Static Friction- Train stops, boxes move?

AI Thread Summary
The discussion revolves around calculating the stopping distance of a train with crates on a flatcar, given a coefficient of static friction of 0.25 and an initial speed of 48 km/h. The user attempts to determine the maximum deceleration and stopping distance but encounters confusion in their calculations. A key point raised is the need to square the initial speed when applying the kinematic equation. The correct approach involves using the maximum static friction to find the deceleration, which is crucial for preventing the crates from sliding. Accurate application of the formulas will yield the correct stopping distance.
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Homework Statement


The floor of a railroad flatcar 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 a distance can the train be stopped at constant acceleration without causing the crates to slide over the floor?

Homework Equations


F = ma
f(s)max= u(s)N
final v^2-init. v^2 = 2a*s(distance)
Newton's second law?

The Attempt at a Solution


I've been cracking my skull on this one... I search and found this https://www.physicsforums.com/showthread.php?t=186380 however, I couldn't figure it out.
Is the maximum deceleration -.25g? But when I plug that into: 0-13.33=2*a*s => -13.33/(-2*.25*9.8) = 2.72m and obviously something is wrong... Thanks in advance.
 
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battlebball said:

Homework Equations


F = ma
f(s)max= u(s)N
final v^2-init. v^2 = 2a*s(distance)
Newton's second law?
Is the maximum deceleration -.25g?

But when I plug that into: 0-13.33=2*a*s => -13.33/(-2*.25*9.8) = 2.72m and obviously something is wrong...

You forgot to square the speed v=13.33 m/s.


ehild
 
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