Why is the net work done equal to the change in K.E. only?

[donaldparida]

Why is it so that 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.

 

[A.T.]

Why is it so that 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?

Quote in context?

 

[Dale]

Why is it so that 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.

The work energy theorem assumes a rigid body, so there is no other form of internal energy.

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

 

[CWatters]

Why is it so that 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?

Shouldn't that read... The net work done on an object by all forces is equal to the change in kinetic energy?

 

[nasu]

The "net work" implies work done by all the forces acting on the object. So the"done by a force on an object" can be removed without any harm.

 

[donaldparida]

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

 

[donaldparida]

@A.T.
I know that
According the law of conservation of energy,

U_i + K_i = U_f +K_f (i-initial, f- final)

=>U_i - U_f =K_f - K_i

=>-(U_f - U_i)=Delta K

=>- Delta U=Delta K
∴Net work done by all forces = Delta K = - Delta U

Now, i read on a website that
W_conservative = ΔK=−ΔU
W_{non−conservative} = ΔK+ΔU = ΔE.
∴ W_net = W_conservative + W_{non−conservative}=-ΔU + ΔK+ΔU=ΔK

Earlier i used to think that the net work done by all forces is equal to the change in kinetic energy but according to the first equation (W_conservative = −ΔU) the net work done by all conservative forces is equal to the change in the kinetic energy.

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?

 

[CAF123]

Why is it so that 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?

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.

 

[A.T.]

Now, i read on a website

Link?

 

[Chandra Prayaga]

Your Q1: It is not true that Change in K is equal to work done by conservative forces only. It is true for all forces together. So the theorem first reads:
ΔK = W_c + W_nc. We will call this equ (1)
W_c in equ (1) can also be represented in terms of potential energy: W_c = - ΔU. In this step, you have included the agents exerting the conservative forces into the system. Then our equ (1) becomes:
ΔK + ΔU = W_nc this is equ (2). This is therefore not different from equ 1. In this form, all conservative forces have been included as part of our system. They are now internal forces. This answers your Q2. Answer to Q3 is also already here. Calling work done by a conservative force as the negative of the change in potential energy is merely a matter of including the agent of the force into our system. It is not really necessary to do that. We may even choose to include some external agents exerting a conservative force into our system, and choose not to so include others.
And our system now has a total mechanical energy E = K + U and
ΔE = W_nc
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_g + W_s + W_f + W_d.
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_s = - ΔU_s, where U_s is the potential energy of the spring. Now take that term to the left in our equation and we have:
ΔK + ΔU_s = W_g + W_f + W_d. In this situation, the system mechanical energy is K + U_s. Gravity is still being considered as an external force.
You can now continue the exercise and include gravity (earth) into our system. Then the same equation becomes:
ΔK + ΔU_s + ΔU_g = W_f + W_d
Our system mechanical energy is now: K + U_s + U_g, and changes in the mechanical energy are because of non-conservative forces only.

 

[donaldparida]

Chandra Prayaga said:

Wc 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

@Chandra Prayaga why is W_conservative equal to -ΔU and why is W_{non-conservative} not equal to -ΔU?

 

[A.T.]

why is W_conservative equal to -ΔU and why is W_{non-conservative} not equal to -ΔU?

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

 

[Chandra Prayaga]

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

 

[donaldparida]

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

 

[donaldparida]

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

 

[Chandra Prayaga]

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

 

[jbriggs444]

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

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.

 

[Chandra Prayaga]

Agreed.

 

[Dale]

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

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.

 

[donaldparida]

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

 

[jbriggs444]

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

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.

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.

 

[sophiecentaur]

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

Then it has internal energy states and the situation is different.

 

[Dale]

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

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.

 

[Shafia Zahin]

@donaldparida Awesome questions really. Never thought like this before in this topic and it's really helpful. Seeming like I'm asking the questions because i tend to fall in problems like this always