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Mass on Vertical Spring homework

  1. Oct 19, 2009 #1
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    A spring with spring constant k = 40 N/m and unstretched length of L0 is attached to the ceiling. A block of mass m = 2 kg is hung gently on the end of the spring.

    a) How far does the spring stretch? dL = 0.49

    Now the block is pulled down until the total amount the spring is stretched is twice the amount found in part (a). The block is then pushed upward with an initial speed vi = 2 m/s.
    |v|max = ?


    F=ma. W=Fd, W = ΔKE, U = mgh, Fs = -Kx


    Part a was easy, weight equals spring force so mg = kx => x=mg/k = 0.49.
    Part b is where I'm really stuck. Since (U + KE)final = (U + KE)initial then
    mgh + 1/2mv^2 = mgh + 1/2mv^2 + Uspring => 2(9.81)(0.981) + 1/2(2)(2^2) = 40(0.981) + 1/2(2)(vmax^2) + 0 so vmax = 4.05 but that's not right. HELP PLEASE!
     
  2. jcsd
  3. Oct 19, 2009 #2

    rl.bhat

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    In the second part, total energy in the system with respect to the unstretched position is
    mg(2δL) + 1/2*k(2δL)^2 + 1/2mv^2
    When it crosses the unstretshed position, it has only KE.
    Equating the above two energies find the vmax.
     
  4. Oct 19, 2009 #3
    Ok so are you saying that: (mg(2δL) + 1/2*k(2δL)^2 + 1/2mv^2) = 1/2m(vmax)^2, cause when I solve that I get 6.51 which is wrong.
     
  5. Oct 20, 2009 #4

    rl.bhat

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    Take mg*2δL negative, because it is the work done against the gravity. It will reduce the KE.
     
  6. Oct 20, 2009 #5
    Ok so then it should be -(mg(2δL) + 1/2*k(2δL)^2 + 1/2mv^2) = 1/2m(vmax)^2 =>
    -2(9.81)(0.98) + .5(40)(.98^2) + .5(2)(2^2) = .5(2)(vmax^2) => vmax = 1.99 but that's wrong too.
     
  7. Oct 20, 2009 #6

    rl.bhat

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    Now the block is pulled down until the total amount the spring is stretched is twice the amount found in part (a)
    So the total stretching from the unstretched position is 3δL.
     
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