- #1
JBP_Ace
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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?
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?