What is the Coefficient of Kinetic Friction for a Box Pushed at an Angle?

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SUMMARY

The coefficient of kinetic friction (uk) for a box weighing 325N, pushed with a force of 425N at an angle of 35.2° below the horizontal, can be determined using the principles of force analysis. The box moves at a constant velocity, indicating that the net force acting on it is zero. The horizontal component of the applied force balances the frictional force, while the vertical component contributes to the normal force acting on the box. This relationship allows for the calculation of uk using the equation: uk = (horizontal force) / (normal force).

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Basic knowledge of force diagrams
  • Familiarity with vector components of forces
  • Concept of friction and its coefficients
NEXT STEPS
  • Study the calculation of friction coefficients in physics problems
  • Learn about vector decomposition of forces
  • Explore Newton's second law and its applications
  • Practice drawing and analyzing force diagrams
USEFUL FOR

Students preparing for physics exams, particularly those focusing on mechanics and forces, as well as educators teaching concepts related to friction and motion.

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A box of books weighing 325N moves with a constant velocity across the floor when it is pushed with a force with 425N exerted downward at an angle of 35.2o below the horizontal. Find uk between the box and the floor.

Please help me I have a test soon and I'm not exactly understanding how to solve that. I skipped a few classes because I traveled for a volleyball tournament.
 
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physicsphail said:
A box of books weighing 325N moves with a constant velocity across the floor when it is pushed with a force with 425N exerted downward at an angle of 35.2o below the horizontal. Find uk between the box and the floor.

Please help me I have a test soon and I'm not exactly understanding how to solve that. I skipped a few classes because I traveled for a volleyball tournament.

Welcome to PF.

Draw a force diagram.

Your box exerts a downward force on the floor.

The force at 35.2° has 2 components - 1 is horizontal and the other is vertical down.

It moves at constant velocity so there is no net force horizontally. The horizontal component of the force is equal then to the weight down and the component down both multiplied by the coefficient of friction.
 

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