# Constant velocity and the coefficient of kinetic friction

"A child rides a wagon down a street with an inclination angle of 38.0 degrees. In order to keep from moving to fast, the child has secured the wheels of the wagon so that they do not turn. The wagon and the child then slide down the hill at constant velocity. What is the coefficient of kinetic friction between the tires of the wagon and the pavement?"

I've thought about this for awhile. I had thought that the coefficient of kinetic friction would equal the coefficient of static friction (because the net force at constant velocity and at rest equals 0), but now I'm not sure. I'm fairly certain that the coefficient of kinetic friction can't equal 0 (there would be no friction at all then), but beyond that I'm not sure. Any ideas?

Doc Al
Mentor
What forces act on the "child + wagon"? Since it moves at constant velocity, what must the net force be in any direction?

Well, the force of kinetic friction and the force that causes the motion would sum to 0 N (the two forces would be equal in magnitude but opposite in direction). The only forces that act on the child and the wagon would be the weight, friction, and applied force...unless you include the normal force, which is the component of the weight perpendicular to the incline. What I don't understand, though, is how to find the coefficient of kinetic friction without a given mass or applied force.

Doc Al
Mentor
There's no applied force--the kid's just sliding down the incline. (Nobody is pushing or pulling, just gravity.) Now analyze the forces acting parallel to the incline.

Mmm...uk (coefficient of kinetic friction) = friction/normal force. Friction = weight(sin 38) and normal force = weight(cos 38). Reducing this fraction to lowest terms would produce sin/cos, which = tan...uk = tan 38 = 0.781. Would this be correct? Good job!! Looks good.