Question

A 8000kg engine pulls a train at 5 wagons,each of 2000kg ,along a horizontal track . If the engine exerts a force of 40000N and the track offers a friction force of 5000N,then calculate:
a) the next accelerating force
b)the acceleration of the train​

Answer 1

$\huge\star{\underline{\mathtt{\red{A}\pink{n}\green{s}\blue{w}\purple{e}\orange{r}}}}$

a) The net accelerating force:-

Sol:- F = 40000N

F1= 5000N

Fa = F - F1

= 40000 - 5000

= 35000N

Hence,the next accelerating force is 35000N.

b) The acceleration of the train:-

Sol:- Acceleration of the train = a

The engine exerts a force of 40000N on all the five wagons.

Mass of the wagon = 2000kg

No, of wagons = 5

Therefore, m = 2000×5

⠀⠀⠀⠀⠀⠀⠀⠀= 10000kg

Total mass (including the mass of engine).

M = m+8000

= 18000kg

Fa = Ma

= a = $\frac{Fa}{M}$ = $\frac{35000}{18000}$ = 1.944m/s^2

Answer 2

Given:

mass of the engine = 8000 kg

mass of one wagon = 2000 kg

mass of 5 wagons = 10,000 kg

Force exerted by the engine = 40,000 N

Friction offered by the track = 5000 N

Solution:

(a) net accelerating force

Net accelerating force = force provided by the engine - frictional force

Net accelerating force = 40,000 - 5000

Net accelerating force = 35, 000 N

Net accelerating force = 35,000 N

_________________________________________________________

(b)the acceleration of the train

Net accelerating force = mass of the train x acceleration

mass of the train = mass of the engine +  mass of 5 wagons

mass of the train = 8000 + 10,000

mass of the train =18,000 kg

$\rightarrow\mathtt {acceleration=\dfrac{net \:accelerating \: force}{mass\: of \:the \:train }}$

$\rightarrow\mathtt {acceleration=\dfrac{35,000}{18,000}}$

$\rightarrow\mathtt {acceleration=1.94\dfrac{m}{s^{2} }}$

The acceleration of the train is 1.94 m/s^2