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