An aeroplane requires for take off a speed of 108 kmph, the run on the ground being 100m. Mass of the plane is 10000kg and coefficient of friction between the plane and the ground is 0.2. Assuming the plane accelerates uniformly, the minimum force required is (g=10m/s)
1) 2*10000N. 2) 2.43*10000N
3) 6.5*10000N. 4)8.86*10000N
Answers
Answer:
Step-by-step explanation:
Given
An aeroplane requires for take off a speed of 108 kmph, the run on the ground being 100 m. Mass of the plane is 10000 kg and coefficient of friction
We know that
Initial velocity u = 0
Final velocity v = 108 km/h = 30 m/s (1080 / 36)
Distance s = 100 m
From equation of motion we have
V^2 = u^2 + 2as
900 = 0 + 2a x 100
200 a = 900
a = 4.5 m / s^2
We know that
Force = m x a
= 10,000 x 4.5
Force = 45,000 N
Frictional force = μmg
= 0.2 x 100000
= 20,000 N
Net force required will be 45000 + 20000 = 65000 N = 6.5 x 10000 N
Step-by-step explanation:
Solution:-
We know that, the required take off speed is 108 kmph
which is equal to 30 m/s over a run of 100m.
We know that,
v^2 - u^2 = 2as
→ v^2 - 0 = 2as
→ a = v^2 /2s
So, we know that a is the acceleration. U is equal to initial velocity and v is final velocity.
We are considering here F is equal to force which should be developed by the engine is given by
F - μmg = ma
→ F = 0.2(10400)(9.81) + 10400[30^2/2(100)]
∴ F = 6.72×10^4 N
- Hope you got it...