A block of mass 6 kg is accelerating at 1.25 m/s ,on the horizontal surface, under the action of a horizontal force of magnitude 22.5 N . Calculate the coefficient of friction between the block and the surface.

Answer 1

Wb = M*g = 6 * 9.8 = 58.8 N. = Wt. of block = Normal(Fn).

Fap-Fk = M*a, 22.5-Fk = 6*1.25 = 7.5, Fk = 15 N. = Force of kinetic friction.

u = Fk/Fn = 15/58.8 = 0.255.