91影视

• The dimension of ice is 400 m by 500 m by 4 mm.
• The density of ice is$917{\text{鈥塳驳/尘}}^{\text{3}}$.
• The air-drag force is:$\mathrm{D}=4{\mathrm{C}}_{\mathrm{ice}}{\mathrm{蚁础}}_{\mathrm{ice}}{\mathrm{v}}^2$.
• The drag coefficient,${\mathrm{C}}_{\mathrm{ice}}=2脳{10}^{鈭�3}$ .
• The air density,$\mathrm{蚁}=1.21{\text{鈥塳驳/尘}}^{\text{3}}$.
• The number of stones is 100.
• The mass of each stone is 20 kg.
• The kinetic friction coefficient between ice and ground, ${\mathrm{渭}}_{\mathrm{k}}=0.10$.

The problem deals with the relation between air-drag force and frictional force. Air-drag force is a force acting opposite to the relative motion of any object moving with respect to a surrounding fluid. The frictional force is the opposing force that is created between two surfaces that try to move in the same direction or that try to move in opposite directions. Here equate the air-drag force to the frictional force to solve this question.

Formula:

Air drag force is given by,

$\mathrm{D}=4{\mathrm{C}}_{\mathrm{ice}}{\mathrm{蚁础}}_{\mathrm{ice}}{\mathrm{v}}^2$

Frictional force is

$\mathrm{F}={\mathrm{渭}}_{\mathrm{k}}\mathrm{mg}$

The mass of ice:

$\begin{array}{l}{\mathrm{m}}_{\mathrm{ice}}=917{\text{鈥塳驳/尘}}^{\text{3}}脳(400\text{鈥尘}脳500\text{鈥尘}脳0.004\text{鈥尘})\\ {\mathrm{m}}_{\mathrm{ice}}=7.34脳{10}^5\text{鈥塳驳}\end{array}$

The mass of ice with 100 stones:

$\begin{array}{l}\mathrm{m}={\mathrm{m}}_{\mathrm{ice}}+100脳20\text{鈥塳驳}\\ \mathrm{m}=7.34脳{10}^5\text{鈥塳驳}+2000\text{鈥塳驳}\\ \mathrm{m}=7.36脳{10}^5\text{鈥塳驳}\end{array}$

To calculate the velocity of wind, equate air-drag force to frictional force:

$\begin{array}{c}\mathrm{D}=\mathrm{F}\\ 4{\mathrm{C}}_{\mathrm{ice}}{\mathrm{蚁础}}_{\mathrm{ice}}{\mathrm{v}}^2={\mathrm{渭}}_{\mathrm{k}}\mathrm{mg}\\ \mathrm{v}=\sqrt{\frac{{\mathrm{渭}}_{\mathrm{k}}\mathrm{mg}}{4{\mathrm{C}}_{\mathrm{ice}}{\mathrm{蚁础}}_{\mathrm{ice}}}}\end{array}$

Substitute the values in the above expression, and we get,

$\begin{array}{l}\mathrm{v}=\sqrt{\frac{\left(0.10\right)脳(7.36脳{10}^5\text{鈥塳驳})脳(9.8{\text{鈥尘/s}}^{\text{2}})}{4脳(2脳{10}^{鈭�3})脳(1.21\text{鈥�}{\text{kg/m}}^{\text{3}})脳(400\text{鈥尘}脳500\text{鈥尘})}}\\ \mathrm{v}=19\text{鈥尘/s}脳\frac{3600\text{鈥塻}}{1\text{hr}}脳\frac{\text{1鈥塳尘}}{\text{1000鈥尘}}\\ \mathrm{v}=\text{69鈥塳尘/丑}\end{array}$

Thus, the required speed is 69 km/h.

Doubling our previous result, we find the required speed to be$139\text{鈥塳尘/丑}$ .

Thus, the speed will be 139 km/h.

The result is reasonable for storm winds. A category-5 hurricane has speeds on the order of $2.6脳{10}^2\text{m/s}$.

Thus, we can say that these values for high-speed winds in a storm are reasonable.