Question

The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction 0.25 ofwith the floor. If the train is initially moving at a speed of 48 km/h, in how short a distance can the train be stopped at constant acceleration without causing the crates to slide over the floor?

Short answer

The distance is 36.23 m

Step-by-step solution

Given

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

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

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

Determining the concept

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:

$\sum _{}{F}_{net}=ma$

Determining the distance

The frictional force acting on the train is given by,

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

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

$$\begin{gathered} \sum _{}F=ma \\ -f_s=ma \\ -{\mu }_s\left(mg\right)=ma \\ a=-{\mu }_{s}g \end{gathered}$$

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.