A75 n box of oranges 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.
calculate the coefficient of kinetic friction between the box and floor?

μk = 0.26885

Explanation:

Conceptual analysis

We apply Newton's second law:

∑Fx = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Data:

a= -0.9  m/s²,  

g = 9.81 m/s² : acceleration due to gravity

W= 75 N :  Block weight

W= m*g  

m =  W/g = 75/9.8= 7.65 kg :  Block mass

Friction force : Ff

Ff= μk*N

μk: coefficient of kinetic friction

N : Normal force (N)  

Problem development

We apply the formula (1)

∑Fy = m*ay    , ay=0

N-W-25 = 0

N = 75 +25

N= 100N

∑Fx = m*ax    

20-Ff= m*ax    

20-μk*100 = 7.65*(-0.90 )

20+7.65*(0.90) = μk*100

μk = ( 20+7.65*(0.90)) / (100)

μk = 0.26885

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