Question

In downhill speed skiing a skier is retarded by both the air drag force on the body and the kinetic frictional force on the skis. (a) Suppose the slope angle is${}^{\mathbf{\theta }\mathbf{=}\mathbf{40}\mathbf{.}{\mathbf{0}}^{\mathbf{\circ }}}$.The snow is dry snow with a coefficient of kinetic friction${}^{{\mathbf{\mu }}_{\mathbf{k}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0400}}$, the mass of the skier and equipment is${}^{\mathbf{m}\mathbf{=}\mathbf{85}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{kg}}$, the cross-sectional area of the(tucked) skier is${}^{\mathbf{A}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{30}\mathbf{}{\mathbf{m}}^{\mathbf{2}}}$, the drag coefficient is${}^{\mathbf{c}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{150}}$, and the air density islocalid="1654148127880" ${}^{\mathbf{1}\mathbf{.}\mathbf{20}\mathbf{}\mathbf{kg}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}}$. (a) What is the terminal speed? (b) If a skier can vary C by a slight amount${}^{\mathbf{d}\mathbf{C}}$by adjusting, say, the hand positions, what is the corresponding variation in the terminal speed?

Short answer

a) The terminal speed is${}^{66\mathrm{m}/\mathrm{s}}$.

b) The variation in the terminal speed is${}^{-2.20\times {10}^2\mathrm{m}/\mathrm{s}}$.

Step-by-step solution

Given

Slope angle is ${}^{\theta =40.0°}$,

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

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

Determining the concept

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:

$F_{net}=\sum _{}ma$

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.

(a) Determining the terminal speed

The force along the slope is given by,

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

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.

(b) Determining the variation in the terminal speed

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

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

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

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