Question

A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and track offers a friction force of 5000 N then, calculate:
(a) The net acceleration force
(b) The acceleration of the train
(c) The force of wagon 1 on wagon 2

Answer 1

We will use formula of net accelerated force which is taken by the frictional force offered by track from the Force exerted by engine, total acceleration which is done by the net acceleration force is divided by mass of train and the force as per the formula to answer this question.

Formula used:
This calculation is taken
$acc{{.}_{net}}={{F}_{engine}}-{{F}_{frict.}},acc{{.}_{train}}=\dfrac{acc{{.}_{net}}}{{{m}_{train}}},acc{{.}_{train}}={{m}_{engine}}+\left( {{m}_{wagons}}\times no{{.}_{wagons}} \right)$ and $4\times \text{mass of wagon }\times acceleration$

Complete answer:
(a) We are given the mass of the engine to be 8000kg and the number of wagons to be 5. We have a mass of a wagon as 2000 kg. Also, the force exerted by the engine is as 40000 N along with a frictional force as 5000 N.
So, the net accelerated force can be taken out as $acc{{.}_{net}}={{F}_{engine}}-{{F}_{frict.}}$ so, we can write this information as 40000 - 5000 = 35000. Thus, the net accelerated force is 35000 N.

(b) Suppose that the train moves with a net acceleration of $m/{{s}^{2}}$. We will use the formula of acceleration to solve it further. We will calculate it by $acc{{.}_{train}}=\dfrac{acc{{.}_{net}}}{{{m}_{train}}}$.
We are not given any information related to the mass of the train. So, we will use formula $acc{{.}_{train}}={{m}_{engine}}+\left( {{m}_{wagons}}\times no{{.}_{wagons}} \right)$ to find it. Therefore, by substituting the values we get $8000+\left( 5\times 2000 \right)=18000$. Now we will put all desired values in the formula $acc{{.}_{train}}=\dfrac{acc{{.}_{net}}}{{{m}_{train}}}$ and get the required acceleration of the train is $\dfrac{35000}{18000}=1.944$.
So, the calculated acceleration of the train is $1.944m/{{s}^{2}}$.

(c) According to the question it is clear that the force is applied only on wagon 1. This results in the net force on the rest of 4 wagons to be equal to the force applied by another wagon. Now, the force applied on wagon 1 on to the wagon 2 is going to be $4\times \text{mass of wagon }\times acceleration=4\times 2000\times 1.944$ which is 15552 N.
Hence, the force of net acceleration is 35000 N, the train’s acceleration is $1.944m/{{s}^{2}}$ and the required force on wagon 2 by wagon 1 is 15552N.

Note:
As we know that frictional force is a force which tries to restrict the speed or acceleration of any object, this is why we have subtracted it from the acceleration of the engine. This is because the frictional force was acting as a restriction to it. Since, we know that the acceleration of any object is the division of net acceleration to the mass of that object this is why we used $acc{{.}_{train}}=\dfrac{acc{{.}_{net}}}{{{m}_{train}}}$. Since, the force is exerted on wagon 1 and the rest are 4 wagons left this is why we used $4\times \text{mass of wagon }\times acceleration$ and not $5\times \text{mass of wagon }\times acceleration$. If these points are not noted while solving the question then, we will get the wrong answer.