Normal Force and Coefficient of kinetic friction (Quiz 4, q15)

[gcombina]

A 250-N force is directed horizontally as shown to push a 29-kg box up an inclined plane at a constant speed. Determine the magnitude of the normal force, FN, and the coefficient of kinetic friction, µk.
FN µk
(a) 330 N 0.31
(b) 310 N 0.33
(c) 250 N 0.27
(d) 290 N 0.30
(e) 370 N 0.26

My attempt:

I know FN = mg
so FN = (29kg) (9.8 m/s^2)
FN = 284

why is it not right?

 

[squelch]

From the diagram and equations you've provided, it doesn't appear that you've accounted for the angle of the incline. Try to account for that.

 

[gcombina]

ok FN = Mg
= 250N sin 27
= 113

still not the right answer

 

[haruspex]

From the diagram and equations you've provided, it doesn't appear that you've accounted for the angle of the incline. Try to account for that.

That's not the only omission.
gcombina, choose two directions, e.g. vertical and horizontal. In each direction, list all the forces that act in that direction. What two equations can you write?
(if you prefer, you can choose parallel to the plane and normal to the plane, or horizontal and normal to the plane, whatever.)

 

[HalfLostAndSearching]

Your calculation for the normal force is correct. The correct answer for the coefficient of kinetic friction can be found by using the formula Ff = µk * FN, where Ff is the force of kinetic friction. Since the box is moving at a constant speed, the force of kinetic friction must be equal and opposite to the horizontal force of 250 N. Therefore, 250 N = µk * 284 N. Solving for µk, we get µk = 0.88. This is not one of the answer choices given, so it is possible that there is a typo in the question or answer choices. Alternatively, it is also possible that the question is asking for the coefficient of static friction instead of kinetic friction, in which case µk would not be applicable. Without more information, it is difficult to determine the correct answer.