Answer:
The kinetic coefficient of friction between box and floor is 0.443.
Explanation:
Let suppose that the box is accelerated uniformly, the acceleration ([tex]a[/tex]), measured in meters per square second, is determined by the following kinematic formula:
[tex]a = \frac{v_{f}-v_{o}}{t}[/tex] (1)
Where:
[tex]v_{o}[/tex], [tex]v_{f}[/tex] - Initial and final speeds of the box, measured in meters per second.
[tex]t[/tex] - Time, measured in seconds.
If we know that [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]v_{f} = 2\,\frac{m}{s}[/tex] and [tex]t = 4\,s[/tex], then the acceleration experimented by the box is:
[tex]a = \frac{2\,\frac{m}{s}-0\,\frac{m}{s} }{4\,s}[/tex]
[tex]a = 0.5\,\frac{m}{s^{2}}[/tex]
Based on the body diagram, we proceed to form the equations of equilibrium:
[tex]\Sigma F_{x} = P\cdot \cos \theta -\mu_{k}\cdot N = m\cdot a[/tex] (1)
[tex]\Sigma F_{y} = N - m\cdot g - P\cdot \sin \theta = 0[/tex] (2)
Where:
[tex]P[/tex] - Magnitude of the force exerted on the box, measured in newtons.
[tex]\theta[/tex] - Direction of the force exerted on the box, measured in sexagesimal degrees.
[tex]\mu_{k}[/tex] - Kinetic coefficient of friction between box and floor, dimensionless.
[tex]N[/tex] - Normal force from the floor to the box, measured in newtons.
[tex]m[/tex] - Mass of the block, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
Now we proceed to solve the system of equation for the kinetic coefficient of friction between box and floor:
By (1)
[tex]\mu_{k}\cdot N = P\cdot \cos \theta -m\cdot a[/tex]
[tex]\mu_{k} = \frac{P\cdot \cos \theta - m\cdot a}{N}[/tex]
By (2)
[tex]N = m\cdot g +P\cdot \sin \theta[/tex]
And then, we get the resulting expression:
[tex]\mu_{k} = \frac{P\cdot \cos \theta - m\cdot a}{m\cdot g + P\cdot \sin \theta}[/tex] (3)
If we know that [tex]P = 400\,N[/tex], [tex]\theta = 50^{\circ}[/tex], [tex]m = 25\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]a = 0.5\,\frac{m}{s^{2}}[/tex], then the kinetic coefficient of friction between box and floor is:
[tex]\mu_{k} = \frac{(400\,N)\cdot \cos 50^{\circ}-(25\,kg)\cdot \left(0.5\,\frac{m}{s^{2}} \right)}{(25\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)+(400\,N)\cdot \sin 50^{\circ}}[/tex]
[tex]\mu_{k} = 0.443[/tex]
The kinetic coefficient of friction between box and floor is 0.443.
