Question

An aeroplane requires taking off the speed of $108{\text{ kmph}}$ the run on the ground being $100{\text{ m}}$. Mass of the plane is ${10^4}{\text{ kg}}$ and the coefficient of friction between the plane and the ground is $0.2$. Assuming the plane accelerates uniformly the minimum force required is $(g = 10{\text{ m}}{{\text{s}}^{ - 2}})$
A. $2 \times {10^4}{\text{N}}$
B. $2.43 \times {10^4}{\text{N}}$
C. $6.5 \times {10^4}{\text{N}}$
D. $8.86 \times {10^4}{\text{N}}$

Answer 1

Hint: This is a kinematics problem for which we need to have a clear concept of initial and final velocities and the various relations between them and also about acceleration and retardation. We will consider the forces as per the free body diagram.

Complete step by step answer:
We know that the required take-off speed is $108{\text{ kmph}}$ which is equal to $30{\text{ m}}{{\text{s}}^{ - 1}}$ over a run of $100{\text{ m}}$.
So using the following relation
${v^2} - {u^2} = 2as$
Substituting the value of initial velocity, we get
$ {v^2} - 0 = 2as$
$ \Rightarrow a = \dfrac{{{v^2}}}{{2s}}$
Where, $u$ is the initial velocity, $v$ is the final velocity, $a$ is the acceleration and $s$ is the distance.

Let the force required be $F$. Now the force which is developed by the engine will be equal to,
$F - \mu mg = ma$
Where $\mu $is the coefficient of friction which is given.
Therefore by substituting the value we get,
$ \Rightarrow F = 0.2 \times {10^4} \times 10 + {10^4} \times \dfrac{{{{30}^2}}}{{2 \times 100}}$
$ \therefore F = 6.5 \times {10^4}{\text{N}}$

Hence option C is correct.

Additional information: Acceleration produced in an aeroplane is described in units of the force called “Gs.” A pilot in a steep turn may experience forces of acceleration equivalent to many times the force of gravity. The ‘g’ in g-force refers to the word gravity, the force currently allowing you to simply sit down and read this. While the force has little to do with gravity, it provides an insight into the measurement of what g-force really is – essentially acceleration.

Note: It should be noted that when acceleration becomes negative then it is called retardation. Retardation is the same thing as acceleration but in the opposite direction. The initial velocity of an object at rest will be zero and it should be kept in mind when substituting the values for initial and final velocities.