Answer:work done on the jet by the catapult =3.31x107j
Explanation:
The kinetic energy of the jet at take off will be equal to the work done by the engines + the work done by the catapult.
First, the work done by the engines = Force x Distance
So the work done by the engines (2.10 X 105J)x 90m = 1.89 x 107J
Having that The total kinetic energy = 5.20x 107J
The work done on the jet by the catapult = the total kinetic energy - the work done by the engines.
Therefore the work done on the jet by the catapult = ( 5.20 X 107 J) -(1.89 X 107 )= 3.31x107j
Answer:
Work done on jet by catapult; W_cp = 33.1 x 10^(6) J
Explanation:
We are given;
Thrust of the Jets engine; F_thr = 2.1 x 10^(5)N
Lift off kinetic energy; K_f = 5.20 10^(7) J
Displacement moved by jet before lift is off; s = 90m
Now, from the work energy theorem, the net work done on an object is equal to the difference between the initial and final kinetic energy of that object.
Thus;
W = KE_f - KE_o
Where, KE_o is initial kinetic energy and KE_f is final kinetic energy.
Now, we the net work done on the jet by both the catapult and the Thrust engine is;
W_n = KE_f = 5.20 10^(7) J
Now, let's find the work done by just the Thrust engine.
The Thrust force is 2.1 x 10^(5)N while the displacement is 90m
Work done = Force x Distance
Thus,
Work done by Thrust engine;
W_thr = 2.1 x 10^(5) x 90 = 18.9 x 10^(6) J
Now, net work done on jet by catapult will be;
W_cp = Wn - W_thr
Where W_cp is the net work done by catapult
Wn is total net work done by both catapult and thrust engine
W_thr is net work done by thrust engine.
Thus,
W_cp = 5.20 10^(7) J - 18.9 x 10^(6) J
W_cp = 33.1 x 10^(6) J
