Asked by Meet
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.
Answers
Answered by
Henry
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.
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.
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