7. A block of mass m = 5.5 kg is pulled up a θ = 22° incline as in the figure with a force of magnitude F = 38 N. (a) Find the acceleration of the block if the incline is frictionless So I got this one right, and here is what I did... From the free body diagram I drew: Fnet = (38 N) + (-mg sin 22) ma = 38 - mg sin 22 a = (38 - mg sin 22)/m a = [ 38 - (5.5)(9.81)(sin 22) ] / (5.5) a = (38 - 20.211) / 5.5 a = 3.24 m/s2 (b) Find the acceleration of the block if the coefficient of kinetic friction between the block and incline is 0.12. So...I know that fk = μkn fk = 0.12(Fappliedsin θ - mg) QUESTION 1: So this is where I'm a little confused...Normal force should be UPWARDS, right? then would that mean that n = Fappliedsin θ + (- mg) after rotating my free body diagram so that I can see the axis better??? Because I almost got the answer... fk = 0.12[38 sin 22 - (5.5)(9.81)] fk = -4.767 (the negative number makes sense since kinetic friction is OPPOSING the motion of the block which is moving up the incline. So then I add it to the equation for acceleration at the top ma = (38 N) + (-mg sin 22) + (-4.767) a = [ 38 - (5.5)(9.81)(sin 22) - 4.767 ] / (5.5) a = 13.022 / 5.5 a = 2.36 m/s2 This answer is wrong...Webassign says the answer is actually a = 2.15 m/s2 QUESTION 2: Where did I go wrong? Was it the normal force equation part?