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But if the resultant force is in vertically up direction it will surely change its potential energy too, so what's the solution here.

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- Thread starter damitr
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But if the resultant force is in vertically up direction it will surely change its potential energy too, so what's the solution here.

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the theorem says that

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damitr said:The work energy theorem says that ''The work done by the net force acting on a body results change only in its kinetic energy. ''

Is that what your textbook says, exactly? It's not the way I would state the work-energy theorem. I would state it simply as "The work done by the net force acting on a body equals the change in its [the body's] kinetic energy." (No "only".) Or, better, "The net work done by all the forces acting on a body equals the change in its kinetic energy."

The work-energy theorem in this form makes no reference to potential energy at all, and has no implication about what happens to the potential energy.

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"The net work done by all the forces acting on a body equals the change in its kinetic energy."

it still does not answer the quastion even here is no mention about the change in potential energy, only kinetic energy is mentioned and that's what the real trouble is.

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If there is a

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I have not understood this statement of yours.

What do you exactly mean by ?

Do we not have to take into consideration the potential energy that has been changed ?

And from what I have learned the work done by the external forces only changes the total energy of the system. In this case it seems that the total energy is changing, kinetic as per the work-energy theorem, and potential due to increase in height.

If I am wrong in reasoning anywhere, if so please point out.

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To put it more precisely, there are two kinds of forces: conservative forces, which have potential energy associated with them, and non-conservative forces, which don't have potential energy associated with them. Define the mechanical energy as the sum of kinetic energy K and the potential energy U:

[tex]E_{mech} = K + U[/tex]

Also define [itex]W_{nc}[/itex] as the net work done by the

[tex]W_{nc} = \Delta E_{mech} = \Delta (K + U)[/tex]

In this version of the work-energy theorem, the effects of gravity are included on the right side of the equation, via U. [itex]W_{nc}[/itex] does not include work done by the graviational force.

In the original version of the work-energy theorem,

[tex]W_{net} = \Delta K[/tex]

the effects of gravity are included on the left side of the equation. That is, the work done by the gravitational force is included in [itex]W_{net}[/itex].

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The purpose of gravitational potential energy is to account for the effect of gravity, so if you consider the object's weight as an external force then you don't also include potential energy--you'd be counting it twice!damitr said:As you have said "the net force includes gravity, so no need to mention potential energy."

I have not understood this statement of yours.

What do you exactly mean by ?

The work-energy theorem is accurate as stated: the work done by theDo we not have to take into consideration the potential energy that has been changed ?

And from what I have learned the work done by the external forces only changes the total energy of the system. In this case it seems that the total energy is changing, kinetic as per the work-energy theorem, and potential due to increase in height.

If I am wrong in reasoning anywhere, if so please point out.

An example may help. Say you lift an object of mass m a height h at constant speed. What's the net force? The force you apply (up) must equal the weight (down), so the net force is zero. Thus, according to the W-E theorem, the change in KE is zero. On the other hand, if you only consider the applied force F=mg (up), then the work done by that force is mgh, which equals the change in "KE + PE" which is mgh. Make sense?

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True, but the work done by theJeff Reid said:

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