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Basic force with pulley question

  1. Nov 2, 2008 #1
    1. The problem statement, all variables and given/known data

    what is the acceleration of the system below? the coefficient of kinetic friction between the block and the table is 0.40. The weigts of the slideing block and hanging mass are 49N and 78.4N respectively.

    2. Relevant equations



    Fnet= Fa-Ff

    3. The attempt at a solution

    I took the 78.4N force as the force pulling the block.

    then i found the Ff.

    Ff= 0.4 * 49 =19.6

    so Fnet = 78.4-19.6 = 58.8N

    using F=ma

    m= 49/9.8= 5kg

    58.8 = 5 * a

    a= 58.8/5= 11.76

    the actual answer given is 4.5m/s^2

    any help to show me where I went wrong would be great. thanks.

    Attached Files:

  2. jcsd
  3. Nov 2, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    The force pulling on the sliding block is the tension in the rope, not the weight of the hanging mass.

    Hint: Set up separate equations for each mass using Newton's 2nd law. Then solve them together to find the acceleration.
  4. Nov 2, 2008 #3
    so since its moving T<mg?
  5. Nov 2, 2008 #4
    would T= mg - Ff?

    so T= 58.8?
  6. Nov 2, 2008 #5

    Doc Al

    User Avatar

    Staff: Mentor

    Why don't you do what I suggest: Set up one equation for the sliding block (what forces act on it?) and another equation for the hanging mass.
  7. Nov 3, 2008 #6
    thanks i think i got it now, i have some consistent results throughout my worksheet. thanks.
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