A 3.60 kg block is pushed along the ceiling with an constant applied force of F = 75.5 N that acts at an angle θ = 61° with the horizontal, as in the figure below. The block accelerates to the right at 5.76 m/s2. Determine the coefficient of kinetic friction between the block and the ceiling.

Answer 1

First get the net force from the acceleration.

Fnet = M*a = 3.6*5.76 = 20.74 N

The net force is the applied horizongtal force component MINUS the friction force.

20.74 = 75.5 cos61 - (75.5sin61-Mg)*U
= 36.60 - (66.03 - 35.28)*U
= 36.60 - 30.75U

Solve for the friction coefficient, U