Answer:
The length of the rough patch is 4.345 meters.
Explanation:
According to the Work-Energy Theorem, change in kinetic energy is equal to the dissipated work due to friction. That is:
[tex]K_{1} = K_{2} + W_{loss}[/tex]
Where:
[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final kinetic energy, measured in joules.
[tex]W_{loss}[/tex] - Work losses due to friction.
By applying definitions of kinetic energy and work, the expression described above is expanded:
[tex]\frac{1}{2}\cdot m \cdot v_{1}^{2} = \frac{1}{2}\cdot m \cdot v_{2}^{2} + f \cdot \Delta s[/tex]
Where:
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speed of the skater, measured in meters per second.
[tex]m[/tex] - Mass of the skater, measured in kilograms.
[tex]\Delta s[/tex] - Length of the rough patch, measured in meters.
[tex]f[/tex] - Friction force, measured in newtons.
According to the statement, friction force is represented by the following expression:
[tex]f = r \cdot m \cdot g[/tex]
Where:
[tex]r[/tex] - Ratio of friction force to weight, dimensionless.
[tex]g[/tex] - Gravitational constant, measured in meters per square second.
Then,
[tex]\frac{1}{2}\cdot m \cdot v_{1}^{2} = \frac{1}{2}\cdot m \cdot v_{2}^{2} + r \cdot m \cdot g \cdot \Delta s[/tex]
The equation is simplified algebraically and patch length is cleared afterwards:
[tex]\frac{1}{2}\cdot (v_{1}^{2}-v_{2}^{2}) = r \cdot g \cdot \Delta s[/tex]
[tex]\Delta s = \frac{v_{1}^{2}-v_{2}^{2}}{2 \cdot r \cdot g }[/tex]
Given that [tex]v_{1} = 5\,\frac{m}{s}[/tex], [tex]v_{2} = 2.5\,\frac{m}{s}[/tex], [tex]r = 0.22[/tex] and [tex]g = 9.807 \,\frac{m}{s^{2}}[/tex], the length of the rough patch is:
[tex]\Delta s = \frac{\left(5\,\frac{m}{s} \right)^{2}-\left(2.5\,\frac{m}{s} \right)^{2}}{2\cdot (0.22)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]
[tex]\Delta s = 4.345\,m[/tex]
The length of the rough patch is 4.345 meters.
