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 $. Let us suppose that the plane accelerates uniformly at the time of the take-off, the minimum force needed by the engine of the plane for take-off will be given as,
$\begin{align}
& A.M\left( \dfrac{{{v}^{2}}}{2s}+\mu g \right) \\
& B.M\left( \dfrac{{{v}^{2}}}{2s}-\mu g \right) \\
& C.M\left( \dfrac{2{{v}^{2}}}{2s}+2\mu g \right) \\
& D.M\left( \dfrac{2{{v}^{2}}}{s}+2\mu g \right) \\
\end{align}$
Answer
610.5k+ views
Hint: The resultant force required for the take-off will be the sum of force because of the frictional force and the acceleration of the plane. And also as the initial velocity is given as zero. The acceleration of the plane is given as the ratio of the square of final velocity to twice the length of the runway. These all will help you to solve this question.
Complete step-by-step answer:
First of all let us calculate the net force required for the take-off of the plane. It is given as the total sum of the forces due to force of friction and the acceleration of the plane. This can be written as,
$F={{F}_{f}}+{{F}_{a}}$
Where ${{F}_{f}}$ be the force of friction and ${{F}_{a}}$ be the force due to acceleration of the plane.
As we all know the force of friction can be given as,
${{F}_{f}}=\mu mg$
Where $\mu $be the coefficient of friction, $m$be the mass of the plane and $g$be the acceleration due to gravity.
And also the acceleration of the plane is given as,
$a=\dfrac{{{v}^{2}}}{2s}$
Therefore the force due to acceleration will be,
${{F}_{a}}=Ma=M\dfrac{{{v}^{2}}}{2s}$
Taking the sum of both will give the minimum force required for the take off.
That is,
${{F}_{a}}=\mu Mg+M\dfrac{{{v}^{2}}}{2s}$
Taking the mass of the plane outside will give,
${{F}_{a}}=M\left( \mu g+\dfrac{{{v}^{2}}}{2s} \right)$
So, the correct answer is “Option A”.
Note: Frictional force is a kind of force which prevents the motion of an object. It is basically seen in the direction opposite to the applied force. They are present when two surfaces get in contact with each other.
Complete step-by-step answer:
First of all let us calculate the net force required for the take-off of the plane. It is given as the total sum of the forces due to force of friction and the acceleration of the plane. This can be written as,
$F={{F}_{f}}+{{F}_{a}}$
Where ${{F}_{f}}$ be the force of friction and ${{F}_{a}}$ be the force due to acceleration of the plane.
As we all know the force of friction can be given as,
${{F}_{f}}=\mu mg$
Where $\mu $be the coefficient of friction, $m$be the mass of the plane and $g$be the acceleration due to gravity.
And also the acceleration of the plane is given as,
$a=\dfrac{{{v}^{2}}}{2s}$
Therefore the force due to acceleration will be,
${{F}_{a}}=Ma=M\dfrac{{{v}^{2}}}{2s}$
Taking the sum of both will give the minimum force required for the take off.
That is,
${{F}_{a}}=\mu Mg+M\dfrac{{{v}^{2}}}{2s}$
Taking the mass of the plane outside will give,
${{F}_{a}}=M\left( \mu g+\dfrac{{{v}^{2}}}{2s} \right)$
So, the correct answer is “Option A”.
Note: Frictional force is a kind of force which prevents the motion of an object. It is basically seen in the direction opposite to the applied force. They are present when two surfaces get in contact with each other.
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