Physics, 06.11.2020 19:40, esmemaluma00

A steam catapult applies a very large force to supplement the engine thrust of a jet fighter as it takes off from an aircraft carrier. The jet has a mass of 4000 kg and must reach its take off speed of 70 m/s at the end of the 220
m flight deck. What is the combined force of the catapult and engines? (Use the work-energy theorem)

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answered: kcdavis318

The combined force of the catapult and engines is 44,545.5 N.

The given parameters;

mass of the jet, = 4000 kgspeed of the jet, v = 70 m/slength of the deck, d = 220 m

The combined force of the catapult and engines is determined by applying work energy theorem as shown below;

$W = \Delta E\\\\Fd = \frac{1}{2}m(v^2 -u^2) \\\\220F = \frac{1}{2}(4000)(70^2 - 0^2) \ \\\\220 F = 9,800,000\\\\F = \frac{9,800,000}{220} \\\\F = 44,545.5 \ N$

Thus, the combined force of the catapult and engines is 44,545.5 N.

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answered: smart1212

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:

$K_{1}+W_{1\rightarrow 2} = K_{2}$ (1)

Where:

$K_{1}$ , $K_{2}$ - Initial and final translational kinetic energies of the jet fighter, measured in joules.

$W_{1\rightarrow 2}$ - 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:

$F\cdot \Delta s = \frac{1}{2}\cdot m\cdot (v_{2}^{2}-v_{1}^{2})$

$F = \frac{m\cdot (v_{2}^{2}-v_{1}^{2})}{2\cdot \Delta s}$ (2)

Where:

- Combined force of the catapult and engines, measured in newtons.

- Mass of the jet fighter, measured in kilograms.

$\Delta s$ - Length of the catapult, measured in meters.

$v_{1}$ , $v_{2}$ - Initial and final speeds of the jet aircraft, measured in meters per second.

If we know that $m = 4000\,kg$ , $v_{1} = 0\,\frac{m}{s}$ , $v_{2} = 70\,\frac{m}{s}$ and $\Delta s = 220\,m$ , then the combined force of the catapult and engines is:

$F = \frac{(4000\,kg)\cdot \left[\left(70\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{2\cdot (220\,m)}$

$F = 44545.455\,N$

The combined force of the catapult and engines is 44545.455 newtons.

answered: Guest

the following is queen elizabeth’s purposes:

- to encourage the parliament that the technique of picking successors to the seat based merely on birthright needs to be reviewed.

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- to persuade the parliament that merely since she has a child does not assure that the kid will be a capable leader.

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answered: Guest

The relationship of both of them

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A steam catapult applies a very large force to supplement the engine thrust of a jet fighter as it t...