# Conservation of energy at terminal velocity

1. Mar 4, 2016

### Jake 7174

1. The problem statement, all variables and given/known data

A spherical object is dropped from an elevation great enough such that it will achieve terminal velocity for some period of time before hitting the ground. Once terminal velocity is achieved what is gravitational potential energy converted to.

2. Relevant equations
Ug = mgh
Ke = mv^2 / 2

3. The attempt at a solution
First off let me say this was not asked as homework. It is a question that occurred to me randomly and made me Realize I am missing some understanding.

Ug is dependent on height which will change until the object is at rest and Ke is dependent on velocity which will continue to change until acceleration is 0 (terminal velocity in this situation). Once terminal velocity is reached Ke is constant but Ug is changing. Ug must be converted to some other form of energy. The only thing i can think of is heat due to the friction of drag but then i ask myself what would happen if the sphere was in a vacuum? Here drag is zero. There must be some other form of energy that we didn't cover. What is it, or am i totally screwed up from the beginning?

2. Mar 4, 2016

### Biker

If you are in a vacuum, There is no terminal velocity.

Terminal velocity is achieved when air resistance balance out gravity perfectly. So as usual if PE changes to KE but KE doesnt change then where did the energy go?

Look at this equation and tell me what do you see
PE1 + KE + W = PE2 + KE
KE is constant doesn't change. But look at W. What does it do? What does it represent?

Work is either giving energy to the system or taking away. Also there is conservative forces which only change the form of energy but doesnt add or take away energy but that is not our point. Try to conclude something out of that equation.

So basically as you mentioned above it is friction and heat

3. Mar 4, 2016

### nrqed

You are correct (about heat and friction). There is no contradiction in case of a vacuum because then there is no such thing as a terminal velocity, the speed keeps increasing following $\Delta U_g + \Delta K =0$ (well, the speed keeps increasing following this equation until relativistic effects become non negligible :-) )

4. Mar 4, 2016

### Jake 7174

Ahh.. of course. Terminal velocity doesn't exist because there is nothing to cancel Ug. Thank you.