Question

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?

Answer 1

The value is
`\mu_ k = 0.249`

Step-by-step explanation:

From the question we are told that

The mass of the crate is
`m = 270 \ kg`

The force applied by the first worker is
`F_1 = 430 \ N`

The force applied by the second worker is
`F_2 = 230 \ N`

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

`F_f = F_1 + F_2`

Here
`F_f` is the frictional force and this is mathematically represented as

`F_f = \mu_ k * m * g`

So

`\mu_ k * m * g = F_1 + F_2`

=>
`\mu_ k * 270 * 9.8 = 430 + 230`

=>
`\mu_ k = 0.249`