First slide

Second law

Question

The aeroplane of mass M requires a speed v for take-off. The length of the run-way is s and the coefficient of friction between the tyres and ground is µ. Assuming that the plane accelerates

Moderate

A

$M (\frac{v^{2}}{2 s} - μg)$
B

$M (\frac{v^{2}}{2 s} + μg)$
C

$M (\frac{2 v^{2}}{s} - 2 μg)$
D

$M (\frac{2 v^{2}}{s} + 2 μg)$

Solution

The acceleration a required to impart a speed v in a distance s is given by
$v^{2} - u^{2} = 2 as or v^{2} = 2 as (∵ u = 0)$
$or a = v^{2} / 2 s$
The force F₁ needed to produce this acceleration is
$F_{1} = M \times a = M (v^{2} / 2 s)$
The force F₂ needed to overcome the friction is
$F_{2} = μMg$
$\begin{aligned} ∴ Total force needed & = F_{1} + F_{2} \\ = M (\frac{v^{2}}{2 s} + μg) \end{aligned}$

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