A box of oranges with a weight of 96 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward. Required:
Calculate the coefficient of kinetic friction between the box and floor.

Explanation:

Please see attached a detailed drawing of the force and line diagram for your reference.

Given data

weight W=96N

Fx=20N

Fy=25N

acceleration a= -0.9m/s^2----slowing down indicate negative

W=mg= 96N

Fk= kinetic friction= μkN

where N= normal reaction

from newtons second law

F=ma

x-----Fx-Fk=ma

Fk=Fx-ma

y----N-Fy-W=0

N=25+96

N=121N

so

Fx-μkN=ma

20-μkN=(96/9.81)-0.9

20-μkN=-86.4/9.81

20-μkN=-8.80

-μkN=-20-8.8

μkN=20+8.8

μk=28.8/N

μk=28.8/121

μk=0.238

μk=0.24

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