Work, Energy and Power Question

AI Thread Summary
The discussion revolves around a physics problem involving a train's work, energy, and power as it travels a 5 km track with a 40 m vertical rise. Participants clarify that the equation Win = Delta K + delta U + Wout applies, with Wout being zero due to negligible friction. There is confusion regarding the significance of the vertical rise and track length, which is clarified to mean the slope distance. To calculate average mechanical power, participants suggest using the formula average power = Win/delta t, with delta t obtainable through kinematic equations. The conversation highlights the need for understanding the relationship between work, energy changes, and time in this context.
ln85
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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
 
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ln85 said:
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.
 
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