Question

A toy train in C.B. garden Kota consist of \[4\] coaches of \[100{\text{ }}kg\] each. Its coaches are joined by a magnetic coupler which can support \[1200{\text{ }}N.\] Mass of engine is \[600{\text{ }}Kg.\] The external horizontal force on engine which is applied by ground for no link to break is (Assume no friction between coaches and rail).
A. \[1200{\text{ }}N\]
B. \[2400{\text{ }}N\]
C. \[3000{\text{ }}N\]
D. \[1000{\text{ }}N\]

Answer 1

Hint: To solve this question, we will first start with taking $f \geqslant ma$, to get the value of acceleration of the toy train. Now after getting the value of a, we will apply the formula $F = Ma$, this is the formula of horizontal force on the engine, by putting all the formulas in this formula, we will get our required answer.

Complete step by step answer:
We have been given that a toy train in C.B. garden Kota consist of \[4\] coaches of \[100{\text{ }}kg\] each. The coaches of toy train are joined by a magnetic coupler which can support \[1200{\text{ }}N.\] It is given that mass of engine is \[600{\text{ }}Kg.\] We need to find the external horizontal force on engine which is applied by ground.
So, the number of coaches of a given toy train \[ = {\text{ }}4\]
Mass of each coach of the given toy train \[ = {\text{ }}100{\text{ }}kg\]
So, the total mass of \[4\] coaches \[ = {\text{ }}4 \times 100{\text{ }} = {\text{ }}400{\text{ }}kg\]
Force supported by magnetic coupler, by which the coaches of toy train are joined \[ = {\text{ }}1200{\text{ }}N\]
We have also been given the mass of engine \[ = {\text{ }}600{\text{ }}Kg\]
Here, $f \geqslant ma$
where, f \[ = \] force supported by magnetic coupler
m \[ = \] total mass of \[4\] coaches
a \[ = \] acceleration of the train
On putting the value in the above equation, we get
$\begin{gathered}
  1200 \geqslant 400a \\
  a \leqslant \dfrac{{1200}}{{400}} \\
  a \leqslant 3m{s^{ - 2}} \\
\end{gathered} $
Now, $F = Ma$
where, F \[ = \] external horizontal force on engine
M \[ = \] mass of engine \[ + \] mass of \[4\] coaches
a \[ = \] acceleration of train
On putting the value in the above equation, we get
$\begin{gathered}
  f = (600 + 400)a \\
   = 1000(3) \\
   = 3000N \\
\end{gathered} $
So, the external horizontal force on engine is \[3000N.\]

So, the correct answer is “Option C”.

Note:
In the question, the word magnetic coupler is mentioned. So, magnetic coupling is a coupling which transfers torque from one shaft to another, but it uses a magnetic field instead of physical mechanical connection, while torque is the product of force and distance and has units of Newton meters.