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Spring constant of a spring

  1. May 13, 2014 #1
    1. The problem statement, all variables and given/known data
    A mass of 50.0 g mass is hung from a spring, spring is elongated to 40 cm, an additional mass of 20g is added, with a new reading of 45cm

    a) what is the spring constant for this spring?
    b) another mass of 30g is added to the spring, what is the new reading?
    2. Relevant equations

    k= mg/x

    3. The attempt at a solution
    not sure if this is the right approach
    a)
    k= change in mass x g / change in height
    0.020x9.8/0.05 =3.92n/m
    b)mg=kx
    0.03x9.8=3.92x
    x=0.075 so it extend another 7.5cm?
     
  2. jcsd
  3. May 13, 2014 #2

    NascentOxygen

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    Staff: Mentor

    Use a capital "N" for newtons.
     
  4. May 13, 2014 #3
    The numbers in your solution don't seem to match the numbers in the problem statement.

    Chet
     
  5. May 13, 2014 #4

    BiGyElLoWhAt

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    Gold Member

    why not just set up a N2L equation for this? If the spring stretches an amount x with a mass m attached to it and comes to rest again, then that means that net force is zero, and mg = kx, you only need 1 weight and 1 distance.
     
  6. May 13, 2014 #5

    Doc Al

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    Staff: Mentor

    They look fine to me.

    I interpret the 40 cm as the total length of the spring, including its unstretched length.
     
  7. May 13, 2014 #6

    BiGyElLoWhAt

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    Gold Member

    Ahhh, I see what you're saying, so you do actually need two masses. Good call Doc.
    elongated to 40 cm
     
    Last edited: May 13, 2014
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