Energy change and work involved in lifting a ball

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
3 replies · 4K views
darkp0tat0
Messages
3
Reaction score
0
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.
 
Physics news on Phys.org
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.