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Pinball Machine

  1. Dec 10, 2006 #1
    1. The problem statement, all variables and given/known data
    The ball launcher in a pinball machine has a spring that has a force constant of 1.2 N/cm. The surface on which the ball moves is inclined at 10° with respect to the horizontal. If the spring is initially compressed 5 cm, find the launching speed of a .1 kg ball when the plunger is released. Friction and the mass of the plunger are negligible.

    2. Relevant equations
    Ue (Elastic potential) = (1/2)kx^2
    K (Kinetic) = (1/2)mv^2
    Ug (Gravitational potential) = mgh

    3. The attempt at a solution

    k = 1.2 N/ cm = .012 N / m
    x = 5 cm = .05 m
    m = .1 kg
    g = 9.8 m/s

    Ue = K + Ug

    .5kx^2 = .5mv^2 + mgh

    .5(.012)(.05^2) = .5(.1)v^2 + .1(9.8)(.05sin(10°))

    0.000015 = .05v^2 + .00851

    At this point, my v^2 will equal a negative number, which makes no sense at all. I'm stuck >.<

    The problem is that the left side of the equation gets exponentially smaller while the right hand side... doesn't.

    Edit: Foolish calculator mistake. I hate how I find this out right after I posted it (I checked my work for 1/2 hr before posting).

    1.2 N / cm = 120 N / m, not .012. I hit * instead of / on my calc.
    Last edited: Dec 10, 2006
  2. jcsd
  3. Dec 10, 2006 #2
    check your conversion from n/cm to n/m
  4. Dec 10, 2006 #3
    I think it's just your conversion for the k value; you have to multiply
    1.2N/cm*100cm/1m = 120N/m

    my bad... didnt see that someone already replied.
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