A dock worker loading crates on a ship

finds that a 18 kg crate, initially at rest on a horizontal surface, requires a 71 N horizontal force to set it in motion. However, after the crate is in motion, a horizontal force of 46 N is required to keep it moving with a constant speed. The acceleration of gravity is 9.8 m/s
2.
Find the coefficient of static friction between crate and floor.

2 answers

Answering my own question here, but...
µs (static friction)
Ffs (peak frictional force)
Fn (Normal force -needs to be in Newtons-)
Ffs = µs * Fn
71 = µs * (18 * 9.8)
71 = µs * 176.4
71 / 176.4 = µs * 176.4 / 176.4 (isolates the variable µs)
.4024943311 = µs

replaces 71 with 46 to find µk ( kinetic friction)
Answering my own question here, but...
µs (static friction)
Ffs (peak frictional force)
Fn (Normal force -needs to be in Newtons-)
Ffs = µs * Fn
71 = µs * (18 * 9.8)
71 = µs * 176.4
71 / 176.4 = µs * 176.4 / 176.4 (isolates the variable µs)
.4024943311 = µs

replace 71 with 46 to find µk ( kinetic friction)