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Spring constant problem

  1. Aug 18, 2009 #1
    1. The problem statement, all variables and given/known data
    Two springs are set up on a table. The longer spring has a spring constant of 225 N/m and an initial length of 42.0 cm. The shorter spring has a spring constant of 675 N/m and an initial length of 28.0 cm. How far above the table is an 8.50-kg ball when it reaches its equilibrium position?


    2. Relevant equations
    f=-kx


    3. The attempt at a solution
    first change cm to m, then would i find the forces of each spring, add them up for find the total force, then use that and the 8.50 kg mass to find x?
     
  2. jcsd
  3. Aug 18, 2009 #2
    Re: Springs

    Yes, the springs combined must be equal to the gravitational force.
     
  4. Aug 18, 2009 #3
    Re: Springs

    how exactly do i use the force that i found to solve for x? because i dont know k
     
  5. Aug 18, 2009 #4
    Re: Springs

    k is the spring constant -- you have two springs and you know both their spring constants.
     
  6. Aug 18, 2009 #5
    Re: Springs

    o! so just add the two up.
     
  7. Aug 18, 2009 #6
    Re: Springs

    does this look right to you then?
    F=-kx
    F=(-225N/m x 0.42m)+(675N/m x 0.28m)
    F=-94.5 + -189N
    F= -283.5N
    X=f/-k
    X=-283.5N / -900N/m
    X= 0.315
     
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