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Physics.

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A 500 g block lies on a horizontal tabletop. The coefficient of kinetic friction between the block and the surface is 0.25. The block is connected by a massless string to the second block with a mass of 300 g. The string passes over a light frictionless pulley as shown above. The system is released from rest.

  • Physics. -

    Start with a free-body diagram (FBD).
    It solves most statics problems automatically when you isolate the system into individual pieces.

    Start with the 500g block, mass m1.
    Normal force=m1g
    coefficient of friction, μk=0.25
    Frictional force = μkm1g
    Let tension in string, T
    Acceleration, a = (T-μkmg)/m
    =T/m1-μkg)

    For the block at the other end of the string.
    Net vertical force
    =m2g-T
    Acceleration, a
    =(m2g-T)/m2
    =g-T/m2

    Since two accelerations must be equal, we have
    T/m1-μkg) = g-T/m2

    Solve for t and substitute into formulas above to get a, acceleration.
    m1=0.500 kg
    m2=0.300 kg
    μk=0.25

    You should get acceleration, a=2.1 m/s² approximately.

  • Physics. -

    2.14

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