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Spring Maximum Length

  1. Mar 26, 2015 #1
    1. The problem statement, all variables and given/known data

    A mass of 0.36kg hangs motionless from a vertical spring whose length is 0.8m and whose unstretched length is 0.51m. Next, the mass is pulled down to where the spring has a length of 1.09m and given an initial speed upwards of 1.2m[PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmmi10/alpha/144/char3D.pngs. [Broken] What is the maximum length of the spring during the motion that follows?

    Maximum length=

    2. Relevant equations

    F = k*y

    3. The attempt at a solution

    k = F/y = m*g/y = 0.36kg*9.8m/s^2/(0.8m - 0.51) = 12.17793N/m

    Energy is conserved

    Let A be the maximum displacement

    So 1/2*k*A^2 = 1/2*k*(1.09 - 0.51)^2 + 1/2*m*v^2

    so A = sqrt((1.09 - 0.51)^2 + m/k*v^2) = sqrt((1.09 -0.51)^2 + 0.36/12.18*1.2^2) = 0.5974m

    So the length will be 0.51 + 0.5974m = 1.11m

    I only have one attempt wrong, Cant find the error..
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Mar 26, 2015 #2


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    You only have two relevant equations -- if I count the second as an equation: spring potential energy = 1/2 k x2.

    You also use a third one: kinetic energy = 1/2 m v2.

    You forget a fourth one. Yet another energy....:wink:

    Oh, and: perhaps you want to check "1/2*k*(1.09 - 0.51)^2" which doesn't look right to me. At what point do you expect there is no energy stored in the spring ? And is that the top position, or is that the middle between top and bottom of the oscillation ?
    Last edited: Mar 26, 2015
  4. Mar 26, 2015 #3
    Would it be 1/2*k*(1.09 - 0.8)^2 and 1/2*k*(.8 - 0.51)^2

    I do you KE=1/2 mv^2 I thought? I'm confused on what else there could be.
  5. Mar 27, 2015 #4


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    Check your energy approach with the following:
    Hold the weight at spring unextended length 0.51 m and let go. How far will it drop until it will go up again (i.e. v = 0)
    Would you do ##{1\over 2} kx^2 = {1\over 2} mv^2= 0 ## to find x ?
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