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

To find the take-off speed (vt), the work-energy principle is used, giving vt ≈ 148.3 m/s. To launch the jet with a spring, the potential energy of the compressed spring should equal the jet's kinetic energy at take-off; you would need to depress the spring by approximately 15.69 m.

Step-by-step explanation:

To solve for the take-off speed of the jet regarding part (a) of the question, we need to apply the work-energy principle. Since the work done by the engine, W, is equal to the force times the distance and also equal to the change in kinetic energy (final kinetic energy - initial kinetic energy, which is 0 since the plane starts from rest), we have:

1. F × d = ½ m × v_t²
2. ½ × 7200 kg × v_t² = 86000 N × 920 m
3. v_t² = (86000 N × 920 m) / (½ × 7200 kg)
4. v_t = √((86000 N × 920 m) / (3600 kg)) ≈ 148.3 m/s

For part (b), the energy stored in the compressed spring, which is potential energy when it is released, must equal the kinetic energy of the jet at take off. Therefore:

1. ½ × k × d_s² = ½ × m × v_t²
2. ½ × 100000 N/m × d_s^2 = ½ × 7200 kg × (148.3 m/s)²
3. d_s² = (7200 kg × (148.3 m/s)²) / (100000 N/m)
4. d_s = √((7200 kg × (148.3 m/s)²) / (100000 N/m))
5. d_s ≈ 15.69 m

Answer 2

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.

Step-by-step 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 =
`(1)/(2)`m
`v^(2)`

Initial velocity is zero hence, PE = 0

Hence W = KE

W =
`(1)/(2)`m
`vt^(2)`................(equation 2)

From equation 1 and 2, we get

F x d =
`(1)/(2)`m
`vt^(2)`

vt =
`\sqrt{(2Fd)/(m) }`

=
`\sqrt{(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 =
`(1)/(2)`m
`vt^(2)`

EPE =
`(1)/(2)`k
`x^(2)` , where k is spring constant and x is displacement (here distance of depression)

KE = EPE

`(1)/(2)`m
`vt^(2)` =
`(1)/(2)`k
`x^(2)`

x =
`\sqrt{(mv^(2) )/(k) }`

=
`\sqrt{(7200 x 148.3^(2) )/(100000) }\\`

= 39.79 m

Hence we need to depress a spring of 39.79 m.