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Kinetic and Potential Energy on a roller coaster

  1. Apr 9, 2013 #1
    1. The problem statement, all variables and given/known data
    A roller coaster starts at the top of hill that is 10 m high. If it is to barely make it to the top of a second hill that is 20 m high, how fast must the initial speed of the roller coaster be? Assume that the roller coaster is frictionless.


    2. Relevant equations
    KE = 0.5mv^2
    PE = mgh

    3. The attempt at a solution
    mgh=0.5mv^2
    gh = 0.5v^2
    v = sqrt(2gh)
    v = sqrt(2*9.81*20)
    v = 20 m/s needed at bottom of second hill.

    mgh=0.5mv^2
    gh = 0.5v^2
    v = sqrt(2gh)
    v = sqrt(2*9.81*10)
    v = 14 m/s at bottom of hill if initial velocity at top of first hill is 0m/s.

    20 m/s - 14 m/s = 6 m/s

    6 m/s is the velocity of the roller coaster at the top of the first hill.

    Does that look right?
     
  2. jcsd
  3. Apr 9, 2013 #2
    No, this is not right. Check it this way: compute KE with v = 6 m/s and PE with h = 10 m; the sum is the total energy initially. It must be equal to the total energy finally, which is KE with v = 0 m/s and PE with h = 20 m.
     
  4. Apr 9, 2013 #3
    Hm, okay. I tried setting the total energy at the top of hill one equal to the total energy at the top of hill two.
    0.5m(v1)^2+mg(h1) = 0.5m(v1)^2+mg(h2)
    0.5(v1)^2 + g(h1) = 0.5(v1)^2 + g(h2)
    v = sqrt(2*g*(h2-h1))
    v = sqrt (2*9.81*(20-10))
    v = 14 m/s

    Would that work?
     
  5. Apr 9, 2013 #4
    The result is correct, but the first two equations are not correct, having the same v1 symbol on both sides.
     
  6. Apr 9, 2013 #5
    Oh, yeah... I meant (v2) on the right side of the equations. That was simply a typo.
     
  7. Apr 9, 2013 #6
    Very well then.
     
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