A 0.5 kg block is being slid up a chalkboard with an applied force of 6.0 N upward and 2.0 N inward towards the board. If the coefficient of kinetic friction is 0.4. a) Calculate the force of friction. b) Calculate the acceleration of the block. F = mju R [ F = u R] 1. F/u = 6/0.4 = 15N ??? 2. F = ma ..... 6/0.5 = a = 12m/s^2 not sure... some1 back me up on this!