Question

Each of the two engines on the F/A-18 aircraft develops a thrust of 100kN. The aircraft has a mass 22,000kg and goes from 0 to 265km/h over a distance of 90 m before taking off. Find the average force exerted by the catapult on the aircraft during the launch. (Answer 462.27 kN)

Answer 1

Here both Engines can apply 200 kN = 200000 N ,

Mass of aircraft = 22,000 kg

Initial velocity(U) = 0 , Final velocity(V) = 265 km/ h = 73.61 m/s   

Traveling Distance (S) = 90 m

Applying motion 3rd Equation = V² - U² = 2 a S , Here U = 0 ,

So a = V² / 2 S = ( 73.61 ) ²   / 2* 90 = 5418.59 / 180 = 30 .1033 m/s²

SO Total Force applied = Mass * Acc (a) = 22000 *30 .1033 = 662277.805 N

Now Force by Engine of aircraft = 200 000 N

SO Force applied by Catapult on aircraft = Total force - force by Engines   

=   662277.805- 200 000 = 462277 N

Force applied by Catapult on aircraft =   462277 N = 462.277 kN