An airliner makes an emergency landing with its nose wheel locked in a position perpendicular to its normal rolling position. The forces acting to stop the airliner arise from friction due to the wheels and from the breaking effort of the engines in reverse thrust mode. The force of the engine on the plane is constant, Fengine=-F0. The sum of the horizontal forces on the airliner (in its forward direction) can be written as
F(t)= -F0+((t/ts)-1)F1, (1)
touchdown at time t=0 to the final stop at time ts=26s (0 ≤t ≤ts). The mass of the plane is M=80 tonnes (one tonne is 1000kg). We have F0=268 kN and F1=42 kN. Neglect all air drag and friction forces, except the one stated in the problem.
(a) Find the speed v0 of the plane at touchdown (in m/s).
v0=
(b) What is the horizontal acceleration of the plane at the time ts ? What is the acceleration at the time of touchdown? (absolute values; in m/s^2)
|a(ts)| =
|a(0)| =
(c) What distance s does the plane go between touchdown and its final stop at time ts ? (in meters)
s=
(d) What work do the engines in reverse thrust mode do during the emergency landing?
(magnitude in Joules; the force due to engines is (-F0))
W=
(e) How much heat energy is absorbed by the wheels during the emergency landing?
(magnitude in Joules; the force due to wheels is (F(t)+F0))
Eheat=
Do no cheat dude
physicsforums com showthread.php?t=720405
but still i am not getting the last two questions
ss01, can you please tell me a,b,c parts?
Oh boy oh boy, don't cheat.
To find the answers to these questions, we will use the given information and formulas related to forces, acceleration, work, and distance. Let's break down each question and go step by step.
(a) Find the speed v0 of the plane at touchdown (in m/s).
The speed of the airplane at touchdown can be found by using the equation of motion:
v = u + at
Where:
v = final velocity (at touchdown)
u = initial velocity (unknown, to be found)
a = acceleration (from equation (1))
t = time taken from touchdown to final stop (ts = 26s)
Since the initial velocity (u) is 0 at touchdown, the equation simplifies to:
v0 = a × t
Plugging in the values:
v0 = (−F0 + ((t/ts) - 1)F1) × t
Substituting the values F0 = 268 kN, F1 = 42 kN, and ts = 26s:
v0 = (−268 × 10^3 N + ((t/26) - 1) × 42 × 10^3 N) × t
Now we can calculate the value of v0 using the given time t.
(b) What is the horizontal acceleration of the plane at the time ts? What is the acceleration at the time of touchdown? (absolute values; in m/s^2)
The horizontal acceleration can be calculated by using the equation:
a = (v - u) / t
Where:
a = acceleration
v = final velocity (0 at touchdown)
u = initial velocity (unknown, to be found)
t = time taken from touchdown to final stop (ts = 26s)
Since the final velocity (v) is 0 at touchdown, the equation simplifies to:
a = -u / t
We need to find the absolute value of the acceleration, so it will be:
|a| = |(-F0 + ((t/ts) - 1)F1) / t|
Again, substituting the values F0 = 268 kN, F1 = 42 kN, and ts = 26s, we can calculate the acceleration at ts.
To find the acceleration at the time of touchdown, t = 0. Plugging this value into the equation for the absolute acceleration:
|a(0)| = |(-F0 + ((0/26) - 1)F1) / 0|
(c) What distance s does the plane go between touchdown and its final stop at time ts? (in meters)
The distance covered by the plane can be calculated using the equation of motion:
s = ut + (1/2)at^2
Where:
s = distance
u = initial velocity (0 at touchdown)
a = acceleration (from equation (1))
t = time taken from touchdown to final stop (ts = 26s)
Simplifying the equation with the known values:
s = 0 × t + (1/2) × (-F0 + ((t/ts) - 1)F1) × t^2
(d) What work do the engines in reverse thrust mode do during the emergency landing? (magnitude in Joules; the force due to engines is (-F0))
Work can be calculated using the formula:
W = F × d
Where:
W = work done
F = force (from the engines)
d = displacement
Since the force from the engines is constant and equal to -F0, the work done by the engines can be calculated as:
W = (-F0) × (s at touchdown)
(e) How much heat energy is absorbed by the wheels during the emergency landing? (magnitude in Joules; the force due to wheels is (F(t) + F0))
The heat energy absorbed by the wheels can be calculated using the same formula as work:
Eheat = F × d
Where:
Eheat = heat energy absorbed
F = force (from the wheels, (F(t) + F0))
d = displacement (s from touchdown to final stop)
Substituting the values and solving for Eheat will give us the answer.
For each question, use the given values and equations provided above to calculate the specific answer.