Question
A worker pulls a 200-N packing crate at constant velocity across a rough floor by exerting a force F = 55.0 N at an angle of 35.0 degrees above the horizontal. What is the coefficient of kinetic friction of the floor?
Answers
Wc = m * g = 200 N.
m * 9.8 = 200
m = 20.41 kg. = Mass of crate.
F = 55N.[35o.]
55*cos35 - Fk = m*a
45.05 - Fk = m*0 = 0
Fk = 45.05 = Force of kinetic friction.
u = Fk/Wc = 45.05/200 = 0.225
m * 9.8 = 200
m = 20.41 kg. = Mass of crate.
F = 55N.[35o.]
55*cos35 - Fk = m*a
45.05 - Fk = m*0 = 0
Fk = 45.05 = Force of kinetic friction.
u = Fk/Wc = 45.05/200 = 0.225
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