# Conservation of energy at terminal velocity

• Jake 7174
In summary, the conversation discusses the concept of terminal velocity and the conversion of gravitational potential energy to other forms of energy once this velocity is reached. It is concluded that in the case of a vacuum, there is no terminal velocity and the speed of the object will continue to increase. The conversion of energy is primarily due to friction and heat.
Jake 7174

## Homework Statement

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.

Ug = mgh
Ke = mv^2 / 2

## 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?

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 doesn't 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 doesn't 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

Jake 7174 said:

## Homework Statement

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.

Ug = mgh
Ke = mv^2 / 2

## 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?
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 :-) )

nrqed said:
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 :-) )
Ahh.. of course. Terminal velocity doesn't exist because there is nothing to cancel Ug. Thank you.

## What is terminal velocity?

Terminal velocity is the maximum speed that an object can reach when falling through a fluid, such as air or water. It occurs when the force of gravity pulling the object down is equal to the force of air resistance pushing against it.

## How does conservation of energy apply to terminal velocity?

Conservation of energy states that in a closed system, energy cannot be created or destroyed, only transformed from one form to another. In the case of terminal velocity, the potential energy of the falling object is converted into kinetic energy as it accelerates due to gravity. Once the object reaches terminal velocity, the kinetic energy is balanced by the energy lost to air resistance, resulting in a constant speed.

## What factors affect the terminal velocity of an object?

The terminal velocity of an object is affected by several factors, including its mass, size, shape, and the density and viscosity of the fluid it is falling through. Objects with a larger surface area or lower density will experience more air resistance, resulting in a lower terminal velocity. The strength of gravity also plays a role, and objects will reach a higher terminal velocity on planets with a higher gravitational pull.

## Can an object reach a terminal velocity in a vacuum?

No, an object cannot reach terminal velocity in a vacuum because there is no fluid present to create air resistance. In a vacuum, an object will continue to accelerate due to gravity until it reaches the maximum speed of light.

## How is terminal velocity used in conservation efforts?

Terminal velocity is an important concept in conservation efforts, particularly in the study of animal flight and the design of wind turbines. By understanding the relationship between an object's mass, shape, and air resistance, scientists can design more efficient structures and devices to reduce energy consumption and promote sustainability.

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