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March 30, 2017

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the 1500 kg automobile is travelling up the 20 degree inclined at a speed of 6 m/s , if the driver wishes to stop his car in a distance of 5m , determine the frictional force at the pavement which must be supplied by the rear wheels.

  • dynamics - ,

    V^2 = Vo^2 + 2ad.
    a = (V^2-Vo^2)/2d.
    a = (0-36)/10 = -3.6 m/s^2.

    Fk = m*a = 1500 * -3.6 = -5400 N. Acting
    in opposite direction of motion.

  • dynamics - ,

    CORRECTION:

    PE = KE.
    h = 5*sin20 = 1.71 m.

    mg*h-Fk*d = 0.5m*V^2.
    1500*9.8*1.71-Fk*5 = 0.5*1500*6^2
    25,137-5Fk = 27,000
    -5Fk = 27000-25,137 = 1863
    Fk = -373 N. = Force of kinetic friction.

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