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?
1 answer:
Answer:
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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