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Rollercoaster Problems

  1. Dec 11, 2008 #1
    Hello!

    I was hoping somebody could help me with these two problems:

    A roller-coaster has a first hill 75.2 meters tall. Immediately after this hill, the roller coaster skims along the ground, then ascends to a second hill 47.8 m tall. If the load limit is 1067.9 kg, how fast is it going at the top of the second hill? As always, no units and 2 decimal places.

    The owners of the amusement park have now decided that having a large first hill is too old-fashioned; they now want to use a spring to start the roller coaster. If the starting gate limits the spring to 3.3 meters of compression, and the next hill is now 44.4 meters, what must the spring constant be to lift a load limit of 2280.9 kg to the top with a minimum speed of 18.4? As always, no units and 2 decimal places.


    Thank you so much in advance!
     
  2. jcsd
  3. Dec 11, 2008 #2

    LowlyPion

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    Welcome to PF.

    How would you think to go about the problems?
     
  4. Dec 11, 2008 #3
    For the first problem, I know how to find the potential energy at either of the hills, but I'm unsure of how exactly to use this knowledge to find velocity. Or, could I find the velocity at the bottom of the first hill, and then find out how much it decelerates as it climbs the second?
     
  5. Dec 11, 2008 #4

    LowlyPion

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    Isn't there a conservation of energy?

    PE becomes KE?

    With no friction won't the KE at any point be just the change from the initial PE?
     
  6. Dec 11, 2008 #5
    Yes! So I can find the potential energy at the top of the hill, set that equal to 1/2mv^2, and just solve for v? How do I factor the first hill into that, though?
     
  7. Dec 11, 2008 #6

    LowlyPion

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    The speed at the top of the second hill is still just the change in PE to that point from the top of the first.

    The load limit then is the max mass of the car and riders?
     
  8. Dec 11, 2008 #7

    LowlyPion

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    The second question is asking you to figure out how much potential energy must be in the spring to propel the car up a 44m hill and still have the KE determined by the max load and the min velocity given.
     
  9. Dec 11, 2008 #8
    Okay, I think I have them both now. Thank you so much!
     
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