Question

When a parachute opens, the air exerts a large drag force on it. This upward force is initially greater than the weight of the sky diver and, thus, slows him down. Suppose the weight of the sky diver is 936 N and the drag force has a magnitude of 1032 N. The mass of the sky diver is 95.5 kg. Take upward to be the positive direction. What is his acceleration, including sign?

Answer 1

Answer:
`1.0052m/s^(2)`

Step-by-step explanation:

Assuming there is only force in the y-component, the total net force
`F_(y)` acting on the parachute and the sky diver is:

`F_(y)=F_(D)-W` (1)

Where:

`F_(D)=1032N` is the drag force acting upwards

`W=936N` is the weight of the sky diver acting downwards, hence with negative sign

Then:

`F_(y)=1032N-936N=96N` (2) This is the total net force excerted on the system parachute-sky diver, and the fact it is positive means is upwards

Now, according Newton's 2nd Law of Motion the force is directly proportional to the mass
`m` and to the acceleration
`a` of a body:

`F_(y)=m.a` (3)

Where
`m=95.5kg` is the mass of the diver.

Substituting the known values and finding
`a`:

`a=(F_(y))/(m)` (4)

`a=(96N)/(95.5kg)` (5)

Finally:

`a=1.0052m/{s^(2)}\approx 1m/s^(2)` This is the acceleration of the sky diver. Note it has a positive sign, which means its direction is upwards.