Answer:
2551.02 m
Explanation:
Given that, the mass of a locomotive is, [tex]m = 50,000 kg[/tex]
coefficient of friction for steel, [tex]\mu= = 0.002[/tex]
Velocity of the locomotive is [tex]v = 10 m/s[/tex]
By conservation of energy it can be written as:
[tex]KE = E[/tex]
Here, E is the energy because of friction.
Now,
[tex]KE = \frac{1}{2}\times m\times v^2[/tex]
Here, m is the mass, v is the velocity of an object.
Now energy of the friction can be written as,
[tex]E= F\times d =\mu\times N = \mu\times mg[/tex]
Here, [tex]\mu[/tex] is the coefficient of friction , m is the mass and g is the acceleration due to gravity.
Therefore
[tex]1/2\times m\times v^2 = \mu\times m\times g\times d\\1/2\times v^2 = \mu\times g\times d\\d=\frac{1 v^2}{2(g\times \mu)}[/tex]
Substitute all the variables.
[tex]d= \frac{10^2}{2(9.8\times 0.002)}\\\\d=2551.02 m[/tex]
The distance traveled by the locomotive ball before stopping is 2551.02 m
