Gravitational potential energy

An object has gravitational potential energy because of its position in a gravitational field. Near Earth’s surface, raising an object increases its gravitational potential energy; lowering it decreases its gravitational potential energy.

Where the gravitational field strength can be treated as constant, the change in energy depends on the object’s mass and its change in vertical height:

\Delta E_p = mg\Delta h

In this equation:

$\Delta E_p$ is the change in gravitational potential energy, in joules ( $\mathrm{J}$ ).
$$$m$$$ is the mass of the object, in kilograms ( $\mathrm{kg}$ ).
$$$g$$$ is the gravitational field strength, in newtons per kilogram ( $\mathrm{N/kg}$ ).
$\Delta h$ is the change in vertical height, in metres ( $\mathrm{m}$ ).

The change in height is the final height minus the initial height. A rise gives a positive $\Delta h$ and therefore a positive energy change. A fall gives a negative $\Delta h$ and therefore a negative energy change.

For example, a $2,\mathrm{kg}$ object raised by $3,\mathrm{m}$ in a field of strength $10,\mathrm{N/kg}$ gains $2 \times 10 \times 3 = 60,\mathrm{J}$ of gravitational potential energy. Lowering the same object through that height gives an energy change of $-60,\mathrm{J}$ .

Gravitational potential energy and reference height

A reference height is a chosen level from which heights are measured. Changing that level changes the stated initial and final heights, but not their difference. The change in gravitational potential energy therefore depends on the vertical height difference, not on the choice of reference height.

Kinetic energy

A moving object has kinetic energy because of its motion. Its kinetic energy depends on its mass and speed:

E_k = \frac{1}{2}mv^2

Here, $$$E_k$$$ is kinetic energy in joules, $$$m$$$ is mass in kilograms and $$$v$$$ is speed in metres per second.

Kinetic energy is proportional to mass: if speed stays the same, doubling the mass doubles the kinetic energy.

Kinetic energy is proportional to the square of speed: if mass stays the same, doubling the speed makes the kinetic energy four times as large. Halving the speed reduces the kinetic energy to one-quarter of its original value.

Energy conservation with resistive forces

Energy cannot be created or destroyed, but it can be transferred between stores or between a system and its surroundings. In a mechanical process, energy may move between the gravitational potential energy store and the kinetic energy store.

If there are no transfers by resistive forces or other processes, a decrease in gravitational potential energy equals the increase in kinetic energy, or vice versa. The total of these two energies remains constant.

Energy transferred by resistance

A resistive force acts against motion. As an object moves against the force, work is done against resistance. Mechanical energy is transferred to other stores, commonly the internal energy stores of the objects and their surroundings. The mechanical energy decreases, but the total energy is still conserved.

When there are no other energy transfers, the changes can be accounted for as:

\Delta E_k + \Delta E_p + W_{\text{res}} = 0

Here, $\Delta E_k$ is the change in kinetic energy, $\Delta E_p$ is the change in gravitational potential energy, and $W_{\text{res}}$ is the positive work done against resistance. Each change is the final energy minus the initial energy, so a decrease is negative.

If a constant resistive force of magnitude $$$F$$$ acts directly opposite to motion over a distance $$$d$$$ , the work done against it is:

W_{\text{res}} = Fd

When energy is measured in joules, force is in newtons and distance is in metres. If there is no resistance, $W_{\text{res}} = 0$ , so the changes in kinetic and gravitational potential energy must add to zero.

Using the energy relationship

The energy equation can be rearranged to find an unknown energy change or resistive work. Near Earth’s surface, kinetic energy is $E_k = \frac{1}{2}mv^2$ and gravitational potential energy is $$$E_p = mgh$$$ . These expressions allow speeds or heights to be found from the energy changes.

For example, a $2,\text{kg}$ object falls vertically from rest through $5,\text{m}$ against a constant resistive force of $4,\text{N}$ . Taking $g = 10,\text{m/s}^2$ , its gravitational potential energy decreases by $100,\text{J}$ . The work done against resistance is $Fd = 4 \times 5 = 20,\text{J}$ . With no other energy transfers, the kinetic energy therefore increases by $80,\text{J}$ , giving a final speed of approximately $8.9,\text{m/s}$ .