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I'm a pure mathematician by trade, and have been trying to teach myself A-level mechanics. (This is not homework, it is purely self-study.)

I've been working through the exercises and have come up against a stubborn problem. Here it is:


A train of mass 150 tonnes is moving up a straight track which is inclined to 2$^{\circ}$ to the horizontal. The resistance to the motion of the train from non-gravitational forces has magnitude 6 kN and the train's engine is working at a constant rate of 350 kW.

Calculate the maximum speed of the train.

The track now becomes horizontal. The engine continues to work at 350 kW and the resistance to motion remains 6 kN

Find the initial acceleration of the train.


Calculate the maximum speed of the train.

This is straightforward. When the train reaches its maximum velocity it will have zero net acceleration, i.e. the force from the engine must neutralise the force from resistance. The gravitational resistance has a magnitude of $150000g\sin 2$ and so the total resistance is $(6 + 150g\sin 2)\times 10^3.$ Using this with the formula $P = Fv$ we get $3.5 \times 10^5 = (6 + 150g\sin 2)v\times 10^3$ and so

$$v = \frac{350}{6 + 150g\sin 2} \approx 6.11 \, \text{m/s} \, .$$

Find the initial acceleration of the train.

This is where I come unstuck. The question asks for the "initial" velocity straight after telling you that the hill levels out. As such I assume it wants the "initial" acceleration right after reaching the peak of the hill.

Since the train is on level ground, the only resistance is the 6 kN non-gravitational resistance. (No value for $\mu$, the coefficient of friction, is given and I assume there is no friction.) We also know that the engine works at a rate of 350 kW. Using the formula $P = Fv$ we get:

$$ 350000 = F \times \frac{350}{6 + 150g \sin 2} \iff F = 1000(6 + 150g\sin 2) \, . $$

Finally, we use the formula $F = ma$ to give $1000(6+150g\sin 2) = 150000a$, and in turn:

$$ a = \frac{6+150g\sin 2}{150} \approx 0.382 \ \text{m/s}^2 \, . $$

This is not the answer in the book. It is very close, but it is slightly out. I feel that either I am misreading the question or the question is under-determined. For example, the maximum possible climbing speed of the train is around $6.11 \ \text{m/s}$, but there is no mention in the question that the train spends a sufficient amount of time climbing in order to reach that speed.

I would be pleased to see your ideas about how to resolve this problem. My misunderstanding is not mathematical, but conceptual; don't worry about using a lot of maths. But please try to explain the physical phenomena in a lucid fashion. Thanks in advance.

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Although self-study and not associated with a class, the homework tag should be added to "exercise"-type problems from textbooks. – tpg2114 Dec 27 '12 at 2:46
Yes, or anything where you're trying to learn the method, rather than get the answer. – David Z Dec 27 '12 at 2:48
By the way, Fly by Night, what exactly are you asking? You've presented the situation and your progress so far very well, but it's not entirely clear what exactly you are looking for an answer to, e.g. is there a formula you're not sure if you're using correctly? Could you try to edit your post and its title to clarify that? It may help to think about how you could phrase your question as an actual question. (Rough rule of thumb: titles with question marks are often better than those without) – David Z Dec 27 '12 at 2:53
@DavidZaslavsky The answer I'm looking for is why my "answer" for the initial acceleration of the train is different to that of the book's. – Fly by Night Dec 27 '12 at 23:16
No, like tpg2114 said, we add the homework tag to questions where it is warranted, and this is definitely one of them. You may have seen something about homework tags on another SE site, but it doesn't apply here. – David Z Dec 27 '12 at 23:37
up vote 2 down vote accepted

You're very nearly there.

You're correct to say that $Power = Fv$, and that the velocity at the moment the train reaches the top of the slope is given by $v_{top} = 350/(6 + 150gsin2)$ so the force at the top of the slope is $F_{top} = 1000(6 + 150gsin2)$.

But the acceleration is the net force divided by the mass, and the net force is $F_{top}$ minus the 6kN frictional force i.e.

$$ a = \frac{F_{top} - 6000}{m} = \frac {1000(6 + 150gsin2) - 6000}{150000} = g sin2$$

which using $g = 9.81m/sec^2$ I get as 0.342.

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That's exactly it! Thanks a lot John. – Fly by Night Dec 27 '12 at 23:22
@John Rennie: I know you're way beyond the level of this type of exercise, but I'd prefer if you be consistent with units (i.e. it'd be nice to write $F_{\text{top}} - 6000 \text{ N}$ instead if $F_{\text{top}} - 6000$). – Vibert Dec 27 '12 at 23:48
Oops, sorry :-) – John Rennie Dec 28 '12 at 7:07

For simplicity imagine you're in the frame where the train is moving at 0 velocity.

When track changes to horizontal, you can picture this as the gravitational force disappearing -- or an additional force acting on the train with magnitude $F_g\sin(\theta)$.

At this instant, this is the net force on the train.

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