Work-Energy-Power Textbook Problem

[suffian]

I am having trouble verifying my answer for a textbook Work-Energy-Power problem. The problem is stated below verbatim from the text. My difficulty appears to arise in part (c), though it might be coming earlier.

Six diesal units in series can provide 13.4 MW of power to the lead car of a freight train. The diesal units have a total mass of 1.10 x 10^6 kg. The average car in the train has a mass of 8.2 x 10^4 kg and requires a horizontal pull of 2.8 kN to move at a constant 27 m/s on level tracks. a) How many cars can be in the train under these conditions? b) However, this would leave no power for accelerating or climbing hills. Show that the extra force needed to accelerate the train is about the same for a 0.10 m/s^2 acceleration or a 1.0% slope (alpha equal to arctan 0.010) c) With the 1.0% slope, show that an extra 2.9 MW of power is needed to maintain the 27 m/s speed of the diesal units. d) With 2.9 MW less power available, how many cars can the diesal units pull up a 1.0% slope at a constant 27 m/s?

a) Okay, so let's denote the power output by the diesal units on the rest of the cars as $$P$$. Then this output should equal the forward force applied to the rest of the cars multiplied by their velocity, or $$P = Fv$$. Now, the forward force supplied will need to be $$n$$ times the force it takes to keep one car steady at 27 m/s. So now the equation becomes $$P = nfv$$. If we isolate $$n$$ then we can get a numerical answer:
$$n = \frac{P}{fv} = \frac{ \left[ 13.4 \times 10^6 \text{ W} \right] }{ \left[ 2800 \text{ N} \right] \left[ 27 \text{ m/s} \right] } \approx 177 \text{ (cars)}$$

b) Allright, when the train is accelerating then Newton's second law is in effect. The net force acting on the train happens to be the extra force needed to maintain the acceleration, $$f = ma$$. The mass of the train is the total mass of the diesal units plus the mass of all the car. Thus we have:
$$f = (m_d + n\, m_c)a = ( \left[1.10 \times 10^6 \text{ kg} \right] + [177][ 8.2 \times 10^4 \text{ kg} ] ) \left[ 0.10 \text{ m/s2} \right]$$
$$f = 1.56 \text{ MN}$$

When the train is going up a slope of angle $$\alpha$$, then it's own weight is pulling it down the incline by a force equal to $$mg \sin \alpha$$. If the train is to be in equilibrium, then it must negate this force by applying an additional force equal to this in the opposite direction (up the incline). Thus we have:
$$f = mg \sin \alpha = (m_d + n\, m_c)g \sin \alpha$$
$$f = ( \left[1.10 \times 10^6 \text{ kg} \right] + [177][ 8.2 \times 10^4 \text{ kg} ] ) [9.80 \text{ m/s2}] \sin\tan^{-1}(0.010))$$
$$f = 1.53 \text{ MN}$$

Close enough to each other.

c) Finally, here is where the problem arises. The book says that the extra power needed when traveling uphill at 1.0% slope is 2.9 MW, but I'm not getting that answer.

I figure that the extra power needed is the additional force that needs to be applied to the train to maintain equilibrium going uphill multiplied by the velocity. The additional force needed was found in part (b) and the velocity is 27 m/s. However, this would yield an extra power of 40 MW, a far cry from 2.9 MW. Can someone see what's going wrong?

 

[Doc Al]

c) Finally, here is where the problem arises. The book says that the extra power needed when traveling uphill at 1.0% slope is 2.9 MW, but I'm not getting that answer.

The question asks for the additional power needed to maintain the speed of diesel units, not the entire train.

 

[suffian]

gah, i always tell myself to read the problem more carefully when i can't figure it out for a while. i guess i was tripped up again though. anyways, many thanks for seeing that, now it's out of my mind.