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13 November, 00:22

A 1500-kg car drives at 30 m/s around a flat circular track 300 m in diameter. Since the net force is equivalent to the force of static friction, your answer to Part D is the magnitude fs. Based on this value, what is the minimum coefficient μs of static friction between the road and the car?

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  1. 13 November, 00:34
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    The minimum coefficient μs of static friction between the road and the car μs is 0.6116

    Explanation:

    Here we have

    Angular acceleration = v²/r = 30²/150 = 6 rad/s²

    Therefore, the force of the car away from its center of motion =

    Force, F = Mass * Acceleration = 1500 kg * 6 rad/s² = 9000 N

    The centripetal force is balanced by the frictional force so the car does not skid off the track

    Frictional force Ff = Normal force (Weight of car) * Coefficient of friction, μs)

    Weight of car = Mass * Acceleration due to gravity = 1500 kg * 9.81 m/s²

    = 14715 N

    Therefore, for equilibrium

    Frictional force Ff of car = Centripetal force on car

    14715 N * μs = 9000 N

    Therefore, μs = 9000 N / (14715 N) = 0.6116

    The minimum coefficient μs of static friction between the road and the car μs = 0.6116.
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