# Simple rolling friciton problem, where did I go wrong?

Word for word:

A50,000kg locomotive is traveling at 10 m/s when its engine and breaks fail. How far will the locomotive roll before it comes to a stop?

Rolling friction for rubber on concrete is 0.02 (by the book)

a= fnet/m = (umg)/m so mass cancels out leaving a = ug = .02(-9.8) = -.196

I use Vf^2 = Vi^2 + 2ad (I'll just say d = distance in meters)
So it works out to
-100 = (-0.392)d
d = 255 m

but wait... the book says the answer's 2550!

I would figure that means the rolling friction is .002 but that's steel on steel (dry)

What am I doing wrong???

## Answers and Replies

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TMM
I'd assume you need the steel on steel coefficient. Don't locomotives with steel wheels run on steel tracks?

I don't see why you're using rubber and concrete.

wow......

I read that and thought because it's a locomotive, then blood drained from my face.

a locomotive, a train........ why the hell am I a physics major? lol

thanks! that's all I needed