Posted by **Jules** on Monday, January 2, 2012 at 9:25pm.

An 89 N box of oranges is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 1.20 m/s each second. The push force has a horizontal component of 25 N and a vertical component of 30 N downward. Calculate the coefficient of kinetic friction between the box and floor.

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**Henry**, Tuesday, January 3, 2012 at 11:02pm
Fb = 89 N. @ 0 deg. = Force of box.

Fp = 89*sin(0) = 0 = Force parallel to

the floor.

Fv = 89*cos(0) = 89 N. = Force perpendicular to the floor.

mg = 89,

m=89/g = 89/9.8 = 9.08 kg=Mass of box.

Fn = Fap-Fp-Fk = ma,

25-0-Fk = 9.08*-1.2 = -10.9,

-Fk = -10.9-25 = -35.9,

Fk = 35.9 N. = Force of kinetic friction.

u(89+30) = 35.9

119u = 35.9,

u = 0.30 = Coefficient of kinetic friction.

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