A 80 N force is needed to slide a 50.0 kg box across a flat surface at a constant velocity. What is the coefficient of kinetic friction between the box and the floor

0.16

Explanation:

Given data

Force F= 80N

Mass m= 50kg

Reaction R= Weight= mg= 50*9.81= 490.5N

We know that

F=UR

Substitute and solve of U

U=F/R

U= 80/490.5

U=0.16

Hence the coefficient of friction is 0.16

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2