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Finding the Spring Constant of a mattress

  1. Nov 13, 2012 #1
    1. The problem statement, all variables and given/known data
    A mattress manufacturer estimates that 20 springs are required to comfortable support a 100kg person. When supporting the person, the 20 springs are compressed 0.035m. Calculate the spring constant for one sprint.

    2. Relevant equations
    The answer is 1400N/M at the back of my textbook. How does one solve it?
    Why does my solution not work?

    3. The attempt at a solution
    (20)Ee = Eg //The gravitational potential energy is transferred to the 20 springs.
    20(0.5)(k)(x^2) = mgh
    10(k)(0.035^2) = (100)(9.8)(0.035)
    k = 2800N/M
    So close yet so far....
  2. jcsd
  3. Nov 13, 2012 #2


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    You have used conservation of energy. Think about the dynamics of that. Person let go from rest on uncompressed springs... what will happen?
    You don't need to think about energy here. Just treat it as a statics question.
  4. Nov 13, 2012 #3
    F = kx
    k = F/x
    k = mg/x
    k = (100)(9.8) / 0.035 x 20
    k = 1400

    I still think my solution should have worked. I don't understand conceptually why energy is not conserved.
  5. Nov 14, 2012 #4


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    As I said, you treated it as though the person was suspended horizontally, just touching the bed, then released. Suppose energy is conserved - what will happen? The person will fall, compressing the springs, but 'overshoot'. When the force from the springs matches the person's weight, the person will stop accelerating, but already has momentum and KE, so will continue to descend. When descent ceases, the springs will be applying a larger force and the person will bounce back up. If energy is conserved they will come all the way back up to starting position before descending again, and bounce forever.
    The question required the point at which there was no acceleration, somewhere between the extremes of the bounce.
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