Asked by omar

Two workers are sliding 310kg crate across the floor. One worker pushes forward on the crate with a force of 440N while the other pulls in the same direction with a force of 230N 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?
Answered by Henry
M*g = 310kg * 9.8N/kg = 3,038 N. = Wt. of the crate = Normal force(Fn).

Fap-Fk = M*a
(440+230)-Fk = M*0 = 0
670 - Fk = 0
Fk = 670 N. = Force of kinetic friction

uk = Fk/Fn = 670/3038 = 0.221

Answered by mat
.221
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