Problem using Newtons second law

Click For Summary
SUMMARY

The discussion focuses on applying Newton's Second Law of Motion to analyze the forces acting on a sky-diver after the parachute opens. The downward gravitational force (Fg) is defined as Fg = mg, while the upward air resistance force (Fr) is modeled as Fr = kv, where k is a constant of proportionality. The net force equation is established as Fnet = Fg - Fr, leading to the acceleration equation a = g - (kv/m). This formulation accurately represents the dynamics of the sky-diver under the influence of gravity and air resistance.

PREREQUISITES
  • Understanding of Newton's Second Law of Motion
  • Knowledge of gravitational force calculations
  • Familiarity with concepts of air resistance and drag
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the derivation of drag force in fluid dynamics
  • Explore the effects of varying the constant of proportionality (k) on motion
  • Learn about terminal velocity and its relation to forces acting on falling objects
  • Investigate numerical methods for solving differential equations in motion analysis
USEFUL FOR

Physics students, educators, and anyone interested in understanding the dynamics of falling objects and the application of Newton's laws in real-world scenarios.

Chosen-wun
Messages
3
Reaction score
0
Once her chute opens, a sky-diver of mass m is acted upon by a downward force Fg due to gravity, and an upward force Fr due to air resistance. If Fg = mg, where g is gravitational acceleration, and Fr is proportional to velocity v, use Newton’s Second Law of Motion to write acceleration a as a function of velocity v.

I don't know if I'm on the right track
Fnet = ma
fnet = Fg+ Fr
Fg = mg

Mg + Fr = ma
 
Physics news on Phys.org
yes on right track except the signs. Now since F_r is proportional to velocity v, we can write

F_r=kv where k is constant of proportionality.

\Rightarrow mg-kv=ma

the negative sign is there since the air drag opposes the downward motion,

\therefore a = g-\frac{kv}{m}
 
IssacNewton said:
yes on right track except the signs. Now since F_r is proportional to velocity v, we can write

F_r=kv where k is constant of proportionality.

\Rightarrow mg-kv=ma

the negative sign is there since the air drag opposes the downward motion,

\therefore a = g-\frac{kv}{m}

That makes a lot more sense, thank you.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
Replies
9
Views
2K
Replies
4
Views
1K
  • · Replies 17 ·
Replies
17
Views
1K
  • · Replies 42 ·
2
Replies
42
Views
6K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
13
Views
3K
Replies
44
Views
4K
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K