The brakes of a 125 kg sled are applied while it is moving at 8.1 m/s, which exerts a force of 261 N to slow the sled down. How
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Answer:
0.2
Explanation:
The given parameters are;
The acceleration of the train, a = 0.2·g
The mass of the person standing on the train = m
Let μ represent the coefficient of static friction, we have;
The force acting on the person, F = m × a = m × 0.2·g
The force of friction acting between the feet and the floor, = m·g·μ
For the person not to slide we have;
The force acting on the person = The force of friction acting between the feet and the floor
F =
∴ m × 0.2·g = m·g·μ
From which we get;
0.2 = μ
The coefficient of static friction that must exist between the feet and the floor if the person is not to slide, μ = 0.2.