Two workers are sliding 330 kg crate across the floor. One worker pushes forward on the crate with a force of 410 N while the other pulls in the same direction with a force of 240 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor?

Answer 1

Wc = M*g = 330 * 9.8 = 3234 N. = Normal force, Fn.

Fap = 410 + 240 = 650 N. = Force applied.

Fap-Fk = M*a.
Fap-Fk = M*0 = 0.
Fk = Fap = 650 N. = Force of kinetic friction.

u = Fk/Fn.

Answer 2

I need answers of find the magnitude of normal force acting on the skateboard on 10 kg àt point A and 10 m at point B