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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.

User Hick
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1 Answer

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Answer:

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

User Logan Reed
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