Calculating Work Done by an Engine to Keep Wagon Moving

AI Thread Summary
The discussion revolves around calculating the work done by an engine to maintain a railway wagon's constant speed while it is loaded with coal. The engine pulls the wagon at a speed of 10 m/s, and the work done is calculated to be 50,000 J. However, participants note that this work is not equal to the kinetic energy imparted to the coal due to the nature of the loading process, which involves a downward impulse on the wagon. The work-energy theorem is referenced to further explain the relationship between work and energy in this context. Clarification is sought on the differences between the work done and the kinetic energy associated with the coal.
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Homework Statement



A railway wagon runs on frictionless rails and is pulled by an engine traveling at
10 ms−1 . The wagon is loaded at constant rate with 1000 kg of coal, dropped vertically
from rest for a time of 2 s. What is the work done by the engine to keep the wagon
moving at constant speed? Is the work done equal to the kinetic energy imparted to
the coal and, if not, explain why not.


Homework Equations





The Attempt at a Solution



So the work done is 50,000J right?

My hench is that this is not exactly equal to the KE imparted to the coal...but I am not sure how to explain this...something to do with downward impulse of coal on wagon?

Thanks
 
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anyone?
 
How did you get 50,000J?
 
KE before = 0.5mv^2

KE after = 1/2(m+1000)v^2..

difference = 50,000J

So please could you help with my second Q?
 
Well, do you know about the work-energy theorem?
 

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