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Spring constant and conservation of energy

  1. Mar 17, 2009 #1
    1. The problem statement, all variables and given/known data
    The ball launcher in a pinbgall machine has a spring that has a force constant of 36 N/cm. THe surface on which the ball moves is inclined theta=10.1 degrees with respect to the horizontal. If the spring is initially conressed 4.25 cm, find the launching speed of a .120 kg bawhen the plunger is released. Friction and the mass of the plunger e negligible.

    2. Relevant equations

    PE=.5k(xf-xi)^2
    KE=.5mv^2

    3. The attempt at a solution
    I'not sure what launching speed is supposed to be, it supposed to be an answer that doesn't require units. Does that mean I need to be finding the spring constant? If so, do I just set the KE=PE?
     
  2. jcsd
  3. Mar 17, 2009 #2

    LowlyPion

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    They give you the spring constant. 36 N/cm.

    So think conservation of energy.

    Spring potential depressed = kinetic energy at release point plus additional increase in gravitational potential energy at release point .
     
  4. Mar 17, 2009 #3
    The force constant is the spring constant. You're looking for a velocity.

    Think about it if we compress a spring at some angle we are storing potential energy in the spring, but we also have to take into account gravity, does this help?
     
  5. Mar 17, 2009 #4
    It can't be simply velocity that I'm looking for because an answer not requiring units is needed. Any other ideas?
     
  6. Mar 17, 2009 #5
    I'm pretty sure it's got units -- it asks for the launching speed. Speed's have units of distance per time.
     
  7. Mar 17, 2009 #6
    I know... it should but, I'm doing homework through lon-capa and it's telling me that it doens't require units, so it can't be asking for velocity.
     
  8. Mar 17, 2009 #7

    LowlyPion

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    They ask for speed.

    Calculate it as m/s.

    Ignore their not asking for units.
     
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