Locomotive, friction, and speed question

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SUMMARY

The discussion revolves around calculating the stopping distance of a locomotive weighing 4.6×104 kg traveling at 14 m/s, with a coefficient of friction of 1.6×10−3. The correct approach involves using the formula for deceleration due to friction, which is derived from Newton's second law. The calculated deceleration is approximately -450800 m/s2, leading to a stopping distance of approximately 2.17×10−4 meters, indicating a significant error in the initial calculations presented by the user.

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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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