Finding the restoring force, Hooke's Law

  1. 1. The problem statement, all variables and given/known data

    Hooke's law describes a certain light spring of unstretched length 35.0cm. When one end is attached to the top of a door frame and a 7.50kg object is hung from the other end, the length of the spring is 41.5cm. a) Find its spring constant. b)The load and spring are taken down. Two people pull in opposite directions on the ends of the spring, each with a force of 190N. Find the length of the spring in this situation.

    2. Relevant equations

    F=-kx

    3. The attempt at a solution
    I found the spring constant in a) and it was 1130N/m which I am unsure if it is the same for b. The real problem is I cant figure out how I am supposed to find the restoring force. My thoughts seem to think that because they are pulling in opposite directions at the same force then the net force would be zero and therefore so would the restoring force?
     
  2. jcsd
  3. Imagine pulling on a spring attached to a wall. Now imagine pulling on the spring in part B.
     
  4. Ok so then it would stretch, and would be double? What about the spring constant would it remain the same?
     
  5. For the purposes of what we're doing, the spring constant is always constant.

    If one pulls on a mounted spring, the spring has a force f=kx. If two people pull on the spring there are now two forces. f1+f2=kx.
     
  6. Okay, thank you I believe I got it :)
     
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