A 5.0-kg box is pulled horizontally with a 45-N force that makes angle of 40 degree with the horizontal. The coefficient of kinetic friction of the surface is 0.20. What is the magnitude of the acceleration of the box?

Answer 1

M*g = 5 * 9.8 = 49 N. = Wt. of box.

Fn = 49 - 45*sin40 = 20.1 N. = Normal
force.

Fk = u*Fn = 0.2 * 20.1 = 4.01 N. = Force
of kinetic friction.

a = (Fx-Fk)/m = (45*Cos40-4.01)/5 = 6.1
m/s^2