Train travelling uphill at s.7s degree incline, find friction coefficient

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Homework Help Overview

The problem involves a train traveling on a 3.73-degree incline, where the last car breaks free and coasts without friction. The discussion focuses on determining the time it takes for the car to come to rest, the distance traveled before stopping, and the coefficient of friction for a real train scenario.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the equations of motion and the relationship between frictional force and the coefficient of friction. There are attempts to calculate time and distance using given parameters, and questions arise about how to approach the third part regarding the friction coefficient.

Discussion Status

Some participants have provided guidance on using momentum and frictional force to find the coefficient of friction. There is an exploration of different approaches, including drawing free body diagrams and considering the implications of mass in calculations. The discussion reflects a mix of understanding and uncertainty among participants.

Contextual Notes

Participants express varying levels of familiarity with physics concepts, with some indicating a lack of confidence in their understanding. The discussion includes references to specific values and assumptions related to the problem setup.

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Homework Statement


A train is traveling at 3.73 degree incline at a speed of 3.25 m/s, when the last car breaks free and begins to coast without friction.

(i) How long does it take for the last car to come to rest momentarily?
(ii) How far did the last car travel before (momentarily) coming to rest?
(iii)For a real train the friction between the car and the track can be described by a friction coefficient. Find this coefficient assuming that the time it takes for the car to come to rest is 3s.


Homework Equations



v=u + at
v^2=u^2 + 2as


The Attempt at a Solution



(i) u=3.25 m/s
v=0m/s
s= ?
t= looking for..
a=9.8sin3.25= -0.55 m/sec^2

t= v-u/a
t= 0-3.25 / -0.55 = 5.909 seconds

(ii) u=3.25 m/s
v=0m/s
s= looking for..
t= 5.909 seconds
a=9.8sin3.25= -0.55 m/sec^2

s= v^2 -u^2 / 2a
s= 0^2 - 3.25^2 / 2(-0.55) = 9.602 metres

I haven't a clue about the third part. can anyone help?
 
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Do you know the expression for frictional force in terms of the coefficient of friction? You can use the fact that the rate of change of momentum is equal to the force applied to obtain the coefficient.
 
I'm afraid i don't. I'm new to physics and just need to pass it for an exam so i don't have much of a grasp of it
 
The frictional force is given by \mu N, where \mu is the coefficient of friction, and N is the normal force between the surface and body. If you draw a free body diagram of the coach, you should see that N=mgcos(\theta) where m is the mass of the coach and \theta is the angle of inclination of the incline. Can you write the change in momentum?
 
nope.. I'm hopeless
 
What is the momentum when the coach breaks free from the train? What is the momentum when it is momentarily at rest?
 
momentum is mass multiplied by the velocity.. there would be no momentum if its at rest, and 3.25 multiplied by the mass when it breaks free?
 
Right. So the rate of change is momentum is the difference of momentum divided by the time takes (3 s). Equate that to the frictional force to obtain the coefficient.

If you find this too tricky, another way would be to use the frictional force formula I showed you to get the deceleration, which you can plug in your normal equation (v=u+at, with v=final velocity=0 m/s, u=initial velocity and t=3s, and a=the acceleration you get from the frictional force)
 
Last edited:
ok... got it but does it matter that i don't actually have the mass?
 
  • #10
The mass will cancel.
 
  • #11
that is perfect.. i get it now.. thank you so so much :D
 

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