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

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.