Question

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.

Answer 1

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.