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The engine of a train is working at a constant rate. The maximum speed of the train up a certain incline is $V_{1}$ and the maximum speed down the same incline is $V_{2}$. Then if the train moves on a level track, its maximum speed will be?

The answer given is:

$P = V_{1}(F+F_{g})$

$P = V_{2}(F-F_{g})$

So we get,

$P = V_{3}F$

Where P is the constant power exerted by the engine, $F_{g}$ is the force of gravity on the train and $F$ is the frictional force on the train. So through this the answer I'm getting is $V_{3} = \frac{2V_{1}V_{2}}{V_{1}+V_{2}}$ which matches with the answer given.

But what I'm concerned about is that won't the value of $F$ be different when the train is level? The $θ$ would also play a role here and hence the question would be incomplete.

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    $\begingroup$ There are two forces on the train: that due to the engine, and gravity. When the train is going down the hill, both are directed downward. The train will accelerate. If you want the train to move down the hill at a constant rate you have to turn off the engine and apply the brakes. The problem doesn't make sense. $\endgroup$ – garyp Nov 19 '17 at 12:47
  • $\begingroup$ @garyp you seem to be neglecting friction? $\endgroup$ – Time4Tea Aug 20 '18 at 14:29
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The frictional force and the gravity force here are most probably the projection of the (vector) force along the velocity. Therefore the angular dependence is already taken into account, albeit hidden in the poor notations. And because the slope is the same up and down (up to a minus sign to the angle), the projections have the same absolute value.

I hope it makes sense to you now ?

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Really, the question should state: “assuming the frictional force on the train (i.e. $F$) is constant”, as the answer they give depends on that.

In reality, F will be composed of two main components:

  1. Mechanical friction from the weight of the train acting on the wheel bearings (typically denoted as $\mu R$, where R is the reaction force from the weight), and
  2. air resistance.

In the case of 1), in reality this will vary with the angle of the incline, since $R = mg \cos\theta$. So, $\mu R$ will be a maximum when the train is on level ground.

In the case of 2), the air resistance is typically modeled as a function depending on $V$ or $V^2$, so this will also vary for each case, if the velocities are different.

In summary, the question is using a simplification that the frictional force is constant, which it should have made clear. Good job to be questioning it though :-)

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Hint: Up and down have no distinction when the train moves on a level surface. What does this say about the relation between $V_1$ and $V_2$?

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  • $\begingroup$ I still don't get it. $\endgroup$ – Aradhye Agarwal Jun 2 '17 at 9:42
  • $\begingroup$ $V_1=V_2$, for the level surface case. What is $V_3$, then? (Also, $F$ remains the same, $F_g$ is zero.) $\endgroup$ – Hritik Narayan Jun 2 '17 at 9:45
  • $\begingroup$ Why does $F$ remain the same? For the level case it would change since the force of gravity vector normal to the surface would be $gcos θ$ whereas in the level case it is just $g$. So $F$ should not remain constant. $\endgroup$ – Aradhye Agarwal Jun 2 '17 at 9:51
  • $\begingroup$ I stand corrected, yes, $F$ won't stay the same. $\endgroup$ – Hritik Narayan Jun 2 '17 at 11:40
  • $\begingroup$ Then how is the solution correct? $\endgroup$ – Aradhye Agarwal Jun 4 '17 at 14:46

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