Static Friction- Train stops, boxes move?

Click For Summary
SUMMARY

The discussion revolves around calculating the stopping distance of a train loaded with crates, given a coefficient of static friction of 0.25. The initial speed of the train is 48 km/h (13.33 m/s). The maximum deceleration is determined to be -0.25g, leading to a calculated stopping distance of 2.72 meters. A critical error identified in the calculations was the failure to square the initial speed when applying the kinematic equation.

PREREQUISITES
  • Understanding of Newton's second law (F = ma)
  • Knowledge of static friction and its coefficient (f(s)max = u(s)N)
  • Familiarity with kinematic equations, specifically final v² - initial v² = 2a*s
  • Basic grasp of acceleration due to gravity (g = 9.8 m/s²)
NEXT STEPS
  • Review kinematic equations in physics for motion analysis
  • Study the concept of static friction and its applications in real-world scenarios
  • Explore advanced problems involving deceleration and friction in physics
  • Learn about the effects of different coefficients of friction on motion dynamics
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of static friction and motion calculations.

battlebball
Messages
4
Reaction score
0

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.
 
Physics news on Phys.org
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
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
3K
Replies
16
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
20
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 5 ·
Replies
5
Views
3K