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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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$M (\frac{2 v^{2}}{s} + 2 μg)$

$M (\frac{2 v^{2}}{s} - 2 μg)$

$M (\frac{v^{2}}{2 s} - μg)$

$M (\frac{v^{2}}{2 s} + μg)$

answer is A.

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Detailed Solution

Net force $F = f_{engine} - f_{friction}$
$\begin{array}{l} F \cdot S = \frac{1}{2} M v^{2} \Rightarrow F = \frac{M v^{2}}{2 s} \\ \Rightarrow f_{engine} = \frac{M v^{2}}{2 s} + μ M g \\ f_{engine} = M (\frac{v^{2}}{2 s} + μg) \end{array}$

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