Coefficient of kinetic friction for 325N box pushed at 35.2°

[logglypop]

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

 

[rl.bhat]

Yes. You are right.

 

[logglypop]

are u sure ?

 

[cristo]

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?

 

[rl.bhat]

Yes.You are right,Cristo. Only 425cos35.2. Sorry.