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The work energy theorem states that for a given object

∆ Kinetic Energy = Work done by all forces ( Conservative, Non Conservative, External ) Here , Suppose if ∆KE = 0 , Potential energy has changed

Ie. For example the ∆ kinetic energy of a body being brought up to a height by a crane , say x , is 0 This means the object doesn't acquire any velocity

The potential energy however is now mgx. Then, what does the given formula suggest?

I think that it means that even though the Kinetic Energy doesn't change , there is a reaction force ( equivalent to that of the Potential Energy one's) acting also. Is it so?

PS : It's an ideal condition.

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    $\begingroup$ You write: "here $\Delta KE = 0$" but you don't describe what that 'here' is referring to. That is, you are not providing enough information to answer the question. According to the work-energy theorem: for any change in state of potential energy there will be a corresponding change in kinetic energy. A mismatch would be an anomaly. That is, a mismatch would be a violation of the laws of motion. $\endgroup$
    – Cleonis
    Commented Feb 10, 2024 at 8:37
  • $\begingroup$ In the given example is what I mean $\endgroup$
    – Krishang Rana
    Commented Feb 10, 2024 at 16:46

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∆KE = 0 , Potential energy has changed , what does the given formula suggest?

It suggests that the net work done by conservative and non conservative forces is zero and that at least one of the forces is conservative since there has been a change in potential energy.

The change in kinetic energy of an object equals the net work done on the object. I emphasize the word net because work can be positive or negative. If $\Delta KE=.0$ it means either no work has been done or the amount of negative work equals the amount of positive work.

For example, if I (a non conservative force) lift an object of mass $m$ initially at rest on the ground and bring it to rest a height $h$ I do positive work on the object of $mgh$. But at the same time gravity (a conservative force), whose force direction is opposite of mine, does an equal amount of negative work $-mgh$, so the net work done on the object is zero. The work done by a conservative force equals the negative of the change in potential energy.

Since the object begins and ends at rest, $\Delta KE=0$ while there is an increase in gravitational PE of $mgh$.In effect, gravity took the energy I gave the object and stored it as gravitational potential energy in the Earth-object system.

Hope this helps.

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Change in kinetic energy of a particular object equals the sum of work done on the kinetic energy associated with a particular object.

This is just a restatement of the conservation of energy plus the definition of work. The amount of energy moved to or from the particle's kinetic energy equals the change of the particle's kinetic energy. Work is what we call the movement of energy from one mathematical object that can hold energy to another mathematical object that can hold energy. Hence the work-energy theorem is a theorem (a statement which can be mathematically proved - in this case the proof is trivial) and not a law (a statement which comports with measurement, like the conservation of energy from which the theorem is trivially derived).

Re: energy conservation. Contrast a non-conserved quantity. The change in the number of flowers in the garden is not equal to the amount of flowers moved to or from the garden.

Re: work done "on an object". Contrast a battery. You do work on the battery to charge it. The battery gets charged but doesn't go anywhere.

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  • $\begingroup$ Ofc it's a theorem. I also know that. But you seem to miss the mark. Your examples make sense , but you didn't really answer my question . $\endgroup$
    – Krishang Rana
    Commented Feb 10, 2024 at 7:14
  • $\begingroup$ About the derivation of the work-energy theorem: Start with $F=ma$ Integrate both sides with respect to position coordinate $$ \int_{s_0}^s F \ ds = \int_{s_0}^s ma \ ds \tag{1} $$ The end result: $$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \tag{2} $$ See: derivation of the work-energy theorem It follows from (2): in any specific case the sum of $E_k$ and $E_p$ will be conserved. We have the option of generalizing that result to a blanket statement: a principle of conservation of energy $\endgroup$
    – Cleonis
    Commented Feb 10, 2024 at 8:58
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The formula suggest that when $\Delta K.E. = 0$, then net work by all forces is zero. In case of the body raised to height $x$, work of the raising force is $mgx$, work of the gravity force is $-mgx$, so net work is zero.

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The work energy theorem states that the change in kinetic energy of a body is the work done on it by the net force, that is the work done by conservative forces plus the work done by the non-conservative forces. The work done by conservative forces can be associated to potential energy changes, and usually, the non-conservative work is lost to heat due to drag, friction, etc. So if $\Delta$KE = 0 for an object, that basically means the changes in the potential energy are dissipated, that is, the agent performing the non conservative force gains/loses energy. (A body falling at terminal velocity in the presence of air drag is a great example to see this.)

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  • $\begingroup$ It's an ideal condition $\endgroup$
    – Krishang Rana
    Commented Feb 10, 2024 at 7:14

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