What is the acceleration (in meters/second^2) of free falling 74kg sky diver, if the air resistance exerts a force of 233n?
To determine the acceleration of the skydiver, we need to use Newton's second law of motion, which states that force equals mass multiplied by acceleration (F = ma). In this case, the force acting on the skydiver is the force of gravity minus the force of air resistance.
The force of gravity can be calculated using the equation F = mg, where m is the mass of the skydiver and g is the acceleration due to gravity, which is approximately 9.8 m/s^2 on Earth.
So, the force of gravity acting on the skydiver is:
F_gravity = m * g
F_gravity = 74 kg * 9.8 m/s^2
F_gravity = 725.2 N
Next, we need to subtract the force of air resistance from the force of gravity to find the net force on the skydiver.
Net force = F_gravity - F_air_resistance
Net force = 725.2 N - 233 N
Net force = 492.2 N
Finally, we can use Newton's second law to find the acceleration of the skydiver.
Acceleration = Net force / mass
Acceleration = 492.2 N / 74 kg
Acceleration ≈ 6.65 m/s^2
Therefore, the acceleration of the 74 kg skydiver, taking into account the force of air resistance, is approximately 6.65 m/s^2.