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

Thread starter ragbash
Start date Sep 24, 2006
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Air Air resistance Force Resistance

AI Thread Summary

The discussion focuses on calculating the force of air resistance acting on a 1.21 g samara falling at a constant speed of 1.1 m/s. The relevant formula used is F=ma, where the mass is converted to kilograms (0.00121 kg). The gravitational force (weight) is calculated as 0.0119 N using the acceleration due to gravity (9.81 m/s²). Since the samara falls at a constant speed, the force of air resistance equals its weight. Thus, the force of air resistance exerted on the samara is 0.0119 N.

Sep 24, 2006

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.

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Sep 24, 2006

skeeter

correction ...

1.21 g = 0.00121 kg

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

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