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Find the spring constant

  1. Nov 21, 2008 #1
    1. The problem statement, all variables and given/known data
    A mass of 50 kg falls 50 cm onto the platform of a spring scale and sticks. The platform comes to rest 10 cm below its initial position. The mass of the platform is 2 kg. Find the spring constant.


    2. Relevant equations



    3. The attempt at a solution
    I thought of the following solution using conservation of energy:
    m*g*h = (k*A^2)/2
    leads to k = 1000g
    But I also thought of the solution using conservation of momentum:
    v0^2=2*a*(ds)=g
    m*v0=M*v1 (M is the total mass)
    m*v0/M=v1=5*g/6
    (m*v1)^2/2=(k*A^2)/2
    k = 5000*g^2/3
    Could it also simply be that the spring comes to rest when forces are at equilibrium
    M*g=k*x
    That yields
    k=100*g
     
  2. jcsd
  3. Nov 21, 2008 #2

    PhanthomJay

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    Using conservation of energy is the correct approach, but you are missing a couple of terms. The initial energy is associated with the potential energy (relative to some reference point) of both masses. And don't forget to consider the distance that the 50 kg mass falls includes both it's initial height and the spring displacement. You can't use conservation of momentum, because it is not conserved after the initial collision.
     
  4. Nov 21, 2008 #3
    Why isn't the momentum conserved? I mean, immediately after collision, m1*v0 should equal (m1+m2)*v1 ??
    About energy I think this is :
    m1*g*h=(k*A^2)/2
    with A = 0.1 + (m2*g)/k
    because you have kx0=m2*g
    and we should get k = 959.58 g
     
    Last edited: Nov 21, 2008
  5. Nov 22, 2008 #4

    PhanthomJay

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    Let me clarify my response, because I didn't state it correctly.
    1. Momentum is conserved only when there is no external force acting on the system.
    2. Mechanical energy is conserved only when conservative forces act (like spring or gravity forces) and when no energy is lost due to heat or sound. etc.
    3. In this problem, the platform comes momentarily to a rest when the spring is extended 10 cm below it's intial position. When mg=kx, it is not at rest; it has no acceleration at that point, but it is still moving.
    To solve this problem, you need to apply conservation of momentum during the collision, and conservation of mechanical energy before and after the collision. Thus, one calculates the speed of the falling block at the instant before it hits the other block, using the kinematic equation of motion or conservation of energy equation, which I believe is what you have designated as Vo. Then during the collision, you use conservation of momentum, (which you can get away with doing even though the external gravity force is still acting, because it acts for such a short distance during the deformation of the 2 objects that you can ignore it), and you can solve for the speed of the combined masses per your correct equation for V1, using conservation of momentum. Finally, you must apply conservation of energy to get the value of the spring constant k, noting that initially after the collision, the combined blocks have both kinetic an potential energies, and at the max compression (10cm) of the spring, there is spring potential energy only.
     
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