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Distance traveled by train using forces and kinetic friction

  • Thread starter Mg53
  • Start date
  • #1
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Homework Statement


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

Homework 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

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:

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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Welcome to PF!

Hi Mg53! Welcome to PF! :smile:

(try using the X2 button just above the Reply box :wink:)
F(net): (4.9*10^7)-(3.9*10^6)= 4.5*10^7 N
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!)
 
  • #3
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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:
  • #4
tiny-tim
Science Advisor
Homework Helper
25,832
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Hi Mg53! :smile:
The reason I did that was because I thought the net force was equal to the normal force minus the force of kinetic friction.
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)​
… I can use that to find the distance using W=Fx but what F do I use? The force of kinetic friction?
yes of course

(i'm going to bed … goodnight! :zzz:)
 

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