1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper

    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


    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


    User Avatar
    Science Advisor
    Homework Helper

    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
    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


    User Avatar
    Science Advisor
    Homework Helper

    (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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook