
During take-off a jet airplane of mass \[10000\text{ kg}\] can produce a maximum thrust of $40\text{ kN}$. At the point of take-off, the plane experiences a drag force of $20\text{ kN}$.
a. Calculate the initial acceleration of the plane.
b. Calculate the acceleration of the plane at the point of take-off.
Answer
519k+ views
Hint: To solve this question we will use our knowledge of Newton’s laws of motion specifically the Newton’s second law of motion. Newton's second law of motion states that the rate of change of momentum of a body is directly proportional to the force applied on it.
Complete step by step answer:
According to Newton’s second law of motion, the force acting on a body is directly proportional to the mass of the body and the acceleration of the body. Hence if we write it in the form of an equation, we would get:
$F=ma$
Where $F$ is the force acting on the body,
$m$ is the mass of the body,
And $a$ is the acceleration of the body.
a. Now to calculate the initial acceleration of the plane we would rearrange the above formula as:
$a=\dfrac{F}{m}$
It is given that the maximum thrust produced by the jet airplane initially is $F=40000\text{ N}$ and the mass of the jet airplane is given as $m=10000\text{ kg}$.
On substituting the values in the equation, we would get:
$\begin{align}
& a=\dfrac{40000}{10000} \\
& \therefore a=4\text{ m}{{\text{s}}^{-2}} \\
\end{align}$
b. Now in the second part we need to calculate the acceleration of the plane at the point we need to take in consideration the drag force experienced by the jet airplane. Thus, the net force experienced by the jet airplane will be as follows:
$\begin{align}
& F=\left( 40000-20000 \right)\text{ N} \\
& \Rightarrow F=20000\text{ N} \\
\end{align}$
The acceleration of the plane at the point of take-off will be:
$\begin{align}
& a=\dfrac{20000}{10000} \\
& \therefore a=2\text{ m}{{\text{s}}^{-2}} \\
\end{align}$
Note: Newton’s second law of motion is the most important law because it quantitatively describes the force. It can be written in different forms. It can be written in terms of rate of change of momentum of a body or the acceleration of the body. Both of these mean the same thing.
Complete step by step answer:
According to Newton’s second law of motion, the force acting on a body is directly proportional to the mass of the body and the acceleration of the body. Hence if we write it in the form of an equation, we would get:
$F=ma$
Where $F$ is the force acting on the body,
$m$ is the mass of the body,
And $a$ is the acceleration of the body.
a. Now to calculate the initial acceleration of the plane we would rearrange the above formula as:
$a=\dfrac{F}{m}$
It is given that the maximum thrust produced by the jet airplane initially is $F=40000\text{ N}$ and the mass of the jet airplane is given as $m=10000\text{ kg}$.
On substituting the values in the equation, we would get:
$\begin{align}
& a=\dfrac{40000}{10000} \\
& \therefore a=4\text{ m}{{\text{s}}^{-2}} \\
\end{align}$
b. Now in the second part we need to calculate the acceleration of the plane at the point we need to take in consideration the drag force experienced by the jet airplane. Thus, the net force experienced by the jet airplane will be as follows:
$\begin{align}
& F=\left( 40000-20000 \right)\text{ N} \\
& \Rightarrow F=20000\text{ N} \\
\end{align}$
The acceleration of the plane at the point of take-off will be:
$\begin{align}
& a=\dfrac{20000}{10000} \\
& \therefore a=2\text{ m}{{\text{s}}^{-2}} \\
\end{align}$
Note: Newton’s second law of motion is the most important law because it quantitatively describes the force. It can be written in different forms. It can be written in terms of rate of change of momentum of a body or the acceleration of the body. Both of these mean the same thing.
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

