an aeroplane requires for a take off speed of 72kmph, the run on the ground being 100m the coefficient of friction between the plane and the ground is 0.2.assume that the plane accelerates uniformlu during the takeoff,the minimum force exerted by the engine of the plane for the takeoff is ( assume,g=10m/s^2 and the mass of the plane is 10^4 kg)
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Answered by
5
Answer:
Net force required will be 45000 + 20000 = 65000 N = 6.5 x 10000 N
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
hope it helps you ♥️♥️
Answered by
0
Answer:
Answer is ma+frictional force
Frictional force is (mu) Mg
mu is coefficient of static friction
acceleration you find out by v2-u2=2as
Where u =0
Substitute every thing in these formula you will get answer
Do on your own
MARK AS BRAINLIST PLEASE
Explanation:
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