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telebender
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Homework Statement
An aircraft of 5000kg mass is diving vertically downwards at a speed of 500 knots. The pilot operates the dive brakes at a height of 10 000m and reduces the speed to 325 knots at 7000m. If the average air resistance of the remainder of the aircraft during the deceleration is 15KN what average force must be exerted by the dive brakes? (assume the engine is throttled back and is not producing any thrust)
Homework Equations
V2=u2+2as
where V= final velocity
u=original velocity
a=acceleration
s=distance
The Attempt at a Solution
First convert velocities into m/s. 1 knot=0.5145 m/s so:
u=500 knots=257.25 m/s
v=325 knots=167.21m/s
transposing above formula for a: a=(V2-u2)/2s
=-6.3696m/s2
The force causing the plane to acelerate downwards is mass*gravity (mg) so to stop the acceleration the drag force would have to be equal to mg. 5000kg * 9.81m/s2=49050N As the drag force acts opposite to acceleration call it -49050N
To acelerate the plane (in this case opposite to direction of travel) there must be an additional force of mass * aceleration (ma). 5000kg*-6.3696m/s2=-31848
The total drag force would be: (force equal to mg)+ma
-49050N+-31848N=-80898N
These figures are -ve because they are based on a negative aceleration so the actual drag force is +80898N
Take away from this the air resistance without dive brakes of 15000N=
65898N
The answer is given as 16.62KN, I have given this a lot of thought but can't work out where I've gone wrong. Any help please?