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A 4.7 kg ball is pushed across a horizontal meter stick track with a constant force of 19.2 N. What is the coefficient of sliding friction between the ball and the meter stick?

A) 0.25
B) 0.40
C) 0.70
D) 0.85

1 Answer

3 votes

Final answer:

The coefficient of sliding friction between the ball and the meter stick is calculated using the formula for the force of friction. With the data provided, it is determined that the coefficient of friction is 0.40, which corresponds to option B.

Step-by-step explanation:

To calculate the coefficient of sliding friction between the ball and the meter stick, we apply Newton's second law, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration (F = ma). However, since the force is constant and there is no mention of acceleration or motion, we can infer that the ball is moving at a constant velocity; hence, the acceleration is zero, and the force applied is balanced by the friction force.

The force of friction is calculated using the formula F_friction = μ * N, where μ is the coefficient of friction and N is the normal force, which in this case, is equal to the weight of the ball due to gravity (N = m*g, where g = 9.8 m/s² is the acceleration due to gravity).

Setting the applied force equal to the friction force gives us 19.2 N = μ * 4.7 kg * 9.8 m/s². Solving for μ yields μ = 19.2 N / (4.7 kg * 9.8 m/s²) = 0.415, which rounds to 0.40, corresponding to option B.

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