Two workers are sliding 490 kg crate across the floor. One worker pushes forward on the crate with a force of 410 N while the ot her pulls in the same direction with a force of 260 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?
2 answers:
Answer:
μ=0.16
Explanation:
Mass of crate = 490 kg
Pushing force = 410 N
Pulling force = 260 N
Lets coefficient of friction =μ
The normal force,N= mg
N= 490 x 9.81 =4806.9 N
We know that friction force(fr)
fr= μ m g
fr =4806.9 μ
The total horizontal force F
F= 410 + 260
F=670 N
For sliding condition the maximum friction force will be equal to total horizontal force
F=fr
670 = 4806.9 μ
μ=0.16
So the coefficient of friction is 0.16.
Answer:
Explanation:
There are two forces acting on the box to slide it over the horizontal surface
1) in forward direction
2) in forward direction
now it is given that box is moving with constant speed so net force on the box must be zero
so we will have
now by the formula of friction force we know that
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time for the change = 2 minutes = 120 seconds
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