Two workers are sliding 470 kg crate across the floor. one worker pushes forward on the crate with a force of 450 n while the other pulls in the same direction with a force of 330 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?

0.169

Explanation:

There are three forces acting on the crate along the horizontal direction:

- The pushing force of the first worker, F1 = 450 N forward

- The pushing force of the second worker, F2 = 330 N forward

- The frictional force acting backward

The crate slides with constant speed, so its acceleration is zero: a = 0. This means that we can write Newton's second law as

The frictional force can be rewritten as

where

is the coefficient of kinetic friction

m = 470 kg is the mass of the crate

g = 9.8 m/s^2 is the acceleration due to gravity

Substituting everything into the previous equation, we find: