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March 29, 2017

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A 13.0 -kg box is released on a 33 degree incline and accelerates down the incline at 0.20 m/s^2

Find the friction force impeding its motion?

Can someone please take me through this problem step by step? Thank you so very much

  • Physics - ,

    ma=mgsinα - F(fr)
    F(fr) =m(gsinα –a)

  • Physics - ,

    Wb = m*g = 13kg * 9.8N/kg = 127.4 N. =
    Wt. of box.

    Fb = 127.4N @ 33 Deg.
    Fp = 127.4*s.n33 = 69.4 N. = Force parallel to the incline.
    Fv = 127.4*cos33 = 106.8 N. = Force perpendicular to plane.

    Fn = Fp - Fk = m*a
    69.4 - Fk = 13*0.20 = 2.6
    -Fk = 2.6 - 69.4 = -66.8
    Fk = 66.8 = Force of kinetic energy.

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