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Spring Gun

  1. Feb 20, 2008 #1
    1. The problem statement, all variables and given/known data
    A spring gun (k = 28 N/m) is used to shoot a 56-g ball horizontally. Initially the spring is compressed by 18 cm. The ball loses contact with the spring and leaves the gun when the spring is still compressed by 12 cm. What is the speed of the ball when it hits the ground, 1.4 m below the spring gun?


    2. Relevant equations

    horizontal velocity, vertical velocity, Kinetic energy
     
  2. jcsd
  3. Feb 20, 2008 #2

    mgb_phys

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    Assuming no energy is lost to friction you have two sources of energy.
    1, The energy from the spring = force * distance.
    2, The gravitational energy of falling the 1.4m = m g h

    Both of these will be converted into kinetic energy of the particle = 1/2 m v^2
    from this you can get the final speed.
     
  4. Feb 20, 2008 #3
    is the distance the change in the compression of the spring for the energy of the spring?
     
  5. Feb 20, 2008 #4

    mgb_phys

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    Yes - the energy stored in a spring is 1/2 K x^2
    where K is the spring constant and X is the distance you compress it.
     
  6. Feb 20, 2008 #5
    i got .0504 for the spring energy and 768.32 for the gravitational potential energy.
    but then what equation or steps do i use to convert those into kinetic energy to get the final velocity
     
  7. Feb 20, 2008 #6

    mgb_phys

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    Check your units!
    You just add the spring energy and potential energy to get the kinetic energy.
     
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