Asked by Rose

A 6.0 kg mass is pulled horizontally along a level surface where the coefficient of friction is
0.20. The mass accelerates at 0.68 m/s2.
a) What is the friction force acting on the mass?
b) What applied force is being used to pull the mass?
Answered by Henry
M*g = 6 * 9.8 = 58.8 N. = Wt. of Mass. = Normal force(Fn).

a. Fk = u*Fn = 0.20 * 58.8 = 11.8 N. - Force of kinetic friction.

b. Fap-Fk = M*a.
Fap = ?.


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