In moving a 35.0 kg desk from one side of a classroom to the other, a professor finds that a horizontal force of 300 N is necessary to set the desk in motion, and a force of 185 N is necessary to keep it in motion at a constant speed.

(a) What is the coefficient of static friction between the desk and the floor?
(b) What is the coefficient of kinetic friction between the desk and the floor?

Could someone please help me? I'm totally lost...I know that
Fsmax = us(mg)...

Nevermind everyone, I've got it.

Of course, I can help you with that! To find the coefficient of static friction (μs) between the desk and the floor, we can use the equation Fsmax = μs(mg), where Fsmax is the maximum static frictional force, μs is the coefficient of static friction, m is the mass of the desk, and g is the acceleration due to gravity.

(a) To find μs, we need to determine the maximum force of static friction. We know that the force required to set the desk in motion is 300 N, so we can say that Fsmax = 300 N. The mass of the desk is given as 35.0 kg, and the acceleration due to gravity is approximately 9.8 m/s^2.

So, we can rearrange the equation as follows:
300 N = μs(35.0 kg)(9.8 m/s^2)

Now, divide both sides of the equation by (35.0 kg)(9.8 m/s^2) to solve for μs:
μs = 300 N / (35.0 kg)(9.8 m/s^2)

Calculating the right-hand side of the equation gives us:
μs ≈ 0.857

Therefore, the coefficient of static friction between the desk and the floor is approximately 0.857.

(b) To find the coefficient of kinetic friction (μk) between the desk and the floor, we can use the equation Fk = μk(mg), where Fk is the force of kinetic friction.

In this case, we are given that a force of 185 N is necessary to keep the desk in motion at a constant speed. So, we can say that Fk = 185 N.

Using the same values for m and g as before, we can rearrange the equation to solve for μk:
185 N = μk(35.0 kg)(9.8 m/s^2)

Divide both sides of the equation by (35.0 kg)(9.8 m/s^2) to isolate μk:
μk = 185 N / (35.0 kg)(9.8 m/s^2)

Calculating the right-hand side of the equation gives us:
μk ≈ 0.529

Therefore, the coefficient of kinetic friction between the desk and the floor is approximately 0.529.

I hope this helps! Let me know if you have any further questions.