# A box of books weighing 325 N moves at a constant velocity across the floor when the box is pushed with a force of 425 N exerted downward at an angle of 35.2 degrees below the horizontal. Fink Mu,k between the box and the floor.

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Each can be solved by writing down Newton's second law. The force "F" in the law F= m a is the net force. In this case, the acceleration a is zero. Therefore the applied force component along the direction of motion equals the friction force. Write that statement in algebraic form and solve for Mu,k.

1. 427

2. 1. The 425 N force needs to be broken into component vectors. 425*sin35.2 will give you the vertical component, 425*cos35.2 will give you the horizontal component.

2. The vertical component gets added to the box's 325 N weight because it is being pushed down on.

3. The horizontal component is equaled by the force of friction backward. Divide this force by the force from step 2 to get the coefficient of friction, which is about 0.61.

3. 20

4. But how did you choose the angle?

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