Does gravitational potential energy exist?

Question

If an object is taken at a height $h$ above the ground, its potential energy is $mgh$. When it is then released from there, it falls freely due to the force of gravity and its kinetic energy increases. Its kinetic energy is maximum at the moment of touching the ground. What about the potential energy? Why do physics books tell us potential energy converts into kinetic energy?

Answer 1

As the object is raised from ground level to height $h$, work is done against gravity. The amount of work done to raise the object by a distance $h$ is $mgh$. Usually we say this energy is stored as potential energy - another way of putting it is that the energy is stored in the gravitational field.

When the object is released from rest it initially has potential energy of $mgh$ and kinetic energy of $0$. When it reaches the ground, after accelerating at a constant acceleration of $g$ for a distance $h$, it has as velocity $v = \sqrt{2gh}$. Therefore it now has kinetic energy $\frac 1 2 mv^2 = mgh$ and potential energy $0$.

Indeed, at any point of its motion between height $h$ and the ground, conservation of energy tells us that kinetic energy + potential energy is always equal to $mgh$. This is why we say that the object's initial potential energy $mgh$ is converted into kinetic energy as the object falls. If the object were thrown upwards instead, we would says that its initial kinetic energy is converted into potential energy as it rises.

Answer 2

Potential energy and kinetic energy are just nice tools to simplify the basic kinematics.

From kinematical definitions, ^dv⁄_dt = a

which when simplified is ^dv⁄_dx ^dx⁄_dt = a;

writing ^dx⁄_dt as v, v ^dv⁄_dx = a and simply solving the differential equation gives us

v²-u²= 2a.dS (v and u are the final and initial velocities on having accelerated an infinitesimal distance dS in which acceleration is almost constant and the component of dS in direction of a is considered for the velocity change (Newton's 2nd Law) only so a dot product is used)

Multiply mass and you will get to see the energy equations of change in ¹⁄₂ mv² = F.dS; based on the situation you can translate F to mg and dS to be the height. Thats where the magic of energy lies. But in reality the energy conservation is not 100% accurate owing to the fact that there are other forces causing acceleration which destroy our ideal equation of change in potential energy (mgh) = change in K.E., however the basic equation of change in ¹⁄₂ mv² = F.dS holds true always.

Answer 3

I'll go through all of the steps, but it's because they are defined as the minus of eachother

$$\vec{F}_{g}= \frac{GMm}{r^2}\hat r$$

$$G_{pe} = -\int_{ref}^{r} \vec{F}_{g} \cdot \vec{dl}$$

The change in $G_{pe}$ is therefore

$$\Delta G_{pe} = -\int_{a}^{b} \vec{F}_{g} \cdot \vec{dl}$$

Now let's consider:

$$\int_{a}^{b} \vec{F}_{g} \cdot \vec{dl} =\int_{a}^{b} \vec{F}_{g} \cdot \vec{dl}$$

$$\int_{a}^{b} \vec{F}_{g} \cdot \vec{dl} =-[-\int_{a}^{b} \vec{F}_{g} \cdot \vec{dl}]$$

$$\int_{a}^{b} \vec{F}_{g} \cdot \vec{dl} =-[ \Delta G_{pe}]$$

Well what IS, $\int_{a}^{b} \vec{F}_{g} \cdot \vec{dl}$?

When this force does work on a mass, $$F=ma$$$$\vec{dl} = \vec{v} dt$$

Substituting this in, We get that this quantity is just:

$$\Delta KE$$

This is called the "derivation of Kinetic energy from the work energy theorem"

So:

$$\Delta KE=- \Delta G_{pe}$$

Meaning that the INCREASE of kinetic energy corresponds to the same DECREASE of gravitational potential energy, hence "gravitational potential energy is being converted into kinetic energy"

From this you also know that:

$$\Delta [KE + G_{pe}] = 0$$

Ie- energy is conserved.