Question

An airplane requires a speed of $80\,km\,{h^{ - 1}}$ for take-off, to run on the ground being $100\,m$. The mass of the plane is $10,000\,kg$ and the coefficient of friction between the plane and the ground is $0.2$. Assuming that the plane accelerates uniformly during the take-off, the minimum force required by the engine for take-off is:
(A) $2 \times {10^4}\,{\text{N}}$
(B) $2.43 \times {10^4}\,{\text{N}}$
(C) $4.43 \times {10^4}\,{\text{N}}$
(D) $8.86 \times {10^4}\,{\text{N}}$

Answer 1

Find the acceleration of the plane using a kinematic equation. Then apply Newton’s second law to the motion of the plane and calculate the force provided by the engine of the plane. The force of kinetic friction is given as, ${f_k} = \mu mg$.

Formula used:
${v^2} = {u^2} + 2as$
Here, v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance.

Complete step by step answer:
We have given the velocity of the plane required to take-off is, $80\,km\,{h^{ - 1}}$. We have to convert it into m/s.
$v = \left( {80\,\dfrac{{km}}{h}} \right)\,\left( {\dfrac{{1000\,m}}{{1\,km}}} \right)\left( {\dfrac{{1\,hr}}{{3600\,s}}} \right)$
$\Rightarrow v = 22.22\,m/s$
Since the initial velocity is zero, we can use kinematic equation to determine the acceleration of the plane as follows,
${v^2} = {u^2} + 2as$
$\Rightarrow {v^2} = 2as$
Here, u is the initial velocity, a is the acceleration and s is the distance.
Substituting $v = 22.22\,m/s$ and $s = 100\,m$ in the above equation, we get,
${\left( {22.22} \right)^2} = 2a\left( {100} \right)$
$\Rightarrow a = \dfrac{{{{\left( {22.22} \right)}^2}}}{{200}}$
$\Rightarrow a = 2.47\,m/{s^2}$
Now, we can apply Newton’s second law on the motion of the plane as follows,
$F - {f_k} = ma$ …… (1)
Here, F is the force due to the engine of the plane, ${f_k}$ is the kinetic friction between the tyres and road and m is the mass of the plane.
We know the kinetic friction is expressed as,
${f_k} = \mu mg$
Here, $\mu$ is the coefficient of kinetic friction and g is the acceleration due to gravity.
Therefore, we can rewrite equation (1) as follows,
$F - \mu mg = ma$
$\Rightarrow F = m\left( {a + \mu g} \right)$
Substituting 10000 kg for m, $a = 2.47\,m/{s^2}$ for a. 0.2 for $\mu$ and $9.8\,m/{s^2}$ for g in the above equation, we get,
$\Rightarrow F = \left( {10000} \right)\left( {2.47 + \left( {0.2 \times 9.8} \right)} \right)$
$\Rightarrow F = \left( {10000} \right)\left( {4.43} \right)$
$\therefore F = 4.43 \times {10^4}\,{\text{N}}$
Therefore, the force due to the engine on the plane is $4.43 \times {10^4}\,{\text{N}}$.

So, the correct answer is option (C).

Note: Students should know that the direction of the frictional force is always opposite to the motion of the body. Therefore, it gains negative sign each time, when we assume the direction of motion is towards the positive axis. We cannot use equation (1) if the acceleration is not uniform.