Air Resistance Force on 1.21 g Samara Falling at 1.1 m/s

[ragbash]

A 1.21 g samara--the winged fruit of a maple tree--falls toward the ground with a constant speed of 1.1 m/s What is the force of air resistance exerted on the samara?
I know that F=ma
I also know I have to account for 9.81 N for g. So by upward force am I looking for W? That would be 11.9N.

 

[skeeter]

correction ...

1.21 g = 0.00121 kg

force of air resistance = mg = (.00121 kg)(9.81 m/s^2) = .0119 N

 

[kyle8921]

I would first like to clarify that the term "constant speed" in this context likely refers to a constant velocity, as speed is a scalar quantity while velocity is a vector quantity. With that being said, the force of air resistance on the samara can be calculated using the formula F = ma, where F is the force, m is the mass of the samara (1.21 g or 0.00121 kg in this case), and a is the acceleration.

In this scenario, the samara is not accelerating, so the net force acting on it must be zero. This means that the force of air resistance, which acts in the opposite direction of the samara's motion, must be equal in magnitude to the force of gravity pulling the samara down. Therefore, we can use the formula F = mg, where g is the acceleration due to gravity (9.81 m/s^2).

Plugging in the values, we get F = (0.00121 kg)(9.81 m/s^2) = 0.0119 N. This is the force of air resistance exerted on the samara as it falls at a constant velocity of 1.1 m/s.

It is important to note that air resistance is a complex force that depends on various factors such as the shape and size of the object, the density of the air, and the speed of the object. Therefore, this calculation is an approximation and the actual force of air resistance on the samara may vary.