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# M2 Power Question watch

1. This is from the OCR Mechanics 2 specification, Q15 on Miscellaneous Exercise 2 of the textbook.

15. If the air resistance to the motion of an airliner at speed v ms-1 is given by kv2 newtons at ground level, then at 6000 metres the corresponding formula is 0.55 kv2, and at 12000 metres it is 0.3 kv2. If an airliner can cruise at 220 ms-1 at 12000 metres, at what speed will it travel at 6000 metres with the same power output from the engines?

Suppose that k=2.5 and that the mass of the airliner is 250 tonnes. As the airliner takes off its speed is 80 ms-1 and it immediately starts to climb with the engines developing three times the cruising power. At what angle to the horizontal does it climb?

I can answer the first part to get an answer of approximately 180 ms-1. This answer is confirmed as right. But how do I go about solving the second part?
But how do I go about solving the second part?
Consider.

What's the power developed when the airliner is crusing?

Hence what's the power developed at takeoff?

This power is consumed in, work done against air resistance plus work done in lifting the plane. This will allow you to work out the vertical velocity.

Then (vertical velocity)/(horizontal velocity) = tan(desired angle)

Edit: See below.
3. (Original post by ghostwalker)
This power is consumed in, work done against air resistance plus work done in lifting the plane. This will allow you to work out the vertical velocity.
But doesn't the power also have to be used in maintaining the cruising? I don't get how it's split into the vertical force and horizontal force (nor do I understand how air resistance is split into these two components).
But doesn't the power also have to be used in maintaining the cruising? I don't get how it's split into the vertical force and horizontal force (nor do I understand how air resistance is split into these two components).
Part of the power maintains the cruising when it's taking off, and that's your horizontal force.

Any unused power goes into the vertical force.

Air resistance is not split into two components.
5. (Original post by ghostwalker)
Part of the power maintains the cruising when it's taking off, and that's your horizontal force.

Any unused power goes into the vertical force.

Air resistance is not split into two components.
So if we call horizontal velocity vx and vertical vy, vertical air resistance = k*vy and vertical = k*vx? I can't think of how else you might model it.
So if we call horizontal velocity vx and vertical vy, vertical air resistance = k*vy and vertical = k*vx? I can't think of how else you might model it.
I think you can safely ignore vertical air resistance.

On reflection my previous post wasn't entirely correct.

The velocity of 80ms^-1 against which we have air resistance would be in the direction the air craft is travelling.

You don't need to deal with vertical/horizontal air resistance, just the resistance against the motion in the direction of travel.

Also angle will be sin^-1 "vertical air speed"/80.
7. (Original post by ghostwalker)
I think you can safely ignore vertical air resistance.

On reflection my previous post wasn't entirely correct.

The velocity of 80ms^-1 against which we have air resistance would be in the direction the air craft is travelling.

You don't need to deal with vertical/horizontal air resistance, just the resistance against the motion in the direction of travel.

Also angle will be sin^-1 "vertical air speed"/80.
If it's travelling at 80 ms-1 in vx then shouldn't the angle be arctan(vy/vx)?

The problem with "resistance against the motion in the direction of travel" is that the direction is at an angle, so there will be two components of motion (vx and vy) so there will be two components of resistance, surely? Unless you're saying I should just take resistance to x motion?
If it's travelling at 80 ms-1 in vx then shouldn't the angle be arctan(vy/vx)?

The problem with "resistance against the motion in the direction of travel" is that the direction is at an angle, so there will be two components of motion (vx and vy) so there will be two components of resistance, surely? Unless you're saying I should just take resistance to x motion?
I think that whatever the angle the resistance and the power required to overcome it remains the same.

IMHO, it's highly debatable what is going on as the plane takes off. Is it 80 m/s horizontally, or at the angle the plane takes off at? I think, but could be wrong, that it's 80 horizontal before it takes off, and becomes 80 at an angle as it takes off.

Either way sin of a small angle is approximately equal to the tan of the same small angle, so hopefully it makes no difference.

Sorry I can't give a definitive answer.

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Updated: March 11, 2013
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