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A toy train has an engine and 5 wagons. The mass of the engine is 0.8 kg and of each wagon 0.2 kg. The engine applies a force of 0.40 N. Assuming a frictionless track, the force applied by the first wagon on the second wagon is:
1. 0.32 N
2. 0.64 N
3. 0.96 N
4. 0.28 N

1 Answer

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Correct Answer - Option 1 : 0.32 N

Ans: Option 1)

CONCEPT:

  • Newton's second law of motion: The acceleration of an object is directly proportional to the net force and inversely proportional to its mass.

i.e. \(F=ma\) 

where F is the force in N, m is mass in kg and a is acceleration in m/s2.

CALCULATION:

Given:

Mass of the engine (\(M\)) = 0.8 kg 

Number of wagons = 5

Mass of the wagons (\(m\)) = 0.2 kg

Force exerted by the engine (\(F\) ) = 0.40 N

Frictional force offered by the track (\(f\)) = 0 N ... as the track is frictionless given.

1) Net accelerating force (\(F_a\)) = \(F-f\)

\(F_a= 0.40-0\)

\(F_a=\) 0.40

2) Consider the acceleration of the train be a m/s2

Acceleration is the ratio of net acceleration force by mass, i.e

\(a=\frac{F_{a}}{m}\)

Where m is the mass of the train

The total mass of the train (\(m\)) = Mass of the engine + (Mass of the wagons  × Number of wagons)

\(m=0.8+(5×0.2)\)

m= 1.8 kg

Acceleration of the train (\(a\)) = 0.40÷ 1.8

a= 0.223 m/s2

3) The net force on the last four wagons is equal to the force applied by the second wagon on the first wagon. 

Let us consider the acceleration of the wagons is \(a'\)

\(0.40= (5m)a'\)  ⇒ \(0.40=(5×0.2)a'\)

a'= 0.40 m/s2

Then the mass of the last four wagons is given as, 

m'=4× 0.2

\(\)m= 0.8 kg

Now let us calculate the net force on the last four wagons

\(F'=m'× a'\)

F'= 0.8× 0.4

F'= 0.32 N 

\(\therefore\) The force applied by the first wagon on the second wagon is \(0.32 N \) .

The correct option is 1)

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