91Ó°ÊÓ

Slope angle is ${}^{\mathrm{θ}=40.0°}$,

Coefficient of kinetic friction is localid="1654148778161" ${}^{{\mathrm{μ}}_k=0.0400}$

Mass of the skier and equipment is ${}^{m=85.0\mathrm{k}}$

Cross-sectional area of the (tucked) skier is ${{}^{A=1.30\mathrm{m}}}^2$

The drag coefficient is ${}^{\mathrm{C}=0.150}$

Air density is ${}^{1.20\mathrm{kg}/{\mathrm{m}}^2}$.

This problem involves Newton’s second law for motion along the slope. According to Newton's 2nd law of motion, a force applied to an object at rest causes it to accelerate in the direction of the force.Also it deals with the drag force and terminal speed which is a type of friction. This is the force acting opposite to the relative motion of an object moving with respect to the surrounding medium.

Formula:

$F_{net}=\underset{}{∑}ma$

where, F is the net force, m is mass and a is an acceleration.

The terminal speed is given by

localid="1654149859961" $v_t=\sqrt{\frac{2F_g}{CpA}}$

Where C is the drag coefficient, ${}^p$is the fluid density, A is the effective cross-sectional area, and ${{}^F}_g$is the gravitational force.

The force along the slope is given by,

$$\begin{gathered} F_g=mg\sin \mathrm{θ}-\mathrm{μ}{F}_N \\ =mg\sin \mathrm{θ}-\mathrm{μ}mg\cos \mathrm{θ} \\ =mg\left(\sin \mathrm{θ}-\mathrm{μ}\cos \mathrm{θ}\right) \\ =\left(85.0\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left[\sin 40.0-\left(0.0400\right)\cos 40.0\right] \\ =510N \end{gathered}$$

Thus, the terminal speed of the skier is,

$$\begin{gathered} v_t=\sqrt{\frac{2F_g}{CpA}} \\ =\sqrt{\frac{2\left(510\mathrm{N}\right)}{\left(0.150\right)\left(1.20\mathrm{kg}/{\mathrm{m}}^3\right)\left(1.30{\mathrm{m}}^2\right)}} \\ =66.0 \end{gathered}$$

Therefore, the terminal speed is 66 m/s.

Differentiating ${}^{v_t}$ with respect to C,

localid="1654588390968" $$\begin{gathered} d{v}_t=-1/2\sqrt{\frac{2F_g}{pA}{C}^{-\frac{3}{2}}}dC \\ =-1/2\sqrt{\frac{2\left(510\mathrm{N}\right)}{\left(1.20\mathrm{kg}/{\mathrm{m}}^3\right)\left(1.30{\mathrm{m}}^3\right)}}{\left(0.150\right)}^{-3/2}dC \end{gathered}$$

${}^{S\frac{d{v}_t}{dC}=-\left(2.20×{10}^2\mathrm{m}/\mathrm{s}\right)}$

Therefore, thevariation in the terminal speed is ${}^{-2.20×{10}^2\mathrm{m}/\mathrm{s}}$.