A 600 N skydiver in free fall is accelerating at -7.6 m/s^2 rather than -9.8 m/s^2 because of the force of air resistance. How much drag force is acting on the skydiver?

a)58.2
b)65.3
c)61.4
d)40.2

I use the equation Σ(F)vertical-ma=0

To find m I did: 600/9.8 = 61.2

I end up getting Fdrag-600-(61.2)(-7.6)=0

I keep getting 135N which is not correct.

force down = 600

force up = Fd

600 - Fd = m a = 61.2 * 7.6
so Fd = 600 - 465
Fd = 135 N
so I agree with you

To find the drag force acting on the skydiver, we can use the equation of motion Σ(F) = ma, where Σ(F) is the sum of all forces acting on the skydiver, m is the mass of the skydiver, and a is the acceleration.

In this case, we are given that the acceleration of the skydiver is -7.6 m/s² instead of the expected acceleration of -9.8 m/s² due to the force of air resistance. We need to find the drag force acting on the skydiver.

Let's use the equation Σ(F)_vertical - mg = ma, but since we are looking for the drag force, we'll rearrange the equation and substitute the known values:

F_drag - mg = ma

First, we need to calculate the mass of the skydiver. You correctly calculated it as 600 N divided by the acceleration due to gravity (9.8 m/s²), which gives us a mass of approximately 61.2 kg.

Substituting m = 61.2 kg and a = -7.6 m/s² into the equation, we have:

F_drag - (61.2 kg)(9.8 m/s²) = (61.2 kg)(-7.6 m/s²)

Simplifying this equation, we get:

F_drag - 599.76 N = -468.72 N

Now, solving for F_drag:

F_drag = -468.72 N + 599.76 N
F_drag = 131.04 N

Therefore, the drag force acting on the skydiver is approximately 131.04 N.

None of the answer choices provided match this result exactly. It's possible that the correct answer was rounded or approximated. However, closest to the calculated result is option a) 58.2 N, so that could be the closest approximation provided in the answer choices.