Coefficient of kinetic friction for 10 kg box on 10° slope

[physlexic]

Homework Statement

Problem:
A 10 kg box slides down a hill and has a constant velocity. The angle of the hill is 10° with respect to the horizontal. What is the coefficient of kinetic friction between the surface of the hill and the box?

Homework Equations

Fn = mg + ma
ƒk = μk * Fn

The Attempt at a Solution

To calculate the coefficient of kinetic friction I used the formula ƒk = μk Fn and isolated μk to be
μk = ƒk / Fn
Where ƒk is kinetic frictional force, μk is the coefficient of kinetic friction, and Fn is the normal force.

I don't have any of these variables, so I drew a free-body diagram and tried to find my normal force by doing:
∑Fy = Fn + mg cos 10° = 0
Fn = mg cos 10°
Fn = (10 kg) (9.80 m/s^2) cos 10° = 96.5 NMy problem is I am not sure on how to determine my kinetic fricitonal force ƒk to plug it into the
μk = ƒk / Fn. Someone please guide me? [/B]

 

[NascentOxygen]

What you can say is that the friction force must exactly balance the opposing force due to that body's weight.

Q: why can you say this?

 

[physlexic]

Newton's third law that states for every action, there is an equal and opposite reaction ?

Therefore does fk = Fn? I'm sorry if this isn't correct, I struggle a lot with this course...

 

[NascentOxygen]

Therefore does fk = Fn?

No, those two forces are not in the same direction. f_k opposes a force acting in the exact opposite direction to f_k. If f_k were not exactly cancelling its opposing force, then the body would be accelerating or decelerating, but we are told it is sliding with a fixed speed.

You draw the diagram and show how the body's weight can be broken into orthogonal components, one along the slope, and the other normal to the slope.