A 15.5-kg sled is being pulled across a horizontal surface at a constant velocity. The pulling force has a magnitude of 76.7 N and is directed at an angle of 27 ° above the horizontal. Determine the coefficient of kinetic friction.

Answer 1

M*g = 15.5 * 9.8 = 151.9 N.=Wt. 0f sled.

Fn = 151.9 - 76.7*sin27 = 117 N.

Fap-Fk = M*a.
76.7*Cos27-Fk = M*0 = 0.
Fk = 68.3 N.=Force of kinetic friction.

u = Fk/Fn 68.3/117 = 0.584.