A clerk moves a box of cans down an aisle by pulling on a strap attached to the box. The clerk pulls with a force of 173.0 N at an angle of 31° with the horizontal. The box has a mass of 40 kg, and the coefficient of kinetic friction between the box and the floor is 0.45. Find the acceleration of the box.

Answer 1

Wb = m*g = 40kg * 9.8N/kg = 392 N. =
Weight of box.

Fk=u*(Wb-F*sin31)=0.45(392-173*sin31) =
136.3 N.= Force of kinetic friction.

Fn=F*CosA-Fk = 173*cos31 - 136.3 = 12 N.
= Net force.

a = Fn/m = 12/40 = 0.30 m/s^2.