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Question

An aeroplane of mass M requires a speed v for take-off. The length of runway is s and the coefficient of friction between the tyres and the ground is μ . Assuming that the plane accelerates uniformly during the take-off, the minimum force required by the engine of the plane for take-off is

A
M(v22s+μg)
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B
M(v22sμg)
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C
M(2v22s+2μg)
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D
M(2v2s+2μg)
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Solution

The correct option is A M(v22s+μg)
Net force, F, required to take off will be the sum of force due to frictional force and acceleration of the plane i.e.
F=Ff+Fa
Ff=μMg
As the initial velocity is zero and final velocity is v, we have acceleration a
a=v22s
Therefore,
Fa=Ma=Mv22s
F=M(v22s+μg)

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