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:
Question
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.
Answers
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.
Best Answer
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 $v_\text{top} = 350/(6 + 150g\sin2)$ so the force at the top of the slope is $F_\text{top} = 1000(6 + 150g\sin2)$.
But the acceleration is the net force divided by the mass, and the net force is $F_\text{top}$ minus the $6$ kN frictional force i.e.
$$ a = \frac{F_{top} - 6000}{m} = \frac {1000(6 + 150g\sin 2) - 6000}{150000} = g \sin2$$
which using $g = 9.81$ m/sec$^2$ I get as $0.342$ m/sec$^2$.