Question

In a recent airline disaster, an airliner ying at 30,000ft,550mi/h, lost power and fell to Earth. The mass of the aircraft was 255,000 lb. If the magnitude of the work done by drag force on the plane during the fall was 2.96 × 106 Btu, estimate the velocity of the aircraft at the time of impact, in mi/h. Let g = 32.08 ft/s2.

Answer 1

To estimate the velocity of the aircraft at the time of impact, use the work-energy principle. The estimated velocity is 256.15 mi/h.

Step-by-step explanation:

To estimate the velocity of the aircraft at the time of impact, we can use the work-energy principle. The work done by the drag force is equal to the change in kinetic energy of the aircraft. The work done by the drag force is given as 2.96x10^6 Btu. We need to convert this energy into foot-pounds, and then use the kinetic energy equation to find the velocity. The equation is:

K = (1/2) mv^2

Where K is the kinetic energy, m is the mass of the aircraft, and v is the velocity. Rearranging the equation, we have:

v = √((2K) / m)

Plugging in the given values, we get:

v = √((2(2.96x10^6 Btu) * (3.968x10^8 ft-lbf/Btu)) / (255,000 lb * 32.08 ft/s^2)) = 256.15 mi/h

Therefore, the estimated velocity of the aircraft at the time of impact is 256.15 mi/h.

Answer 2

To develop this problem we will apply the energy conservation theorem and the principle of work. Basically we will have that the kinetic, potential and work energy must be equivalent to the kinetic energy just before the impact. We will start converting the units given to the British system and then proceed with the Calculations.

Remember that according to the energy balance in this case it would be balanced like this

`T_1 +\sum U_(1-2) = T_2`

`(1)/(2) mv_1^2+mgh -W_(drag) = (1)/(2) mv_2^2`

Here

m = mass

`v_(1,2)`= Velocity at each moment

`W_(drag)`= Work by drag

h = Height

g = Acceleration due to gravity

Mass

`m =255000 lb ((1slug)/(32.174lb))`

`m = 7925.654slug`

Initial Velocity

`v_1 = 550 (mi)/(h) ((5280ft)/(1mi))((1hr)/(3600s))`

`v_1 = 806.667ft/s`

Work by drag

`W_(drag) = (2.96*10^6BTU)((778.169lb\cdot ft)/(1BTU))`

`W_(drag) = 2303380240lb\cdot ft`

By work energy principle

`(1)/(2) mv_1^2+mgh -W_(drag) = (1)/(2) mv_2^2`

Replacing,

`(1)/(2) (7925.654slug)(806.667ft/s)^2 +(7925.654slug)(32.08ft/s^2)(30000ft) -2303380240lb\cdot ft = (1)/(2) (7925.654slug)v_2^2`

Solving for
`v_2`, we have that

`v_2 = 1412.2 ft/s`

Converting this value,

`v_2 = 1412.2 ft/s ((1mi)/(5280ft))((3600s)/(1h))`

`v_2 = 962.85mi/h`

Therefore the velocity of the aircraft at the time of impact is 962.85mi/h