Question

Ans.
opposes the motion of stone.
7. 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 the track offers a friction force of 5000 N, then calculate :
(a) the net accelerating force,
(b) the acceleration of the train, and
(c) the force of wagon 1 on wagon 2-
​

Answer 1

AnswEr :

• The net accelerating force is 35000 N

• Acceleration of the train is 1.944 m/s²

• Force on wagon 2 by wagon 1 is 17774.22 N.

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Given :

• Mass of engine = 8000 kg .

• Mass of each wagon = 2000 kg

• Force acting in backward direction(f) = 5000 N .

• Force exerted by engine(f) = 40000 N

• Total number of wagons = 5

To Find :

(a) The net accelerating force(f')= ?

(b) The acceleration of the train = ?

(c) The force of wagon 1 on wagon 2 = ?

Solution :

• (a) The net accelerating force ,

$\sf\dashrightarrow\ \ \ f' = F-f \\ \\ \\ \dashrightarrow \sf \ \ \ 40000-5000 \\ \\ \\ \dashrightarrow\sf \ \ \ {\underline{\boxed{\sf{\blue{35000 \ N}}}}} \ \bigstar$

Hence, the net accelerating force is 35000 N.

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• (b) The acceleration of the train,

Let us assume that the acceleration be a.

$\sf\dashrightarrow \ \ \ F=(5m+M)a\\ \\ \\ \sf\dashrightarrow \ \ \ 35000=(5 \times 2000+8000)a \\ \\ \\ \sf\dashrightarrow \ \ \ 35000=(10000+8000)a \\ \\ \\ \sf\dashrightarrow \ \ \ 35000=18000a \\ \\ \\ \sf\dashrightarrow \ \ \ a=\dfrac{35000}{18000} \\ \\ \\ \sf\dashrightarrow \ \ \ {\underline{\boxed{\sf{\orange{ \ a = 1.944 \ {m/s}^{2} }}}}} \ \bigstar$

Thus, acceleration of the train is 1.944 m/s².

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• (c)The force of wagon 1 on wagon 2,

Let us assume acceleration of both the wagons be a' .

Total force of 4 wagons :

$\sf\dashrightarrow \ \ \ \dfrac{5000 \times 2000 \times 4}{18000} \\ \\ \\ \sf\dashrightarrow \ \ \ \dfrac{5000 \times 8}{18} \\ \\ \\ \sf\dashrightarrow \ \ \ 2222 .22 N$

Now,

Force accelerating on 4 wagons :

$\sf\dashrightarrow \ \ \ 2000 \times 4 \times 1.944 \\ \\ \\ \sf\dashrightarrow \ \ \ 8000 \times 1.944 \\ \\ \\\sf\dashrightarrow \ \ \ 15552 N$

Total force exertred by wagon 1 on wagon 2 = Frictional force + Accelerating force

$\sf\dashrightarrow \ \ 15552+2222 .22 \\ \\ \\ \sf \dashrightarrow \ \ { \underline{ \boxed{ \sf{ \red{17774.22 N}}}}} \ \bigstar$

Therefore, force on wagon 2 by wagon 1 is 17774.22 N.

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