# Relation between potential energy and forces

Can you also explain why can't we define potential energy corresponding to a non-conservative internal force? Non-conservative forces are those which don't depend on the initial and final states but on the path taken. If such a force act as in a system as internal force why can't we define potential energy?

Relation between Forces and Potential Energy

Can you explain why can't we define potential energy corresponding to a non-conservative internal force?

In order to examine the relation between two terms we must consider the definitions of each term:

# a) forces:

• 1) internal forces are those that act inside a body (note that in engineering also a structure is considered a body), they interact between the parts of a body and keep it together. If a body is elastic it can have an internal force when it is compressed or stretched beyond the line of natural equilibrium.
• 2) contact (or applied) forces are those that act from outside and are in contact with a part of the body. A push or a pull do positive work, while friction and drag do negative work on a body
• 3) non-contact forces (gravity, electric and magnetic) can accelerate a body without any contact. We can consider these forces as elastic if we connect, for example, B and the ground with an ideal spring that stretches out when we separate them as in the bottom sketch:

# b) Potential Energy

Mechanical energy (ME) is the ability of a body to do [mechanical work]. A body has ME because of:

• 1) motion: kinetic energy is defined as the ability of a body to do work. If a massive body A impacts on another body B it will give some KE to B and do work. If KE is lost by a body B because of a conservative force (2,3) it is conserved PE
• 2) position: if a body is distant from the source of non-contact force has PE and it will acquire the KE lost
• 3) condition, (compressed/stretched): if a body is elastic it can have PE, it will tend to reach the position of natural equilibrium and do work on another body

Potential energy is associated only with elastic, conservative forces, that act on a body in a way that depends only on the body's position in space. These forces can be represented by a vector at every point in space forming what is known as a vector field or force field

An elastic force is conservative because it conserves the KE it subtracts to a body as potential energy. In the bottom sketch when body B is shot up in the air, it has PE = 0 and KE (mgh) = 10 (mg) * h, when it reaches h/2 has KE = 5 * h and PE = 5 *h, and at height h has KE = 0 and PE = 10 * h: ME is costant = mgh.

mechanical energy is the sum of potential energy and kinetic energy ($ME = KE +PE$. It is the energy associated with the motion and position of an object. The principle of conservation of mechanical energy states that in an isolated system that is only subject to conservative forces the mechanical energy is constant. If an object is moved in the opposite direction of a conservative net force, the potential energy will increase.

This you have learned in another answer. You ask now:

Non-conservative forces are those which don't depend on the initial and final states but on the path taken. If such a force act as in a system as internal force why can't we define potential energy?

• Probably you realize by now that PE cannot be associated to a non-conservative force, it would be a contradiction in terms, since PE is the conserved energy
• Besides that, no internal force is known apart from the spring force. If other non conservative force exist or existed inside a body we could never define a PE associated to them. The only forces associated with PE are the non-contact forces and the internal spring force. If you are interested you can find here details on how PE is stored in a spring.

Potential Energy is negative of the work done by conservative forces. The reason why we define potential energy is so that we can get the result::

Work done on a body = change in potential energy of the body (if only conservative forces act on the body) this is proved by the very definition of potential energy.

However let us say we were to define potential energy for a non conservative force. Let at a point x potential energy be U. Let at a point y potential energy be U2. We defined potential energy but there is no use for it. Difference in potential energy will not give us work done. There is no way to equate potential energy to work done. It does not simplify calculation of work done and is of no use. That is why we do not define potential energy for non conservative forces. The concept of work is to simplify complicated systems where newtons laws of mechanics is insufficient.

If you are wondering why i considered potential energy at a point x, it is because potential energy is defined as the energy a body contains by virtue of its position, orientation or state. Generally potential energy is by virtue of its position (gravitational and spring). I could have considered the other factors and still there would be no known way to equate potential energy to, work done by non conservative forces.

• Why potential negative of work done? Isn't work done by on any force be change in kinetic energy then why have you written it is change in potential energy? – pcforgeek Nov 30 '14 at 12:44
• its because P.E when defined was mainly for spring forces, gravitational forces. When a body is lifted up a height h, gravity does -ve work. So rather than having -ve potential energy, the p.e is positive. Same for spring forces. It is a matter of definition and is helpful generally in solving big equations wothout a calculator (at that time all equations were soved by hand) – Sashurocks Nov 30 '14 at 12:53

If you have an example in mind I could point out what is the problem. But you can prove mathematically that if a force is not conservative then it cannot be written as the gradient of a potential. And the reverse is true too. You can show mathematically as a theoren that is a force is conservative, it can be written as the gradient of some potential.

Work: $$W = \int_\gamma\mathbf F\cdot\mathbf{dr} = \int_{t_0}^{t_1}\mathbf F(\mathbf r(t))\cdot\mathbf{r}'(t)dt$$

If it is conservative then $\mathbf F = -\nabla U$. Therefore: $$W = \int_{t_0}^{t_1}\mathbf F(\mathbf r(t))\cdot\mathbf{r}'(t)dt = \int_{t_0}^{t_1} -\nabla U(\mathbf r(t))\cdot\mathbf{r}'(t)dt = -\int_{t_0}^{t_1}\left[U(\mathbf r(t))\right]'dt$$

Since we have a mathematical equality: $\nabla U(\mathbf r(t))\cdot\mathbf{r}'(t) = \left[U(\mathbf r(t))\right]'$. Therefore, the work: $W = U(\mathbf r(t_0)) - U(\mathbf r(t_1))$.

We can describe a force field in terms of potentials. Mathematically, we have to separate a divergence-free and a rotational-free parts of the force field. If we want a force field $\mathbf F$ to be written, will be: $$\mathbf F = \nabla\phi + \nabla\times\mathbf A$$

Which means, there are two potentials: A scalar potential $\phi$ and a vector potential $\mathbf A$ who can describe any force field. However, if the force field is conservative, then $\nabla\times\mathbf F = \mathbf 0$ and therefore $\mathbf F = \nabla\phi$. And then we can identify the potential energy being $U = -\phi$.

The relationship between the potential energy and what you call "conservative" force is:

$${\bf F} = \nabla U = {\bf n} \frac{\partial U}{{\partial\bf n}}\tag{1}$$

where ${\bf n}$ is the direction normal to the equipotential surface passing through some point $P$. Then, the mechanical work $dW$ done by the force ${\bf F}$ along WHATEVER path $dL$ that begins at the point P, is

$$dW = {\bf n}\cdot dL \frac{\partial U}{{\partial\bf n}},\tag{2}$$

where ${\bf n}\cdot dL$ is the projection of the vector $dL$ on the unit vector ${\bf n}$.

Now, let's denote by $S$ the equipotential surface that passes through $P$, and by $S'$, that which passes through $P'$. As says Sashurocks, the difference of potential between the two surfaces is

$$dU = -dW.$$

Now, any path you choose between P and P' has the same projection on the vector n. Thus, the difference in potential in FIXED, look at the equality (2), because ∂U/∂n is fixed and n∙dL is fixed. So, you can't make dU dependent on the path taken between P and P'.

Good luck!