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Moving two blocks connected by a spring

  1. Nov 30, 2008 #1
    1. The problem statement, all variables and given/known data
    m_1, v_1f
    m_2, v_2f
    m_2 = m_1 = m

    2. Relevant equations

    KE = .5mv^2
    W = F*d
    spring: F= -kx
    pot energy = PE = -.5kx^2

    3. The attempt at a solution

    A) K + U = .5m(v_1f)^2 + .5m(v_2f)^2 - .5kx^2
    .5(.1kg)(v_1f)^2 + .5(.1kg)(v_2f)^2 -.5(100N/m)(.02m)^2

    B) ??
    translational KE - going straight?
    isnt it
    .5(.1kg)(v_1f)^2 + .5(.1kg)(v_2f)^2

    C) ??
    V_cm....
    ??
    am i looking for how fast the center of the spring is moving?
    the center of the spring moved from .03 m to .1m

    and so

    ?? im lost

    or do I get the average speed of the two boxes which would be

    V_cm = (V_1f+V_2f)/2

    or

    .5(.1kg)(v_1f)^2 + .5(.1kg)(v_2f)^2 -.5(100N/m)(.02m)^2 = Wd = (5N)(.08m)
    (v_1f)^2 + (v_2f)^2 = 8.4 (m^2)/(s^2)

    does that equal V_cm?

    V_cm = ( 8.4 (m^2)/(s^2) ) ^ (1/2) = 2.90 m/s

    D) ??

    vibrational?
     

    Attached Files:

  2. jcsd
  3. Nov 30, 2008 #2

    Redbelly98

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    (A)
    How much work has been done on this system?
     
  4. Nov 30, 2008 #3
    Wd = (5N)(.08m)
     
  5. Dec 1, 2008 #4

    Redbelly98

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    Okay, so the energy system has increased by that amount. That helps with (A), getting the total energy K+U. Do you see why?
     
  6. Dec 1, 2008 #5
    what are you trying to say?

    the initial energy = 0
    becuz the objects are at rest and spring is at its natural lenght

    so therefore

    the final energy is K + U which is equal to Wd

    that?

    and so

    .5m(v_1f)^2+.5m(v_2f)^2-.5kx^2 = (100N/m)(.08m)
    KE m1 KE m2 PEspr Wd

    so my answer is??

    8 Joules
     
  7. Dec 1, 2008 #6

    Redbelly98

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    Yes, correct.

    Almost. K+U is equal to the work done, W=Fd:

    Well, this is really W or Fd, but not Wd. At any rate, multiplying (5N) times (0.08m) will give you the answer to (A).

    No, see above comments.
     
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