Force, Friction, Feel like I am not given sufficient info

[1MileCrash]

Homework Statement

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

Homework Equations

The Attempt at a Solution

I've worked the entirety of this chapter's problems and this is literally number 5 out of ten dozen. Can someone explain to me how I am supposed to go about this without mass or a normal force? I don't know what to do with just a coefficient of static friction..

 

[PeterO]

Homework Statement

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

Homework Equations

The Attempt at a Solution

I've worked the entirety of this chapter's problems and this is literally number 5 out of ten dozen. Can someone explain to me how I am supposed to go about this without mass or a normal force? I don't know what to do with just a coefficient of static friction..

Method #1. Since you are not given the mass of the crate(s), it clearly doesn't matter - so assume a nice convenient figure like 14 kg. [I always like any assumed mass to be even, so that fractions are not introduced when you calculate Kinetic Energy. 2 x 7 is good, since 7 is a prime number so you are not likely to get an incorrect answer by co-incidence. [note that 2+2 = 2x2 so you can get the correct answer by coincidence when working with just a 2]
As you go through the problem you will find you alternately multiply and divide by this mass showing it didn't matter what value you chose.

Method #2. Let the mass of the crate(s) be M. As you work through the problem you will find that M cancels out - so you will get a "number only" answer in the end.

 

[1MileCrash]

I got it, I just called normal force mg, so acceleration needed to overcome max static friction was equal to ug.

Seems like weight matters in this case though. Wouldn't those crates be less likely to slide if I put 500kg of mass in them?

 

[PeterO]

I got it, I just called normal force mg, so acceleration needed to overcome max static friction was equal to ug.

Seems like weight matters in this case though. Wouldn't those crates be less likely to slide if I put 500kg of mass in them?

I might seem like the mas mattered, but your physics calculation just showed it doesn't. have faith! Trust the Force [calculation] Luke!