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 25°, the box starts to slip, and it slides 2.6 m down the plank in 3.6 s at constant acceleration.

Answer 1

The angle A at which slipping starts can be used to determine the coefficient of static friction, u_s.

M*g*sinA = u_s*cosA*M*g
M and g cancel, leaving
u_s = tanA = 0.466

The rate at which it accelerates after that can be used to determine the coefficient of kinetic friction, u_k.

In your case, the accleration rate is given by
X = (1/2) a t^2
a = 2X/t^2 = 0.401 m/s^2
F' = M a can be used get the kinetic friction coefficent u_k, since F' is the component of the weight in the direction of motion, MINUS the friction force.

M*a = M*g*sinA - M*g*cosA*u_k

u_k = tanA - (a/g)*secA
= 0.466 - 0.045 = 0.421