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

  1. Aug 15, 2011 #1
    1. The problem statement, all variables and given/known data
    A train is travelling 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.


    2. Relevant equations

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


    3. 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?
     
  2. jcsd
  3. Aug 15, 2011 #2
    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.
     
  4. Aug 15, 2011 #3
    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
     
  5. Aug 15, 2011 #4
    The frictional force is given by [itex]\mu N[/itex], where [itex]\mu[/itex] 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 [itex]N=mgcos(\theta)[/itex] where m is the mass of the coach and [itex]\theta[/itex] is the angle of inclination of the incline. Can you write the change in momentum?
     
  6. Aug 15, 2011 #5
    nope.. I'm hopeless
     
  7. Aug 15, 2011 #6
    What is the momentum when the coach breaks free from the train? What is the momentum when it is momentarily at rest?
     
  8. Aug 15, 2011 #7
    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?
     
  9. Aug 15, 2011 #8
    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: Aug 15, 2011
  10. Aug 15, 2011 #9
    ok... got it but does it matter that i don't actually have the mass?
     
  11. Aug 15, 2011 #10
    The mass will cancel.
     
  12. Aug 15, 2011 #11
    that is perfect.. i get it now.. thank you so so much :D
     
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