Two workers are sliding 450 kg kg crate across the floor. One worker pushes forward on the crate with a force of 380 NN while the other pulls in the same direction with a force of 230 NN 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

The coefficient of kinetic friction on the floor is 0.138

Explanation:

Given;

mass of the crate, m = 450 kg

force applied by the first worker, F₁ = 380 N

force applied by the second worker in the same direction as the first worker, F₁ = 230 N

frictional force opposing the motion of the box = -

Apply Newton's second law of motion;

∑F = ma

If the crate slides with constant speed, acceleration is zero (0).

Therefore, the coefficient of kinetic friction on the floor is 0.138

true tan speed = ang vel x radius

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer