Calculating how to overcome drag

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AI Thread Summary
The discussion focuses on calculating the drag force and energy required for a model car overcoming air resistance during a journey. The drag force is determined to be proportional to both the cross-sectional area and the velocity of the car. When the cross-sectional area is doubled and the speed is increased to 20 m/s, the drag force increases to 40 N. Consequently, the energy required to overcome this drag over a distance of 1000 m is calculated to be 40,000 joules. The importance of accurately applying the relationship between drag force, area, and velocity is emphasized.
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Homework Statement



Studetns examine the aerodynamics of a model car which has a drag force in air of 10 N at 10 m/s.

They wish to estimate what the drag will be for a full scale model, and the energy lost on a given trip. The drag coefficient C remains constant C= 0.6. The cross-sectional area A of the car is increased to 2 times its original area, and the speed is now increased to 20 m/s.
How much work does this car do in overcoming air resistance in a journey of 1000 m. Answer in joule.

Homework Equations



new drag force = (C/0.5) * (area/original area)* original drag force
then Energy to overcome drag = Work = force x distance.

The Attempt at a Solution



(.6*.5)(20/10)(10)=24

24*1000=2400

but answer is 40000
 
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Drag force is proportional to the cross-sectional area and velocity of the object. Since the drag coefficient remains the same, we can effectively ignore it.
Thus it is clear that when I double cross-sectional area or velocity, the magnitude of the drag force would be doubled as well. When both are doubled simultaneously, the drag force is quadrupled to 40N. The relation between old and new drag force would be:
\frac{F_{1}}{F_{2}} = \frac{A_{1} v_{1}}{A_{2} v_{2}}
The drag coefficient cancels out, so there is no need to include it.
And, 24*1000=2400?? Do take more effort in putting forth your attempt here...
 
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