Question 38: Laws of Motion

Question

A plane requires for take off a speed of 72 kmh$$-$$1, the run on the ground being 50 m. The mass of the plane is 10000 kg and the coefficient of friction between the plane and the ground is 0.2. Assume that the plane accelerates uniformly during take off. The minimum force required by the engine of the plane for take off is

Accepted Answer

Given, mass of plane, M = 10000 kg
Coefficient of friction, $$\mu$$ = 0.2
Distance, s = 50 m
Initial speed, u = 0
Final speed, v = 72 kmh
$$-$$1
= 20 ms
$$-$$1
Let a be the acceleration of the plane.
From third equation of motion,
v
2
$$-$$ u
2
= 2as
(20)
2
$$-$$ (0)
2
= 2 $$	imes$$ a $$	imes$$ 50
$$	herefore$$ $$a = {{20 	imes 20} \over {100}} = 4$$ ms
$$-$$2
Force required to provide necessary acceleration to the plane,
F
1
= ma = 10000 $$	imes$$ 4 = 4 $$	imes$$ 10
4
N
Force required to overcome friction by engine,
F
2
= $$\mu$$R = $$\mu$$mg = 0.2 $$	imes$$ 10000 $$	imes$$ 9.8
= 1.96 $$	imes$$ 10
4
N
Therefore, the minimum force required by the engine,
F
1
+ F
2
= 4 $$	imes$$ 10
4
+ 1.96 $$	imes$$ 10
4
= 5.96 $$	imes$$ 10
4
N

Answer

4.43 $$	imes$$ 104 N

Answer

5.96 $$	imes$$ 104 N

Answer

2.25 $$	imes$$ 104 N

Answer

3.45 $$	imes$$ 104 N

Laws of Motion

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