Question

A 65kg teacher (including the parachute) is skydiving! As the parachute opens, the system experiences a 1200N drag.
How many forces are acting on the system?

What is the magnitude of the net force on the system?

What is the magnitude of the system's acceleration?

Describe the system's speed:

Answer 1

1) 2 Forces

2) 563 N, upward

3) $8.66 m/s^2$, upward

4) Speed is decreasing

Explanation:

1)

There are two forces acting on teacher during his skydiving:

- The force of gravity (also known as weight), acting downward (towards the Earth's centre), of magnitude

$W=mg$

where

m is the mass of the teacher

g is the acceleration due to gravity

- The air resistance, of magnitude R = 1200 N, acting upward (air resistance always acts in the direction opposite to the direction of motion)

There are no other forces, therefore the correct answer is:

2 Forces

2)

The net force on the system is given by the vector resultant of the two forces described in part 1). Since the two forces act along the same line but in opposite directions, the net force will be equal to the difference between the two forces, so:

$F_{net}=W-R=mg-R$

where we have chosen downward as positive direction, and where

m = 65 kg is the mass of the teacher

$g=9.8 m/s^2$ is the acceleration due to gravity

R = 1200 N is the air drag

Therefore, the net force on the teacher is

$F_{net}=(65)(9.8)-1200=-563 N$

And the negative sign means that the direction of the net force is upward.

3)

The acceleration of the system can be found by using Newton's second law of motion, which can be written as

$a=\frac{F_{net}}{m}$

where

a is the acceleration

$F_{net}$ is the net force

m is the mass of the teacher

Here we have

$F_{net}=-563 N$

m = 65 kg

Therefore, the acceleration is

$a=\frac{-563}{65}=-8.66 m/s^2$

And the negative sign indicates that the direction of the acceleration is opposite to the motion (so, upward).

4)

From part 3), we observed that:

- The motion of the teacher is downward (he is moving downward)

- However, the direction of his acceleration is upward

This means that the velocity and the acceleration of the system have opposite directions.

As we know, the velocity of an accelerating system can be written as

$v=u+at$

where

v is the velocity after time t

u is the initial velocity

a is the acceleration

Since a is negative, we see that $at<0$, so $v<u$: this means that the velocity of the teacher is decreasing with time, so, the teacher is slowing down, and therefore his speed is decreasing.

Answer 2

(a) Two forces are acting on the system.

(b) The magnitude of net force on the system is 592.4 N.

(c) The magnitude of the acceleration is 9.11 m/s².

(d) The speed of the system decreases with time

The given parameters;

• mass of the teacher, m = 65 kg
• drag force on the teacher, $F_f$ = 1200 N

The forces acting on the teacher are downward force due to the teachers weight and drag force acting upward opposing the downward motion.

The magnitude of net force on the system is calculated as follows;

$F_{net}= W - F_f\\\\F_{net}= (65 \times 9.8) - (1200)\\\\F_{net} = -592.4 \ N$

The magnitude of the acceleration is calculated as follows;

$a = \frac{-592.4}{65} \\\\a = -9.11 \ m/s^2\\\\|a| = 9.11 \ m/s^2$

The speed of the system is calculated as follows;

$v= u + at\\\\v= 0 + (-9.11)t\\\\v = -9.11t$

The speed of the system decreases with time.

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