Calculate Power Used to Climb Small Airplane - 2 Sig Fig %

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The discussion focuses on calculating the fraction of engine power used by a small single-engine airplane to climb. The airplane, with a mass of 680 kg and an engine power of 80 kW, climbs at a rate of 2.1 m/s. By applying Newton's second law and the formula for power, the lifting force is determined to be 6664 N, resulting in a power usage of 13994.4 W for climbing. Consequently, 17% of the engine's power is utilized for altitude gain.

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When its engine of power 80 kW is generating full power, a small single-engine airplane with mass 680 kg gains altitude at a rate of 2.1 m/s.

What fraction of the engine power is being used to make the airplane climb? (The remainder is used to overcome the effects of air resistance and of inefficiencies in the propeller and engine.)

Express your answer as a percentage using two significant figures.


So I'm stumped here. I need some hints on how to get started. Any help at all would be greatly appreciated. Thanks.
 
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Power is the rate of doing work. What is the work the engine is doing to gain height compared to the work of overcoming resistance and other effects? That is what do you know about the work done to move a mass from one height to another?
 
I couldn't quite follow your post, but I believe I have a solution now. Thanks.

P=\vec{F}\cdot\vec{v}

I'm given v=2.1\ m/s

I need to find F which I do by using Newton's 2nd law:

\Sigma F_{y}=-mg+L=0 where L is the lifting force.

So

L=mg

m=680\ kg
g=9.8\ m/s^{2}

Therefore,

F=L=6664\ N

P=\left(6664\ N\right)\left(2.1\ m/s\right)=13994.4\ W

Now I simply need to find what fraction of the full power this is:

\frac{13994.4\ W}{80000\ W}=0.17493

Hence, the percentage of engine power being used to make the airplane climb is 17\%.
 
Last edited:

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