Maintaining constant Potential Energy

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Discussion Overview

The discussion revolves around the concept of maintaining constant potential energy for an object at a fixed height, particularly in the context of gravitational force and the energy expenditure of helicopters. Participants explore the relationship between work, force, and potential energy, questioning why energy is required to keep an object at a constant height despite the notion that no work is done when displacement is zero.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant notes that to maintain an object at a constant height, an upward force (upthrust) is necessary to counteract gravity, which helicopters achieve by forcing air downwards.
  • Another participant emphasizes that while work is defined as force times distance, energy is still expended to maintain position, as seen in biological processes even when standing still.
  • A different perspective suggests that mixing energy and force concepts can lead to confusion, highlighting that lifting an object requires equal force to its weight and that potential energy is stored when lifting against gravity.
  • This participant also explains that if the lifting force is removed, the object will fall, converting potential energy to kinetic energy, and that helicopters must continuously do work against air resistance, which requires fuel consumption.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between work, energy, and maintaining constant potential energy. While there is some agreement on the necessity of an upward force to counteract gravity, the implications of energy expenditure and the definitions of work remain contested.

Contextual Notes

Participants discuss the assumptions involved in lifting objects, the conditions under which work is considered done, and the implications of energy expenditure in biological systems and mechanical systems like helicopters. The discussion highlights the complexity of these concepts without resolving the nuances involved.

Who May Find This Useful

This discussion may be of interest to individuals studying physics, engineering, or those curious about the principles of energy, force, and motion in both theoretical and practical contexts.

cool0crash
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Hi all,

While thinking about potential energy and force, I couldn't figure out how to maintain an object at a constant height above the ground i.e to maintain a constant potential energy.

I know that potential energy will be equal to the amount of workdone to lift the object to a particular height (considering gravity) but W = F x displacement. But to maintain an object at a constant potential energy (height), displacement = 0 & hence no work or energy is required.
If that is right, why does helicopters expend energy (fuel), to stay at a particular height?Also consider an object on ground, what is the exact action taking place if the same magnitude of force as that of gravity is applied in the upward direction?
 
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cool0crash said:
. But to maintain an object at a constant potential energy (height), displacement = 0 & hence no work or energy is required.
An upthrust is required to counteract gravity.Helicopters do that by forcing air, but air being a fluid constantly moves and is replaced so constant power is to be delivered
 
Work is defined as force times distance, W = fd, so NO work is done. Energy and power, via the consumption of fuel, IS expended.

When you stand still, or sit, for example, no work is done to the extent you are not undergoing translation (no movement). But you are burning calories,energy, just to maintain body functions and stay alive. Of course some movement IS taking place inside you as blood circulates, your lungs expand and contract, and your heart beats...
 
I think it is not good to mix up the energy and the force.. say that you are lifting a stone against the gravity.. then as you and I have heard we can use E=Fx to find the energy.. but here it has assumed that the lifting process has done in uniform velocity.. that is why E=Fx and not E=Fx + 1/2 m*v*v ,(here the v is the final velocity which the stone takes), anyway what I am pointing out is you are lifting the stone against the gravity and using the same force as the weight of the stone.. that same force is applied by your hand.. so the two fores are equal and ni resultant force.. so the uniform velocity assumption is there (here the motionless situation can't be assumed , because the stone is moving). anyway mg=F, F is the force by your hand, then E=Fx , which gives the energy you wasted (give) for the lifting and stored in the stone..
E=Fx=mg*x, then you will have the known equation 'E=mgh'... this only gives which amount of energy that stone stored.. let's come to your point... say that you
took your hand away, what happens now.. the stone will face to the force field by the Earth's (gravitational force).. if there is no a hand to give an equal reaction as the weight
(or reaction by the wings of helicopter), the stone will fell (accelerate) to the down and it will be a free fall.. which convert its potential energy to the kinetic energy.. and in your case the helicopter spend fuel to provide that equal force, .. not like the hand which keeps the stone in stable situation(when the stone is in the final place).. helicopter have to do a work(against the air) for rotating the wings.. that is why it has to spend fuel...I think that this will be a support for your problem...
 

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