91Ó°ÊÓ

Coefficient of static friction,${\mathrm{μ}}_s=0.25$

Initial speed of the train,$v_0=48\mathrm{km}/\mathrm{h}=13.33\mathrm{m}/\mathrm{s}$

Final speed of the train,$v_f=0\mathrm{m}/\mathrm{s}$

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. Use the Newton's 2nd law of motion to find the acceleration of the train and then by using the Kinematic equations of motion, find the distance travelled by train.

Formula:

$\underset{}{∑}{F}_{net}=ma$

The frictional force acting on the train is given by,

$$\begin{gathered} f_s={\mathrm{μ}}_{s}N \\ ={\mathrm{μ}}_s\left(mg\right) \end{gathered}$$

By using Newton's 2nd law along the horizontal direction,

$$\begin{gathered} \underset{}{∑}F=ma \\ -f_s=ma \\ -{\mathrm{μ}}_s\left(mg\right)=ma \\ a=-{\mathrm{μ}}_{s}g \end{gathered}$$

This is the acceleration of the train.

Now, by using the Kinematic equation of motion,

$$\begin{gathered} v^2=v_0^2+2as \\ s=\frac{v^2-v_0^2}{2a} \\ =\frac{-{\left(13.33\right)}^2}{2\left(\left(-0.25\right)\left(9.81\right)\right)} \\ =36.23m \end{gathered}$$

Therefore, the distance is 36.23 m.