Answer
Please see the work below.
Work Step by Step
We know that the acceleration of the train is given as
$a=\mu g$
We plug in the known values to obtain:
$a=(0.58)(9.8)=5.684\frac{m}{s^2}$
The required distance is given as
$d=\frac{v^2}{2a}$
$d=\frac{(\frac{140}{3.6})^2}{(2)(5.684)}=130m$
Hence, the train will stop in time.