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Two workers are sliding 270 kg k g crate across the floor. One worker pushes forward on the crate with a force of 430 N N while the other pulls in the same direction with a force of 230 N 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?

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okpalawalter8

Answer:

The value is [tex]\mu_ k = 0.249[/tex]

Explanation:

From the question we are told that

   The mass of the crate is  [tex]m = 270 \ kg[/tex]

   The force applied by the first worker is  [tex]F_1 = 430 \ N[/tex]

   The force applied by the second worker is [tex]F_2 = 230 \ N[/tex]

   

Generally the kinetic frictional force between the crate and the floor is equal to the force applied by both worker , this is mathematically  represented as

        [tex]F_f = F_1 + F_2[/tex]

Here  [tex]F_f[/tex] is the frictional force and this is mathematically represented as

      [tex]F_f = \mu_ k * m * g[/tex]

So

     [tex]\mu_ k * m * g = F_1 + F_2[/tex]

=>  [tex]\mu_ k * 270 * 9.8 = 430 + 230[/tex]

=>  [tex]\mu_ k = 0.249[/tex]

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