. Ancilla (mass 57 kg) goes skydiving. At one point in her descent, the force of air

Question

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]