Work, Energy and Power Question

[ln85]

1. A train of mass 2.4*10^6 kg enters a 5 km stretch of track with a vertical rise of 40 m at a speed of 1.0 m/s and leaves at 3.0 m/s. Assuming that frictional drag is negligible, find:

A) Each term of the equation Win = Delta K + delta U + Wout

B) Average Mechanical Power delivered by the engine for the climb if it is done at constant acceleration.



2. A) Win = Delta K + delta U + Wout
K= 1/2*mv^2
U=mgh


3. A)I can't seem to get the picture of what is happening. I know that Wout is zero since there is no frictional drag, but I don't get the 40m rise and 5 km stretch part.
B) average power = Win/delta t, how do you get delta t?

Please help, I'm really confused. Thanks.

Lisa

 

[PhanthomJay]

I can't seem to get the picture of what is happening. I know that Wout is zero since there is no frictional drag, but I don't get the 40m rise and 5 km stretch part.

I believe it means that the track length measured along the slope is 5000 m; and the track rises 40 m above it's start point in that stretch.

B) average power = Win/delta t, how do you get delta t?

You can use one of the kinematic equations.