(a) No, because the mechanical energy is not conserved
Explanation:
The work-energy theorem states that the work done by the engine on the airplane is equal to the gain in kinetic energy of the plane:
[tex]W=\Delta K[/tex] (1)
However, this theorem is only valid if there are no non-conservative forces acting on the plane. However, in this case there is air resistance acting on the plane: this means that the work-energy theorem is no longer valid, because the mechanical energy is not conserved.
Therefore, eq. (1) can be rewritten as
[tex]W=\Delta K + E_{lost}[/tex]
which means that the work done by the engine (W) is used partially to increase the kinetic energy of the airplane ([tex]\Delta K[/tex]) and part is lost because of the air resistance ([tex]E_{lost}[/tex]).
(b) 77.8 m/s
First of all, we need to calculate the net force acting on the plane, which is equal to the difference between the thrust force and the air resistance:
[tex]F=7.70\cdot 10^4 N - 5.00 \cdot 10^4 N=2.70\cdot 10^4 N[/tex]
Now we can calculate the acceleration of the plane, by using Newton's second law:
[tex]a=\frac{F}{m}=\frac{2.70\cdot 10^4 N}{1.60\cdot 10^4 kg}=1.69 m/s^2[/tex]
where m is the mass of the plane.
Finally, we can calculate the final speed of the plane by using the equation:
[tex]v^2- u^2 = 2aS[/tex]
where
[tex]v=?[/tex] is the final velocity
[tex]u=66.0 m/s[/tex] is the initial velocity
[tex]a=1.69 m/s^2[/tex] is the acceleration
[tex]S=5.00 \cdot 10^2 m[/tex] is the distance travelled
Solving for v, we find
[tex]v=\sqrt{u^2+2aS}=\sqrt{(66.0 m/s)^2+2(1.69 m/s^2)(5.00\cdot 10^2 m)}=77.8 m/s[/tex]
