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

Answer:

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

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

$F_1+F_2-\mu mg=0\\\\\mu=\dfrac{F_1+F_2}{mg}\\\\\mu=\dfrac{400+380}{400\times 10}\\\\\mu=0.195$

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