# Work Done/Power

1. Jun 5, 2010

### bon

1. The problem statement, all variables and given/known data

A railway wagon runs on frictionless rails and is pulled by an engine travelling 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.

2. Relevant equations

3. The attempt at a solution

Done the first part fine..Just unsure about the second part...I know the work-energy theorem..But am i right in thinking that the WD =/= KE imparted to coal as some KE of coal is lost as heat as it collides with the wagon on its descent?

2. Jun 5, 2010

### Staff: Mentor

You can just compare the work done with the gain in KE of the coal and see if they're equal. You're right that mechanical energy is not conserved as the coal is brought up to speed--think of the wagon as inelastically colliding with the coal. (It's not the KE due the coal's falling that matters here; you can just assume that the coal is dropped into the wagon from a low height with small vertical speed.)

3. Jun 5, 2010

### ehild

The work-energy theorem is valid only to a rigid body of constant mass.You have variable mass here. Newton's second law is valid in the form dp/dt =F. The momentum is mv, so
dp/dt= v dm/dt +m dv/dt.
v is constant, so F=v*dm/dt. Multiplying both sides with v, and integrating with respect to time you get the relation between work and the change of the KE in this case.

As Doc Al said, you can not expect energy conservation, as this is kind of inelastic collision.

ehild