Energy change and work involved in lifting a ball

AI Thread Summary
The discussion centers on the work-energy theorem, specifically regarding the lifting of a ball and the associated work done. It highlights that when a ball is lifted, the net work done appears to be zero due to no change in kinetic energy, leading to confusion about the forces involved. The conversation emphasizes that while gravitational force and the upward force balance out, the work done against gravity must be considered, which contributes to potential energy. The net work is defined strictly as the change in kinetic energy, without incorporating potential energy terms. Ultimately, the total work done in lifting the ball translates into kinetic energy when it falls back to its original position.
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I was thinking about the work-energy theorem today and how it states that:

Wnet = ΔEkinetic

If this is true, then when a ball is moved upward a distance of d, the net work done is equal to zero because there is no change in kinetic energy.

Because: Work = Force x Displacement

for every infinitely small distance, dr, that the ball moves in the upward direction, the work done by the upward force is equal to F * dr and the work done by gravitational force is equal to -mg * dr.

However, according to the previous statement, net work done when the ball moves a distance of dr is zero, which means that F*dr = mg * dr and F = mg

Because Fnet = F - mg, there is no net force. If there is no net force, why does the ball move up?

I feel like I am missing a very crucial part of logic, but I can't seem to figure it out. Any help would be greatly appreciated.
 
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For total energy, you need to include a potential term.

W = ΔEkinetic - ΔEpotential
 
rcgldr said:
For total energy, you need to include a potential term.

W = ΔEkinetic - ΔEpotential

This is wrong. The net work will always be the change in kinetic energy. No potential energy
terms are involved.
 
Acut said:
This is wrong. The net work will always be the change in kinetic energy. No potential energy terms are involved.
Note I dropped "net" from the work term in the equation I posted. I was relating total work done to total energy. If a 1kg ball is raised 1 meter, than 9.8 Newton meters of total work is done. If I then release the ball, it's kinetic energy at the moment it falls back to it's original position will be 9.8 Newton meters.
 
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