1. Okay, so this is a homework problem, just so that everybody knows it.

A train with the mass m=260 metric tons accelerates (from velocity=0) along horizontal, straight rail tracks. The moment the velocity is 36 km/h the power is 650 kW. The resistance against the movement is 0,5 % of the weight of the train.

a)Find the force with which the engine pulls the train forwards and the acceleration when the velocity is 36 km/h.

b)How does the acceleration change if the power of the engine is the same for a while (the acceleration after the one we calculated in a))?

Ok, so my physics teacher says that P=W/t. To this point I agree.
Furthermore he says that W=F*d (d is the distance) and thus P=F*d/t=F*v, and I don't agree. To says that W=F*d, the force must be constant, which it isn't.
So I reckon I'm right and I don't know how I should solve it. Any suggestions?

:-D

2.

3. Originally Posted by thyristor
Ok, so my physics teacher says that P=W/t. To this point I agree.
Furthermore he says that W=F*d (d is the distance) and thus P=F*d/t=F*v, and I don't agree. To says that W=F*d, the force must be constant, which it isn't.
So I reckon I'm right and I don't know how I should solve it. Any suggestions?

:-D
Do it the way your teacher says.

You are right that the work is the an integral of the area under the force-distance curve, but you are just looking for the power at a certain instant. For this purpose the force can be considered a constant.

4. Ok, thanks a lot!
Though, why can the force be considered a constant?

5. Originally Posted by thyristor
Ok, thanks a lot!
Though, why can the force be considered a constant?
The assumption that the force is constant is just that, an assumption. But since the formulas that you quoted earlier, such as P=W/t also assume that quantities like work and power are constant (you are not using differential calculus) it is a reasonable assumption. ( in reality.) The assumption lets you solve the problem, and without calculus.

The point of this problem is for you to show that you understand what energy, power and work are and the relationships among them. Apparently the level of the class is such that calculus is not available as a tool. Therefore to address the basic physics you are having to assume that many things are constant. If you do not do this you wind having to deal with differential equations, and you also need more information than what is given.

6. Originally Posted by thyristor
Ok, thanks a lot!
Though, why can the force be considered a constant?
You are solving for the force at the instant that the velocity is 36 km/h. That's just one point on the curve, so there is just one value of force to consider.

7. Originally Posted by thyristor
Okay, so this is a homework problem, just so that everybody knows it.

A train with the mass m=260 metric tons accelerates (from velocity=0) along horizontal, straight rail tracks. The moment the velocity is 36 km/h the power is 650 kW. The resistance against the movement is 0,5 % of the weight of the train.

a)Find the force with which the engine pulls the train forwards and the acceleration when the velocity is 36 km/h.

b)How does the acceleration change if the power of the engine is the same for a while (the acceleration after the one we calculated in a))?

Ok, so my physics teacher says that P=W/t. To this point I agree.
Furthermore he says that W=F*d (d is the distance) and thus P=F*d/t=F*v, and I don't agree. To says that W=F*d, the force must be constant, which it isn't.
So I reckon I'm right and I don't know how I should solve it. Any suggestions?

:-D
Ok, lets do this using calculus

and

So

Where is the work done on a general object by the total force applied on the object.

Let be the resisting force. Then the total force on the train is , where is the force applied by the engine, so that which shows that

Now plug in the specific numbers for your problem.

Note that here you do not need to assume that anything is constant in either time or space, but you no longer have the simple relation that .[/tex]

8. Ok, but isn't ?
If so, how do we get that ?

9. Originally Posted by thyristor
Ok, but isn't ?
If so, how do we get that ?
is the same thing as

And I did not say that but rather that

10. Originally Posted by thyristor
Ok, but isn't ?
If so, how do we get that ?
We don't. P=fv

11. Ok, sorry my mistake, I made a misprint.
But why is the same as ?
If I have understood it right, is for as in power, so you must be referring to the time derivative.
If so, I dont understand how you get that the same as .

12. Originally Posted by thyristor
Ok, sorry my mistake, I made a misprint.
But why is the same as ?
If I have understood it right, is for as in power, so you must be referring to the time derivative.
If so, I dont understand how you get that the same as .

becomes on differentiating with respect to

or

Here is distance and I used for the same thing earlier.

What may be throwing you is the use of "differentials" like and . There is a way to do this rigorously, but you can simply think of it as a notational convenience. It is pretty common in physics texts.

13. I'm so sorry if I am understanding slowly, and thanks a lot for your help.
I don't understand how you can just multiply with .
I mean, actually we have and I don't understand how we now can write .

14. Originally Posted by thyristor
I'm so sorry if I am understanding slowly, and thanks a lot for your help.
I don't understand how you can just multiply with .
I mean, actually we have and I don't understand how we now can write .
You can either interpret it in the language of differential forms or simply take the equation as really meaning that

15. Ok, thanks a lot! :-D

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