Calculating Energy Dissipation in Free Fall with Drag Force

Click For Summary

Homework Help Overview

The discussion revolves around calculating energy dissipation due to drag force on a ball in free fall, specifically when it reaches terminal velocity before hitting the ground. The drag force is defined as bv², where b is the drag coefficient and v is the instantaneous speed of the ball.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between energy dissipation and the equations for potential and kinetic energy, questioning the validity of the equation Ediss=mgh. There is also consideration of conservation of energy principles and the need for height in relation to terminal velocity.

Discussion Status

The discussion is active, with participants exploring different interpretations of energy dissipation and the equations involved. Some guidance has been offered regarding the conservation of energy approach, but there is no explicit consensus on the next steps.

Contextual Notes

There is a noted uncertainty regarding the distance required to reach terminal velocity and how it relates to the overall energy dissipation calculation.

impendingChaos
Messages
24
Reaction score
0

Homework Statement


Need to energy dissipation by the drap force on a ball in free fall from height h and when it has reached terminal speed, assuming it happens before it hits thr ground. Drag force has magnitude bv^2, where b is drag coefficient and v is instantaneous speed of the ball.


Homework Equations



I have calculated in an earlier part of the problem that terminal velocity:
Vt=root(mg/b)

Also, isn't the energy dissipation equation:
Ediss=mgh

The Attempt at a Solution



I have the above equations but am unclear where to move next.
 
Physics news on Phys.org
Not sure what you mean by Ediss=mgh. If chosing conservation of energy, one could simply assume that the drag energy(work) is the difference between the final and initial sums of PE and KE. The problem I see, is that the eqn you posted for Vt says nothing about height.

In other words if we knew the distance required to reach terminal velocity, you're home free.
 
Ok will check that, Ediss is Energy dissipated

thx!
 
no sweat, either way problem is soluble:

mv'=-b*v^2+mg.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
9
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 31 ·
2
Replies
31
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
4K
Replies
1
Views
2K
Replies
13
Views
3K