1 simple word problems.. physics experts please help.
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1 simple word problems.. physics experts please help.

[From: ] [author: ] [Date: 11-08-27] [Hit: ]
How much power is required to overcome aerodynamic drag if p = 0,0002378 slugs/ft^3?Please help me with the steps as well.Thank you so much guys.FD is the force of drag, which is by definition the force component in the direction of the flow velocity,......
A new sport car has a drag coefficient of 0.29 and a frontal area of 20ft^2, and is traveling at 100 mi/h.

How much power is required to overcome aerodynamic drag if p = 0,0002378 slugs/ft^3?

Please help me with the steps as well. I really need help
Thank you so much guys.

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Fd = 1/2 * rho * V^2 * A * CD


where

FD is the force of drag, which is by definition the force component in the direction of the flow velocity,
ρ is the mass density of the fluid,
v is the velocity of the object relative to the fluid,
A is the reference area, and
CD is the drag coefficient — a dimensionless constant, e.g. 0.25 to 0.45 for a car.

100 miles * 5280 feet / mile = 5280 * 100 = 528000 feet
v = 528000 feet/3600 second = 146.7 feet/sec

F = 1/2 0.0002378 slugs/ft^3 * (146.7 ft/sec)^2 * 20 ft^2 * 0.29
F = 14.84 slugs ft / sec^2 = 14.84 pounds (believe it or not).

Work = F * d
Work = 14.84 * 528000 = 7835520 foot pounds

Power = W / T = 7835520/3600 = 2176 foot pounds / second

Power = 2176 * 1.36 = 2959 watts

Power = 2959/746 = 3.966 Hp

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[Edit: are you sure that's the right value for the density of air? It looks kinda low..]

Let's start with the formula and see what happens:

C-drag = 2*F-drag/(rho*v^2*A), rho = fluid density, v = velocity relative to fluid, A = reference area.

Do the substitutions:
(100mph = 146.67 ft/s)

0.29 = 2*F-drag / (0.0002378 slugs/ft^3 * (146.7ft/s)^2 * 20ft^2)

Solve for drag force:

F-drag = 0.29 * 2.378x10-4 slugs/ft^3 * 2.15ft^2/s^2 * 20ft^2 / 2
= 14.83 slug-ft/s^2 (pounds)

Work = Force x Distance
Power = Work / Time = Force * Velocity
P = 14.83 slug-ft/s^2 * 146.67 ft/s = 2176 slug-ft^2/s^3

You can convert that to ft-lbs/s then to horsepower or whatever else you like.

[Edit #2: I'm more comfortable with metric units, so I'm going to try it again converted to metric to see if I can get the same answer.]

[Edit #3: Yeah, there's definitely something fishy here. I used 1.2256 kg/m^3 for the density of air and got a value for power = 32.76kW which is about 44 horsepower. Using your value of rho, it came out as only 4 horsepower. I think there's an extra zero in your air-density number..]
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