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Two workers are sliding 290 kg crate across the floor. One worker pushes forward on the crate with a force of 430 N while the other pulls in the same direction with a force of 360 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?

User BTMPL
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1 Answer

6 votes

Answer:0.27

Step-by-step explanation:

Given

One worker Pushes with force
F_1=430 N

other Pulls it with a rope of rope
F_2=360 N

mass of crate
m=290 kg

both forces are horizontal and crate slides with a constant speed

Both forces are in the same direction so Friction will oppose the forces and will be equal in magnitude of sum of two forces because crate is moving with constant speed i.e. net force is zero on it


f_r=F_1+F_2

where
f_r is the friction force


f_r=430+360


f_r=790 N


f_r=\mu N

where
\muis the coefficient of static friction


N=mg


790=\mu 29\cdot 9.8


\mu =0.27

User Jishan
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