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?

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,

Force applied by the second worker,

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 is frictional force

Here,

is the crate's coefficient of kinetic friction on the floor

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

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west

it's a b and