f a 20.8 N horizontal force must be applied to slide a 10.1 kg box along the floor at constant velocity what is the coefficient of sliding friction between the two surfaces?

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  1. Wb = m*g = 10.1kg * 9.8N/kg = 98.98 N.=
    Wt. of box.

    Fb = 98.98N @ 0 Deg. = Force of box.
    Fp = 98.98*sin(0) = 0 = Force parallel to floor.
    Fv = 98.98*cos(0) = 98.98 N. = Force perpendicular to floor.

    u = Coefficient of friction.

    Fn = !@#$%^& - Fp - u*Fv = ma. a = 0.
    20.8 - 0 - u*98.98 = 0. a = 0.
    -98.98u = -20.8
    u = -20.8 / -98.98 = 0.21.

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