Question
A 225-kg box is pushed horizontally with a force of 620N. If the coefficient of friction is 0.20, calculate the acceleration of the box.
Answers
Henry
Fb = m*g = 225kg * 9.8N/kg = 2205 N. =
Force of box. = Normal force.
Fk = u*Fn = 0.20 * 2205 = 441 N. = Force of kinetic friction.
a=(Fap-Fk)/m = (2205-441)/225=7.84 m/s^2
Force of box. = Normal force.
Fk = u*Fn = 0.20 * 2205 = 441 N. = Force of kinetic friction.
a=(Fap-Fk)/m = (2205-441)/225=7.84 m/s^2