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 $\mu$ . 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. $M(\frac{v^2}{2s}+\mu g)$
• B. $M(\frac{v^2}{2s}-\mu g)$
• C. $M(\frac{2v^2}{2s}+2\mu g)$
• D. $M(\frac{2v^2}{s}+2\mu g)$

Answer 1

The correct option is A $M(\frac{v^2}{2s}+\mu g)$

Net force, $F$, required to take off will be the sum of force due to frictional force and acceleration of the plane i.e.
$F=F_f+F_a$
$F_f=\mu Mg$

As the initial velocity is zero and final velocity is $v$, we have acceleration a
$a=\frac{v^2}{2s}$

Therefore,
$F_a=Ma=M\frac{v^2}{2s}$
$F=M(\frac{v^2}{2s}+\mu g)$