Answer
(a) The weight of the plane is $10^5~N$
(b) The air speed above the wings is $85.1~m/s$
Work Step by Step
(a) We can find the total force that the pressure difference exerts on the wings:
$F = \Delta P~A$
$F = (500~N/m^2)(2)(100~m^2)$
$F = 10^5~N$
Since the airplane flies on a level path, this upward force is equal in magnitude to the plane's weight. Therefore, the weight of the plane is $10^5~N$
(b) Let $P_1$ be the pressure below the wings and let $P_2$ be the pressure above the wings. We can use Bernoulli's equation to find the air speed $v_2$ above the wings:
$P_2 + \frac{1}{2}\rho~v_2^2 = P_1+\frac{1}{2}\rho~v_1^2$
$\frac{1}{2}\rho~v_2^2 = P_1-P_2 +\frac{1}{2}\rho~v_1^2$
$v_2^2 = \frac{2(P_1-P_2)}{\rho} +v_1^2$
$v_2 = \sqrt{\frac{2(P_1-P_2)}{\rho} +v_1^2}$
$v_2 = \sqrt{\frac{2(500~N/m^2)}{1.3~kg/m^3} +(80.5~m/s)^2}$
$v_2 = 85.1~m/s$
The air speed above the wings is $85.1~m/s$
