Question

The force of the wind blowing on a vertical surface varies jointly as the area of the surface and the square of the velocity. If a wind of 50 mph exerts a force of 20 lb on a surface of 1/5thft2, how much force will a wind of 125 mph place on a surface of 2 ft2?

Answer 1

Answer:

a force of 1250lb is exerted

Step-by-step explanation:

since the force of the wind varies jointly as the area of the surface and the square of the velocity,

let f = force

a = area

velocity =v

from the above statement, we find out that

$f \alpha a*v^2$----1

that is $f= k* a* v^2$     -----2

where k is a coefficent of proportionality

since velocity of wind in mph,v =50

and force in lb = 20

and surface area =$1/5 th ft^2$

from eqn 2

we have that

$20 =k*\frac{1}{5}* 50^2$

making k subject of the formula

we have that

$k =\frac{20}{\frac{1}{5} *50^2 } =\frac{1}{25} (lb.mph^-1.ft^-2)$

that means that $f =\frac{1}{25} *a *v^2$  ---3

however, when velocity = 125mph

surface area =$2ft^2$

putting values in eqn 3 we have

$f= \frac{1}{25}*2*125^2 = 1250 lb$

hence a force of 1250lb is exerted