91Ó°ÊÓ

Here, M is the mass of the engines plus the cars, F is the force exerted on the track, f is the friction force, and a is the acceleration of the train.

• Mass of the two engines = 8.00×10⁴ kg
• average masses of 45 cars = 5.50×10⁴ kg.
• acceleration of the train = 5.00×10^–2 m/s² .
• the force of friction is 7.50×10⁵ N.

Calculate the mass of the engines and cars as:

$\begin{array}{c}M=\left\{2×\left(8×{10}^4\right)+45×\left(5.5×{10}^4\right)\right\}\text{ k²µ}\\ =\left\{\left(16×{10}^4\right)+\left(247.5×{10}^4\right)\right\}\text{ k²µ}\\ =263.5×{10}^4\text{ k²µ}\end{array}$

From the Newton’s second law of motion, we get the folllowing:

${\mathit{F}}_{\mathbf{n}\mathbf{e}\mathbf{t}}\mathbf{=}\mathit{M}\mathit{a}$

$\mathbf{2}\mathit{F}\mathbf{−}\mathit{f}\mathbf{=}\mathit{M}\mathit{a}$ ……………………… (i)

Here, F_net is the net force acting on the system.

Substitute 263.5x10⁴ kg for M, 7.5x10⁵ N for f, and 5x10^-2 m/s² for a in equation (i) and we get,

$\begin{array}{c}2F−7.5×{10}^5\text{ N}=263.5×{10}^4\text{ k²µ}×5×{10}^{−2}{\text{ m/²õ}}^{\text{2}}\\ 2F−7.5×{10}^5\text{ N}=131750\text{ N}\\ F=\frac{\left(131750+750000\right)\text{ N}}{2}\\ F=4.41×{10}^5\text{ N}\end{array}$

Hence, the force exerted on the ground is 4.41x10⁵ N.

There are eight cars behind the 37^th car. The force of friction is equally divided between 45 cars and two engines.

Calculate the force of friction between the 37^th and 38^th cars as it is evenly distributed between cars and engines as, we get

$\mathbf{\text{f'=}}\frac{\mathbf{8}\mathbf{f}}{\mathbf{47}}$

Here, ´Ú’ is the force of friction between the 37th and 38th cars.

Substitute 7.5x10⁵ N for f in the above expression, and we get,

$\begin{array}{c}f\text{'}=\frac{8×7.5×{10}^5\text{ N}}{47}\\ =1.277×{10}^5\text{ N}\end{array}$

Calculate the force exerted in the coupling between 37^th and 38^th car as:

$F\text{'}=8ma+f\text{'}$

Here, m is the mass of the car.

Substitute 5.5x10⁴ kg for m, 1.277x10⁵ N for ´Ú’, and 5x10^-2 m/s² for a in the above expression, and we get,

$\begin{array}{l}F\text{'}=8×5.5×{10}^4\text{ k²µ}×5×{10}^{−2}{\text{ m/²õ}}^2+1.277×{10}^5\text{ N}\\ F\text{'}=\left(0.22×{10}^5+1.277×{10}^5\right)\text{ N}\\ F\text{'}=1.497×{10}^5\text{ N}\end{array}$

Hence, the force exerted in the coupling between 37^th and 38^th car is 1.497x10⁵ N.