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Frictional Forces without coefficient

  1. Oct 2, 2008 #1
    1. The problem statement, all variables and given/known data
    Block A (4.13kg) is on a table with a massless chord attached and draped over a frictionless pulley and the other end of the chord is attached to Block B (2.17kg). The system
    starts from rest. When the 2.71 kg mass has fallen through 0.388 m, its downward speed is
    1.28 m/s.
    The acceleration of gravity is 9.8 m/s2 .

    What is the frictional force between the 4.13 kg mass and the table? Answer in units
    of N.


    2. Relevant equations
    F=ma V=d/t a=v/t
    Not sure what other equations to use.



    3. The attempt at a solution
    I assume the speed to be constant, so there is no acceleration. I calculate the force of tension on block A to be equal to the force of gravity on block B: Ft=-mg=(2.71kg)*(-9.8m/s)=-26.56N

    I assumed that 26.56N would be the force of tension on Block A and the force of friction would simply be the opposite of that: -26.56N. That definitely isn't the answer and I pretty much knew that because I didn't use all the info given. I'm stuck at this point and any help would be appreciated. Thanks
     
  2. jcsd
  3. Oct 2, 2008 #2

    mgb_phys

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    Science Advisor
    Homework Helper

    Block A is falling - why would you assume it's a constant speed?
     
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