91影视

• Mass of each car $=5.0脳{10}^4\mathrm{kg}$.
• 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鈥檚 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鈥檚 second law to obtain the tension T and the angle of inclination$\mathit{胃}$

Formula:

$F=ma$

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

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

Along the level track, this object experiences a 鈥渇orward鈥� force T exerted by the locomotive, so that Newton鈥檚 second law leads to:

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

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

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

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

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

Thus,

$T-f-mg\sin 胃=ma$

Where, a = 0 (according to given condition)

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

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

$$\begin{gathered} 3.0脳{10}^{5}N-5.2脳{10}^{4}N=\left(1.25脳{10}^{6}kg\right)\left(9.8m/s^2\right)\sin 胃 \\ 2.48脳{10}^5=12.25脳{10}^6脳\sin 胃 \\ \sin 胃=0.02 \\ 胃={\sin }^{-1}\left(0.02\right) \\ 胃=1.16掳 \\ 胃鈮�1.2掳 \end{gathered}$$

Thus, the steepest grade up is 1.2 degrees.