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Caleb asked in Science & MathematicsPhysics · 2 months ago

physics kinetic friction?

A 15.9 kg box rests on a horizontal surface. The coefficient of kinetic friction between the surface and the box is 0.33. The box is pulled to the right by a force of 109 N.

What is the acceleration in m/s^2?

3 Answers

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  • Ash
    Lv 7
    2 months ago
    Favorite Answer

    F(net) = F(applied) - F(friction)

    ma = F(applied) - μmg

    a = [F(applied) - μmg]/m

    a = [(109 N) - (0.33)*(15.9 kg)(9.8 m/s²)]/(15.9 kg)

    a = 3.6 m/s²

  • oubaas
    Lv 7
    2 months ago

    acceleration a = (F-FF)/m = (109-15.9*9.806*0.33)/15.9 = 3.6 m/sec^2 (2 sign. digits)

  • Whome
    Lv 7
    2 months ago

                                     F = ma

    109 - 0.33(15.9)(9.81) = 15.9a

    a = 3.6180... ≈ 3.62 m/s² ◄

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