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Homework Help: Conservation of Energy with a spring

  1. Feb 27, 2012 #1
    1. The problem statement, all variables and given/known data

    A spring is attached to two blocks. The smaller block is 4.0 kg and the larger block is 8.0 kg. Initially the spring is in equilibrium and the blocks are separated by a d = 25 cm. The spring constant k = 400 N/m. No friction is present.

    I now push on the two blocks compressing the spring. The separation between the two blocks is now 15 cm.

    a) What is the speed of each block when they get back to being separated by 20 cm?
    b) What will the maximum separation between the blocks be?

    2. Relevant equations

    Spring PE = .5*k*x^2
    Kinetic Energy = .5*m*v^2

    3. The attempt at a solution

    So for a, I managed to solve for the potential energy of the system, with the spring compressed .1 meters.

    PE = .5kx^2 -> .5 * 400 * (.1)^2 = 2 J.

    Then I split that in half, and applied conservation of energy to solve for the kinetic energy at equilibrium for each block.

    So for smaller block -> KE = .5*m*v^2 = 1, 1 = .5*4*v^2, 1/2 = v^2, v = .707 m/s or so.
    Larger block 1 = .5 * 8*v^2, v = .5 m/s

    And part b, I assume the total stretch will be equal to the total compression, so stretch = 10 cm, thus the max separation should be 35 cm.

    Are my answers correct?
  2. jcsd
  3. Feb 27, 2012 #2


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    Staff: Mentor

    Hi navydrgn, welcome to PF.

    Can you confirm for part (a) that the required separation is 20 cm, and not the original 25 cm?

    Note that the kinetic energy won't split evenly between the blocks unless both blocks are of the same mass. What applies here is conservation of momentum -- the momentum of both blocks must be equal (and opposite in sign) in order for conservation to hold, which it must.
    This will also cause the center of mass of the system to remain stationary, which must be true since no external forces are operating on it.
  4. Feb 27, 2012 #3
    Yes, thank you, it should be 25 cm. That was a typo.

    So in order to solve the system, I'd have to determine a Pi (initial momentum) and a final momentum (Pf) using P = mv, where m is mass and is v is velocity? And the solve for the final velocity of both blocks?

    Why can't this problem be solved with conservation of energy?

    Thanks for the swift response, by the way.
  5. Feb 27, 2012 #4


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    Staff: Mentor

    You need both conservation of energy and conservation of momentum for this problem. Otherwise how can you decide how to apportion the kinetic energy between the blocks?

    For the conservation of momentum part it is sufficient to know that to begin with the momentum is zero, and so it must remain zero. It means that at any given time the momentum must sum to zero, so that the momenta of the two blocks are equal and opposite in sign. In other words, m1*v1 = m2*v2.
  6. Feb 27, 2012 #5
    Okay, I oversimplified the problem, then, by assuming I could apportion the kinetic energy just based on ratios of masses. In a way that's what I was doing, (after your initial advice) but I didn't have a name for it.

    Thus (using your notation, where m1 = 4.0 kg, and m2 = 8.0 kg), we have:

    m1*v1 = m2*v2, 4v1 = 8v2, v1 = 2v2, and then plugging that in to the kinetic energy equation (modified to include both blocks) we have

    .5*m1*v1^2 + .5*m2*v2^2 = 2 J
    2(2v2)^2 + 4*v2^2 = 2 J
    12v2^2 = 2J
    v2 = .408 m/s

    And then:
    v1 = 2v2 = .816 m/s

    Assuming I understood you correctly, these would be the correct values.

    Thank you
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