Question

Two workers are sliding 400 kg crate across the floor. One worker pushes forward on the crate with a force of 400 N while the other pulls in the same direction with a force of 380 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed.

a. What is the crate's coefficient of kinetic friction on the floor?

Answer 1

The crate's coefficient of kinetic friction on the floor is 0.195.

Step-by-step explanation:

Given that,

Mass of crate, m = 400 kg

Force applied by the first worker,
`F_1=400\ N`

Force applied by the second worker,
`F_2=380\ N`

Both forces are horizontal, and the crate slides with a constant speed. It means that the acceleration is equal to 0. As a result, net force is equal to 0. So,

`F_1+F_2-f=0`

f is frictional force

`f=\mu mg`

Here,

`\mu` is the crate's coefficient of kinetic friction on the floor

`F_1+F_2-\mu mg=0\\\\\mu=(F_1+F_2)/(mg)\\\\\mu=(400+380)/(400* 10)\\\\\mu=0.195`

So, the crate's coefficient of kinetic friction on the floor is 0.195.