Two workers are sliding 350 kgkg crate across the floor. One worker pushes forward on the crate with a force of 390 NN while the other pulls in the same direction with a force of 220 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

Explanation:

First, we write the equations of motion for each axis. Since the crate is sliding with constant speed, its acceleration is zero. Then, we have:

Where T is the tension in the rope, F is the force exerted by the first worker, f_k is the frictional force, N is the normal force and mg is the weight of the crate.

Since and , we can rewrite the first equation as:

Now, we solve for and calculate it:

This means that the crate's coefficient of kinetic friction on the floor is 0.18.

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2

i already answered it by 34/s