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Conservation of mechanical energy (low difficulty)

  1. Feb 11, 2009 #1
    1. The problem statement, all variables and given/known data

    There is a rollercoaster. A car starts 35m "high" and descends to the ground. A.) What is the velocity at this point? The car continues an goes 28m above the ground. B.) calculate its velocity at this point. C.) If the car descends another 13m from this point, what is the velocity

    Ignore Friction
    2. Relevant equations

    mgy=1/2mv^2

    3. The attempt at a solution

    using mgy= 1/2mv^2 i got the answer to a.), which is 26 m/s. But after this I became flustered and I do not know what to do to get its velocity if it goes 28m higher. I understand c. but I need the velocity at point B to get it!

    Please help
     
  2. jcsd
  3. Feb 11, 2009 #2
    I'm sorry I just figured it out for B.

    I set 1/2mv^2=1/2mv^2+mgy and found my second velocity, which comes out to 11.7 roughly when using g=9.8

    I would appreciate if someone would confirm my findings

    I then set up the (square root of(1/2mv^2 + mgy-mgy)/.5)=v and came out with 20 m/s for C.
     
  4. Feb 11, 2009 #3
    You're on the right track, think about conservation of energy
    [tex] \Delta U = m g \Delta y[/tex]
    [tex] \Delta K = \frac{1}{2} m(v_f ^2-v_i^2) [/tex]
    Thus [tex] \Delta K + \Delta U = 0[/tex]
    Thus [tex] -m g \Delta y +\frac{1}{2} m v_i^2= \frac{1}{2} m v_f^2[/tex]
     
  5. Feb 12, 2009 #4
    I also got 11.7 for part B.
     
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