Discover the Coefficient of Kinetic Friction for a 325N Box on the Floor

AI Thread Summary
The discussion focuses on calculating the coefficient of kinetic friction for a 325N box being pushed with a 425N force at a 35.2-degree angle. The coefficient is determined using the formula that divides the frictional force by the normal force. The initial calculation mistakenly included the full 425N in the numerator instead of just the horizontal component. After clarification, the correct frictional force is identified as 425N multiplied by the cosine of 35.2 degrees. The final coefficient of kinetic friction is confirmed to be approximately 0.136.
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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 of 425N exerted downward at an angle of 35.2 degree below the horizontal . find the coefficient of kinetic friction between the box and the floor.

what i do :please see check it

the coefficientof kinetic friction equal to the force of kinetic divided by the normal force = 425-425xcos35.2 / 325+425xsin35.2 = 0.136
 
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Yes. You are right.
 
are u sure ?
 
logglypop said:
a box of books weighing 325N moves with a constant velocity across the floor when it is pushed with a force of 425N exerted downward at an angle of 35.2 degree below the horizontal . find the coefficient of kinetic friction between the box and the floor.

what i do :please see check it

the coefficientof kinetic friction equal to the force of kinetic divided by the normal force = 425-425xcos35.2 / 325+425xsin35.2 = 0.136

Where does the first 425 come from on the numerator? What much the frictional force be equal to?
 
Yes.You are right,Cristo. Only 425cos35.2. Sorry.
 

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