Asked by Anonymous

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?

Answers

Answered by Henry
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

There are no AI answers yet. The ability to request AI answers is coming soon!

Related Questions