A 9.0 kg potted plant slides down a 25.0º incline with an acceleration of 2.4 m/s2. What is the coefficient of kinetic friction between the potted plant and the incline?

0.196

Explanation:

According to Newton's second law

\sumFx = ma

Fm-Ff = ma

Fm is the moving force

Ff is the frictional force

m is the mass = 9kg

a is the acceleration = 2.4m/s²

Fm = Wsin theta

Fm = 9(9.8)sin25°

Fm = 37.28N

Ff is the frictional force = nmgcos theta

Ff = n(9)(9.8)cos25°

Ff = 79.94n

Substitute the given values into the formula

37.28-79.94n = 9(2.4)

-79.94n = 21.6-37.28

-79.94n = -15.68

n = 15.68/79.94

n = 0.196

Hence the coefficient of kinetic friction between the potted plant and the incline is 0.196

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