(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A train has a mass of 5 million kilograms and is traveling south at 22 m/s. If the train driver has to initiate a panic stop, how far will the train go before it comes to a halt. Use 0.08 as the coefficient of kinetic friction.

m=5*10^{6}kg

v=-22 m/s

μ-kinetic friction=0.08

g= 9.8 m/s^{2}

2. Relevant equations

Horizontal surface: F(normal)=F(weight)

F-weight=mg

F-kinetic = (F-normal)*(μ-kinetic friction)

v= sqrt(2*(F-net/m)*x)→x= v^{2}/2*(F-net/m) I'm not sure if it's F(net)/m or g that's supposed to used. My book has both written in it.

F(net)=F(normal)-F(kinetic)

** where F(net)/m comes from the equation F(net)=ma

3. The attempt at a solution

F(weight): (5*10^{6})(9.8) = 4.9*10^{7}N

F(kinetic): (4.9*10^{7})(0.08)= 3.9*10^{6}N

F(net): (4.9*10^{7})-(3.9*10^{6})= 4.5*10^{7}N

F(net)/m: (4.5*10^{7})/(5*10^{6}) = a = 9 m/s^{2}

x=(-22^2)/(2*9)

= 54 m

I've tried this problem multiple times but I keep getting the wrong answer. 54 meters doesn't make sense but none of my other answers have either. I think the problem is somewhere in the F(net)/m step though. Is it suppose to be a instead where a=9.8? Any help would be appreciated!

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# Homework Help: Distance traveled by train using forces and kinetic friction

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