Locomotive coefficient of friction?

Question

A 5.6×104 locomotive, with steel wheels, is traveling at 16 m/s on steel rails when its engine and brakes both fail. The coefficient of friction is 1.7×10−3.

How far will the locomotive roll before it comes to a stop?

Answer 1

solve this using primitive methods only, as wind resistance, moment of inertia and other factors are omitted in youre question.

Mass = 5,6 x10^4 kg

initial velocity = 16m/s

final velocity = 0m/s

x(distance) = ???

coefficient of friction = 1,7 x10 ^-3

Newtons second law and the equations of kinematics are to be used.

Go for it!

Answer 2

You will need only two kinematic equations to solve this:

Vf² = Vi² + 2ad.

For your needs, the equation will break down to:

d = -Vi² / 2a. remembering that the a is negative

You will find the a, by just using F=ma

Remember, there are 9.8 Newtons per kg.

Your answer will be just under 7800 m