The mechanical energy of the plane when it is on the runway is zero, because its altitude from the ground is zero (so, no gravitational potential energy) and its speed is zero (so, no kinetic energy).
At the final location instead, the mechanical energy of the plane is the sum of the kinetic energy K and the potential energy U of the plane, so:
[tex]E=K+U= \frac{1}{2}mv^2+mgh= [/tex]
[tex]= \frac{1}{2}(865 kg)(90m/s)^2+(865kg)(9.81 m/s^2)(2230m) =2.24 \cdot 10^7 J [/tex]
So, the change in mechanical energy of the plane is [tex]2.24 \cdot 10^7 J [/tex].
