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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

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$M (\frac{v^{2}}{2 s} - μ g)$

$M (\frac{2 v^{2}}{s} + 2 μ g)$

$M (\frac{2 v^{2}}{s} - 2 μ g)$

$M (\frac{v^{2}}{2 s} + μ g)$

answer is A.

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Detailed Solution

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 = F_f+ F_a
F_f = μMg

As the initial velocity is zero and the final velocity is v, we have an acceleration a
a = v²/2s

Therefore,
F_a= M_a= M(v²/2s)

F = M(v²/2s + μg)

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