Question

A 50,000 kg locomotive is traveling at 10 m/s when its engine and brakes both fail. How far will the locomotive roll before it comes to a stop? Assume the track is level.

Answer 1

2551.02 m

Step-by-step 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 = (1)/(2)* m* 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* d =\mu* N = \mu* mg`

Here,
`\mu` is the coefficient of friction , m is the mass and g is the acceleration due to gravity.

Therefore

`1/2* m* v^2 = \mu* m* g* d\\1/2* v^2 = \mu* g* d\\d=(1 v^2)/(2(g* \mu))`

Substitute all the variables.

`d= (10^2)/(2(9.8* 0.002))\\\\d=2551.02 m`

The distance traveled by the locomotive ball before stopping is 2551.02 m