A $\mathbf{40}\mathbf{}\mathbf{KG}$skier skis directly down a frictionless slope angled at ${\mathbf{10}}^{\mathbf{0}}$to the horizontal. Assume the skier moves in the negative direction of an xaxis along the slope. A wind force with component role="math" localid="1657169142011" ${\mathbf{F}}_{\mathbf{x}}$acts on the skier. (a) What is ${\mathbf{F}}_{\mathbf{x}}$if the magnitude of the skier’s velocity is constant, (b) What isif the magnitude of the skier’s velocity is increasing at a rate of $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$, and (c) What is ${\mathbf{F}}_{\mathbf{x}}$if the magnitude of the skier’s velocity is increasing at a rate of$\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$?

• The mass of the skier is, $m=40kg.$
• The inclination angle with the horizontal is, $\theta ={10}^0.$
• The acceleration of a skier is, $a=1.0m/s^2.$
• The acceleration of the skier is,$a=2.0m/s^{2.}$

As per Newton’s second law, the force acting on the object is directly proportional to the acceleration of the body. The force is equal to the product of mass and acceleration of the body. A free-body diagram shows the forces that are acting on the body along with their directions.

By using Newton’s second law x component of wind force when skier moves with constant velocity and with acceleration can be found.

Formulae:

$\overrightarrow{F_{}}net=m\overrightarrow{a}$

Here,$\overrightarrow{F_{}}net$ is the net force, m is mass and$\overrightarrow{a}$ is acceleration.

By applying Newton’s second law along x axis as,

$\overrightarrow{F_{net}}=m\overrightarrow{a}$

The skier is moving with constant velocity hence it has zero acceleration. Consider the sign convention according to the direction of skier

$$\begin{gathered} F_x-mg.\sin \theta =0 \\ {F}_x=mg.\sin \theta \\ =40kg\times 9.8m/s^2\times \sin \left({10}^0\right) \\ =68N\left(uphill\right) \end{gathered}$$

Hence, the value of role="math" localid="1657170901844" $F_x$when skier is moving with constant velocity is 68N .

The skier is going down with some acceleration, hence according to the Newton’s second law,

$$\begin{gathered} F_x-mg.\sin \theta =-ma \\ {F}_x=mg.\sin \theta -ma \\ =40kg\times 9.8m/s^2\times \sin \left({10}^0\right)-40kg\times 1.0m/s^2 \\ =28N\left(uphill\right) \end{gathered}$$

Hence, the value of $F_x$when skier is going down with $1.0m/s^2$ the acceleration is$28N$.

The skier rate is increasing in downward direction, hence according to the Newton’s second law,

$$\begin{gathered} F_x-mg.\sin \theta =-ma \\ {F}_x=mg.\sin \theta -ma \\ {F}_x=(40kg\times 9.8m/s^2\times \sin ({10}^0\left)\right)-(40kg\times 1.0m/s^2) \\ {F}_x=-12N\left(downhill\right) \end{gathered}$$

Hence, the value of $F_x$when skier is going down with the $2.0m/s^2$acceleration is -12 N

The negative sign indicates that wind is also flowing in the downward direction, hence skier rate increases.