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A Breelyn and Tina pull on opposite ends of a 5.0 Kg box resting on a flat ground.Breelyn pulls to the left with a force of 65 N. Tina pulls to the right with a force of 38 N. the box accelerates to the left at 2.0 m/s? What is the coefficient of friction between the box and the ground

User Holtorf
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Final answer:

To calculate the coefficient of friction for the box being pulled on the ground, we find the net force acting on the box, then use the relationship between friction, normal force, and the coefficient of friction to solve for the latter, giving us an approximate value of 0.55.

Step-by-step explanation:

To find the coefficient of friction between the box and the ground, we need to consider the net force acting on the box and the acceleration. Given that Breelyn pulls to the left with a force of 65 N and Tina pulls to the right with a force of 38 N, we can determine the net force on the box:

Net Force = Force by Breelyn - Force by Tina = 65 N - 38 N = 27 N.

According to Newton's second law, the net force is the product of the mass and the acceleration (Net Force = mass × acceleration).

Therefore, we can set up the equation:

27 N = 5.0 Kg × 2.0 m/s².

This gives us the net force acting on the box which is already known from the forces applied by Breelyn and Tina. The difference between this net force and the net force calculated from these two forces gives us the force of friction. To find the coefficient of friction (μ), we use the formula:

Friction = μ × Normal force.

The normal force for an object resting on a flat surface is equal to its weight (mg), which is:

Normal force = 5.0 Kg × 9.8 m/s² = 49 N.

Since we know the frictional force (Net Force) and the normal force, we can rearrange the equation to solve for μ:

μ = Friction / Normal Force = 27 N / 49 N.

Therefore, the coefficient of friction is:

μ = 0.55 (approximately).

User Bishwarup Das
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