A locomotive accelerates a 25-car train along a level track. Every car has a mass of $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{}\mathbf{kg}$and is subject to a frictional force $\mathit{f}\mathbf{=}\mathbf{250}\mathit{v}$, where the speed vis in meters per second and the force fis in newtons. At the instant when the speed of the train is 30km/h, the magnitude of its acceleration is 0.20 m/s².

(a) What is the tension in the coupling between the first car and the locomotive? (b) If this tension is equal to the maximum force the locomotive can exert on the train, what is the steepest grade up which the locomotive can pull the train at 30 km/h?

• Mass of each car $=5.0\times {10}^4\mathrm{kg}$.
• The frictional force on each car $f=250v$ here ,v is the speed in m/s .
• Speed of the train, $v=30\mathrm{km}/\mathrm{h}=8.3\mathrm{m}/\mathrm{s}$.
• The magnitude of Acceleration$a=0.20\mathrm{m}/{\mathrm{s}}^2$ .

The problem deals with Newton’s second law of motion, which states that the acceleration a, of an object is dependent upon the net force F, acting upon the object and the mass m, of the object. Apply Newton’s second law to obtain the tension T and the angle of inclination$\mathit{\theta }$

Formula:

$F=ma$

Treat all 25 cars as a single object of mass$m=25\times 5.0\times {10}^4\mathrm{kg}$ and (when the speed is 30 km/h = 8.3 m/s) subject to a friction force equal to:

$$\begin{gathered} f=25\times 250\times 8.3 \\ =5.2\times {10}^4\mathrm{N} \end{gathered}$$

Along the level track, this object experiences a “forward” force T exerted by the locomotive, so that Newton’s second law leads to:

$$\begin{gathered} T-f=ma \\ T=f+ma \end{gathered}$$

Substitute the values in the above expression, and we get,

$$\begin{gathered} T=5.2\times {10}^4\mathrm{N}+\left(25\times 5.0\times {10}^4\mathrm{kg}\right)\times \left(0.2\mathrm{m}/{\mathrm{s}}^2\right) \\ T=5.2\times {10}^4\mathrm{N}+25\times {10}^4\mathrm{N} \\ T=30.2\times {10}^4\mathrm{N} \\ T\approx 3.0\times {10}^5\mathrm{N} \end{gathered}$$

Thus,the tension in the coupling between the first car and the locomotive is $3.0\times {10}^5\mathrm{N}$.

The +x direction (which is the only direction in which we will be applying Newton’s second law) is uphill (to the upper right in our sketch).

Thus,

$T-f-mg\sin \theta =ma$

Where, a = 0 (according to given condition)

$$\begin{gathered} T-f-mg\sin \theta =ma \\ T-f=mg\sin \theta \end{gathered}$$

Substitute the values in the above expression, and we get,

$$\begin{gathered} 3.0\times {10}^{5}N-5.2\times {10}^{4}N=\left(1.25\times {10}^{6}kg\right)\left(9.8m/s^2\right)\sin \theta \\ 2.48\times {10}^5=12.25\times {10}^6\times \sin \theta \\ \sin \theta =0.02 \\ \theta ={\sin }^{-1}\left(0.02\right) \\ \theta =1.16° \\ \theta \approx 1.2° \end{gathered}$$

Thus, the steepest grade up is 1.2 degrees.