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Kinetic Energy including a pulley

  1. May 21, 2012 #1
    1. The problem statement, all variables and given/known data
    Two blocks, of masses M = 2.0 kg and 2M are connected to a spring of spring constant k = 200 N/m that has one end fixed, as shown in the figure below. The horizontal surface and the pulley are frictionless, and the pulley has a moment of inertia I=0.005 kg.m^2 with radius R=0.050 m. The blocks are released from rest with the spring relaxed.i.e in its equilibrium position).

    (a) What is the combined(total) kinetic energy of the two blocks when the hanging two-block-pulley system has fallen a distance of 0.090 m?
    J

    (b) What is the rotational kinetic energy of the pulley when the hanging blockhas fallen 0.090 m?



    http://www.webassign.net/hrw/W0155-N.jpg


    ]2. Relevant equations[/b]
    Mass 1
    Fnet(X)= [T[/1] -[F][/sp] =[M][/1][a][/x]= [T][/1] -KΔs

    Pulley
    [T][/net] =[T][/1]R -[T][/2]R= -IΩ
    Hanging Mass
    [F][/net]=[M][/2]-[T][/2]=[M][/2]



    3. The attempt at a solution

    Three equations Four unknowns ,I do not know how to proceed from this point.
     
    Last edited: May 21, 2012
  2. jcsd
  3. May 21, 2012 #2

    tiny-tim

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    Hi Lydia! :smile:

    Why forces? :confused:

    Just call the speed "v", and calculate all the energies. :wink:
     
  4. May 21, 2012 #3
    I was trying to get the acceleration the lead to velocity
     
  5. May 21, 2012 #4

    tiny-tim

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    i know! :smile:

    you don't need to
     
  6. May 21, 2012 #5
    If the velocity is final 'v'
    K =0.5M[V][/2] + 0.5I[ω][/2] + 0.5{2M}[V][/2] + mgh
    =0.5M[V][/2] + 0.5(0.5M[R][/2])[V][/2][R][/2] + 0.5{2M}[V][/2] - 2Mgh
     
  7. May 21, 2012 #6
    But how do I know i don't need to? won't this be in terms of unknowns ?? #swimming in the pool of confused (EXAM in a few hours) :cry:
     
  8. May 21, 2012 #7

    tiny-tim

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    (try using the X2 button just above the Reply box :wink:)
    Yes, but

    i] I is given in the question

    ii] you've left out the springy part! :wink:
     
  9. May 21, 2012 #8
    If the velocity is final 'v'
    K =0.5M[V][/2] + 0.5I[ω][/2] + 0.5{2M}[V][/2] + mgh
    =0.5M[V][/2] + 0.5(0.005[V][/2][0.050][/2] + 0.5{2M}[V][/2] - 2(9.8)(0.090) + 0.5(200)(ΔS)
     
  10. May 21, 2012 #9

    tiny-tim

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    a bit difficult to read :redface:

    i think it's ok, except that the spring KE should be 1/2 k h2 :wink:
     
  11. May 21, 2012 #10
    Sorry i cannot use those fancy buttons.Thank you for the little light
     
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