A dockworker loading crates on a ship finds that a 22-kg crate, initially at rest on a horizontal surface, requires a 80-N horizontal force to set it in motion. However, after the crate is in motion, a horizontal force of 59 N is required to keep it moving with a constant speed. Find the coefficients of static and kinetic friction between crate and floor.

To find the coefficients of static and kinetic friction between the crate and the floor, we need to analyze the forces acting on the crate.

Let's denote the coefficient of static friction as μs and the coefficient of kinetic friction as μk.

1. When the crate is at rest:
In this case, the applied force of 80 N is equal to the maximum force of static friction that prevents the crate from moving. Therefore, we can write the equation:
Static friction force (Fs) = Applied force (Fapplied)
μs * Normal force (N) = Fapplied
μs * m * g = Fapplied (where m is the mass of the crate and g is the acceleration due to gravity)

2. When the crate is in motion:
Once the crate is set in motion, the force required to keep it moving with a constant speed is the force of kinetic friction. Therefore, we can write the equation:
Kinetic friction force (Fk) = Applied force (Fapplied)
μk * Normal force (N) = Fapplied
μk * m * g = Fapplied

Now, let's solve these equations to find the values of μs and μk.

Given:
Mass of crate (m) = 22 kg
Force required to set the crate in motion (Fapplied) = 80 N
Force required to keep the crate moving (Fapplied) = 59 N

First, we'll find the normal force:
Normal force (N) = m * g
Normal force (N) = 22 kg * 9.8 m/s^2
Normal force (N) = 215.6 N

1. Coefficient of static friction (μs):
μs * N = Fapplied
μs = Fapplied / N
μs = 80 N / 215.6 N
μs = 0.37

2. Coefficient of kinetic friction (μk):
μk * N = Fapplied
μk = Fapplied / N
μk = 59 N / 215.6 N
μk = 0.27

Therefore, the coefficients of static and kinetic friction between the crate and the floor are 0.37 and 0.27, respectively.