91Ó°ÊÓ

The given data can be listed below,

• The initial velocity of the skater is,${\mathrm{v}}_{\mathrm{i}}=3\mathrm{m}/\mathrm{s}$
• The final velocity of the skater is,${\mathrm{v}}_{\mathrm{f}}=1.65\mathrm{m}/\mathrm{s}$
• The percentage of frictional force is,of weight.${\mathrm{f}}_{\mathrm{k}}=25\%\mathrm{of}\mathrm{weight}$

Energy transfer from one body to another is referred to as work. When a force moves an object, work is said to be done.

The length of the rough patch is obtained by the relation-

$$\begin{gathered} W_{tot}=-\mathrm{μ}mgl \\ -\mathrm{μ}mgl=\frac{1}{2}m\left(v_i^2-v_f^2\right) \\ l=\frac{1}{2\mathrm{μ}g}\left(v_i^2-v_f^2\right) \end{gathered}$$

Here,$\mathrm{μ}$ is the coefficient of friction,g is the acceleration due to gravity,$v_i$ is the initial velocity of the skater and is the final velocity of the skater.

For the given values, the above equation becomes-

role="math" localid="1663759592364" $$\begin{gathered} \mathrm{l}=\frac{1}{2\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(0.25\right)}\left(3^2-{\left(1.65\right)}^2\right){\left(\mathrm{m}/\mathrm{s}\right)}^2 \\ =1.28\mathrm{m} \end{gathered}$$

Thus, the length of the rough patch is 1.28 m.