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3 years ago

7

Two workers are sliding 300 kg crate across the floor. One worker pushes forward on the crate with a force of 390 N while the ot

her pulls in the same direction with a force of 320 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

1 answer:

romanna [79]3 years ago

6 0

Answer:

The crate's coefficient of kinetic friction on the floor is 0.23.

Explanation:

Given that,

Mass of the crate, m = 300 kg

One worker pushes forward on the crate with a force of 390 N while the other pulls in the same direction with a force of 320 N using a rope connected to the crate.

The crate slides with a constant speed. It means that the net force acting on it is 0. Net force acting on it is given by :

$F_1+F_2=\mu N\\\\\mu=\dfrac{F_1+F_2}{mg}\\\\\mu=\dfrac{390+320}{300\times 10}\\\\\mu=0.23$

So, the crate's coefficient of kinetic friction on the floor is 0.23.

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