Calculating Work Done by an Engine to Keep Wagon Moving

[bon]

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

 

[bon]

anyone?

 

[diazona]

How did you get 50,000J?

 

[bon]

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?

 

[diazona]

Well, do you know about the work-energy theorem?