Question

The rolling resistance for steel on steel is quite low; the coefficient of rolling friction is typically ? r =0.002 . Suppose a 180,000 kg ${\rm kg}$ locomotive is rolling at 25 m/s ${\rm m/s}$ on level rails.

The rolling resistance for steel on steel is quite low; the coefficient of rolling friction is typically ? r =0.002 mu_{rm r}=0.002. Suppose a 180,000 kg {rm kg}locomotive is rolling at 25 m/s on level rails. Part A If the engineer disengages the engine, how much time will it take the locomotive to coast to a stop? Express your answer using two significant figures. Part B How far will the locomotive move during this time? Express your answer to two significant figures and include the appropriate units.

Part A

If the engineer disengages the engine, how much time will it take the locomotive to coast to a stop?

Express your answer using two significant figures.

Part B

How far will the locomotive move during this time?

Express your answer to two significant figures and include the appropriate units.

Answer 1

Concepts and reason

The concepts used to solve this problem are Newton’s second law and the fundamental equations of kinematics.

Initially, the acceleration of the object can be calculated from the equation of motion of the object. Then, the time taken to stop the object can be calculated by using the equations of kinematics. Finally, the travelled distance by the object can be calculated using the kinematics equation.

Fundamentals

From Newton’s second law, the equation of motion of the object is,

$F=ma $

Here, F is the force acting on the engine, m is the mass of the engine, and a is the acceleration of the engine.

The expression for the friction force is,

Here, is the frictional force acting on the engine.

The kinematics equation which is used to calculate the time taken by the object is,

Here, is the final velocity, is the initial velocity, andis the time.

The equation for the travelled distance from kinematics is,

Here, is the travelled distance.

(A)

The expression for the friction force is,

Here, is the frictional force acting on the engine.

The engine is disengaged. On the way to stop the only force acting on the engine is the friction force.

Here, is the friction coefficient.

The expression for the acceleration is,

$a=4,8 $

Substitute for, forin the above expression.

The negative indicates, the engine is slowing down.

The expression for the time taken by the object is,

Rearrange the above expression for t.

Substitute for, 25m/s for , and for in the above expression.

(B)

The expression for the travelled distance is,

Rearrange the above expression for s.

Substitute for, 25m/s forand for in the above expression.

Ans: Part A

The time taken by the object to stop is

Part B

The travelled distance by the object is.