A man exerts a horizontal force of 141 N on a crate with a mass of 35.7 kg. What is the acceleration of the crate?

Question

10-07-2025
Physics

High School

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A man exerts a horizontal force of 141 N on a crate with a mass of 35.7 kg. What is the acceleration of the crate?

Answer :

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astha8579 astha8579 · Answer 1 · 2024-10-04T10:21:05-04:00

Final answer:

The magnitude of the static friction force is 125 N. The minimum possible value of the coefficient of static friction is approximately 0.425.

Explanation:

In order to determine the magnitude of the static friction force, we need to calculate the force of friction when the crate doesn't move. The force of friction is equal to the force exerted by the man, which is 125 N.

Therefore, the magnitude of the static friction force is 125 N.

The minimum possible value of the coefficient of static friction can be found by dividing the magnitude of the static friction force by the normal force. The normal force is equal to the weight of the crate, which is given by the equation N = mg, where m is the mass of the crate (30.0 kg) and g is the acceleration due to gravity (9.8 m/s²).

Therefore, the normal force is 294 N. Dividing the magnitude of the static friction force (125 N) by the normal force (294 N), we find that the minimum possible value of the coefficient of static friction is approximately 0.425.

Learn more about static friction here:

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Complete question:
A man exerts a horizontal force of 125 N on a crate with a mass of 30.0 kg. (a) If the crate doesn’t move, what’s the magnitude of the static friction force? (b) What is the minimum possible value of the coefficient of static friction between the crate and the floor?