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A fighter plane is pulling out for a div...

A fighter plane is pulling out for a dive at a speed of 900 km/hr. 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=9.8m/s^2`.

Text Solution

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At the lowest point the path the acceleration is vertically upward (towards the centre) and its magnitude is `v^2/r`.
The forces on the plane are
a. weight Mg downward and
b. force F by the air upward.
Hence, Newton's second law of motion gives
F-Mg=Mv^2/r`
or `F=M(g+v^2/r)`
Here `v=900 km/hr theta=(9xx10^5)/3600 m/s =250 m/s`
or `F=16000(62500/2000)N=6.56xx10^5N(upward).`
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