Q4.6-27PE - College Physics (Urone)

Question

A freight train consists of two 8.00×10⁴ -kg engines and 45 cars with average masses of 5.50×10⁴ kg.

(a) What force must each engine exert backward on the track to accelerate the train at a rate of 5.00×10–2 m/s² if the force of friction is 7.50×10⁵ N, assuming the engines exert identical forces? This is not a large frictional force for such a massive system. Rolling friction for trains is small, and consequently trains are very energy-efficient transportation systems.

(b) What is the force in the coupling between the 37th and 38th cars (this is the force each exerts on the other), assuming all cars have the same mass and that friction is evenly distributed among all of the cars and engines?

Step-by-Step Solution

Answer

(a) The force exerted on the ground is 4.41x10⁵ N.

(b) The force exerted in the coupling between 37^th and 38^th car is 1.497x10⁵ N.

Step 1: Draw the free body diagram of the system of interest

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.

Step 2: Given data

• 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.

Step 3: Determine the known and the unknown elements

Calculate the mass of the engines and cars as:

 

$\begin{array}{c}M=\left\{2\times \left(8\times {10}^4\right)+45\times \left(5.5\times {10}^4\right)\right\}\text{\hspace{0.33em}kg}\\ =\left\{\left(16\times {10}^4\right)+\left(247.5\times {10}^4\right)\right\}\text{\hspace{0.33em}kg}\\ =263.5\times {10}^4\text{\hspace{0.33em}kg}\end{array}$

Step 4: (a) Calculate the force exerted by the train

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,

 

$\begin{array}{c}2F-7.5\times {10}^5\text{\hspace{0.33em}N}=263.5\times {10}^4\text{\hspace{0.33em}kg}\times 5\times {10}^{-2}{\text{\hspace{0.33em}m/s}}^{\text{2}}\\ 2F-7.5\times {10}^5\text{\hspace{0.33em}N}=131750\text{\hspace{0.33em}N}\\ F=\frac{\left(131750+750000\right)\text{\hspace{0.33em}N}}{2}\\ F=4.41\times {10}^5\text{\hspace{0.33em}N}\end{array}$

 

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

Step 5: (b) Determine the force in the coupling between the 37 th and 38 th cars

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, f’ is the force of friction between the 37th and 38th cars.

 

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

 

$\begin{array}{c}f\text{'}=\frac{8\times 7.5\times {10}^5\text{\hspace{0.33em}N}}{47}\\ =1.277\times {10}^5\text{\hspace{0.33em}N}\end{array}$

 

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 f’, and 5x10^-2 m/s² for a in the above expression, and we get,

$\begin{array}{l}F\text{'}=8\times 5.5\times {10}^4\text{\hspace{0.33em}kg}\times 5\times {10}^{-2}{\text{\hspace{0.33em}m/s}}^2+1.277\times {10}^5\text{\hspace{0.33em}N}\\ F\text{'}=\left(0.22\times {10}^5+1.277\times {10}^5\right)\text{\hspace{0.33em}N}\\ F\text{'}=1.497\times {10}^5\text{\hspace{0.33em}N}\end{array}$ 

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