A dockworker loading crates on a ship finds that a 19-kg crate, initially at rest on a horizontal surface, requires a 83-N horizontal force to set it in motion. However, after the crate is in motion, a horizontal force of 53 N is required to keep it moving with a constant speed. Find the coefficients of static and kinetic friction between crate and floor.

Wc = m*g = 19kg * 9.8N/kg = 186.2 N. =

Weight of crate.

Fap-Fs = m*a
83-u*mg = m*0
83-u*186.2 = 0
186.2u = 83
us = 0.446

uk*186.2 = 53
uk = 0.285

To find the coefficients of static and kinetic friction between the crate and the floor, we can use the given information about the applied forces.

1. First, let's calculate the coefficient of static friction (μs).

The applied force required to set the crate in motion is the maximum force of static friction. We can set up the equation:

F_applied = μs * N

Where F_applied is the applied force, μs is the coefficient of static friction, and N is the normal force (equal to the weight of the crate).

The weight of the crate can be calculated as:

Weight = mass * gravity

Given that the mass of the crate is 19 kg and the acceleration due to gravity is approximately 9.8 m/s^2:

Weight = 19 kg * 9.8 m/s^2 = 186.2 N

Substituting the values into the equation, we have:

83 N = μs * 186.2 N

Solving for μs:

μs = 83 N / 186.2 N ≈ 0.446

So, the coefficient of static friction between the crate and the floor is approximately 0.446.

2. Now, let's calculate the coefficient of kinetic friction (μk).

The applied force required to keep the crate moving at a constant speed is equal to the force of kinetic friction. Using the same equation as before:

F_applied = μk * N

Substituting the known values:

53 N = μk * 186.2 N

Solving for μk:

μk = 53 N / 186.2 N ≈ 0.284

Therefore, the coefficient of kinetic friction between the crate and the floor is approximately 0.284.