A jet aircraft is traveling at 233 m/s in horizontal flight. The engine takes in air at a rate of 122 kg/s and burns fuel at a rate of 3.36 kg/s. The exhaust gases are ejected at 487 m/s relative to the aircraft. Find the thrust of the jet engine. Answer in units of N.

33624.32 N

Explanation:

From second law of motion, thrust of engine can be computed as

F= Δp/ Δt

= m1 * Δv + m2 v(o)

F= m1 *[v(o) - v] + m2 *v(o)eqn(1)

Where

v(o) = initial velocity of the gas when ejected= 487 m/s

v= final velocity=233 m/s

m1= initial rate The engine takes in air at =122 kg/s

m2= rate of burning fuel= 3.36 kg/s.

If we input the given values into eqn (1) we have

F= 122×[ 487-233] + [3.36×487]

F=30988+1636.33

=33624.32 N

Hence, the thrust of the jet engine is 33624.32 N

+15-35=-20

going in opposite directions