Question

9.
An engine pulls a train of 30 compartments, each of mass 103 kg, with a force of 1.5x 105 N. Calculate
the force in the coupling between 21st and 22nd compartment.
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Answer 1

Answer:

The answer is 60 x 10^5 N .

Explanation:

Acceleration of the wagons = a

Force = total mass x acceleration

a = Force / total mass

a = 1.5 x 10^5 / 30 x 10^3 = 0.05 x 10^2 = 5 m/s^2  

Number of compartments from 22 to 30 = 8  

Mass of 8 compartments = 8 X 1.5 x 10^5 = 12 x 10^5 kg  

acceleration of 8 compartments = 5 m/s^2

Tension between 31-22 coupling = force required to produce an acceleration of 1 ms-2 in 8 compartments

T = 12 x 10^5 kg   x 5 m/s-2 = 60 x 10^5 N

Answer 2

Given :

Force=$1.5*10^{5} \N$

Mass =$10^{3} \ kg$

To find:

Force coupling between 21st and 22nd compartment=?

Solution :

We know that

$Force = Mass\ * \ Acceleration\\\\Acceleration = \frac{Force}{Mass}$

Total Mass =$30 * 10^{3}$

Acceleration =$1.5*10^{5} \ N$

Putting these value into the acceleration equation we get

$a = \frac{1.5 * 10^5 }{ 30 *10^3 }\\ =5\ ms^{2}$ 

$Total\ number\ of \ compartment\ between\ 22\ to\ 30=8$

$Mass\ of \ the\ 8\ compartments\ \\ =\ 8 * 1.5 * 10^5 \\=\ 12 * 10^5 \ kg\ \\Acceleration = 5 m/s^2$

Therefore tension  between 31-22 coupling  can  be  determined  by  

= Mass of  the 8 compartments * Acceleration

Tension = 12 x 10^5 kg   * 5 m/s-2

$= 60 x 10^5 N$