91Ó°ÊÓ

The coefficient of kinetic friction on a rough surface is ${\mathrm{μ}}_k=0.300$

The length of the rough patch is: l = 4.20 m

The height of the hill is: h = 2.50 m

The mass of the skier is m = 62 kg

The initial speed of a skier is u = 6.5 m/s

The kinetic energy of the skier changed in rough patches from the frictionless icy surface due to the frictional work. The kinetic energy of the skier at the top of the hill changed to the potential energy of the skier.

The speed of the skier at the bottom of the hill is given as:

$v=\sqrt{u^2-2{\mathrm{μ}}_{k}gl+2gh}$

Here, g is the gravitational acceleration and its value is $9.8m/s^2$, h is the height of the hill. u is the initial speed of the skier and l is the length of a rough patch.

Substitute all the values in the above equation and we get,

$$\begin{gathered} v=\sqrt{{(6.5 m/s)}^2-2(0.300)(9.8 m/s^2)(4.20 m)+2(9.8 m/s^2)(2.50 m))} \\ v=8.16 m/s \end{gathered}$$

Therefore, the speed of the skier at the bottom of the hill is 8.16 m/s .

The generated internal energy is calculated as:

$E={\mathrm{μ}}_{k}mgl$

Here, E is the generated internal energy, ${\mathrm{μ}}_k$is the coefficient of kinetic friction on the rough surface.

Substitute all the values in the above equation, and we get,

$$\begin{gathered} E=(0.300)(62 kg)(9.8 m/s^2)(4.20 m) \\ E=766 J \end{gathered}$$

Therefore, the generated internal energy is 766 J .