Answer:
The kinetic coefficient of friction between the box and the floor is 1.020.
Explanation:
Let suppose that box is on horizontal ground. According to the Newton's Laws, an object has a net acceleration of zero when either is at rest or moves at constant velocity. Friction is a reaction to the external force that moves the box. Hence, the equation of equilibrium for the 20-Kg box is:
[tex]\Sigma F = F-f = 0[/tex] (Eq. 1)
Where:
[tex]F[/tex] - External force, measured in newtons.
[tex]f[/tex] - Friction force, measured in newtons.
If we know that [tex]F = 200\,N[/tex], then the magnitude of the kinetic friction force is:
[tex]f = F[/tex]
[tex]f = 200\,N[/tex]
In addition, friction force is represented by the following formula:
[tex]f = \mu_{k}\cdot m\cdot g[/tex] (Eq. 2)
Where:
[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, dimensionless.
[tex]m[/tex] - Mass, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
Now we clear the kinetic coefficient of friction:
[tex]\mu_{k} = \frac{f}{m\cdot g}[/tex]
If we know that [tex]f = 200\,N[/tex], [tex]m = 20\,kg[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the kinetic coefficient of friction is:
[tex]\mu_{k} = \frac{200\,N}{(20\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]
[tex]\mu_{k} = 1.020[/tex]
The kinetic coefficient of friction between the box and the floor is 1.020.
