Work Energy Theorem and Potential Energy violation?

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Discussion Overview

The discussion revolves around the Work Energy Theorem and its application to potential energy when lifting a particle. Participants explore the relationship between work done, kinetic energy, and potential energy, particularly in scenarios where a particle is lifted to a height.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about the application of the Work Energy Theorem, noting that if the initial and final kinetic energy of a particle is zero, then the net work done should also be zero, leading to a contradiction regarding the change in potential energy.
  • Another participant clarifies that the work done by all forces, including gravity, should be considered, and that the work done by the hand lifting the particle is equal to the increase in potential energy, while the work done by gravity is negative.
  • A different viewpoint suggests that the kinetic energy gained while lifting the object transforms into potential energy when the object reaches a certain height.
  • One participant questions the logic of equating kinetic energy to potential energy when the object remains at a height, emphasizing that work must be done against gravity to achieve this height.
  • Another participant presents a scenario involving throwing a rock, explaining the transition of kinetic energy to potential energy as the rock is lifted, and requests a mathematical representation of this idea.
  • Several participants provide a detailed breakdown of energy states during the lifting process, illustrating how work done translates into changes in potential and kinetic energy.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of the Work Energy Theorem and the relationship between kinetic and potential energy. There is no consensus on how to reconcile the apparent contradictions in the application of these concepts.

Contextual Notes

Some participants' arguments depend on specific assumptions about the system, such as the negligible change in height while throwing the rock. The discussion also highlights the complexity of energy transformations and the need for careful consideration of work done by different forces.

easwar2641993
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I came across a rather confusing topic about Work Energy Theorem and Potential Energy applied in lifting a particle.I will be glad if anyone clears it for me.

Consider a particle at a height =0.Potential Energy is considered as zero at height=0.Now it is lifted to a position where height is h so that Potential Energy of the particle is increased by a value mgh where m is the mass of it,g is the gravitational acceleration.Or in other words energy possessed by the particle is increased from zero to mgh.Now the problem begins.
Net work done on the system W(net)=Change in Kinetic energy.But here initial and final kinetic energy of the particle is considered to be zero since the particle is stationary at initial and final positions.This means there is no change in energy of the particle.Because energy is something required to do work.But since net work is zero ,then energy of the particle should be constant.And I just can't apply energy function for this case
Potential energy U(final)-U(initial)=-W=K(initial)-K(final)
But W=0.Then how can I say potential energy change is zero since it is obvious that it possesses mgh energy?
 
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The work-energy theorem says that if you consider the work done by all forces (including gravity) it will equal the change in the kinetic energy (not total energy). And that's just what happens.

If you consider the work done by all forces except gravity, then the work done will equal ΔK + ΔU. (The work done by gravity is already included in the potential energy term.)
 
The work that is equal to mgh is the work of the hand which lift the particle. The work done by the gravitational force is -mgh. The net work is zero as it should be since no kinetic energy was gained.

Best wishes,

DaTario
 
Maybe we should view the situation in this way?

since the object was moved a certain distance by a force work was done on it so its kinetic energy has increased but since it stays at that height the kinetic energy becomes potential energy?
 
mihirviveka said:
Maybe we should view the situation in this way?

since the object was moved a certain distance by a force work was done on it so its kinetic energy has increased but since it stays at that height the kinetic energy becomes potential energy?

That doesn't make any sense, since the height in the problem state by the OP has clearly changed! Otherwise, there's no work done in the first place against gravity.

Zz.
 
this thing has been bothering me for a long time

what I had in mind was this situation:

If we throw a rock from the ground to the top of a building then it's kinetic energy has clearly increased but it just stays on the roof of the building so the increase in kinetic energy has become increase in potential energy.

What we need is the mathematical representation of this idea, can anyone please do that?
 
Suppose that the rock has zero potential energy at ground level and 100 J of potential energy when it's on the roof of the building.

You (standing on the ground) throw the rock upwards and do 100 J of work on it. Before you start to throw it, it has 0 J PE and 0 J KE, for a total of 0 J mechanical energy. When it leaves your hand, it has 0 J PE and 100 J KE, for a total of 100 J mechanical energy which came from the work you did on it. (I'm assuming the rock doesn't change height significantly while it's in your hand, so its PE doesn't change while you throw it.) When it reaches the rooftop and comes to rest again, it has 100 J PE and 0 J KE, which is still a total of 100 J of mechanical energy.
 
jtbell said:
Suppose that the rock has zero potential energy at ground level and 100 J of potential energy when it's on the roof of the building.

You (standing on the ground) throw the rock upwards and do 100 J of work on it. Before you start to throw it, it has 0 J PE and 0 J KE, for a total of 0 J mechanical energy. When it leaves your hand, it has 0 J PE and 100 J KE, for a total of 100 J mechanical energy which came from the work you did on it. (I'm assuming the rock doesn't change height significantly while it's in your hand, so its PE doesn't change while you throw it.) When it reaches the rooftop and comes to rest again, it has 100 J PE and 0 J KE, which is still a total of 100 J of mechanical energy.

great! I think this settles the issue
 

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