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Velocity for a puck up a ramp

  1. May 22, 2009 #1
    1. The problem statement, all variables and given/known data
    What minimum speed does a 136.4 g puck need to make it to the top of a 3.31-m-long, 24° frictionless ramp?


    2. Relevant equations
    K=mV^2/2 U=mgY



    3. The attempt at a solution
    PHysics9.jpg
     
  2. jcsd
  3. May 22, 2009 #2

    rock.freak667

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    the 'y' in your equation is the vertical distance, not horizontal. So the cos24 should be sin24
     
  4. May 22, 2009 #3
    wrong its cos.

    cos angle = b/c

    f * s + potential energy = mv^2 / 2

    edit: mistake if you want height then its sin.
     
  5. May 22, 2009 #4
    if the equation is mgh = mv^2/2 then you can cancel out the mass.

    gh = v^2/2

    9.82 * 1.35m = v^2 / 2

    26.5 =v^2

    v = 5.15m/s

    but it think this is totally wrong.
     
  6. May 22, 2009 #5

    Cyosis

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    You have the angle, hypotenuse and want to know the opposite side. Therefore you need to use the sine.
     
  7. May 23, 2009 #6
    ahhhhh. Thanks guys i see where i went wrong you guys are awesome!!
     
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