Anonymous
Anonymous asked in Science & MathematicsPhysics · 4 weeks ago

physics and forces?

Two blocks, m1 = 4.4 kg and m2 = 2 kg, are in contact with each other and are on a horizontal surface for which the coefficient of kinetic friction is 0.1. A 27 N horizontal force acts on block 1.What is the the force exerted by m2 on m1.? 

I have tried every possible answer and none of them work, please help.

5 Answers

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  • 4 weeks ago
    Best answer

    friction force

    f = µ(m1+m2)g = 0.1 * 6.4kg * 9.8m/s² = 6.272 N

    and so the net motive force

    F = (27 - 6.272)N = 20.728 N

    results in a system acceleration

    a = F / (m1 + m2) = 20.728N / 6.4kg = 3.24 m/s²

    In order to accelerate m2 at this rate, the applied force must be

    F = m1*a + µ*m1*g = 2kg * (3.24 + 0.1*9.8)m/s² = 8.4 N

    If the direction of motion is considered "positive," then the force that m2 exerts on m1 is equal and opposite, so you might need a negative sign.

  • 4 weeks ago

    Assuming that the 27 N is pushing m1 toward m2 then for the system as a whole:  27 - (.1)(6.4)(9.8) = 6.4 a  then a = 3.23875 m/s*s.  With m1 pushing on m2:  P - (.1)(2)(9.8) = 2 a = 2(3.23875)  Giving P = 8.4375 N.

  • Anonymous
    4 weeks ago

    Post your work.................

  • oubaas
    Lv 7
    4 weeks ago

    F(2→1) = -27*2/(6.4) = -8.44 N

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  • 4 weeks ago

    Well, unbalanced force results in acceleration.

    The frictional force is (4.4+2)×.1×9.8 = 6.272N

    So, the combined system should accelerate at

    (27-6.272)÷6.6 = 2.181 m/s^2

    2.181×2 + 2×.1×9.8 = 6.322 Newtons 

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