Power Problem: Calculating the Power Required to Push a Train Up an Incline

In summary: Yes, this is correct. From here, you could use the second equation to calculate the time it takes for the train to reach a point 1 m high: t= 21582000/9.81=2.42 seconds.
  • #1
adk
8
0

Homework Statement



What power is required to push a train weighing 110,000 kilograms up an incline of 1 in 100 at 20 kilometers per hour, all frictional resistances being neglected?

Homework Equations


P = ΔE/ ΔT


The Attempt at a Solution



This is the only power equation we discussed:
P= ΔE/ ΔT = (m)(g)(h)/t
This doesn't really fit with the data given b/c there's a velocity. Also
I don't know how to express that ratio of 1 in 100 in the equation.
 
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  • #2
adk said:

Homework Statement



What power is required to push a train weighing 110,000 kilograms up an incline of 1 in 100 at 20 kilometers per hour, all frictional resistances being neglected?

Homework Equations


P = ΔE/ ΔT


The Attempt at a Solution



This is the only power equation we discussed:
P= ΔE/ ΔT = (m)(g)(h)/t
This doesn't really fit with the data given b/c there's a velocity. Also
I don't know how to express that ratio of 1 in 100 in the equation.
the power required to push the train at constant speed may be determined from Power = Force times velocity, where the force is the pushing force which you can calculate from equilibrium considerations. Or you can calculate that force and use P= work/time, where work is the pushing force times the distance traveled in the direction of the force (letting h=1 m, find the distance traveled up the incline to reach the 1 m high mark using the 1:100 slope), then you can calculate the time it takes for the train to reach a point 1 m high using kinematics. Or else use your listed equation, finding t in the same manner...it all leads to the same answer.
 
  • #3
PhanthomJay said:
the power required to push the train at constant speed may be determined from Power = Force times velocity, where the force is the pushing force which you can calculate from equilibrium considerations.

If we use the first equation:

F=m*a = (110,000)(9.81) = 1079100
P=1079100*20km/hr=21582000 N

Is this correct? Where do you go from here?
 

1. What is the "Power Problem Pushing a Train?"

The "Power Problem Pushing a Train" is a common issue in transportation engineering where a train experiences difficulty in moving due to insufficient power from the locomotive.

2. What causes the "Power Problem Pushing a Train?"

The most common cause of the "Power Problem Pushing a Train" is an imbalance between the power output of the locomotive and the weight and resistance of the train. Other factors such as track conditions, weather, and mechanical issues can also contribute to this problem.

3. How is the "Power Problem Pushing a Train" addressed?

The "Power Problem Pushing a Train" can be addressed by adjusting the locomotive's power output, redistributing the weight of the train, or improving the track conditions. In some cases, a secondary locomotive may be added to provide additional power.

4. Is the "Power Problem Pushing a Train" a safety concern?

While the "Power Problem Pushing a Train" can cause delays and inconvenience, it is not typically considered a safety concern. However, if the issue is not addressed, it can lead to mechanical failures and potential accidents.

5. How can the "Power Problem Pushing a Train" be prevented?

The "Power Problem Pushing a Train" can be prevented by regularly maintaining and upgrading the locomotives and tracks, properly balancing the weight and resistance of the train, and ensuring that the power output of the locomotive is sufficient for the load it is carrying.

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