Answer:
The combined force of the catapult and engines is 44545.455 newtons.
Explanation:
Let suppose that platform of the aircraft carrier is horizontal, such that changes in gravitational potential energy can be neglected. In addition, effects of non-conservative forces are neglected. By using Principle of Energy Conservation and Work-Energy Theorem we find that steam catapult-jet aircraft is represented by:
[tex]K_{1}+W_{1\rightarrow 2} = K_{2}[/tex] (1)
Where:
[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final translational kinetic energies of the jet fighter, measured in joules.
[tex]W_{1\rightarrow 2}[/tex] - Work done on the jet fighter due to the combined force of the catapult and engines, measured in joules.
By applying definitions of translational kinetic energy and work, we expand and simplify the equation above as follows:
[tex]F\cdot \Delta s = \frac{1}{2}\cdot m\cdot (v_{2}^{2}-v_{1}^{2})[/tex]
[tex]F = \frac{m\cdot (v_{2}^{2}-v_{1}^{2})}{2\cdot \Delta s}[/tex] (2)
Where:
[tex]F[/tex] - Combined force of the catapult and engines, measured in newtons.
[tex]m[/tex] - Mass of the jet fighter, measured in kilograms.
[tex]\Delta s[/tex] - Length of the catapult, measured in meters.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speeds of the jet aircraft, measured in meters per second.
If we know that [tex]m = 4000\,kg[/tex], [tex]v_{1} = 0\,\frac{m}{s}[/tex], [tex]v_{2} = 70\,\frac{m}{s}[/tex] and [tex]\Delta s = 220\,m[/tex], then the combined force of the catapult and engines is:
[tex]F = \frac{(4000\,kg)\cdot \left[\left(70\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{2\cdot (220\,m)}[/tex]
[tex]F = 44545.455\,N[/tex]
The combined force of the catapult and engines is 44545.455 newtons.
