Q:

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?

Accepted Solution

A:
Answer:a force of 1250lb is exertedStep-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 = forcea = area velocity =v from the above statement, we find out that [tex]f \alpha a*v^2[/tex]----1that is [tex]f= k* a* v^2[/tex]     -----2where k is a coefficent of proportionalitysince velocity of wind in mph,v =50and force in lb = 20and surface area =[tex]1/5 th ft^2[/tex]from eqn 2we have that[tex]20 =k*\frac{1}{5}* 50^2[/tex]making k subject of the formulawe have that [tex]k =\frac{20}{\frac{1}{5} *50^2 } =\frac{1}{25} (lb.mph^-1.ft^-2)[/tex]that means that [tex]f =\frac{1}{25} *a *v^2[/tex]  ---3however, when velocity = 125mphsurface area =[tex]2ft^2[/tex]putting values in eqn 3 we have[tex]f= \frac{1}{25}*2*125^2 = 1250 lb[/tex] hence a force of 1250lb is exerted