Question

The thrust produced by a single jet engine creates a force of F = 86000 N. It takes the jet (with a mass of m = 7200 kg) a distance of d = 0.92 km to take off. show answer Correct Answer 50% Part (a) What is the take-off speed of the jet vt in m/s? vt = 148.25 vt = 148.3 ✔ Correct! show answer No Attempt 50% Part (b) How far in meters would you need to depress a giant spring k = 100,000 N/m in order to launch the jet at the same speed without help from the engine? ds = | sin() cos() tan() cotan() asin() acos() atan() acotan() sinh() cosh() tanh() cotanh() Degrees Radians π ( ) 7 8 9 HOME E ↑^ ^↓ 4 5 6 ← / * 1 2 3 → + - 0 . END √() BACKSPACE DEL CLEAR

Answer 1

Answer:

a) take off speed of jet in m/s =  148.3 m/s

b) we need to depress a spring for distance of 39.79 m.

Explanation:

Given:

F = 86000 N

m =7200 kg

d = 0.92 Km= 920 m

a) To find take off speed of jet in m/s

We know that work done is given by

W = F x d ...........(equation 1)

F is force and d is displacement

Also work done in terms of kinetic energy KE and potential energy PE is given by

W = PE + KE

PE = $\frac{1}{2}$m$v^{2}$

Initial velocity is zero hence, PE = 0

Hence W = KE

W = $\frac{1}{2}$m$vt^{2}$................(equation 2)

From equation 1 and 2, we get

F x d =  $\frac{1}{2}$m$vt^{2}$

vt = $\sqrt{\frac{2Fd}{m} }$

   = $\sqrt{\frac{2 x 86000 x 920}{7200} }$

   =148.25

 vt  = 148.3 m/s

Hence proved

b) To find how far giant spring must be depressed (in meters)

Given

spring constant K = 100000 N/m

We know that for a spring, kinetic energy KE is equal to elastic potential energy EPE

KE =   $\frac{1}{2}$m$vt^{2}$  

EPE =  $\frac{1}{2}$k$x^{2}$ , where k is spring constant and x is displacement (here distance of depression)

KE = EPE

$\frac{1}{2}$m$vt^{2}$  = $\frac{1}{2}$k$x^{2}$

x = $\sqrt{\frac{mv^{2} }{k} }$

  = $\sqrt{\frac{7200 x 148.3^{2} }{100000} }$

  = 39.79 m

Hence we need to depress a spring of 39.79 m.

Answer 2

a) The take-off speed of the jet is approximately 148.249 meters per second.

b) The spring deformation is approximately 39.779 meters.

Let suppose that the jet accelerates on a horizontal ground.

a) The final speed of the jet ($v$), in meters per second, is determined by Work-Energy Theorem:

$\frac{1}{2}\cdot m\cdot v^{2} = F\cdot s$ (1)

Where:

• $m$ - Mass of the jet, in kilograms.
• $F$ - Thrust of the jet engine, in newtons.
• $s$ - Take-off distance, in meters.

If we know that $m = 7200\,kg$, $F = 86000\,N$ and $s = 920\,m$, then the take-off speed is:

$v = \sqrt{\frac{2\cdot F\cdot s}{m} }$

$v = \sqrt{\frac{2\cdot (86000\,N)\cdot (920\,m)}{7200\,kg} }$

$v \approx 148.249\,\frac{m}{s}$

The take-off speed of the jet is approximately 148.249 meters per second.

b) By principle of energy conservation, the change in translational kinetic energy ($\Delta K$), in joules, is equal to the change in elastic potential energy ($\Delta U$), in joules. By applying the definition of translational kinetic energy and elastic potential energy, we get the resulting formula:

$\frac{1}{2}\cdot m \cdot v^{2} = \frac{1}{2}\cdot k\cdot x^{2}$ (2)

Where:

• $k$ - Spring constant, in newtons per meter.
• $x$ - Spring deformation, in meters.

Now, we clear $k$ within the expression:

$x = \sqrt{\frac{m}{k} }\cdot v$ (3)

If we know that $m = 7200\,kg$, $k = 100000\,\frac{N}{m}$ and $v \approx 148.249\,\frac{m}{s}$, then the spring deformation is:

$x = \sqrt{\frac{7200\,kg}{100000\,\frac{N}{m} } }\cdot \left(148.249\,\frac{m}{s} \right)$

$x \approx 39.779\,m$

The spring deformation is approximately 39.779 meters.

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