Question

1. a few seconds after a 55 kg sky - diver jumps out of an airplane, she notices that the air friction blocking upward on her has a force of 300 n.

a. what is the net force on her?
b. what is her acceleration?
c. which way will she accelerate?
i. upward
ii. downward
iii. zero acceleration

1. a person pushes a 3000 kg car forward. he pushes with a force of 600 n against a friction force of 100 n.

a. how big are all the forces in this situation?

• active force =
• friction force =
• gravity force =
• normal force =

b. how much does the car accelerate?
equations:
$f_g=m \times 10$
$a = f_{net}/m$
equations:
$f = m \times a$
$f_{net}=m \times a$
$f_g=m \times 10$
$a = f_{net}/m$
$f_d = f_a$
$f_f=\mu \times f_n$

Explanation:

Step1: Calculate forces for the car - active and friction

The active force is given as 600 N and the friction force is 100 N.

• active force = 600 N
• friction force = 100 N

Step2: Calculate gravity force

Use the formula $F_g = m\times g$, where $m = 3000$ kg and $g = 10$ m/s².
$F_g=3000\times10 = 30000$ N

Step3: Determine normal force

On a flat - surface, the normal force $F_N$ equals the gravity force. So $F_N = 30000$ N.

Step4: Calculate car's acceleration

First, find the net force $F_{net}=F_{active}-F_{friction}=600 - 100=500$ N.
Then use the formula $a=\frac{F_{net}}{m}$, where $m = 3000$ kg.
$a=\frac{500}{3000}=\frac{1}{6}\approx0.17$ m/s²

Answer:

• active force = 600 N
• friction force = 100 N
• gravity force = 30000 N
• normal force = 30000 N
• acceleration of the car: $a=\frac{1}{6}\text{ m/s}^2\approx0.17$ m/s²