Question regarding Work

[rslancer]

Hi I am a lil bit confused about work and energy conservation. So I understand that Work= F(delta x) and Initial E + Work = Final E; however I'm not sure I understand the work done by gravity so much.

So if a still object starts at height of 0 and is lifted up to a height of H and continues to be still, then Initial E is = to 0 but Final E= mgH. So this would imply that the person lifting it did work of mgH but gravity must have also done -mgH. This would mean that the system actually hasn't gained energy??? So where did the Final E of mgH come from???

Sorry if this is obvious. I'm not too hot with physics

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[RogueUnicorn]

pushing & pulling are not conservative forces. the increase in the potential energy in this system came not from gravity but from an external source putting it in the higher h.

 

[rslancer]

hmm i'm not sure i entirely understand. so since the increase in PE came from the work done by lifting it up to the new height shoudn't

PE = to -mgh + work done? in otherwords work done= 2mgh???

 

[RogueUnicorn]

work done by lifting object up to height h is simply force times distance, which of course will equal mgh. we're dealing with nonconservative forces, so it's best not to be concerned with a direct input of work into potential energy. this only works with conservative forces

remember, work-energy theorem applies to kinetic energy only. it's possible to put in an infinite amount of work without changing the potential energy one bit, and, as we see here, it's possible to put in NO work and gain potential energy.

 

[kentavr]

rslancer, I don't think bleargh is right. Energy-work relationship is correct for total mechanical energy which is a sum of kinetic and potential energy. The trick is that if the only conservative forces(gravity,electrostatic etc) act on the system, then the total energy will not change at all. (Conserved). Drop your ball and it will get kinetic energy. But the total PE+KE=const. So in your case, work of gravity is irrelevant, it does not change the total energy. What does? You. Outside force, its origin is the transformation of chemical energy to mechanical. It does work against gravity and change total energy to mgh. If the speed of the boll at h is not zero, then the additional kinetic part should be included too. In order for speed at h to be non zero, the acceleration(a) has to be involved. and the work will be (mg+ma)*S. And the change in total energy will be equal to this value.

 

[RogueUnicorn]

rslancer, I don't think bleargh is right. Energy-work relationship is correct for total mechanical energy which is a sum of kinetic and potential energy. The trick is that if the only conservative forces(gravity,electrostatic etc) act on the system, then the total energy will not change at all. (Conserved). Drop your ball and it will get kinetic energy. But the total PE+KE=const. So in your case, work of gravity is irrelevant, it does not change the total energy. What does? You. Outside force, its origin is the transformation of chemical energy to mechanical. It does work against gravity and change total energy to mgh. If the speed of the boll at h is not zero, then the additional kinetic part should be included too. In order for speed at h to be non zero, the acceleration(a) has to be involved. and the work will be (mg+ma)*S. And the change in total energy will be equal to this value.

thank you. this is what i was trying to say but with some difficulty