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a 50kg box is pushed up a 30 degree plane which is 100m long. what is the speed of the box after being pushed the 100m? Mk=0.1

  • physics -

    m*g = 50kg * 9.8N./kg = 490 N. = Wt. of
    the box.

    Fp = 490*sin30 = 245 N. = Force parallel
    to the incine.

    Fn = 490*cos30 = 424.4 N. = Force
    perpendicular to the incline.

    Fk = u*Fn = 0.1 * 424.4 = 42.44 N. =
    Force of kinetic friction.

    At bottom of incline:
    KE + PE = mg*h - Fk*L
    0 + PE = 490*100*sin30-42.44*100
    PE = 24,500-4244 = 20,256 J.

    At top of incline:
    KE + PE = 20,256
    KE + 0 = 20,256
    0.5m*V^2 + 0 = 20,256
    0.5*50*V^2 = 20,256
    25V^2 = 20,256
    V^2 = 810.24
    V = 28.5 m/s.

    Note: h = 100*sin30

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