What Is the Frictional Force Acting on a Skydiver at Constant Velocity?

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SUMMARY

The discussion focuses on calculating the frictional force acting on a skydiver descending at a constant velocity of 2 m/s with a mass of 70 kg. The gravitational force acting on the skydiver is calculated as 686 N using the formula F = mass x gravity (F = 70 kg x 9.8 N/kg). Since the skydiver is at constant velocity, the net force (Fnet) is zero, indicating that the air's frictional force (Ff) equals the gravitational force, thus Ff is also 686 N. Participants emphasize the importance of recognizing that constant velocity implies no acceleration, simplifying the calculation of forces.

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frigid
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A skydiver jumps from an airplane and descends at a constant velocity of 2m/s. The mass of the skydiver and his equipment is 70kg. Calculate the magnitude of the air's frictional force acting on the skydiver and his equipment.

F = 70kg x 9,8N/kg = 686N

I'm lost
 
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Can anyone please help me with this one real quick.

I know that Fnet is Fa - Ff = MA
His mass is 70kg and velocity 2m/s. But since i don't have the distance of his fall or the time i can't calculate the acceleration. So how do I figure out the magnitude of the air's frictional force which would be Ff?

:cry: :cry: :cry: :cry: :cry:
 
frigid said:
But since i don't have the distance of his fall or the time i can't calculate the acceleration.
Reread the problem statement carefully: the acceleration is given.
 

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