# Problem concerning energy

1. Apr 12, 2004

### JamesL

Problem concerning energy....

Heres the problem:

A train with total mass of 2.16 x 10^6 kg rises 750 m while traveling a distance of 51.4 km at an average speed of 9.12 m/s. The frictional force is .8 percent of the weight.

Given: acceleration of gravity is 9.8 m/s.

*Find the kinetic energy of the train in MJ.
*Find the total change in its potential energy in J.
*Find the energy dissipated by friction in J.
*Find the power output of the trains engines in MW.

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On the first part i tried using K = .5mv^2, but that is apparently not the way to do it. Im assuming i have to factor in the work done by friction as well?

That brings me to another question. In trying to find the energy dissipated by friction i did the following:

(mass)(.8 percent)(distance) = Work done by friction
(2.16 x 10^6)(.8/100)(51400) = 8.8819 x 10^8

this is not correct either... so im puzzled as to what im doing wrong?

Can anyone point me in the right direction for this problem? Thanks!

2. Apr 12, 2004

The kinetic energy at what point? All that's given is the average velocity, so all you can find is the average kinetic energy, and the equation you tried to use is the equation you want to use.

Change in potential energy. What's the formula for potential energy? U = mgh, so $\Delta U = mg \Delta h$. Plug and chug.

You know the force of friction, so how do you get energy from force?

Once again we seem to be looking at average power. P = F/t.

3. Apr 13, 2004

### JamesL

Thanks for the response!

Maybe we are missing something for the first part of the problem. I tried using K = .5mv^2

so... K = (.5)(2.16 x 10^6)(9.12^2) = 8.9828 x 10^7 J

into MJ ------> (8.9828 x 10^7)/(10^6) = 89.828 MJ. However when i submit this answer to my online homework service, it says that it is incorrect.

I got the second part about the change in potential energy correct.

I think i am missing something about the 3rd part. When they ask for the energy dissipated by friction, im assuming they are talking about the work done by friction. Is this correct?
If so, then ive tried doing this:

(mass)(.8 percent)(distance) = Work done by friction
(2.16 x 10^6)(.8/100)(51400) = 8.8819 x 10^8 J

But that is not counted as correct by my homework service. The energy dissipated by friction should be the work done by it, right? And then it should just be (force of friction) x (distance)...???

For the last part, im not sure how to find the "Force" in the equation for power. Perhaps this is how you solve it: if we know the change in kinetic energy (from part 1) that would be equal to the net Work. Since work = (force)(distance)... we should be able to find the force, and then the power. Am I headed in the right direction?

4. Apr 13, 2004

### Staff: Mentor

I'm not clear on the statement of the problem. Is the train moving with constant speed? Uniformly accelerated from rest? What?

The KE is simply 1/2mv^2, but as cookiemonster pointed out, at what point? I assume at the top of the hill, but we need to know how it's moving. If I had to guess, I would assume uniform acceleration: so calculate the final speed based on that assumption. (But you shouldn't have to guess!)

Yes, but the friction force is 0.8 percent of the weight, not the mass!

5. Apr 13, 2004

### JamesL

Ahh! Thanks for the clarification about the force of friction. That had complelely slipped my mind.

I dont know what else to say about the first part. I wrote the problem out exactly as it is on my worksheet. I do believe we are supposed to assume uniform acceleration. Im really not sure what im doing wrong...

6. Apr 14, 2004

### Chen

Ehh, almost... Your answer is correct however, because you forgot two minus signs - firstly, the energy dissipated by any force is minus the work done by that force; secondly, the work done by friction is always going to be negative because it always acts against the motion of the object, so the angle between the force and the displacement vectors is 180.