The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.32 with the floor. If the train is initially moving at a speed of 54 km/h, in how short a distance can the train be stopped at constant acceleration without causing the crates to slide over the floor?

The shortest braking distance is 35.8 m

Explanation:

To solve this problem we must use Newton's second law applied to the boxes, on the vertical axis we have the norm up and the weight vertically down

On the horizontal axis we fear the force of friction (fr) that opposes the movement and acceleration of the train, write the equation for each axis

Y axis

N - W = 0

N = W = mg

X axis

-Fr = m a

-μ N = m a

-μ mg = ma

a = μ g

a = - 0.32 9.8

a = - 3.14 m/s²

We calculate the distance using the kinematics equations

Vf² = Vo² + 2 a x

x = (Vf² - Vo²) / 2 a

When the train stops the speed is zero (Vf = 0)

Vo = 54 km/h (1000m/1km) (1 h/3600s) = 15 m/s

x = (0 - 15²) / 2 (-3.14)

x = 35.8 m

The shortest braking distance is 35.8 m