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A heavy-duty stapling gun!

  1. Nov 2, 2009 #1
    1. The problem statement, all variables and given/known data
    A heavy-duty stapling gun uses a 0.151-kg metal rod that rams against the staple to eject it. The rod is pushed by a stiff spring called a "ram spring" (k = 35007 N/m). The mass of this spring may be ignored. Squeezing the handle of the gun first compresses the ram spring by 3.9 multiplied by 10-2m from its unstrained length and then releases it. Assuming that the ram spring is oriented vertically and is still compressed by 1.0 multiplied by 10-2m when the downward-moving ram hits the staple, find the speed of the ram at the instant of contact.


    2. Relevant equations
    U=(1/2)kx^2
    (1/2)kx^2=(1/2)mv^2

    3. The attempt at a solution
    I have no idea what to do. My answer was 13.96 m/s but it was wrong.
     
    Last edited: Nov 2, 2009
  2. jcsd
  3. Nov 2, 2009 #2

    Delphi51

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    Homework Helper

    I don't get 13.96. (1/2)kx^2=(1/2)mv^2 is not quite right because the spring still has some energy left (x = .01) when it hits.
     
  4. Nov 3, 2009 #3
    Conservation of energy. U_0 + KE_0 = U_f + KE_f. Think about each of these terms and what their values are initially and finally for this situation.
     
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