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Spring constant, should be easy?

  1. Nov 14, 2007 #1
    1. The problem statement, all variables and given/known data
    A spring is suspended vertically parallel to a meterstick. When a 170-g mass is attached to the bottom of the spring, the spring stretches until its bottom is adjacent to the 80-cm mark on the meterstick. When a 760-g mass is then used to replace the 170-g mass, the bottom of the spring is measured to be at the 30-cm mark. Based on these measurements, the value of the spring constant is


    2. Relevant equations
    k=mg/x


    3. The attempt at a solution
    Seemed simple enough, (0.17*9.8)/.2 = 8.33 N/m
    then (0.76*9.8)/.7=10.64 N/m

    But neither of these are correct, and if i take the sum of the masses and distance i get:
    (.93*9.8)/0.9=10.126~

    None of these can possibly be the correct answer.. and im sort of lost =/
     
  2. jcsd
  3. Nov 14, 2007 #2
    New question: A spring is suspended vertically parallel to a meterstick. When a 130-g mass is attached to the bottom of the spring, the spring stretches until its bottom is adjacent to the 20-cm mark on the meterstick. Based on this measurement, the value of the spring constant is
    6.37 kg/s2.
    637 kg/s2.
    0.00637 kg/s2.
    unknown.
    0.650 kg/s2.
    ---------------------
    Where i thought it was 6.37; apparantly it is unknown because you dont know that the spring's bottom was at 0cm when it started =/ Still not sure how it will help me with the first one though [the first question doesnt state that either]
     
  4. Nov 14, 2007 #3
    For the first one:

    Take the two points (use the weight) and plot them on a graph..... your gradient is the spring constant.

    Alternatively k = (m2 - m1)g / (x2 - x1)
     
  5. Nov 14, 2007 #4
    Ugh, so it was the difference in the mass and distance; no idea why i took the sum. Thank you.
     
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