The coefficient of kinetic friction between the box and the floor is 0.323.
By Newton's Law of Motion we understand that an object is at equilibrium if and only if it is at rest or it is moving at constant velocity. If a horizontal force is applied on the box and it is moving, then a force with equal magnitude and opposed to that force must exists, which corresponds to the kinetic friction. Let suppose that the box is moving on a horizontal ground.
The equation of equilibrium for the box is described below:
[tex]\Sigma F = P - \mu_{k}\cdot m\cdot g = 0[/tex] (1)
Where:
- [tex]P[/tex] - External force, in newtons.
- [tex]\mu_{k}[/tex] - Kinetic coefficient of friction, no unit.
- [tex]m[/tex] - Mass, in kilograms.
- [tex]g[/tex] - Gravitational constant, in meters per square second.
An expression for the kinetic coefficient of friction is derived by clearing the variable in (1):
[tex]\mu_{k} = \frac{P}{m\cdot g}[/tex] (2)
If we know that [tex]P = 38\,N[/tex], [tex]m = 12\,kg[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the kinetic coefficient of friction is:
[tex]\mu_{k} = \frac{38\,N}{(12\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]
[tex]\mu_{k} = 0.323[/tex]
The coefficient of kinetic friction between the box and the floor is 0.323.
We kindly invite to see this question on friction: https://brainly.com/question/18332986
