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An aircraft is designed to have landing weight of 2500N and stall speed V of 9.8m/s. How much kinetic energy should be absorbed by wheel? Consider number of wheels with brake is 2.

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An aircraft is designed to have landing weight of 2500N and stall speed V of 9.8m/s. How much kinetic energy should be absorbed by wheel? Consider number of wheels with brake is 2.

(a) 6125J

(b) 5.98 KJ

(c) 2100

(d) 235 KJ

The question was asked in unit test.

The doubt is from Landing Gear in division Landing Gear and Subsystems of Aircraft Design

1 Answer

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Best answer
The correct option is (a) 6125J

Easy explanation: Given, landing weight W = 2500N, stall speed v = 9.8m/s. Number of wheels n = 2.

Kinetic energy K.E. = 0.5*W*v^2/g*n = 0.5*2500*9.8^2/9.81*2 = 6125 J.

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