Physics help! (forces, 10 points for best answer)?

Question

a 4 kg block is pushed along the celing with a constant applied force of 85N that acts at an angle of 55 degrees. The block accelerates to the right at 6 m/s2. determine the coefficient of the kenetic friction between the block and the ceiling.

Answer 1

I'm assuming that the applied force is 55° to the horizontal.

I recommend drawing a free body diagram for these types of problems. Doing so, you'll notice that the vertical component of the applied force points upward, force or gravity points downward, normal force points downwards, the horizontal component of applied force points to the right, and friction points to the left.

(1) The force of gravity exerted on the box,

G = mg = (4 kg)(9.81 N/kg) = 39.24 N [down]

(2) The horizontal component of the applied force,

Ax = 85cos(55°) N = 48.75 N [right]

(3) The vertical component of the applied force,

Ay = 85sin(55°) N = 69.63 N [up]

(4) Normal Force, N, and Force of Friction, F, are unknown.

Let upwards and to the right be positive.

Since the box is not moving vertically, the sum of the vertical forces exerted on the box equal zero:

Ay - G - N = 0

=> N = Ay - G

=> N = 69.63N - 39.24N = 30.39 N [down]

We know that the box is accelerating the right at 6 m/s². This means that the sum of all the horizontal forces = the mass of the box * 6 m/s². Let μ be the coefficient of kinetic friction:

Ax - F = ma

=> 48.75N - μ(30.39N)  = (4 kg)(6 m/s²)

=> -μ(30.39N) = 24N - 48.75N

=> μ = -24.75N / -30.39N

=> μ = 0.814

Therefore, the coefficient of kinetic friction is: 0.814