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Posted by on Thursday, October 13, 2011 at 12:50am.

you are driving a 2520.0-kg car at a constant speed of 14.0 m/s along a wet, but straight, level road. As you approach an intersection, the traffic light turns red. You slam on the brakes. The car's wheels lock, the tires begin skidding, and the car slides to a halt in a distance of 25.8 m. What is the coefficient of kinetic friction between your tires and the wet road?

  • physics - , Friday, October 14, 2011 at 7:45pm

    Wc = ma = 2520kg * 9.8N/kg = 24,696N =
    Weight of car.

    Fc = (24,696N.,0 deg.).

    Fv = 24,696cos(0) = 24,696N. = Force perpendicular to road.

    a = (Vf^2 - Vo^2) / 2d,
    a = (0 - (14)^2) / 51.6 = -3.8m/s^2.

    Fn = ma = 2520 * (-3.8) = -9572N.

    Ff = u*Fv = -(-9572),
    24,696u = 9572,
    u = 0.388. = Coefficient of friction.

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