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: Distance traveled by train using forces and kinetic friction

  1. Jul 4, 2012 #1
    1. The problem statement, all variables and given/known data
    A train has a mass of 5 million kilograms and is traveling south at 22 m/s. If the train driver has to initiate a panic stop, how far will the train go before it comes to a halt. Use 0.08 as the coefficient of kinetic friction.

    m=5*106 kg
    v=-22 m/s
    μ-kinetic friction=0.08
    g= 9.8 m/s2

    2. Relevant equations
    Horizontal surface: F(normal)=F(weight)
    F-weight=mg
    F-kinetic = (F-normal)*(μ-kinetic friction)
    v= sqrt(2*(F-net/m)*x) x= v2/2*(F-net/m) I'm not sure if it's F(net)/m or g that's supposed to used. My book has both written in it.
    F(net)=F(normal)-F(kinetic)
    ** where F(net)/m comes from the equation F(net)=ma

    3. The attempt at a solution
    F(weight): (5*106)(9.8) = 4.9*107 N
    F(kinetic): (4.9*107)(0.08)= 3.9*106 N
    F(net): (4.9*107)-(3.9*106)= 4.5*107 N

    F(net)/m: (4.5*107)/(5*106) = a = 9 m/s2

    x=(-22^2)/(2*9)
    = 54 m

    I've tried this problem multiple times but I keep getting the wrong answer. 54 meters doesn't make sense but none of my other answers have either. I think the problem is somewhere in the F(net)/m step though. Is it suppose to be a instead where a=9.8? Any help would be appreciated!
     
    Last edited: Jul 4, 2012
  2. jcsd
  3. Jul 4, 2012 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi Mg53! Welcome to PF! :smile:

    (try using the X2 button just above the Reply box :wink:)
    i don't understand this …

    you seem to be subtracting the horiztonal friction from the vertical weight

    i] you can't do that

    ii] why did you want to? :confused:

    just do work done = change in kinetic energy :smile:

    (and don't waste space writing in the m, it makes no difference!)
     
  4. Jul 4, 2012 #3
    Hi, thanks for the help :)

    The reason I did that was because I thought the net force was equal to the normal force minus the force of kinetic friction. Like I said though, I had a feeling that's where I was messing up in the problem but I didn't really know what else to do.

    Okay, I'm still a little confused by one part of that.

    Work done=change in kinetic energy
    =(1/2)(5*106)(222)
    = 1.21*109

    I can use that to find the distance using W=Fx but what F do I use? The force of kinetic friction?

    I did that and got x=310 m which I guess might make sense but I'm still not sure if that's right because I don't have an answer key :/
     
    Last edited: Jul 4, 2012
  5. Jul 4, 2012 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi Mg53! :smile:
    you're completely confused :confused:

    the net vertical force is zero (because the vertical acceleration is zero)

    the net horizontal force is the friction (because it's the only horizontal force) ​
    yes of course

    (i'm going to bed … goodnight! :zzz:)
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook