Does work-energy theorem involve potentials?

Question

According to khan academy,

"Net work done on an object equals the object’s change in kinetic energy. Also called the work-energy principle."

Now, here we see that there is no mention of 'potential energy' but in a physics libre text on fluids, they mention work energy as the sum of the change in kinetic energy plus the sum of the change in potential energy. So which is it? Also, how do we derive this 'other' work energy theorem?

edit: In the libre text article they equate $\Delta K + \Delta U = \Delta p V$ where p is the pressure, how did they get get left side of equation?( $W = \Delta K + \Delta U$)

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References:

Libre Text

Khan Academy

Answer 1

The work energy theorem does not include potential energy because for a change in potential energy only, the net work done on an object equals zero.

For example lift an object a height $h$ beginning at rest and ending at rest. You do positive work transferring energy to the object equal to $mgh$. At the same time since the direction of the gravitational force is opposite the displacement of the object, gravity does an equal amount of negative work $-mgh$, for a net work of zero. Gravity takes the energy you gave the object and stores it as gravitational potential energy of the object-earth system. Since the object starts and ends at rest the change in kinetic energy is zero.

edit: In the libre text article they equate $\Delta KE + \Delta PE = V\Delta p$ where p is the pressure, how did they get get left side of equation? that is $W = \Delta KE + \Delta PE$. [ more context]

If only internal conservative forces (e.g., gravity) are involved in a system, the total mechanical energy (KE+PE) is conserved (constant) and we have

$$\Delta PE + \Delta KE = 0$$

If an external force does net external work $W_{ext}$ on the system, then there will be a change in its total mechanical energy

$$\Delta PE + \Delta KE=W_{ext}$$

Rearranging

$$W_{ext}-\Delta PE=\Delta KE$$

Applying the work energy theorem,

$$W_{net}=W_{ext}-\Delta PE=\Delta KE$$

In the case Bernoulli's equation, the external work done on the fluid is the flow work $V\Delta p$ and is due to the difference in pressures at the input and output and is positive if $p_{1}>p_{2}$. The internal work done on the system is the work done by gravity resulting in a change in potential energy. The change in potential energy $\Delta PE$ is negative when $y_{1}>y_{2}$.

Putting it all together we have

$$V(p_{2}-p_{1})-(-mg(y_{2}-y_{1})=W_{net}=\frac{1}{2} mv_{2}^{2} - \frac{1}{2} mv_{1}^{2} \ldotp$$

When applying the work energy principle, the source of the energy for performing $W_{net}$ is irrelevant. It can come from flow work, work done by gravity, or a combination of both. Bernoulli's equation allows for the possibility of both.

The above is consistent with the work-energy theorem which only addresses the effect of the net work done on an object, namely to change its kinetic energy. The source(s) of the energy for performing the net work is irrelevant.

Hope this helps.

Answer 2

BobD's answer is long, Charles's answer is short, so I will go in the middle :)

The key is the word net. The net work done on an object is equal to its change in kinetic energy. $$W_\text{net}=\Delta K$$ You can easily break the net work done on an object into work done by conservative forces internal to the system and work done by external forces.

$$W_\text{net}=W_\text{cons.}+W_\text{ext.}=\Delta K$$

By the definition of potential energy, the work done by conservative forces is equal to the negative change in potential energy associated with those conservative forces $$W_\text{net}=-\Delta U+W_\text{ext.}=\Delta K$$

This lets us arrive at the second expression you are confused about: $$W_\text{ext.}=\Delta K+\Delta U$$

i.e. the total mechanical energy changes when an external force does work.

So, both expressions say exactly the same thing, it is just that the latter breaks the net work into other classifications depending on how you define your system.

Answer 3

My approach will be a bit more systematic. In systems where energy is conserved it states : $$K+U = \text {const} \tag 1,$$

Differentiating this equation we arrive at : $$dK + dU = 0 \tag 2,$$

This means that work-energy theorem can be expressed equally well in terms of potential energy :

$$ W = \Delta K = - \Delta U. \tag 3$$

So to say, for object to increase it's kinetic energy it must consume some sort of potential energy. For example, body falling in Earth gravity consumes gravitational potential energy $mgh$. Mass speeding up on uncompressing spring - consumes spring potential energy $1/2 kx^2$ and so on.

Answer 4

The work energy theorem does not itself involve potentials, but in the case of a conservative force, in which work is independent of path, it is used to define the potential.