Distance Problem: Find distance of 50,000 kg locomotive w/ 0 m/s Vf

  • Thread starter vac
  • Start date
If the train has lots of friction, it will not go very far. If it is well oiled and the track is excellent, it will go a considerable distance before stopping. So no answer unless you know the coefficient of friction or something related like the deceleration or the time to stop. In summary, a 50,000 kg locomotive is traveling at 10 m/s when its engine and brakes both fail. Known: Vi=10 m/s, Vf=0 m/s, mass=50,000 kg. Unknown: how far will the locomotive roll before it comes to a stop? Assume the track is level.
  • #1

vac

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Homework Statement


A 50,000 kg locomotive is traveling at 10 m/s when its engine and brakes both fail.

Known
-----------------
Vi= 10 m/s
Vf = 0 m/s
mass = 50,000 kg
_______________
Find distance



Homework Equations


How far will the locomotive roll before it comes to a stop? Assume the track is level.


The Attempt at a Solution


F = ma
[tex]a = \frac{0-10}{0-t}[/tex]
[tex]v_f = at+v_i t[/tex]
[tex]0 = \frac{0-10}{0-t}t+10 m/s t[/tex]

after using distance formula, my answer is 29.8 m
 
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  • #2
Welcome to PF, vac!
The problem is not solvable without further information. If the train has lots of friction, it will not go very far. If it is well oiled and the track is excellent, it will go a considerable distance before stopping. So no answer unless you know the coefficient of friction or something related like the deceleration or the time to stop.
 
  • #3
This is the question that supposed to have all the information: "A 50,000 kg locomotive is traveling at 10 m/s when its engine and brakes both fail. How far will the locomotive roll before it comes to a stop? Assume the track is level."

Also the answer in the back of the book is 2.6 km
I don't know how they got that answer. Now I am thinking they used [itex]F_k = u_k F_N[/itex]
[itex]F_k = ({rubber~~on~~~~concrete~~} u_k = 0.80 ) F_N[/itex]
[itex]F_k = 0.80 F_N[/itex]
 
  • #4
vac said:
This is the question that supposed to have all the information: "A 50,000 kg locomotive is traveling at 10 m/s when its engine and brakes both fail. How far will the locomotive roll before it comes to a stop? Assume the track is level."

Also the answer in the back of the book is 2.6 km
I don't know how they got that answer. Now I am thinking they used [itex]F_k = u_k F_N[/itex]
[itex]F_k = ({rubber~~on~~~~concrete~~} u_k = 0.80 ) F_N[/itex]
[itex]F_k = 0.80 F_N[/itex]
Doesn't locomotives usually run on rails?
 
  • #5
Ah, you can look up the coefficient of friction! Wouldn't it be steel wheel on steel track? The friction force causes the train to decelerate. Calculate the acceleration and then you just have an accelerated motion problem to solve for the distance.

I must add that this thinking is not really correct; the kinetic coefficient of friction between track and wheel would only apply if the wheel was locked and sliding on the track. When rolling with no braking, the friction involved would be air resistance and friction of the moving parts in the bearings which you have no way to calculate. However, the writer of the problem may be thinking of it just the way you are - work it out and see if you get the "correct" answer.

If you haven't posted the complete wording of the problem, please do so!
 
  • #6
lep11 said:
Doesn't locomotives usually run on rails?

And they use steel wheels to boot.
 
  • #7
  • #8
SteamKing said:
And they use steel wheels to boot.
Indeed. I thought it'd be obvious since I already mentioned the rails . But anyway the coefficient will be much smaller than 0.8.
 
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  • #9
lep11, does your marker appreciate students going a little further than necessary?
If so, you could begin with the given answer and work backward to find the coefficient of friction.
Then use that coefficient to work the problem through to find the stopping distance.
 
  • #10
I think it is steel on steel so [itex]F_k = u_k F_N[/itex]
[itex]F_k = ({steel~~on~~steel~~(dry)~~} u_k = 0.60 ) F_N[/itex]
[itex]F_k = 0.60 F_N[/itex]

I still get the wrong answer though. Any ideas?
 
  • #11
0.6 is a tremendous coefficient of friction! I calculate that the train will stop in 1.7 seconds, as it goes distance 8.49 meters. I guess that is not what the creator of the question had in mind.

Steam King's link has 0.010...0.035 for a car on a road; something in that range would be reasonable.
You might take a guess somewhere in there and run it through again. Or work backward from the correct stopping distance to see what the coefficient needs to be to get the right answer.

Actually, if most of the marks are for the work, we should concentrate on getting your calculation right regardless of the value for u or the distance. Did we agree on 8.49 meters for the u = 0.6?
 
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  • #12
The coefficient of resistance for a steel wheel on a rail is 1 to 2 orders of magnitude smaller than the resistance of a car on a road, according to the link in post #7.
 
  • #13
The thing is, it ISN'T steel SLIDING on steel. It is steel rolling on steel plus air resistance plus bearing and gear frictions. The question writer should have put it in the question (or if it is from a book, the coefficient should be in the chapter preceding the question).

See slide 21 here for a description of how rolling friction occurs with train wheels
This problem was taken from slide 23. It assumes you saw slides 21 and 22 where the u is given.
http://www.docstoc.com/docs/134983793/PowerPoint-Presentation
 
  • #14
I understand this question is very old but the coefficient of static friction for rolling friction for steel on steel is 0.002 (given in the book).

Solution:

Friction = 0.002*(490,000 N) = 980 N
F = m*a -->

-980 N = 50,000kg * a
a = -0.0196 m/s/s
Solve for distance:
V^2-Vo^2 = 2a(X-Xo)

X-Xo = 2.6 km
 

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