# What exactly is potential energy?

Consider a ball falling from a height $$h$$. It gains velocity because of the work done by gravity on it. I don't quite understand the role of potential energy here. What does the potential energy exactly do in this scenario; i.e. what is the role of potential energy in the falling of an object?

• Potential energy change is just a scalar defined for a special kind of force , usually called as conservative force ; usually for solving problems easily. Commented Mar 10, 2023 at 7:51
• Potential energy of body is converted to kinetic energy by gravity field. Commented Mar 10, 2023 at 10:32
• Potential energy can mostly be usedfor the transformation of energy to kinetic energy by the law of conservation of energy. Intuitively it is useful to think as some gained up energy present in the object which converts into kinetic energy in required situations Commented Mar 10, 2023 at 16:11
• Why is this question closed? Surely it's a conceptual question! Commented Jul 28, 2023 at 3:05

The potential energy of the system does nothing in itself since it is a mathematical construct representing the work associated to a force from one object to another. In your problem, it is the force of gravity of the earth that pulls the ball into an accelerated falling motion. The potential energy of the Earth-ball system is the work (within a constant, depending on the reference point) that must be done to bring the ball to the height h. It can be interpreted as "how hard it is" for the Earth-ball system to have the ball at height $$h$$.

In traditional teaching, mechanics courses start by working with forces acting on different bodies, and later introduce energy. You can switch from a force representation of your problem to an energy representation (1.) and vice-versa (2.) by

1. Integrating a force in its path corresponds to the potential energy associated with it.
2. Differentiating a potential energy with respect to space coordinates corresponds to calculating the force associated with this potential.

We can therefore move from a representation in terms of force to a representation in terms of potential energy without losing any information about the problem, due to mathematical properties of derivation and integration. This means that a mechanical problem can be solved (if solvable) in both framework.

1. When solving a problem using in the force "framework", the second law of Newton has to be solved.
2. When solving a problem using the energy "framework", an integrated version of the second law of Newton has to be solved.

The reason why potential energy is introduced is that if it can be determined, powerful analytical tools can be used to find the time equation of motion (analytical mechanics). It can also simplify equation solving when using Newton's mechanics: in the case of your problem, the equation to solve is $$\underbrace{mgh}_{\text{Potential energy at height }h}=\underbrace{\frac{1}{2}mv(0)^2}_{\text{Kinetic energy at height } 0}-\underbrace{\frac{1}{2}mv(h)^2}_{\text{Kinetic energy at height } h}$$ in an energy "framework".

• The potential energy of the ball is . . . . . rather it is the potential energy of the ball and Earth system. The ball of its own cannot have potential energy as it must be part of a system of at least two masses. Commented Mar 10, 2023 at 15:11

As others have effectively pointed out Gravitational Potential Energy is

the energy possessed by an object because of its position in the gravitational field.

"...It gains velocity because of the work done by gravity on it..."

...What does the potential energy exactly do in this scenario?...

This is the confusion. It is not that both of them affect the object separately, GPE and work done by gravity both describe the gain in velocity of the object as well as by how much.

If you look closely, you'll realise that both of them involve the same expressions and equations.

### Using the concept of work

Let $$h$$ be the change in height. Now, since the object moves through $$h$$ by a constant force $$mg$$, $$W_g= mgh$$

With that being said, you can easily calculate the velocity using the kinetic energy formula.

### Using GPE

Since the change in height is $$h$$, change is potential energy of the object is $$mgh$$, all of which has to convert into kinetic energy.

Now, plug the values and use kinetic energy formula to get the velocity.

## Intuition

You can think of potential energy at a height being the total work which could be done on the object in future (if the object is released) by the conservative force in whose field the object is.
• . . . . . change is potential energy of the object is . . . . ., a single object cannot possess potential energy. Commented Mar 10, 2023 at 16:06
• @Farcher It's important to remember that when you're using the approximation that $m_o << m_e$, the potential energy is only in the position of the object. Commented Mar 10, 2023 at 16:30
• @SeñorO I entirely agree with your comment but to me an important idea is that the potential energy is stored in a system of at least two masses. Commented Mar 10, 2023 at 23:17

Consider a ball falling from a height $$h$$. It gains velocity because of the work done by gravity on it. I don't quite understand the role of potential energy here.

If your system is the ball alone gravitational potential energy has no role to play at all.
To define a quantity called gravitational potential energy of a system the system must contain at least two masses.
If the system that you are considering is the ball alone then that system has one external force acting on in - the gravitational attraction of the Earth, $$mg$$.
When the ball falls through a height $$h$$ the work done by the external force is $$mg\times h = mgh$$ and that amount of work done is equal to the change in the kinetic energy of the ball.
The confusing part of this analysis is probably because you equate $$mgh$$ to a change in the gravitational potential energy but as I have pointed out the idea that the ball alone has gravitational potential energy is flawed.

If the system under consideration is the ball and the Earth then there are no external forces but there are two internal forces, the gravitational attraction the Earth has for the ball and the gravitational attraction the ball has for the Earth, and they equal in magnitude and opposite in direction.
When the ball is accelerating towards the Earth (and the Earth accelerating towards the ball) the internal forces do work.
As the mass of the Earth is so much bigger than the ball the work done by the gravitational force on the Earth due to ball is usually neglected because the Earth moves hardly at all compared to the distance moved by the ball.
The work done by the internal force, gravitational attraction on the ball due to the Earth, $$mg$$ when the ball falls a distance $$h$$ is $$mg\times h=mgh$$ and again this is equal to the change in kinetic energy of the ball.
Note that the motion of the Earth has been neglected in this calculation as in reality if the distance between the ball and the Earth changes by $$h$$ the ball has actually moved very slightly less than $$h$$.
So everything has been done in terms of forces but if the idea of gravitational potential energy is introduced the situation of the ball falling can be described as follows.
When the separation of the ball and the Earth changes by a distance $$\Delta h$$ the gravitational potential energy of the ball and Earth system changes by $$mg\Delta h$$.
Suppose the ball mass, $$m$$, started at a height $$h_{\rm i}$$ above the Earth an initial velocity $$v_{\rm i}$$ with an initial.
Now, what is the gravitational potential energy of the ball and Earth system?
Well, it can be anything you like so we could define the zero of the gravitational potential of the ball and Earth system when the ball is at the surface of the Earth.
Thus at a height $$h_{\rm i}$$ the ball and Earth system has a gravitational potential energy is $$mgh_{\rm i}$$ and at the final separation $$h_{\rm f}$$ the system has a gravitational potential energy $$mgh_{\rm f}$$.

Conservation of energy tells us that the sum of the kinetic energy and gravitational potential energy must be constant, thus, $$\frac 12 mv_{\rm i}^2 + mgh_{\rm i} = \frac12 mv_{\rm f}^2 + mgh_{\rm f}$$.
This can be put another way $$mgh_{\rm i}-mgh_{\rm f} = \frac 12 mv_{\rm f}^2 - \frac 12 mv_{\rm i}^2$$ and if $$h = h_{\rm i}-h_{\rm f}$$ you get the equation that you are familar? with $$mgh = \frac 12 mv_{\rm f}^2 - \frac 12 mv_{\rm i}^2$$, or in words, the loss in gravitational potential energy of the ball (and Earth) is equal to the gain in kinetic energy of the ball (and Earth), noting again that usually only the ball is mentioned.

Thus the falling ball can be analysed via forces or via energies (by the introduction of the concept of potential energy).
You can pick and choose which approach to use noting that forces are vectors so you have both magnitude and direction to deal with whereas energy is a scalar and hence only the magnitude needs to be considered making energies easier to add when compared with the addition of forces.

Potential energy is the energy possessed by it by the virtue of its position, shape or configuration.

This simply means that work is done on an object and it gains some energy. But this energy, instead of initiating any change in the velocity of the object, simply gets stored.

Consider a ball falling from a height h

At this height h, the ball (more precisely, the ball-earth system) has some potential energy because work has been done on it against the force of gravity, which is precisely defined as $$W_{net} = F_{net}*displacement$$ = mgh. When it is released, it falls and gains velocity since the total mechanical energy of the ball remains conserved i.e., Potential Energy + Kinetic Energy = constant and as the ball loses height, its potential energy decreases while it gains kinetic energy.

Thus, the potential energy of the ball slowly gets converted into kinetic energy.

• The ball doesn’t have potential energy. The ball-Earth system does. Commented Mar 10, 2023 at 11:18

It comes from the second law of Newton, according to which the net force acting on a body equals its mass $$\times$$ acceleration. Multiplying both sides by a infinitesimal vertical displacement $$dh$$:$$Fdh = madh = m\frac{dv}{dt}dh = mdv\frac{dh}{dt} = mdv.v = d\left(\frac{1}{2}mv^2\right)$$ When we consider the gravitational force ($$F = -mg$$) constant for small displacements, it is possible to integrate the expression:$$-mg\Delta h = \Delta \left(\frac{1}{2}mv^2\right)$$ So this approximation is useful because the variation of the right side of the equation involving velocities can be known by the information of the difference of height.

A model that works well is to consider the potential energy as energy "within the conservative force" between two objects. Kinetic energy has been removed from either or both of the objects by the force between them, but the force has not yet transferred the energy back to either or both of the two objects. This is easiest to imagine with a spring between two objects, considering the potential energy as energy stored within the spring. This also works for a system of several objects, provided the force between each pair of objects is conservative. You need a kinetic energy for each object. You need one potential energy for each pair of objects, or for each one force between each pair of objects.