Avatar
    Homework Helper

    At the bottom of the hill, the height above the ground is zero. So the PE is 0.

    Hence your KE at C = KE+PE (at B)
     
  20. May 25, 2010 #19
    But I am solving for speed at b. That is velocity right?
     
  21. May 25, 2010 #20

    rock.freak667

    User Avatar
    Homework Helper

    ah sorry, I did not read your entire question. I thought the bottom of the first hill was at the bottom, I forgot the first hill ended at B. So your answer was correct. Sorry about that.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Roller Coaster Physics and Energy
  1. Roller Coaster Energy (Replies: 1)

Loading...