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

AI Thread Summary
To find the coefficient of kinetic friction (uk) for a box weighing 325N pushed with a 425N force at a 35.2° angle, a force diagram is essential. The force has horizontal and vertical components, with the box moving at constant velocity indicating no net horizontal force. The horizontal component of the applied force equals the frictional force, which is the product of uk and the normal force. The vertical component of the applied force increases the normal force acting on the box. Solving these equations will yield the value of uk between the box and the floor.
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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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