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LenaWriter [7]
3 years ago
11

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?
Physics
1 answer:
mezya [45]3 years ago
7 0

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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Answer:

a = 2.72 [m/s2]

Explanation:

To solve this problem we must use the following kinematics equation:

v_{f} =v_{o} + a*t

where:

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if we convert these units to units of meters per second squared

35250.35[\frac{km}{h^{2} }]*(\frac{1}{3600^{2} })*[\frac{h^{2} }{s^{2} } ]*(\frac{1000}{1} )*[\frac{m}{km} ] = 2.72 [\frac{m}{s^{2} } ]

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I would pick C

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