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Stopping distance help please

  1. Oct 9, 2009 #1
    a 25000kg train travel down a track at 18m/s. a cat wander onto the track 45m ahead of train, causing the conductor to slam on the brakes. The train skids to a stop. If the brakes can provide 75,000 newtons of friction, will the conductor have enough stopping distance to avoid hitting the cat?
     
  2. jcsd
  3. Oct 10, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi xbebegirlx! Welcome to PF! :smile:
    Is this a lab experiment?

    (I blame Schrödinger :rolleyes:)

    What results did you get? :smile:

    Show us what you've tried, and where you're stuck, and then we'll know how to help! :wink:

    (Hint: you have the force and the distance, so the obvious thing to calculate would be the … ? :smile:)
     
  4. Oct 10, 2009 #3
    I have the Mass=25000kg
    friction force = 75000n
    time 18m/s
    I dont know where the 45m fit in
    I was confuse, to get stopping distance dont I need the coefficient of friction between the train and the rail? usually it giving but in this problem it not

    I think I need to use this equation
    change KE=1/2mvf^2-mvi^2
    normal force:mg 25000kg(9.8m/s gravity) =245000n
    friction force =coefficient * Normal force: 75000 =245000x; x =.306 this is my coefficient

    vi^2=2*coefficient*gravity*Distance
    18^2=2(.306)(9.8)D
    324=5.998D
    D=54.018m

    so yes it have enough distance to avoid hitting the cat

    Can someone help me check see if I got it right ?????????
     
  5. Oct 10, 2009 #4

    tiny-tim

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    Hi xbebegirlx! :smile:
    It doesn't really! … you find D, and then right at the end you check whether D ≤ 45. :wink:
    ah, you're not reading the question properly :redface:
    … the question doesn't bother to give you µ, it gives you µmg (= 75000N) all at once.

    Try again. :smile:
     
  6. Oct 10, 2009 #5
    change KE =WF
    ke=1/2mv
    1/2*25000kg*18m/s^2 =75000x
    d = 54M
    no conduction dont have enough time to stoo

    I hope this one right...because if not then I am stuck
     
  7. Oct 10, 2009 #6

    tiny-tim

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    (what does WF stand for? :confused:)

    I think you meant …
    ke=1/2mv2 = work done
    1/2*25000kg*182m/s^2 =75000xd
    so d = 54m
    no the conductor doesn't have enough time to stop :wink:

    Yes, that's right. :smile:
     
  8. Oct 12, 2009 #7
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