Jet fighter planes are launched from aircraft carriers with the aid of their own engines and a catapult. If in the process of being launched from rest, the thrust of a jet's engines is 2.10 105 N and after moving through a distance of 90 m the plane lifts off with a kinetic energy of 5.20 107 J, what is the work done (in J) on the jet by the catapult

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

unbalanced force

explanation:

false.

explanation:

this statement is false. after falling for 5 seconds, its speed will not be 100 m/s.