Answer:
Explanation:
Given
coefficient of kinetic friction [tex]\mu _k=0.002[/tex]
mass of locomotive=180,000 kg
v=10 m/s
deceleration provided by friction [tex]a=\mu _kg[/tex]
[tex]a=0.002\times 9.8=0.0196 m/s^2[/tex]
Time taken to stop completely is
v=u+at
where v=final velocity
u=initial velocity
a=acceleration
[tex]0=10-0.0196\times t[/tex]
t=510.20 s
Distance traveled during this point
[tex]v^2-u^2=2 a s[/tex]
[tex]0-10^2=2(-0.0196)s[/tex]
[tex]s=\frac{100}{0.0392}[/tex]
s=2551.02 m
