Solving for Coefficient of Friction with 30N Force

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SUMMARY

The discussion focuses on calculating the coefficient of kinetic friction for a wooden crate being pushed with a horizontal force of 30.0 N. The crate has a mass of 12.0 kg, and the gravitational acceleration is 9.8 m/s². The force of friction is confirmed to be equal to the applied force of 30.0 N, as the crate moves at a constant velocity, indicating that the net force is zero. The coefficient of kinetic friction (μk) can be calculated using the formula Ff = μk * FN, where FN is the normal force.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with the concept of friction and its coefficients
  • Basic algebra for solving equations
  • Knowledge of gravitational force calculations (g = 9.8 m/s²)
NEXT STEPS
  • Calculate the coefficient of kinetic friction using the formula μk = Ff / FN
  • Explore the effects of different surface materials on friction coefficients
  • Learn about static vs. kinetic friction and their applications
  • Investigate real-world applications of friction in engineering and physics
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and friction, as well as educators looking for practical examples of force and motion concepts.

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Homework Statement



If you use a horizontal force of 30.0N to slide a 12.0-kg wooden crate across a floor at a constant velocity, what is the coefficient of kinetic friction between the crate and the floor?

Homework Equations



Ff=[tex]\mu[/tex]kFN

The Attempt at a Solution



I thought maybe it would be 30.0N=[tex]\mu[/tex]k(12.0kg)(9.8m/s290, but I don't know if that's right. Is 30N the force of friction? Please help me!
 
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You are doing fine. Yes, the 30N force = force of friction. (Since the box moves with constant velocity, what must be the net force on it? That's what should convince you that the applied force and the friction must be equal and opposite.)
 

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