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Question: A railway engine of mass \(100\,tons\) is pulling a wagon of mass \(80\,tons\) with a force of \(450...

A railway engine of mass 100tons100\,tons is pulling a wagon of mass 80tons80\,tons with a force of 4500N4500N. The resistance force acting is 1N1N per ton. The tension in the coupling between the engine and the wagon is:
(1). 1600N1600N
(2). 2000N2000N
(3). 200N200N
(4). 1500N1500N

Explanation

Solution

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:
1ton=1000kg1\,ton=1000kg
F=maF=ma

Complete step-by-step solution:
Given, an engine of mass 100tons100\,tons is pulling a wagon of mass 80tons80\,tons.
1ton=1000kg 80tons=8×104kg 100tons=105kg \begin{aligned} & 1\,ton=1000kg \\\ & \Rightarrow 80tons=8\times {{10}^{4}}kg \\\ & \Rightarrow 100tons={{10}^{5}}kg \\\ \end{aligned}
The force applied to pull the wagon by the engine is 4500N4500N.From Newton’s second law, the force is given by-
F=maF=ma - (1)
Here,
FF is the force
mm is the mass
aa is acceleration

The whole system of engine and wagon are moving with acceleration aa, therefore,
a=Fm1+m2 a=45008×104+10×104 a=450018×104ms2 a=0.025ms2 \begin{aligned} & 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{aligned}

Therefore, the system of engine and wagon are moving with acceleration of 0.025ms20.025m{{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=mwaF={{m}_{w}}a
Here, mw{{m}_{w}} is the mass of the wagon on which tension acts

We substitute values in the above equation to get,
F=8×104×0.025 F=2000N \begin{aligned} & F=8\times {{10}^{4}}\times 0.025 \\\ & \therefore F=2000N \\\ \end{aligned}

Therefore, the tension in the coupling between the engine and the wagon is 2000N2000N.

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.