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**the net work done by a force on an object is equal to change in kinetic energy only and not any other form of energy like potential energy?**Also is the work energy theorem valid for both conservative and non conservative forces.

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A.T.

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Quote in context?Why is it so thatthe net work done by a force on an object is equal to change in kinetic energy only and not any other form of energy like potential energy?

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The work energy theorem assumes a rigid body, so there is no other form of internal energy.the net work done by a force on an object is equal to change in kinetic energy only and not any other form of energy like potential energy?Also is the work energy theorem valid for both conservative and non conservative forces.

It is always important to remember the assumptions or constraints on any formula.

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CWatters

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Shouldn't that read... The net work done on an object byWhy is it so that

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nasu

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@CWatters , Yes that should be 'all' instead of 'a'.

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I know that

According the law of conservation of energy,

=>U

=>-(U

=>- Delta U=Delta K

∴Net work done by all forces = Delta K =

Now, i read on a website that

W

∴ W

Earlier i used to think that the

1.How to prove that for conservative forces only the net work done is equal to the change in kinetic energy(or negative of the change in potential energy) and not for non conservative forces?

2.Why is work done by a non conservative force equal to the sum of change in kinetic and potential energy as in the second equation?

3.Also why is the work-energy theorem valid for kinetic energy only and not for any other forms of energy?

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CAF123

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The statement is that ##T_2 - T_1 = \Delta K = \int \mathbf F \cdot \mathbf{dr}##, ie that the net work done on an object is equal to its change in kinetic energy. The right hand side encodes that work done by both conservative and non conservative forces. The former type can always be written as the gradient of some scalar function, the potential energy. Viz. $$\int \mathbf F \cdot \mathbf{dr} = \int (\mathbf F_{\text{cons}} + \mathbf F_{\text{non-cons}}) \cdot \mathbf{dr} = \int \mathbf F_{\text{cons}} \cdot \mathbf{dr} + \int \mathbf F_{\text{non-cons}} \cdot \mathbf{dr}.$$ Now, ##\mathbf F_{\text{cons}} = - \nabla V##, where V is the scalar function alluded to previously. Then $$\Delta K = -\int \nabla V \cdot \mathbf{dr} + \int \mathbf F_{\text{non-cons}} \cdot \mathbf{dr} = - \Delta V + \int \mathbf F_{\text{non-cons}} \cdot \mathbf{dr} $$ Rearranging gives $$\Delta K + \Delta V = \int \mathbf F_{\text{non-cons}} \cdot \mathbf{dr} $$ If there are no non-conservative forces at play, then the right hand side is identically zero. If there are, it is not zero and corresponds to the ##\Delta E## in your equations. C.f, if friction is present in your system (a non conservative force) then it will dissipate kinetic energy to heat say so that the change in kinetic energy is not the change in potential energy, as ##\Delta K = - \Delta V## would imply.Why is it so that

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A.T.

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Link?Now, i read on a website

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Chandra Prayaga

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ΔK = W

W

ΔK + ΔU = W

And our system now has a total mechanical energy E = K + U and

ΔE = W

Here is an example: A block hanging from a vertical massless spring. List of forces:

Conservative: gravity, spring

Non-conservative: Friction, drag

Work energy theorem: Consider the block alone as our system. All four forces listed above are external to the system. The theorem (equ (1)) states:

ΔK = W

Subscripts g, s, f, and d stand for gravity, spring, friction and drag respectively. The first two terms on the right in the equation are work by conservative forces. The last two are by non-conservative forces. This is correct. No potential energies are involved here.

Now, think of our system as ball + spring. To do this, we replace the work done by the spring with the corresponding spring potential energy, W

ΔK + ΔU

You can now continue the exercise and include gravity (earth) into our system. Then the same equation becomes:

ΔK + ΔU

Our system mechanical energy is now: K + U

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@Chandra Prayaga why is WWc in equ (1) can also be represented in terms of potential energy: Wc = - ΔU. In this step, you have included the agents exerting the conservative forces into the system

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A.T.

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Per definition:why is W_{conservative }equal to -ΔU and why is W_{non-conservative}not equal to -ΔU?

https://en.wikipedia.org/wiki/Conservative_force#Mathematical_description

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Chandra Prayaga

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Thank you A.T. That reference does give a complete answer to the question.

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@A.T. , can you please explain in simple terms since i havenot taken a calculus course yet.

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@Dale why does a rigid body has no other form of energy other than K.E.?

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Chandra Prayaga

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That statement is not correct. A rigid body + the earth can have gravitational potential energy.

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jbriggs444

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A system consisting of a rigid body plus the earth can indeed have internal energy such as gravitational potential energy. But then a system consisting of a rigid body plus the earth is not a rigid body.That statement is not correct. A rigid body + the earth can have gravitational potential energy.

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Chandra Prayaga

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Agreed.

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A rigid body has no other forms of internal energy because it has no internal degrees of freedom. More exactly, there can be no change in internal energy because there are no internal degrees of freedom available to be changed.@Dale why does a rigid body has no other form of energy other than K.E.?

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@Dale What if the body is non-rigid?What will be the relation between work and Energy in that case?

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jbriggs444

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In the case of a non-rigid object (or a rotating object) one can distinguish between the work computed by force times parallel distance moved by the point of application and "pseudo-work" or "center of mass work" computed as force times parallel distance moved by the center of mass of the object. The latter will give you the increment to the object's bulk kinetic energy of linear motion. The former will give you the increment to total energy of all forms.@Dale What if the body is non-rigid?What will be the relation between work and Energy in that case?

So, for instance, if you stretch a rubber band by pulling both ends, the "center of mass" work is zero. The center of mass is not moving and the net force is zero. But if you compute the work done by your right hand on the right end of the rubber band, that's positive. If you compute the work done by your left hand on the left end of the rubber band, that's also positive. Work has been done and internal energy has been stored in the rubber band.

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sophiecentaur

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Then it has internal energy states and the situation is different.@Dale What if the body is non-rigid?What will be the relation between work and Energy in that case?

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The relation is more complicated. For a non rigid body it is possible to have a net external force that does no work and instead the body changes some internal energy into kinetic energy. An example is a car.@Dale What if the body is non-rigid?What will be the relation between work and Energy in that case?

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