A “kulbit” (Russian for “summersault”) is an aerial maneuver were a friendly air

Question

A “kulbit” (Russian for “summersault”) is an aerial maneuver were a friendly aircraft performs a tight loop to get directly behind an enemy aircraft. Immediately before beginning the maneuver, the pilot reduces the aircraft’s thrust to zero. The aircraft as two engines capable of 75.22 kN of thrust each. If the 16,380 kg aircraft’s initial speed is 650 km/h at the beginning of the loop of radius 27.4 m, what is the aircraft’s speed at the top of the loop? THE PLANE STARTS HORIZONTALLY HEADING TO THE RIGHT AND DOES A LOOP AND CONTINUES TO MOVE IN THE RIGHT DIRECTION.

PLEASE SHOW ALL DETAILED WORK. THIS IS AN EXAM QUESTION.

Explanation / Answer

initial speed, v1 = 650 km/h

= 650*5/18

= 180.6 m/s

Workdone by engines on the plane, W_engines = F*d

= 2*75.22*10^3*pi*27.4

= 1.295*10^7 J

Workdone by gravity, Wg = -m*g*(2*r)

= -16380*9.8*2*27.4

= -8.797*10^6 J

net workdone, Wnet = W_engines + W_g

= 1.295*10^7 -8.797*10^6

= 4.153*10^6 J

now use work-energy theorem

Wnet = change in kinetic energy

Wnet = (1/2)*m*(v2^2 - v1^2)

v2^2 - v1^2 = 2*Wnet/m

v2 = sqrt(v1^2 + 2*Wnet/m)

= sqrt(180.6^2 + 2*4.153*10^6/16380)

= 182 m/s

= 182*18/5

= 655.2 km/h

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