# Energy change and work involved in lifting a ball

1. Dec 18, 2011

### darkp0tat0

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.

2. Dec 18, 2011

### rcgldr

For total energy, you need to include a potential term.

W = ΔEkinetic - ΔEpotential

3. Dec 18, 2011

### Acut

This is wrong. The net work will always be the change in kinetic energy. No potential energy
terms are involved.

4. Dec 18, 2011

### rcgldr

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.