Locomotive, friction, and speed question

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The discussion revolves around calculating the stopping distance of a locomotive weighing 4.6×10^4 kg traveling at 14 m/s with a coefficient of friction of 1.6×10−3. The user attempted to apply the friction force and kinematic equations but expressed confusion over their calculations and the units involved. They calculated the friction force as 721 N and derived an acceleration of -450800 m/s², but their subsequent steps led to an incorrect stopping distance. The user seeks clarification on their approach and the correct application of the equations. The conversation highlights the challenges of applying physics concepts to real-world problems.
xgoddess210
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Homework Statement



A 4.6×104 kg locomotive, with steel wheels, is traveling at 14 m/s on steel rails when its engine and brakes both fail. The coefficient of friction is 1.6×10−3.

1.) How far will the locomotive roll before it comes to a stop?

Answer in meters using two significant figures.

Homework Equations




The Attempt at a Solution



I tried to do something here and I thought I had the solution but it was wrong. Please check my work and help me understand this problem.

Here is what I did:

1.6*10^-3 x4.6 *10^4 x9.8 =721 N

-721.28/1.6*10^-3 = -450800 m/s^2

v2-u2/2a ----> 0-14^2/2*-450800 ---> -196/-901600

d=2.17*10 ^-4
 
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Your second equation makes no sense with the units. What equation are you trying to use for step 2?
 
Honestly, I had no idea what I was doing. I tried looking at how someone solved the same problem with different numbers and just plugged my numbers in but it didn't work.
 
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