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A 2.6 kg block slides down a 25° inclined plane with constant acceleration. The block starts from rest at the top. At the bottom, its velocity is 0.75 m/s. The incline is 1.6 m long.
(a) What is the acceleration of the block?

(b) What is the coefficient of friction?

  • physics -

    a. Wb = M*g = 2.6kg * 9.8N/kg=25.48 N. =
    Weight of block.

    Fp = 25.48*sin25 = 10.77 N. = Force
    parallel to the inclined plane.
    Fn = 25.48*cos25 = 23.09 N. = Normal Force = Force perpendicular to the
    incline.

    a=V^2-Vo^2)/2d
    a = (0.75^2-0)/2.6 = 0.2163 m/s^2.

    b. Fp-Fk = m*a
    10.77-Fk = 2.6 * 0.2163 = 0.5625
    Fk = 10.77 - 0.5625 = 10.21 N. = Force
    of kinetic friction.

    u = Fk/Fn = 10.21/23.09 = 0.4421

  • physics -

    Correction:
    a = (0.75^2-0)/3.2 = 0.1758 m/s^2.

    b. Fp-Fk = m*a
    10.77-Fk = 2.6*0.1758 = 0.4570
    Fk = 10.77 - 0.4570 = 10.31 N

    u = Fk/Fn = 10.31/23.09 = 0.4466

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