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A 1460-kg car is being driven up a 10.0 ° hill. The frictional force is directed opposite to the motion of the car and has a magnitude of 497 N. A force F is applied to the car by the road and propels the car forward. In addition to these two forces, two other forces act on the car: its weight W and the normal force FN directed perpendicular to the road surface. The length of the road up the hill is 209 m. What should be the magnitude of F, so that the net work done by all the forces acting on the car is 149 kJ?

  • physics -

    Fc=mg=1460kg * 9.8N/kg=14308N @ 10deg.

    Fp = 14308*sin10 = 2484.56N = Force parallel to the plane downward.

    Fv = 14308*cos10 = Ff = .63N = Force perpendicular to the plane and acting downward.

    Ff = 497N = Force of friction acting
    opposite to the motion.

    W = Fn*d = 149000J.
    Fn * 209 = 149000,
    Fn = 149000 / 209 = 712.92 = Net force.

    Fn = Fap - Fp - Ff = 712.92,
    Fap - 2484.56 - 497 = 712.92,
    Fap - 2981.56 = 712.92,
    Fap = 712.92 + 2981.56 = 3695N = Force
    applied.

  • physics -

    CORRECTION:

    Fv = 14308*cos10 = 14090.63N = Force perpendicular to the plane and acting
    downward.

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