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Energy with springs on an angle

  1. Feb 7, 2008 #1
    1. The problem statement, all variables and given/known data
    The ball launcher in a pinball machine has a spring that has a constant of 1.2 N/cm. The surface on which the ball moves is inclined 10 degrees with respect to the horizontal. If the spring is initially compressed 5.0 cm, find the launching speed of a 100g ball when the plunger is released.

    2. Relevant equations
    W = .5kx^2
    mgh + .5kx^2 + .5mv^2 = mghf + .5kxf^2 + .5mvf^2

    3. The attempt at a solution
    first i converted the spring constant
    1.2 N/cm * 100cm/m = 120 N/m

    next i found PE(spring)
    .5kx^2 = .5(120N/m)(.05m)^2 = .15 J
    then i used the mechanical energy eq...i canceled the unnecessary terms and got:
    .5kx^2 + .5mv^2 = mgh(f)

    and when i solve for V(initial) i get:
    ((mgh(f) - .5kx^2 )/(.5m))^.5

    but then i dont know the distance which the ball travels so i cant use this equation
    and also i know the 10 degrees has to factor in somewhere, but where? is it part of the potential energy of the string? or does it affect the initial velocity? or both?
     
  2. jcsd
  3. Feb 7, 2008 #2
    Please Help If You Can Thanks
     
  4. Feb 7, 2008 #3
    first, W =! .5kx^2
    Work = change in energy (not just energy)

    I would use the conservation of energy to solve this problem
    Energy stored in the spring = the kinetic energy of the ball right when its launched
     
  5. Oct 20, 2009 #4
    The angle goes into the final height.
     
  6. Oct 20, 2009 #5
    Your initial energy is the potential energy in the srping. Your final energy, right when the ball lleaves the spring, is kinetic and gravitational potentiall energy. You know everytheing but it's velocity. (draw a triangle to find h in mgh)
     
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