A 3.40 kg block is pushed along the ceiling with a constant applied force of 85.0 N that acts at an angle of 55.0° with the horizontal, as in the figure below. The block accelerates to the right at 6.00 m/s2. Determine the coefficient of kinetic friction between block and ceiling. Ugghhh so... this is what I did... F = 85.0 N m = 3.40 kg a = 6.00 m/s2 fk = μkn 1) Forces in the x-direction: F cos θ fk (kinetic friction) Forces in the y-direction: F sin θ n (normal force) 2) Sum of forces ƩFx = F cos θ + fk = 0 ƩFy = F sin θ + n = 0 ƩFy = F sin θ + n = 0 ƩFy = n = -F sin θ 3) ƩFx = F cos θ + μkn = 0 3) ƩFx = F cos θ + μk(-F sin θ) = 0 3) ƩFx = F cos θ + μk(-F sin θ) = 0 3) ƩFx = 85 cos 55 - μk(85 sin 55) = 0 3) ƩFx = 85 cos 55 = μk(85 sin 55) 3) ƩFx = (85 cos 55)/(85 sin 55) = μk 3) ƩFx = 0.700 = μk So I got this wrong, the answer is actually 0.781 (???) I don't know what to do!