The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.320 with the floor. If the train is initially moving at a speed of 49.0 km/h, in how short a distance can the train be stopped at constant acceleration without causing the crates to slide over the floor?

Answer: 29.50 m

Explanation: In order to calculate the higher accelation to stop a train without moving the crates inside the wagon which is traveling at constat speed we have to use the second Newton law so that:

f=μ*N the friction force is equal to coefficient of static friction multiply the normal force (m*g).

f=m. a=μ*N = m*a = μ*m*g = m*a

then

a=μ*g=0.32*9.8m/s^2 = 3.14 m/s^2

With this value we can determine the short distance to stop the train

as follows:

x = vo*t - (a/2) * t^2

Vf=0 = vo-a*t then t=vo/a

Finally; x=vo*vo/a-a/2 * (vo/a) ^2=vo^2/2a = (49*1000/3600) ^2 / (2*3.14) = 29.50 m