1. Oct 9, 2009

### xbebegirlx

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. Oct 10, 2009

### tiny-tim

Welcome to PF!

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

(I blame Schrödinger )

What results did you get?

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

(Hint: you have the force and the distance, so the obvious thing to calculate would be the … ? )

3. Oct 10, 2009

### xbebegirlx

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 ?????????

4. Oct 10, 2009

### tiny-tim

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

Try again.

5. Oct 10, 2009

### xbebegirlx

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

6. Oct 10, 2009

### tiny-tim

(what does WF stand for? )

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

Yes, that's right.

7. Oct 12, 2009

thanks