Question

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

Answer 1

Answer:

$\mu_k=0.18$

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:

$x: T+F-f_k=0\\\\y:N-mg=0$

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 $f_k=\mu_k N$ and $N=mg$, we can rewrite the first equation as:

$T+F-\mu_k mg=0$

Now, we solve for $\mu_k$ and calculate it:

$\mu_k=\frac{T+F}{mg}\\ \\\mu_k =\frac{220N+390N}{(350kg)(9.8m/s^{2})} =0.18$

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