Question

. Ancilla (mass 57 kg) goes skydiving. At one point in her descent, the force of air
resistance on Ancilla and her parachute is 670 N [up].
(a) What is the force of gravity on Ancilla?
(b) What is the net force on Ancilla?
(c) What is Ancilla’s acceleration?

Answer 1

To solve the problems, we'll use the following formulas:

1. The force of gravity (weight) acting on an object can be calculated using the formula:
\[
F_{\text{gravity}} = m \cdot g
\]
where \( m \) is the mass and \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

2. The net force acting on an object can be calculated using Newton's second law:
\[
F_{\text{net}} = F_{\text{gravity}} - F_{\text{air resistance}}
\]
(Here, air resistance acts in the opposite direction of gravity.)

3. Finally, the acceleration can be calculated using:
\[
a = \frac{F_{\text{net}}}{m}
\]

Now, let's calculate each part:

### (a) What is the force of gravity on Ancilla?
Using the mass of Ancilla \( m = 57 \, \text{kg} \):
\[
F_{\text{gravity}} = m \cdot g = 57 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 \approx 559.17 \, \text{N}
\]
So, the force of gravity on Ancilla is approximately **559 N [down]**.

### (b) What is the net force on Ancilla?
The force of air resistance \( F_{\text{air resistance}} \) is given as \( 670 \, \text{N} \) [up]. Now calculate the net force:
\[
F_{\text{net}} = F_{\text{gravity}} - F_{\text{air resistance}} = 559 \, \text{N} - 670 \, \text{N} = -111 \, \text{N}
\]
The negative sign indicates that the net force is directed upward. So, the net force on Ancilla is **111 N [up]**.

### (c) What is Ancilla’s acceleration?
Now using the net force to find the acceleration:
\[
a = \frac{F_{\text{net}}}{m} = \frac{-111 \, \text{N}}{57 \, \text{kg}} \approx -1.95 \, \text{m/s}^2
\]
The negative acceleration indicates that Ancilla is decelerating (moving upward), so her acceleration is approximately **1.95 m/s² [up]**.

### Summary:
(a) Force of gravity on Ancilla: **559 N [down]**
(b) Net force on Ancilla: **111 N [up]**
(c) Ancilla’s acceleration: **1.95 m/s² [up]**