A 4.0×104kg locomotive, with steel wheels, is traveling at 9.0m/s on steel rails when its engine and brakes both fail. The coefficient of friction is 1.7×10−3.

Question

anonymous · Answer

how far will it go before it stops?

anonymous · Answer

any help would greatly be appreciated...

anonymous · Answer

UncleRhakus can you help

unklerhaukus · Answer

@UnkleRhaukus

unklerhaukus · Answer

$$ m=4.0\times10^4[\text{kg}]\v=9.0 [\text{m/s}]\\mu=1.7\times10^{−3}$$

anonymous · Answer

ok, as for the formula i think i'll be using two different ones. Can you confirm that?

unklerhaukus · Answer

Yeah what are the relevant forces?

anonymous · Answer

Can you elaborate on your question? Are you referring to the Newtons or m coefficient of friciton of 
[1.7 x 10^{3}\]

unklerhaukus · Answer

isn't it -3  ?

anonymous · Answer

yes sorry. i'm new to the equation thing

anonymous · Answer

forces will be drag, friction

unklerhaukus · Answer

type like this 

```
$$1.7 \times 10^{-3}$$
```

unklerhaukus · Answer

what about gravity?

anonymous · Answer

ok yea that also. -g

unklerhaukus · Answer

and the normal force

anonymous · Answer

perpendicular plane, so weight also of $$\4.0[ x \10^{4}]$$

anonymous · Answer

so force would be weight x -g

anonymous · Answer

Force weight=mass*gravity ?

unklerhaukus · Answer

The weight force $$F_w=-mg$$if we take up to be positive

unklerhaukus · Answer

yeah and the normal force will be
$$F_N=+mg$$

unklerhaukus · Answer

(assuming the train track is not on a hill)

anonymous · Answer

no its flat

anonymous · Answer

I don't think I'm putting together the concepts very well.

unklerhaukus · Answer

the friction force will be $$F_f=\mu N$$ and will point in the opposite direction to the motion of the train

anonymous · Answer

ok, i'll try to crunch it out. thanks

unklerhaukus · Answer

What is the question asking for/?

unklerhaukus · Answer

ops $F_N=N$

unklerhaukus · Answer

maybe the the friction force makes more sense as
$$F_f=-μN$$