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March 28, 2017

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An object with a mass of 3.7 kg slides in the positive x direction across a horizontal surface with a coefficient of kinetic friction of 0.24. During its slide a force of 364 Newtons acts on it in the positive x direction while a force of 44 Newtons acts on it in the negative x direction. If the object gains 548 Joules of kinetic energy during its slide, how far did the object slide in meters?
Help me solve please .

  • PHYSICS PLEASE HELP !!!!! - ,

    Wo = mg = 3.7kg * 9.8N/kg = 36.26N. =
    Weight of object.

    Fo = 36.26N@0deg.
    Fp=36.26sin(0) =0=Force perpendicular to surface.
    Fv = 36.26cos(0) = 36.26N. = Force
    perpendicular to surface.

    Ff = u*Fv = 0.24 * 36.26 = 8.70N. =
    Force of friction.

    Fn = (F1 + F2) - Fp - Ff,
    Fn = (324 - 44) - 8.70 = 271.3N.

    Fn = ma,
    a = Fn / m = 271.3 / 3.7 = 73.3m/s^2.

    KE = 0.5mV^2 = 548J.
    1.85V^2 = 548,
    V^2 = 296.2,
    V = 17.2m/s.

    d = (Vf^2 - Vo^2) / 2a,
    d = ((17.2)^2 - 0) / 146.6 = 2.02m.

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