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Spring Loaded Cork Gun

  1. Oct 9, 2012 #1
    1. The problem statement, all variables and given/known data
    A cork gun contains a spring whose spring constant is 20.0 N/m. The spring is compressed by a distance ∆X = 8.2 cm and used to propel a cork of mass 3.76 g from the gun. Assuming the cork is released when the spring passes through its equilibrium position (Xeq).
    (a) What is the speed of the cork as it is released from the spring?
    (b) Suppose now that the cork temporarily sticks to the spring, causing the spring to extend 4.0 cm beyond its equilibrium position before separation occurs. What is the speed of the cork as it is released from the spring in this case?


    2. Relevant equations
    Initial Mechanical Energy (Ei) = Final Mechanical Energy (Ef)
    (a)Ei=KEi+PEi; Ef=KEf+PEf; KEf+PEf=KEi+PEi ⇔ 0+1/2kx^2=1/2mv^2+0 (solve for "v") v=(√(k)x)/√(m) ⇔ (√(20.0)*0.082)/√(0.00376) ⇔ v=5.98m/s (this is correct according to the lon-capa homework thing)

    (b)Ei=KEi+PEi; Ef=KEf+PEf; KEf+PEf=KEi+PEi ⇔ 0+1/2kx^2=1/2mv^2+0 (this is where I'm stuck, I can't figure out where to add the extra distance due to the spring. I think you subtract it because it will slow down the cork. Am I heading the right direction? Or am I totally off?)

    Thanks a lot :)
    Allyn
     
  2. jcsd
  3. Oct 9, 2012 #2
    Sorry I forgot to post the picture.....
     

    Attached Files:

  4. Oct 9, 2012 #3
    Ok. I figured it out!you add the energy from the spring and get this equation:
    1/2kx^2=1/2mv^2+1/2ky^2 Where y= the distance the spring travels with the cork i.e. after the equilibrium
    Then solve for v and get:
    v=(√(20)*√(.082^2-.04^2))/√(.00376) and you get v=5.22 m/s.
     
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