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Spring Problem, conservation of energy equation

  1. Oct 16, 2011 #1
    Write the conservation of energy equation for this situation and solve it for the speed of the mass as it passes equilibrium.

    A horizontal spring attached to a wall has a force constant of 900 N/m. A block of mass 1.20 kg is attached to the spring and oscillates freely on a horizontal, frictionless surface as in the figure below. The initial goal of this problem is to find the velocity at the equilibrium point after the block is released.


    The points of interest are where the mass is released from rest (at x = 5.40 cm) and the equilibrium point, x = 0.

    x= 5.40 it is 1.31 J
    x= 0 it is 0 J.

    I am getting suck on write the conservation of energy equation for this situation.
    the answer is 1.48 M/S but the only equation I can come up with is 1/2kx^2=1/2mv^2

    Nothing is adding up here.
     
  2. jcsd
  3. Oct 16, 2011 #2

    Doc Al

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    Staff: Mentor

    What does 'it' represent?

    Note that total energy = KE + PE
     
  4. Oct 16, 2011 #3
    "it" is energy stored in the spring at those moments.
     
  5. Oct 16, 2011 #4
    KE+PE = total Energy, but how do I convert energy to a velocity?
     
  6. Oct 16, 2011 #5

    Doc Al

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    Staff: Mentor

    Good. So what's the total energy at each moment? (That's where conservation of energy comes in.)
     
  7. Oct 16, 2011 #6
    I dont quite understand, throw a dog a bone here!
    PE=1/2kx^2 so that is energy stored technically.
    KE= 1/2mv^2 I can't find the velocity this is where I am stuck.

    I am assuming I can take the PE and convert it to KE *some how*

    then take the known variables of KE and plug it into KE=1/2mv^2 and solve for v^2.

    So I need to figure out how to convert PE into KE.
     
  8. Oct 16, 2011 #7
    Put a fork in it, this ham is done boys.

    I wasn't converting CM to M...so naturally it was making my numbers super high.

    Incase anyone comes up with this problem a nice lil formula to use is.


    Vf= squareroot (k/m (Xi^2-Xf^2)

    in other words you take the Sprint Constant, Divide it by the mass. You then take this number and multiply it by (Xi^2-Xf^2), and square root all of it.
     
  9. Oct 16, 2011 #8

    Doc Al

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    Staff: Mentor

    Good. But rather than memorize some formula, just think of it as energy conservation:

    KE1 + PE1 = KE2 + PE2

    Where KE = 1/2mv2 and PE = 1/2kx2.

    This way you can solve for whatever you need, no matter what they throw at you.

    In this case, things get easy:

    Since the mass is released from rest, you know that KE1 = 0.
    And since the second point is at the equilibrium position, you know that PE2 = 0.
     
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