Please help I can't figure this one out to save my life!?

Question

A student wants to determine the coefficients of static friction and kinetic friction between a box and a plank. She places the box on the plank and gradually raises one end of the plank. When the angle of inclination with the horizontal reaches 34o, the box starts to slip, and it then slides 2.7 m down the plank in 1.7 s at constant acceleration. What are (a) the coefficient of static friction and (b) the coefficient of kinetic friction between the box and the plank?

Answer 1

To find the static coefficient, you know that at equilibrium:
down-slope gravity = mg sin theta
=
up-slope friction = mu * normal force = mu mg cos theta
Therefore:
mu = tan (theta) --that's your static

To find the kinetic coefficient:
net force = mg sin theta - mu mg cos theta
= ma

mu = (g sin theta - a) / g cos (theta)

distance fallen from rest is given by:
d = 1/2 at^2
so a = 2d / t^2

so plug that back in to get your kinetic coefficient:
mu = tan(theta) - 2d / (gt^2 cos(theta))

Answer 2

Without loss of generality, you can assume any mass you like for the box. So make it easy on yourself and choose 1 kg.
Calculate the gravitational forces on the box that are normal to the plane and parallel to the plane. (If you're already lost at this point, go back and take trig again.) Since you know the normal force, you can now calculate the coefficient of static friction since the force parallel to the plank is known and it's just enough to start the box sliding (so it's equal to the frictional force on the box)
Since you know how far it moved(2.7 m) and how long it took (1.7 s) you can calculate the acceleration. Since you now know the acceleration you can calculate the required force. Since you already know the parallel force to the plank, the difference is the frictional force. And since you know the normal force, you can now calculate the coefficient of kinetic friction.

Doug