Question

A fighter plane is pulling out for a dive at a speed of 900 km//h. Assuming its path to be a vertical circle of radius 2000 m and its mass to be 16000 kg, find the force exerted by the air on it at the lowest point. Take g=9.8 m//s^(2)

Answer 1

Given as :

The speed of a fighter plane = s = 900 km/h  = $\dfrac{900000}{3600}$ = 250 m/s

The  path to be a vertical circle of radius = r = 2000 m

The mass of the plane = m = 16000 kg

The value of g = 9.8 m/s²

To Find :

The force exerted by the air on plane at the lowest point

Solution :

Let The force exerted by the air on plane = F Newton

So, At lowest point a centripetal force is exerted on it

And , upward force applied on plane = F

        downward force applied on plane = mg

Thus, At equilibrium position

       F - mg = $\dfrac{mv^{2} }{r}$

Or,   F - 16000 kg × 9.8  m/s²  = $\dfrac{16000 kg \times (250 m/s)^{2}}{2000 m}$

Or,  F - 156800  kg m/s² = $\dfrac{16000 kg \times 62500 (m/s)^{2} }{2000 m}$

Or,   F - 156800  = $\dfrac{10^{9} }{2000}$

Or,   F - 156800  = 5 × $10^{5}$

∴      F = 5 × $10^{5}$  + 156800

i.e    F = 656800

Or, Force = F = 6.56 × $10^{5}$   Newton  upward

Hence, The force exerted by the air on it at the lowest point is  6.56 × $10^{5}$ Newton ( upward )  . Answer