Calculating Energy Dissipation in Free Fall with Drag Force

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SUMMARY

The discussion focuses on calculating energy dissipation due to drag force on a ball in free fall, specifically when it reaches terminal velocity. The drag force is defined as bv², where b is the drag coefficient and v is the instantaneous speed. The terminal velocity is determined using the formula Vt = √(mg/b). The energy dissipation is represented by Ediss = mgh, which is the difference between potential energy and kinetic energy when considering conservation of energy principles.

PREREQUISITES
  • Understanding of drag force and its mathematical representation (bv²).
  • Knowledge of terminal velocity calculation (Vt = √(mg/b)).
  • Familiarity with energy conservation principles in physics.
  • Basic calculus for solving differential equations related to motion.
NEXT STEPS
  • Explore the derivation of the drag force equation in fluid dynamics.
  • Study the relationship between height and time to reach terminal velocity.
  • Learn about differential equations in the context of motion under drag forces.
  • Investigate energy conservation in non-conservative systems.
USEFUL FOR

Physics students, educators, and anyone interested in understanding the dynamics of free fall and the effects of drag force on motion.

impendingChaos
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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.
 
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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.
 

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