Answer
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Hint: The system of wagon alone is not isolated as the engine exerts a force on it. So, by using Newton’s second law, its acceleration can be calculated. The tension in the coupling acts as the pulling force on the wagon, so again using Newton’s second law pulling force can be calculated.
Formulae used:
$1\,ton=1000kg$
$F=ma$
Complete step-by-step solution:
Given, an engine of mass $100\,tons$ is pulling a wagon of mass $80\,tons$.
$\begin{align}
& 1\,ton=1000kg \\
& \Rightarrow 80tons=8\times {{10}^{4}}kg \\
& \Rightarrow 100tons={{10}^{5}}kg \\
\end{align}$
The force applied to pull the wagon by the engine is $4500N$.From Newton’s second law, the force is given by-
$F=ma$ - (1)
Here,
$F$ is the force
$m$ is the mass
$a$ is acceleration
The whole system of engine and wagon are moving with acceleration $a$, therefore,
$\begin{align}
& a=\dfrac{F}{{{m}_{1}}+{{m}_{2}}} \\
& \Rightarrow a=\dfrac{4500}{8\times {{10}^{4}}+10\times {{10}^{4}}} \\
& \Rightarrow a=\dfrac{4500}{18\times {{10}^{4}}}m{{s}^{-2}} \\
& \therefore a=0.025m{{s}^{-2}} \\
\end{align}$
Therefore, the system of engine and wagon are moving with acceleration of $0.025m{{s}^{-2}}$
The tension in the coupling between engine and wagon act as the pulling force on the wagon.
From eq (1), the tension is calculated as-
$F={{m}_{w}}a$
Here, ${{m}_{w}}$ is the mass of the wagon on which tension acts
We substitute values in the above equation to get,
$\begin{align}
& F=8\times {{10}^{4}}\times 0.025 \\
& \therefore F=2000N \\
\end{align}$
Therefore, the tension in the coupling between the engine and the wagon is $2000N$.
Hence, the correct option is (2).
Note:
The Newton’s Second law says that force is required to change the state of rest or motion of a body, this means if a body is accelerating then a force acts on it. Tension is a pulling force which acts along the axis of continuous objects like string, rod etc. By Newton’s third law, it acts equal and opposite on both the objects at its ends. Convert the units as required.
Formulae used:
$1\,ton=1000kg$
$F=ma$
Complete step-by-step solution:
Given, an engine of mass $100\,tons$ is pulling a wagon of mass $80\,tons$.
$\begin{align}
& 1\,ton=1000kg \\
& \Rightarrow 80tons=8\times {{10}^{4}}kg \\
& \Rightarrow 100tons={{10}^{5}}kg \\
\end{align}$
The force applied to pull the wagon by the engine is $4500N$.From Newton’s second law, the force is given by-
$F=ma$ - (1)
Here,
$F$ is the force
$m$ is the mass
$a$ is acceleration
The whole system of engine and wagon are moving with acceleration $a$, therefore,
$\begin{align}
& a=\dfrac{F}{{{m}_{1}}+{{m}_{2}}} \\
& \Rightarrow a=\dfrac{4500}{8\times {{10}^{4}}+10\times {{10}^{4}}} \\
& \Rightarrow a=\dfrac{4500}{18\times {{10}^{4}}}m{{s}^{-2}} \\
& \therefore a=0.025m{{s}^{-2}} \\
\end{align}$
Therefore, the system of engine and wagon are moving with acceleration of $0.025m{{s}^{-2}}$
The tension in the coupling between engine and wagon act as the pulling force on the wagon.
From eq (1), the tension is calculated as-
$F={{m}_{w}}a$
Here, ${{m}_{w}}$ is the mass of the wagon on which tension acts
We substitute values in the above equation to get,
$\begin{align}
& F=8\times {{10}^{4}}\times 0.025 \\
& \therefore F=2000N \\
\end{align}$
Therefore, the tension in the coupling between the engine and the wagon is $2000N$.
Hence, the correct option is (2).
Note:
The Newton’s Second law says that force is required to change the state of rest or motion of a body, this means if a body is accelerating then a force acts on it. Tension is a pulling force which acts along the axis of continuous objects like string, rod etc. By Newton’s third law, it acts equal and opposite on both the objects at its ends. Convert the units as required.
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