91Ó°ÊÓ

Mass of the stone,$M=20\mathrm{kg}$

Area of cross section,$A=0.040{\mathrm{m}}^2$

Drag coefficient,$\mathrm{C}=0.8$

Air density,$p=1.21\mathrm{kg}/{\mathrm{m}}^3$

Coefficient of kinetic friction,${\mathrm{μ}}_k=0.80.$

This problem is based on the drag force 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:

The terminal speed is given by

$v_t=\sqrt{\frac{2F_g}{CpA}}$

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.

Free body diagram of the stone:

Kinetic frictional force must act on the stone during its motion.

To find its magnitude, use Newton’s 2nd law along y direction.

Stone is not moving along y.So, the acceleration is 0, it gives,

$$\begin{gathered} \underset{}{∑}{F}_y=0 \\ N-mg=0 \\ N=mg \end{gathered}$$

Kinetic frictional force is defined as,

$$\begin{gathered} f_k={\mathrm{μ}}_{k}N \\ ={\mathrm{μ}}_k\left(mg\right) \\ =\left(0.80\right)\left(\left(20\right)\left(9.81\right)\right) \\ =156.96\mathrm{N} \end{gathered}$$

In this case,${}^{F_k}$is nothing but the drag force. Thus,

Where D is the drag force. Then, by using the equation,

$D=\frac{1}{2}CpA{V}^2,$

expression for speed of the wind can be written as,

$$\begin{gathered} v=\sqrt{\frac{2F}{CpA}} \\ =\sqrt{\frac{2\left(157\mathrm{N}\right)}{\left(0.80\right)\left(1.21\mathrm{kg}/{\mathrm{m}}^3\right)\left(0.040{\mathrm{m}}^2\right)}} \\ =90\mathrm{m}/\mathrm{s} \\ =3.2×{10}^2\mathrm{km}/\mathrm{h} \end{gathered}$$

Hence,the wind speed V along the ground is needed to maintain the stone’s motion once it has started moving is${}^{3.2×{10}^2\mathrm{km}/\mathrm{h}}$.

Doubling our previous result, the reported speed is found to be${}^{6.5×{10}^2\mathrm{km}/\mathrm{h}}$.

The result is not reasonable for a terrestrial storm. A category 5 hurricane has speeds

on the order of${}^{2.6×{10}^2\mathrm{m}/\mathrm{s}}$.