If you use a horizontal force of 31.0 N to slide a 10.5 kg wooden crate across a floor at a constant velocity, what is the coefficient of kinetic friction between the crate and the floor?

I tried 3.32 but it said that was wrong ?

To find the coefficient of kinetic friction between the crate and the floor, we need to use the formula:

Coefficient of kinetic friction (μk) = Force of kinetic friction (Fk) / Normal force (N)

First, let's calculate the force of kinetic friction acting on the crate.

Since the crate is moving at a constant velocity, we know that the force applied (horizontal force) must be equal and opposite to the force of kinetic friction.

Therefore, the force of kinetic friction (Fk) is 31.0 N.

Next, we need to find the normal force (N), which is the perpendicular force exerted by the floor on the crate.

The normal force is given by:

Normal force (N) = Mass (m) x Acceleration due to gravity (g)

The mass of the crate is given as 10.5 kg, and the acceleration due to gravity is approximately 9.8 m/s^2.

So, the normal force (N) is 10.5 kg x 9.8 m/s^2 = 102.9 N.

Now we can calculate the coefficient of kinetic friction:

Coefficient of kinetic friction (μk) = 31.0 N / 102.9 N

μk ≈ 0.301

Therefore, the coefficient of kinetic friction between the crate and the floor is approximately 0.301.

It seems like you tried a different value of 3.32, which might have been a mistake. Please make sure to follow the calculations step by step to get the correct answer.