Conservation of energy of released ball

AI Thread Summary
A ball released from a height experiences a conversion of gravitational potential energy (GPE) to kinetic energy (KE) as it falls. When a ball falls at terminal velocity, it maintains constant KE while GPE decreases, as the gravitational force is balanced by drag force from air resistance. The energy 'lost' is not entirely converted to heat; instead, it is used to do work against viscous drag. This interaction explains why the ball can lose GPE without an increase in KE. Understanding terminal velocity is key to grasping these energy transformations.
O.J.
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suppose we have a ball held at a certain height above the ground where it has 0 KE and x potential enegy. as its release it fall down with uniform acceleration so its KE starts increasing while its GPE starts decreasing. Now suppose a ball is acted upon by a force (say air resistance) causin it to fall with constant speed where it has constant KE while its GPE is decreasing. Where is this energy missing going to?> heat due to friction with air? doesn't seem really convincing to me that all the lost GPE is converted to heat.

a question in one of my excercises asked: explain how can a sphere falling through a liquid have a constant KE but decreasing GPE/? i assumed this is relevant so I am putting it across too.
 
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Why isn't it convincing that the energy 'lost' is converted to heat?

As for you exercise question, have you heard of the term 'terminal velocity'?

Regards,
~Hoot
 
i've heard of it. but question is, how come the sphere loses GPE but maintains a consant KE? is it because it's doing work against viscous drag?
 
O.J. said:
is it because it's doing work against viscous drag?

You've got it in one. At terminal velocity the gravitational force is equal the the drag force created by the fluid throught which it is travelling. The air resistance does negative work on the ball becuase it is acting in the opposite direction to the velocity.

Regards,
~Hoot
 
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