quick homework question?

Question

A dockworker loading crates on a ship finds that a 20kg crate, initially at rest on a horizontal surface, requires a 75N horizontal force to set it in motion. However, after the crate is in motion, a horizontal force of 60N is required to keep it moving with a constant speed. find the coefficients of static and kinetic friction between crate and floor

Answer 1

Since it requires 75N horizontal force to set it in motion. therefore, terminal static friction Fs = 75N

The reaction force of the surface on the crate = weight of the crate = (20 x 9.8)N =196N
ms = coefficient of static friction between crate and floor

now, Fs = ms x R
so,ms = Fs / R = 75 / 196 = 0.3826

Since 60N horizontal force is required to keep it moving in const speed so the net force acting on the body is zero.

therefore, kinetic friction,Fk = applied force =60N

mk = coefficient of kinetic friction between crate and floor

Fk = mk x R
mk = Fk / R = 60 / 196 = 0.3061

Answer 2

Coefficient of static friction = 75 / 20g = 3.75/g
Coefficient of kinetic friction = 60/20g = 3/g

If we take g as 10m/sec^2 we can write coefficient of static friction = 0.375 and coefficient of kinetic friction = 0.3