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Work and energy with velocity

  1. Apr 21, 2009 #1
    1. The problem statement, all variables and given/known data

    Each of the sliders A and B has a mass of 2 kg and moves with negligible friction in its
    respective guide, with y being in the vertical direction (see Figure 3). A 20 N horizontal force
    is applied to the midpoint of the connecting link of negligible mass, and the assembly is
    released from rest with θ = 0°. Determine the velocity vA with which slider A strikes the
    horizontal guide when θ = 90°.
    [vA = 3.44 m/s]

    http://http://img.photobucket.com/albums/v242/021003/tutorial12.jpg [Broken]


    2. Relevant equations

    1/2 mv^2

    F = ma

    Wp = mgh

    SUVAT


    3. The attempt at a solution

    When at 0 degrees

    W=0J

    At 90

    F=20N

    W = 20xd = 8J

    Work from cart A = 0.5mv^2

    Therefore 16 = mv^2

    v = 2 rt2

    Or...

    do i need to add the energy from 20n force and from cart b...

    0.5mv^2 (b) + 8J = 0.5mv^2 (A)

    with F = ma, 20/10 a = 10 therefore v (b) = 2 rt 2

    sub this into above eq.

    8 + 8 = 0.5mv^2

    v = 4

    Help!

    Iv been goin round in circles, clearly im wrong lol can someone explain how i could work this out please
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 21, 2009 #2

    tiny-tim

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    Hi dietwater! :smile:

    hmm … your d in your work done is wrong.

    And what happened to mgh?
     
  4. Apr 24, 2009 #3
    lol il start again with mgh added in! lol

    Fd + mgh + o.5mv^2 (b) = 0.5mv^2 (a)

    I dont know how to get d, I just assumed its 0.4 as its the distance it has to travel?
     
  5. Apr 24, 2009 #4

    tiny-tim

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    No … the point of application of the force is only moving 0.2 :wink:
     
  6. Apr 25, 2009 #5
    am i right in thinking v(b) is still 2rt2?

    If so..

    Fd + mgh + o.5mv^2 (b) = 0.5mv^2 (a)

    20x0.2 + 2x9.81x0.4 + 0.5x2x(2rt2)^2 = 0.5x2x(v^2)

    so v = 3.98

    Im not sure where Iv gone wrong
     
  7. Apr 25, 2009 #6

    tiny-tim

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    Hi dietwater! :smile:

    (have a square-root: √ and try using the X2 tag just above the Reply box :wink:)

    (and your picture didn't show up because you put too many http//s in it :rolleyes:)

    Nooo :redface:

    imagine it's a cross-roads instead of a T-junction …

    which way would B be going as A sails across? :wink:
     
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