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Homework Help: Double spring question

  1. Oct 25, 2007 #1
    1. The problem statement, all variables and given/known data
    A block of mass 20kg with a spring with a stiffness of 500N/m attached to the bottom is dropped .5m from rest onto another spring of stiffness 800N/m. The weight of the springs can be neglected. What is the maximum deformation of each spring due to the collision.

    2. Relevant equations
    work = change in kinetic energy + change in potential energy + change in elastic energy

    3. The attempt at a solution
    I have assumed that I can add the different spring coefficients, giving me an energy equation of:

    work=0= -.5mv^2 + -mg(.5+x) + .5k(x^2)

    here k is the sum of the spring coefficents and x is the total deformation of the spring.
    I solved the equation, got two answers for x one positive and one negative.

    Can I use the ratio of the spring coefficients to work out the individual reformation, and also do i use the positive value calculated for x?

    Sorry for the long question, your help is appreciated.
    Last edited: Oct 25, 2007
  2. jcsd
  3. Oct 25, 2007 #2
    Ermm, if the springs are inline with each other (on top of each other) you add the reciprocals of the constants, so that:

    [tex]{k_T}^{-1} = {k_1}^{-1} + {k_2}^{-1}[/tex]

    The system adds (becomes stiffer) when the springs are used adjacent to each other.

    So I think, from the wording of the question, you need to use this equation maybe:

    [tex]k_T = \frac{k_1 k_2}{k_1 + k_2}[/tex] - Your thoughts

    But yeah - the kinetic energy term would become [itex]mg\Delta h[/itex] in your equation (because the kinetic energy came about from falling the distance [itex]\Delta h[/itex], cause when it hits the spring, your [itex]mgh = 0[/itex]. Apart from that, logic seems fine...

    Let me know how you get on :P
  4. Oct 25, 2007 #3
    I do agree with Sam.. Keq = k.k'/(k + k') , is the one to be used.
    Since, the two springs are adjacent, force in them must be equal => k.x = k'.x'
    Equivalent spring, Keq.X = k.x = k'.x', where X = x + x'.

    @ sumeer@dinsum.c
    I posted this, just to urge one thing: please REASON OUT your assumptions! Do not make blind assumptions. The assumptions you made (that Keq = k+ k') would have been true, if elongations in both the springs would have been same! x = x', X = x = x', and Ftot = Keq.X = k.x + k'.x'
    I hope you get the idea.
  5. Oct 25, 2007 #4
    Hi Sam,

    Thanks for your information, I've managed to work through the problem.
    Last edited: Oct 25, 2007
  6. Oct 26, 2007 #5


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    Homework Helper

    Hi Sameer, I don't how you solved the problem. But see whether it matches with this. Loss of PE of block = mg(0.5 +X). This energy is shared by the two springs and it is equal to 1/2*kx^2 + 1/2*k'x'^2. During the compression force in each spring is kx and k'x'. Hence down ward force =mg + kx and upward force = k'x'. In equilibrium position mg +kx = k'x' and X = x+x', write x' and X in terms of x and substitute it in the above energy equation. Then solve for x, and x'
    Last edited: Oct 26, 2007
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