An aeroplane of mass M requires a speed v for take off. The length of the 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 given by

The required force is to (i) accelerate the plane from rest to a speed v over a distance s and (ii) to overcome the force of friction (=μR=μMg). The acceleration a required to impart a speed v in a distance s is given by v2−u2=2as. Since, u = 0, we have v2=2 as or a=v2/2s. The force needed to produce this acceleration isF1= mass × acceleration =Mv22sThe force needed to overcome the force of friction is F2=μMg∴Total force needed =F1+F2=Mv22s+μg