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Question: Spring constant

  1. Jan 16, 2009 #1
    1. The problem statement, all variables and given/known data
    The left side of the figure shows a light (`massless') spring of length 0.340 m in its relaxed position. It is compressed to 67.0 percent of its relaxed length, and a mass M= 0.250 kg is placed on top and released from rest (shown on the right).

    The mass then travels vertically and it takes 1.50 s for the mass to reach the top of its trajectory. Calculate the spring constant, in N/m. (Use g=9.81 m/s2). Assume that the time required for the spring to reach its full extension is negligible.

    2. Relevant equations
    EK=1/2mv2=1/2 * 0.25 * V2

    3. The attempt at a solution
    I plugged in the variables, but it still isn't right.
    Last edited: Jan 16, 2009
  2. jcsd
  3. Jan 16, 2009 #2


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    The mechanical energy of the spring will be converted into gravitational potential energy at the top of the trajectory. You have used kinetic energy instead.
  4. Feb 11, 2009 #3
    im really confused by this same type of question...
    i found U(y) now how do i use that to find K?
  5. Feb 11, 2009 #4
    As someone above said, you should be converting mechanical energy into gravitational potential energy, rather than using kinetic. 1/2kx^2=mgy is the proper conversion of energy. Hope that helps
  6. Feb 11, 2009 #5
    is x = .2278 m like above? and y = the distance the mass traveled in the air? cuz i have been using those and gettting the wrong answer
  7. Feb 11, 2009 #6

    The x i used in my equation is the distance that the spring moves from its "position of relaxation". You need to find the distance that the object goes vertically, and plug that in for y in the equation

    1/2kx^2=mgy >>>>> k= (mgy)/(.5x^2)

    Once you have y you can solve
  8. Feb 11, 2009 #7
    If you can confirm that the asnwer is roughly 3.1E3 N/m , I will tell you how i got it.

    I believe in the equation x= x(initial) + Velocity(Initial)*time + .5gt^2 you forgot to account for the .5gt^2, which equals roughly 22 m. Added to the 11 m you found using the middle term of the equation above, you have solved for the correct height. You were only a step off from getting it.
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