What is the acceleration (in meters/second^2) of a freely falling 74.0 kg sky-diver, if air resistance exerts a force of 264 N?
Fs = 74kg * 9.8Nkg = 725.2N,
a=Fn/m = (725.2 - 264) / 74 = 6.2m/s^2
To find the acceleration of a freely falling skydiver with air resistance, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
In this case, the net force is the force of gravity minus the force of air resistance. The force of gravity can be calculated using the formula: Fgravity = mass × acceleration due to gravity.
Let's assume the acceleration due to gravity is approximately 9.8 m/s². The force of gravity can be calculated as follows:
Fgravity = mass × acceleration due to gravity
Fgravity = 74.0 kg × 9.8 m/s²
Next, we subtract the force of air resistance from the force of gravity to find the net force:
Net force = Fgravity - Force of air resistance
Net force = (74.0 kg × 9.8 m/s²) - 264 N
Finally, we can use Newton's second law to find the acceleration of the skydiver:
Net force = mass × acceleration
(74.0 kg × 9.8 m/s²) - 264 N = 74.0 kg × acceleration
Rearranging the equation, we can solve for acceleration:
acceleration = ((74.0 kg × 9.8 m/s²) - 264 N) / 74.0 kg
Calculating the values, we get:
acceleration = (725.2 N - 264 N) / 74.0 kg
acceleration = 461.2 N / 74.0 kg
acceleration ≈ 6.23 m/s²
Therefore, the acceleration of the skydiver is approximately 6.23 m/s².