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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)`A. `3.28xx10^(5)N`B. `6.56xx10^(5)N`C. `9.28xx10^(5)N`D. `12.56xx10^(5)N` |
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Answer» Correct Answer - A Given , radius of circular path = 2000 m Mass of plane = 16000 kg and speed = 900 `kmh^(-1)` `=(900xx10^(3))/(3600)ms^(-1)` = 250 `ms^(-1)` The required centripetal force, `f_(c)=(mv^(2))/(r)` (in upward directon) `=(16000xx250xx250)/(2000)=5xx10^(5)N` The gravitational force, `f_(g)=mg=16000xx9.8` `=1.568xx10^(5)N` Thus, force will act as reaction force in upward direction. ltbtgt `therefore` Net upward force `=5xx10^(5)-1568xx10^(5)` `=(5-16)xx10^(5)N=3.4xx10^(5)N` |
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