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.
g= 9.8 m/s2
Horizontal surface: F(normal)=F(weight)
F-kinetic = (F-normal)*(μ-kinetic friction)
v= sqrt(2*(F-net/m)*x) → x= v2/2*(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.
** where F(net)/m comes from the equation F(net)=ma
The Attempt at a Solution
F(weight): (5*106)(9.8) = 4.9*107 N
F(kinetic): (4.9*107)(0.08)= 3.9*106 N
F(net): (4.9*107)-(3.9*106)= 4.5*107 N
F(net)/m: (4.5*107)/(5*106) = a = 9 m/s2
= 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!