91Ó°ÊÓ

The work done by nonconservative force is,

$W_{net}=W_c+W_{nc}$ (1.1)

Here,$W_{net}$is the net work done,$W_c$ is the work done by all conservative force, and$W_{nc}$is the work done by all nonconservative force.

According to work-energy theorem, the net work done is,

$$\begin{gathered} W_{net}=\mathrm{Δ}\text{KE} \\ =\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2 \end{gathered}$$ (1.2)

Here, m is the mass of the skier,$v_f$is the final velocity of the skier, and$v_i$is the initial velocity of the skier$\left(v_i=12.0\text{m}/\text{s}\right)$.

The work done by conservative force is,

$$\begin{gathered} W_c=-\mathrm{Δ}\text{PE} \\ =mg{h}_i-mg{h}_f \end{gathered}$$ (1.3)

Here, m is the mass of the skier, g is the acceleration due to gravity$\left(g=9.8\text{m}/{\text{s}}^2\right)$,$h_i$is the initial height ($h_i =0$as the skier is initially at the ground), and$h_f$is the final height of the skier or height of the slope$\left(h_f=2.50\text{m}\right)$.

The work done by nonconservative force is the work done by the frictional force,

$$\begin{gathered} W_{nc}=W_f \\ =-\mathrm{μ}Nd \\ =-\mathrm{μ}mg\cos \left(30°\right)d \end{gathered}$$ (1.4)

Here,$\mathrm{μ}$is the coefficient of friction$\left(\mathrm{μ}=0.0800\right)$, Here, m is the mass of the skier, g is the acceleration due to gravity$\left(g=9.8\text{m}/{\text{s}}^2\right)$, and d is the distance travelled by the skier.

From equations (1.1), (1.2), (1.3) and (1.4),

$\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2=mg{h}_i-mg{h}_f-\mathrm{μ}mg\cos \left(30°\right)d$

Rearranging the above equation to get an expression for the final velocity,

$v_f=\sqrt{v_i^2+2g\left(h_i-h_f\right)-2\mathrm{μ}gd\cos \left(30°\right)}$ (1.5)

The length of the slope is,

$d=\frac{h_f}{\sin \left(30°\right)}$

Putting all known values,

$$\begin{gathered} d=\frac{2.50\text{m}}{\sin \left(30°\right)} \\ =5\text{m} \end{gathered}$$

The final velocity of the skier can be calculated using equation (1.5).

Putting all known values in equation (1.5),

$$\begin{gathered} v_f=\sqrt{\left\{{\left(12.0\text{m}/\text{s}\right)}^2+2×\left(9.8\text{m}/{\text{s}}^2\right)×\left[\left(0\text{m}\right)-\left(2.5\text{m}\right)\right] \\ -2×0.0800×\left(9.8\text{m}/{\text{s}}^2\right)×\left(5\text{m}\right)×\cos \left(30°\right)\right\}} \\ =9.36\text{m}/\text{s} \end{gathered}$$

Therefore, the required final speed at the top is $9.36\text{m}/\text{s}$.