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.

 

[nlightner]

, it seems like you are having trouble understanding the problem. Let me break it down for you.

First, we have a train with a mass of 2.4*10^6 kg. This train enters a 5 km stretch of track with a vertical rise of 40 m. This means that the train is going uphill and will increase in height by 40 m over the course of 5 km.

The train enters this stretch of track at a speed of 1.0 m/s and leaves at 3.0 m/s. This means that the train is accelerating, since its speed has increased from 1.0 m/s to 3.0 m/s.

Now, let's look at the equation Win = Delta K + delta U + Wout. Win represents the work done on the train, while delta K represents the change in kinetic energy and delta U represents the change in potential energy. Wout represents the work done by the train.

Since there is no frictional drag, Wout is equal to 0. This means that the work done on the train is equal to the change in kinetic energy plus the change in potential energy.

To find the average mechanical power delivered by the engine for the climb, we can use the equation P = W/delta t, where P represents power, W represents work, and delta t represents time.

We can find the work done on the train by using the equation W = F*d, where F represents force and d represents distance. In this case, the force is equal to the mass of the train multiplied by its acceleration, which we can find by using the equation a = (vf-vi)/t, where vf represents final velocity, vi represents initial velocity, and t represents time.

Once we have the work done on the train, we can divide it by the time it took for the train to climb the 40 m (which we can find by using d=1/2*at^2) to find the average mechanical power delivered by the engine for the climb.

I hope this helps clarify the problem for you. Remember, it's important to break down the problem into smaller parts and use the appropriate equations to solve it. Keep practicing and don't be afraid to ask for help if you're still confused. Good luck!