can anyone help me please ? physics problem ?

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3 Answers

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  • 2 months ago

    The weight of the 12kg block on the 9kg block = 12*9.8 = 117.6N

    The friction force opposing motion = 117.6*0.3 = 35.28N

    Fnet = 5*9.8 - 35.28 = 13.72N

    a = Fnet/(9+5) = 0.98m/s² <<<<<

    OR we can use the tension T:

    (5*9.8 - T)/5kg = a = (T - 12*9.8*0.3)/9kg

    Cross multiply

    9(5*9.8 - T) = 5(T - 12*9.8*0.3)

    441 - 9T = 5T - 176.4

    441+176.4 = 617.4 = 14T

    T = 44.1

    a = (49 - 44.1)/5 = 0.98m/s² same answer.

    a = (44.1 - 12*9.8*0.3)/9kg = 0.98m/s²

    That rounds to 1m/s² the first choice.

  • 2 months ago

    What is the friction between the 9Kg block and the table?

  • 2 months ago

    weight of 12 kg block is 12•9.8 N

    force of friction is 0.3•12•9.8 = 4•9.8 N

    force due to weight is 5•9.8 N

    net force is the difference, 1•9.8 N

    a = F/m = 9.8 / (5+9) = 0.7 m/s²

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