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Plz could you check this working for me?

  1. Feb 6, 2013 #1
    Plz could you check this working for me??

    1. The problem statement, all variables and given/known data
    Two blocks, A and B (with mass 50 kg and 100 kg, respectively), are connected by a string, as
    shown in figure.

    The pulley is frictionless and of negligible mass. The coefficient of kinetic friction between
    block A and the incline is μk = 0.25. Determine the change in the kinetic energy of block A as
    it moves from (C) to (D), a distance of 20 m up the incline (and block B drops down a
    distance of 20 m) if the system starts from rest.



    2. Relevant equations

    Block A
    T-μR+mg sinθ= ma

    Block B

    mg-T=ma




    3. The attempt at a solution


    acceleration in both blocks is the same since they are connected directly with no elastic strings

    Block A
    T=ma+μR-mg sinθ

    Block B
    T=mg-ma

    Substitute for T

    mg-ma=ma+μR-mg sinθ
    (100 x 9.8) - (100a) = (50a) + (0.25 x 50 x 9.8 x cos 37) - (50 x 9.8 x sin 37)
    980-100a-50a=97.83-294.89
    -100a-50a=97.83-294.89-980
    -150a=-1177.057
    a=7.8


    is this correct????
     

    Attached Files:

  2. jcsd
  3. Feb 6, 2013 #2

    PhanthomJay

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    Re: Plz could you check this working for me??

    You have a signage error here...which way does the weight component act along the plane? Once you find the acceleration, then you need to find the speed of the blocks to get the KE change. Alternatively, using work-energy methods, you will get the same result.
     
  4. Feb 6, 2013 #3
    Re: Plz could you check this working for me??

    oh i see so are you saying that it should be more like

    T-μR-mg sinθ= ma

    because the component of weight is opposite to tension?
     
  5. Feb 6, 2013 #4

    PhanthomJay

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    Re: Plz could you check this working for me??

    Yes, that is correct, and i think the best way to proceed to then find the speed of the block A and its KE change.
     
  6. Feb 6, 2013 #5
    Re: Plz could you check this working for me??

    Thank you very much i feel that i have learnt a lot since i joined here
     
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