Physics · Energy

Potential energy in a uniform gravitational field

Q: How do you calculate potential energy in a gravitational field?

Multiply the mass in kilograms by the local factor $g \approx{}\frac{ 9.81 m}{s^{2}}$ and by the height h in metres: $E_{\mathrm{pot}}=mgh$. The result is the positional energy in joules. A body of 50 kg at 4 m height has $E_{\mathrm{pot}}=(50\,\mathrm{kg})(9.81\,\mathrm{m/s^2})(4\,\mathrm m)=1962\,\mathrm J$. This assigns $U=0$ at $h=0$. The height h is always understood relative to a freely chosen zero level, usually the ground or a table top. Only height differences are physically meaningful, since the absolute height depends on the chosen reference point. Insert g as 9.81 or, rounded, $\frac{10 m}{s^{2}}$ depending on the task.

In a uniform gravitational field, $E_{\mathrm{pot}}=mgh$ gives gravitational potential energy relative to the chosen zero level. Changes depend on vertical height differences.

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Formula

Calculate

E_{p o t} = m \cdot g \cdot h

E in J · m in kg · g in m/s² (often 9.81 near Earth) · h in m

Variables & units

Variables & units
Symbol	Meaning	Unit
$E_{\mathrm{pot}}$	Potential energy	J
$m$	Mass	kg
$g$	Gravitational acceleration in the chosen field	m/s²
$h$	Vertical centre-of-mass height relative to the zero level	m

LaTeX

E_{pot} = m \cdot g \cdot h

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Diagram: a body of mass m sits at height h above the ground, a dimension arrow marks its centre-of-mass height h, an arrow g points straight down. — In a uniform gravitational field, Epot = mgh. h measures centre-of-mass height above the chosen zero level; g points downward.

Potential energy calculator

Find the change in gravitational potential energy, the height change or the mass in a uniform field, such as near the Earth’s surface.

Find

Mass m

Mass in kg, not the weight in N, e.g. 50 kg for a heavy crate.

Gravitational acceleration g

Gravitational acceleration, 9.81 m/s² on Earth, about 1.62 m/s² on the Moon.

m/s²

Height change Δh (up is positive)

Vertical height change, positive upwards, negative downwards, e.g. 2 m. On a ramp only the height counts.

Filled with example values. Type your own and the result updates straight away.

Result: Change in potential energy ΔE

ΔE = 981 J

With m = 50 kg, g = 9.81 m/s² and Δh = 2 m, ΔE = 981 J.

Working · rearrange for ΔE, substitute, evaluate

1. Rearrange for ΔE
$\Delta E=mg\Delta h$
2. Substitute values (in SI units)
$\Delta E=50\cdot 9.81\cdot 2=981$
3. Result
ΔE = 981 J

Quick tips

Estimate with g ≈ 10 m/s²: about 10 J per kilogram and metre of height. Lifting 50 kg by 2 m therefore takes about 1000 J.
On a ramp or staircase only the vertical height change counts, not the length of the path.

Assumptions and limits

Uniform gravitational field with g > 0 and m > 0. Signed height change Δh is positive upward; ΔE = mgΔh is a change, not an absolute universal energy.

Derivation & background

With height positive upward, gravity performs $W_g=-mg\Delta h$ and potential energy changes by $\Delta U=-W_g=mg\Delta h$ . $U=mgh$ sets zero at $h = 0$ . For larger distance changes in a spherically symmetric central field, $U=-\frac{GMm}{r}$ sets zero at infinity. Mechanical energy is constant only without additional nonconservative work.

Exam blueprint

Validity range

Applies in an approximately uniform gravitational field near Earth and relative to a chosen reference height.

Derivation steps

Potential-energy change is the negative of gravitational work.

1Choose a uniform field, constant mass and upward-positive height.
2Weight has component $Fz = -mg$ .
3Integrate over vertical displacement: $Wg = -mg(h_{2} - h_{1})$ .
4Using $\Delta{}U = -Wg$ gives ΔU = mgΔh.
5Set zero energy at $h = 0$ ; then $U=mgh$ .
6External lifting work may also include acceleration and energy transfers.

Rearrangements

Height from energy

h = \frac{E_{pot}}{m g}

The zero height is arbitrary; energy differences are physically decisive.

Task variant

A 50 kg body is lifted vertically by 4 m in a constant field with $g =\frac{ 9.81 m}{s^{2}}$ . How much gravitational potential energy does it gain?

Approach: $\Delta U=mg\Delta h$ .
Substitution: $\Delta U=(50\,\mathrm{kg})(9.81\,\mathrm{m/s^2})(4\,\mathrm m)=1962\,\mathrm J$ .
This increase depends on height change and requires no absolute zero level.

Common mistakes

Interpreting h as absolute rather than relative to the chosen reference level.

Define the zero level before calculating.

Exam context

Often asked with falling motion, energy conservation and efficiency.

Formula cluster

Mechanical energy

Complements kinetic energy and gravitational force models.

Ekin = ½mv²

Kinetic Energy

F = G·m₁m₂/r²

Newton's Law of Universal Gravitation

ΣF_ext = F_res = ma

Newton's Second Law

E = ½·D·s²

Elastic Energy of a Spring

Worked example

An 80 kg mass rising vertically by 30 m with $g =\frac{ 9.81 m}{s^{2}}$ gains $\Delta{}U = 80\cdot{}9.81\cdot{}30 = 23544 J \approx{} 23.54 kJ$ . Use height change, not automatically elevation above sea level.

Applications

Pumped-storage power plants, drop-tower experiments, bridge statics, spaceflight

How do you calculate with Potential Energy (Gravitational)?

Here is how to work through a typical Potential Energy (Gravitational) (Epot = mgh) task step by step:

1
Task
A 50 kg body is lifted vertically by 4 m in a constant field with $g =\frac{ 9.81 m}{s^{2}}$ . How much gravitational potential energy does it gain?
Solution path
1. Approach: $\Delta U=mg\Delta h$ .
2. Substitution: $\Delta U=(50\,\mathrm{kg})(9.81\,\mathrm{m/s^2})(4\,\mathrm m)=1962\,\mathrm J$ .
3. This increase depends on height change and requires no absolute zero level.

Understand and apply the equation

Lifting a crate stores energy in the crate–Earth system. Vertical height change matters, rather than the length of the lifting path. Learn to choose a zero level deliberately, interpret work signs and connect potential energy to kinetic energy or lifting efficiency.

Getting an intuitive picture

A crate can be lifted vertically or moved up a longer ramp. Without friction, the same height gain requires the same energy: the ramp spreads the work over a longer path with a smaller force. Gravitational potential energy belongs to the interaction between crate and Earth. It is not a material stored exclusively inside the crate.[1][2]

Prerequisite: weight has magnitude mg. Mass in kilograms and weight in newtons are different quantities.
For a constant parallel force, work is force times displacement. Opposite force and displacement directions give negative work.
Choose a zero level. Negative energy below it is allowed; energy change depends only on height difference.

Which zero level applies to gravitational potential energy?

The zero level of potential energy can be chosen freely; for the same field and mass, energy differences between the same positions are independent of that choice. Near Earth’s surface, weight can be treated as constant over small height ranges. Potential energy is then U = mgh when we assign U = 0 at h = 0. Height h is the vertical coordinate relative to that chosen level. Between two positions use ΔU = mgΔh. A longer sloping path does not automatically mean a larger height difference.[1]

\Delta E_{\mathrm{pot}}=mg(h_2-h_1)

Use the change in height, not automatically elevation above sea level.

Original zero-level check: a 4 kg body rises from 1 m to 3 m. With zero at the floor, energies are 39.24 J and 117.72 J. With zero at 2 m, they are −39.24 J and +39.24 J. Both choices give an increase of 78.48 J. Negative potential energy is therefore consistent.

Different zero, same energy change

QUANTA STUDYquanta-study.de

For 4 kg, raising the object from 1 m to 3 m always gains 78.48 J. Choosing zero at 2 m makes values below it negative; slope and energy difference stay unchanged.[1][2]

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From weight to energy

Lift the object upward with unchanged speed and neglect other forces. Your lifting force balances downward weight and has magnitude mg. Over height gain Δh it performs positive work mgΔh. Gravity opposes displacement and performs −mgΔh. We describe gravity’s negative work as a gain in potential energy.[1][2]

\begin{aligned}W_g&=-mg\Delta h\\\Delta U&=-W_g=mg\Delta h\\U(h)-U(0)&=mgh\\U(0)=0&\;\Rightarrow\;U(h)=mgh\end{aligned}

Potential energy is defined from conservative gravitational work. Equating it to external lifting work additionally requires unchanged kinetic energy and no other work contributions.[1][2]

The integral expression below is the general notation for the same work. In a constant field it reduces to force times height change. If you have not studied integrals, that product is sufficient for the following examples.

\begin{aligned}W_{\mathrm g}&=\int_{z_1}^{z_2}(-mg)\,\mathrm dz=-mg(z_2-z_1)\\\Delta U&=-W_{\mathrm g}\end{aligned}

Constant g lets mg be taken outside the integral.

Setting U = 0 at z = 0 gives U = mgz. External lifting work equals this increase when initial and final speeds are equal and there are no other energy transfers. Additional acceleration also changes kinetic energy.

View diagram: forces during steady lifting

Lifting without a speed change

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During idealized constant-speed lifting, lifting force and weight have equal magnitudes. Net force is zero, but the object can still move. Lifting force performs positive work and gravity negative work; potential energy increases.[1][2]

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Units and useful rearrangements

\begin{aligned}[mgh]&=\mathrm{kg}\cdot\mathrm{m/s^2}\cdot\mathrm m=\mathrm J\\h&=\frac{E_{\mathrm{pot}}}{mg}\end{aligned}

The rearrangement uses the same zero level as the energy value.

m is mass in kilograms, not weight in newtons. g is measured in m/s² or N/kg. The familiar 9.81 m/s² is an approximate near-surface Earth value, not an exact constant for every location. Use the problem’s value. Height is in metres; ramp length is not directly height.

For an increase, Δh = ΔU/(mg) requires m > 0 and g > 0. Doubling either mass or height change doubles the energy change; doubling both multiplies it by four. Zero net height change gives ΔU = 0 even after a long journey.

Two examples with an energy balance

Lifting a crate with losses

Given: 35 kg crate; rise 2.4 m; g = 9.81 m/s²; equal initial and final speed. Lifting-system efficiency 80%.
Find: Find potential-energy increase and input energy.

Approach: height gain determines useful energy change. With equal initial and final speeds there is no kinetic-energy change; efficiency relates useful energy to total input.
$\begin{aligned}\Delta U&=mg\Delta h\\E_{\mathrm{in}}&=\frac{\Delta U}{\eta}\end{aligned}$
Multiply mass, gravity and height change.
$\Delta U=35\cdot9.81\cdot2.4=824.04\,\mathrm J$
Use decimal efficiency.
$\eta=0.80=\frac{\Delta U}{E_{\mathrm{in}}}$
Solve for input energy.
$E_{\mathrm{in}}=\frac{824.04}{0.80}=1030.05\,\mathrm J$

Result: Potential energy increases by 824.04 J; the modelled lifting system requires 1030.05 J.

Check: The difference, 206.01 J, is transferred to other forms. Input must exceed useful output when efficiency is below 100%.

Falling from rest without drag

Given: A 2.5 kg particle falls 3.2 m from rest; g = 9.81 m/s². No drag or rotation.
Find: Find potential-energy decrease and speed after falling.

Approach: in an ideal drop from rest, lost potential energy becomes translational kinetic energy. Height difference is specified as positive distance fallen.
$\begin{aligned}U_1&=mgh\\\frac12 mv^2&=mgh\\v&=\sqrt{2gh}\end{aligned}$
Set zero at the endpoint and calculate initial energy.
$U_1=2.5\cdot9.81\cdot3.2=78.48\,\mathrm J$
Balance energy changes.
$\tfrac12mv^2=mgh$
Cancel mass and take the positive root.
$v=\sqrt{2gh}=\sqrt{2\cdot9.81\cdot3.2}\approx7.92\,\mathrm{m/s}$

Result: Potential energy decreases by 78.48 J and kinetic energy increases equally. Speed is about 7.92 m/s.

Check: Mass cancels from speed but not from energy. Motion is downward; the positive result gives speed only.

Does a ramp’s length or height difference determine potential-energy change?

Gravitational energy change depends only on endpoint heights. For ideal slow lifting along a frictionless ramp of length L and angle α, h = L sin α. Required force along the ramp is mg sin α, giving mechanical work mg sin α·L = mgh. A longer route reduces required force, not ideal lifting work.[2]

Original ramp check: 10 kg on a 4 m ramp at 30° gives h = 2 m, lifting force 49.05 N and work 196.2 J. Friction requires extra work, so equal endpoint heights alone do not imply equal external work on real paths.

Mechanical energy and energy transfers

E_{\mathrm{kin},1}+U_1+W_{\mathrm{other}}=E_{\mathrm{kin},2}+U_2

For a particle in the chosen fixed gravitational field, Wother is work by other forces.[3]

Mechanical energy is conserved when only conservative forces act and their potentials are included. Braking or drag can reduce mechanical energy while total energy of an appropriately larger system remains conserved. A motor can increase it. Do not automatically set K + mgh constant when other forces perform work. For a powered lift, efficiency relates the gained gravitational potential energy as useful output to the total supplied energy; the crate example illustrates this balance.[3][5]

A body at rest high above the zero level still has potential energy. Below that level it can have negative U and positive kinetic energy. These signs have different origins: U depends on reference choice; kinetic energy depends on squared speed.

When is m·g·h no longer sufficient?

When distance changes are no longer small compared with Earth’s radius, g varies appreciably. Outside a spherically symmetric attracting mass, gravitational potential energy is U(r) = −GMm/r with zero at infinity. r is distance from its centre, not altitude above the surface.[4]

\Delta U=GMm\left(\frac1{r_1}-\frac1{r_2}\right)

For r₂ > r₁, energy increases. Small height changes relative to radius recover approximately mgΔh.

Do not equate the two absolute potential expressions with different zero choices. Compare differences instead. On the Moon or Mars, use the appropriate local g. A height alone, without field and reference level, does not fully specify an energy problem.

Five problems with solutions

Calculate · Foundation

An 8 kg mass rises 1.5 m with g = 9.81 m/s². Find its potential-energy increase.

Show hint

Use vertical height change.

Learning goal: Calculate potential-energy change from vertical lift.

Show solution and calculation

Approach: the required quantity is an increase, so use $\Delta U=mg\Delta h$ .
Multiply mass, g and the vertical height gain.
For 8 kg and a rise of 1.5 m, the increase is 117.72 J.

Check: weight is 78.48 N; multiplying it by the height gain gives the same energy. No zero level is required to find this change. A reference would be needed to assign an absolute potential-energy value. Any consistently chosen zero gives the same increase between the same positions.

\begin{aligned}\Delta U&=mg\Delta h\\&=(8\,\mathrm{kg})(9.81\,\mathrm{m/s^2})(1.5\,\mathrm m)\\&=117.72\,\mathrm J\end{aligned}

[1][2]

Rearrange · Foundation

A 12 kg body gains 588.6 J potential energy. Find Δh for g = 9.81 m/s².

Show hint

Rearrange for height change.

Learning goal: Infer height change from energy gain.

Show solution and calculation

Approach: a given energy change determines height gain through Δh = ΔU/(mg).
Weight is 117.72 N, so 588.6 J corresponds to a 5 m height gain.

Check: weight times that height returns the specified energy. This calculation determines height difference, not automatically absolute height above the ground.

\begin{aligned}\Delta h&=\frac{\Delta U}{mg}\\&=\frac{588.6\,\mathrm J}{(12\,\mathrm{kg})(9.81\,\mathrm{m/s^2})}\\&=5\,\mathrm m\end{aligned}

[1][2]

Calculate · Application

A 3 kg mass descends from h = 4 m to h = 1 m. Find ΔU and gravitational work.

Show hint

Height change is negative; gravitational work has the opposite sign.

Learning goal: Compare signs for descent and gravitational work.

Show solution and calculation

Approach: subtract initial height from final height.
Height change is −3 m, so potential energy decreases by 88.29 J.
Gravity performs work with the opposite sign, +88.29 J.

Check: during downward motion weight and displacement point the same way, so gravitational work is positive.

\begin{aligned}\Delta h&=1\,\mathrm m-4\,\mathrm m=-3\,\mathrm m\\\Delta U&=mg\Delta h=(3\,\mathrm{kg})(9.81\,\mathrm{m/s^2})(-3\,\mathrm m)\\&=-88.29\,\mathrm J\\W_g&=-\Delta U=+88.29\,\mathrm J\end{aligned}

[1][2]

Interpret · Application

A 2 kg mass is moved from rest to rest to a point 3 m higher, with constant g=9.81 m/s². Path A is frictionless; path B has additional friction work −10 J. Compare potential-energy change and external work for both paths.

Show hint

ΔU depends only on endpoint heights; ΔK is zero.

Learning goal: Separate path-independent gravitational work from friction work.

Show solution and calculation

Approach: both paths have $\Delta U=mg\Delta h=58.86\,\mathrm J$ .
From $\Delta K=W_{\mathrm{ext}}+W_g+W_{\mathrm{friction}}=0$ , external work is 58.86 J for A and 68.86 J for B.

Check: the extra external work exactly offsets the 10 J friction loss. Path-independent potential change does not make every work contribution path-independent; inclined path length cannot replace vertical height change.

\begin{aligned}\Delta U&=2\cdot9.81\cdot3=58.86\,\mathrm J\\W_{\mathrm{ext}}&=\Delta U-W_{\mathrm{friction}}\end{aligned}

[2][3]

Error analysis · Foundation

The potential-energy zero is 10 m above a floor. A 1 kg mass falls frictionlessly from rest 2 m above the floor to it. A solution refuses calculation because both U values are negative. Find ΔU and kinetic energy just before contact, g=9.81 m/s².

Show hint

The fall balance requires only the energy difference.

Learning goal: Interpret negative potential energy as a reference choice rather than a motion prohibition.

Show solution and calculation

Approach: heights relative to the zero are −8 and −10 m, so initial U=−78.48 J and final U=−98.10 J; ΔU=−19.62 J.
The loss becomes 19.62 J kinetic energy.

Check: choosing the floor as zero gives the same change. Negative U is allowed and does not forbid motion; impact itself requires a separate contact model.

\begin{aligned}\Delta U&=mg(-10\,\mathrm m+8\,\mathrm m)=-19.62\,\mathrm J\\K_f&=-\Delta U=19.62\,\mathrm J\end{aligned}

[1][3]

What to take away

Potential energy belongs to the body–gravitational-source system. In a constant field, changes depend only on vertical height difference.
The zero level is arbitrary but must remain consistent within a calculation. Negative heights and energies below it are normal.
Lifting work equals potential-energy gain only under the appropriate energy balance. Account separately for acceleration, friction and efficiency where needed.

Sources for this explanation

Quanta · Updated October 1, 2026

Editorial responsibility: Quanta Study editorial team

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This explanation, 2 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

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An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 1, 2026Made zero-level choice, ramp height and the limit of the m·g·h approximation accessible through direct answers and headings; added contextual links to mechanical work, efficiency and central-field energy.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Added prerequisites and an intuitive introduction; clarified the derivation, worked-example approaches and exercise solutions, and added a summary.

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Quanta. Potential Energy (Gravitational). Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/energie-potentiell

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Frequently asked questions about Potential Energy (Gravitational)

How do you calculate potential energy in a gravitational field?+

Multiply the mass in kilograms by the local factor $g \approx{}\frac{ 9.81 m}{s^{2}}$ and by the height h in metres: $E_{\mathrm{pot}}=mgh$ . The result is the positional energy in joules. A body of 50 kg at 4 m height has $E_{\mathrm{pot}}=(50\,\mathrm{kg})(9.81\,\mathrm{m/s^2})(4\,\mathrm m)=1962\,\mathrm J$ . This assigns $U=0$ at $h=0$ . The height h is always understood relative to a freely chosen zero level, usually the ground or a table top. Only height differences are physically meaningful, since the absolute height depends on the chosen reference point. Insert g as 9.81 or, rounded, $\frac{10 m}{s^{2}}$ depending on the task.

Why can you freely choose the zero level of height?+

Because only energy differences act physically, not the absolute energy value. When a body falls from one height to another, only the height difference matters for the energy converted. Whether you place the zero level at the ground or at the table edge changes the numerical value of E_pot, but the difference between start and finish stays the same. Therefore you deliberately fix the zero level before each calculation and keep it. A common mistake is to interpret h as absolute height above sea level, although the problem means the height above the ground.

How are potential and kinetic energy linked in free fall?+

When gravity alone acts in the constant field, total mechanical energy stays constant, so $E_{\mathrm{pot}}+E_{\mathrm{kin}}=\mathrm{constant}$ . During the fall the body loses height and thus potential energy, but gains exactly the same amount of kinetic energy. For a drop from rest with pure translation, use $mgh=\frac12mv^2$ , where h is distance fallen. This cancels the mass and gives the impact speed $v=\sqrt{2gh}$ . From 5 m height, for example, $v=\sqrt{2(9.81\,\mathrm{m/s^2})(5\,\mathrm m)}\approx9.9\,\mathrm{m/s}$ , independent of the mass. This energy approach is often faster than the equations of motion, because you do not even need to know the fall time.

When does

E_{\mathrm{pot}} = m\cdot{}g\cdot{}h

no longer apply?+

The formula assumes a nearly uniform gravitational field, that is a constant local factor g. This holds well near the ground as long as the height is small compared with the Earth radius. If you rise to several hundred kilometres, for example into a satellite orbit, g decreases noticeably and you need the general gravitational potential energy, $E_{\mathrm{pot}}=-\frac{GMm}{r}$ outside a spherically symmetric source. Here r is distance from its centre and zero energy is assigned at infinity. The local and central-field expressions may use different zeros; compare energy differences. On other celestial bodies you must also insert their own local factor, on the Moon about $\frac{1.62 m}{s^{2}}$ . For everyday heights from metres to a few kilometres, however, $E_{\mathrm{pot}}=mgh$ gives very accurate values.

How do you rearrange

E_{\mathrm{pot}} = m\cdot{}g\cdot{}h

for the height?+

Divide the potential energy by the product of mass and local factor: $h=\frac{E_{\mathrm{pot}}}{mg}$ . The height is therefore the ratio of positional energy to weight, since $mg$ is exactly the weight $F_{\mathrm G}$ . Insert the energy in joules, the mass in kilograms and g in $\frac{m}{s^{2}}$ , then h comes out in metres. Example: 1962 J at 50 kg give $h=\frac{1962\,\mathrm J}{(50\,\mathrm{kg})(9.81\,\mathrm{m/s^2})}=4\,\mathrm m$ . If you want the mass instead, rearrange to $m=\frac{E_{\mathrm{pot}}}{gh}$ . Always check the result for plausibility; negative heights are allowed below the chosen zero level. Only disagreement with that reference or the initial data indicates an error. Given an energy change, the rearrangement gives a height change; a negative value then means descent.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Vol. 1, §8.1 Potential Energy of a System

More formulas: Mechanics

F = k \cdot x

Hooke's Law (Spring Force)

s = \frac{1}{2} a t^2 + v_0 t

Uniformly Accelerated Motion (Position–Time Equation)

v = a \cdot t + v_0

Velocity-Time Law

h = \frac{1}{2} g t^2

Free Fall

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Potential energy in a uniform gravitational field

In a uniform gravitational field,

E_{\mathrm{pot}}=mgh

gives gravitational potential energy relative to the chosen zero level. Changes depend on vertical height differences.

BasicExam-relevantLower secondaryUpper secondary

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Symbol

Meaning

Unit

E_{\mathrm{pot}}

Potential energy

m

Mass

g

Gravitational acceleration in the chosen field

m/s²

h

Vertical centre-of-mass height relative to the zero level

Derivation & background

Exam blueprint

Validity range

Applies in an approximately uniform gravitational field near Earth and relative to a chosen reference height.

Derivation steps

Potential-energy change is the negative of gravitational work.

1Choose a uniform field, constant mass and upward-positive height.
2Weight has component $Fz = -mg$ .
3Integrate over vertical displacement: $Wg = -mg(h_{2} - h_{1})$ .
4Using $\Delta{}U = -Wg$ gives ΔU = mgΔh.
5Set zero energy at $h = 0$ ; then $U=mgh$ .
6External lifting work may also include acceleration and energy transfers.

How do you calculate with Potential Energy (Gravitational)?

Here is how to work through a typical Potential Energy (Gravitational) (Epot = mgh) task step by step:

Task

A 50 kg body is lifted vertically by 4 m in a constant field with $g =\frac{ 9.81 m}{s^{2}}$ . How much gravitational potential energy does it gain?

Solution path

Approach: $\Delta U=mg\Delta h$ .
Substitution: $\Delta U=(50\,\mathrm{kg})(9.81\,\mathrm{m/s^2})(4\,\mathrm m)=1962\,\mathrm J$ .
This increase depends on height change and requires no absolute zero level.

Getting an intuitive picture

Prerequisite: weight has magnitude mg. Mass in kilograms and weight in newtons are different quantities.
For a constant parallel force, work is force times displacement. Opposite force and displacement directions give negative work.
Choose a zero level. Negative energy below it is allowed; energy change depends only on height difference.

Which zero level applies to gravitational potential energy?

\Delta E_{\mathrm{pot}}=mg(h_2-h_1)

Use the change in height, not automatically elevation above sea level.

Different zero, same energy change

QUANTA STUDYquanta-study.de

For 4 kg, raising the object from 1 m to 3 m always gains 78.48 J. Choosing zero at 2 m makes values below it negative; slope and energy difference stay unchanged.[1][2]

Free with a Quanta account. Your download stays ready after registration.

From weight to energy

\begin{aligned}W_g&=-mg\Delta h\\\Delta U&=-W_g=mg\Delta h\\U(h)-U(0)&=mgh\\U(0)=0&\;\Rightarrow\;U(h)=mgh\end{aligned}

Potential energy is defined from conservative gravitational work. Equating it to external lifting work additionally requires unchanged kinetic energy and no other work contributions.[1][2]

\begin{aligned}W_{\mathrm g}&=\int_{z_1}^{z_2}(-mg)\,\mathrm dz=-mg(z_2-z_1)\\\Delta U&=-W_{\mathrm g}\end{aligned}

Constant g lets mg be taken outside the integral.

View diagram: forces during steady lifting

Lifting without a speed change

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Units and useful rearrangements

\begin{aligned}[mgh]&=\mathrm{kg}\cdot\mathrm{m/s^2}\cdot\mathrm m=\mathrm J\\h&=\frac{E_{\mathrm{pot}}}{mg}\end{aligned}

The rearrangement uses the same zero level as the energy value.

Two examples with an energy balance

Lifting a crate with losses

Given: 35 kg crate; rise 2.4 m; g = 9.81 m/s²; equal initial and final speed. Lifting-system efficiency 80%.
Find: Find potential-energy increase and input energy.

Approach: height gain determines useful energy change. With equal initial and final speeds there is no kinetic-energy change; efficiency relates useful energy to total input.
$\begin{aligned}\Delta U&=mg\Delta h\\E_{\mathrm{in}}&=\frac{\Delta U}{\eta}\end{aligned}$
Multiply mass, gravity and height change.
$\Delta U=35\cdot9.81\cdot2.4=824.04\,\mathrm J$
Use decimal efficiency.
$\eta=0.80=\frac{\Delta U}{E_{\mathrm{in}}}$
Solve for input energy.
$E_{\mathrm{in}}=\frac{824.04}{0.80}=1030.05\,\mathrm J$

Result: Potential energy increases by 824.04 J; the modelled lifting system requires 1030.05 J.

Check: The difference, 206.01 J, is transferred to other forms. Input must exceed useful output when efficiency is below 100%.

Falling from rest without drag

Given: A 2.5 kg particle falls 3.2 m from rest; g = 9.81 m/s². No drag or rotation.
Find: Find potential-energy decrease and speed after falling.

Approach: in an ideal drop from rest, lost potential energy becomes translational kinetic energy. Height difference is specified as positive distance fallen.
$\begin{aligned}U_1&=mgh\\\frac12 mv^2&=mgh\\v&=\sqrt{2gh}\end{aligned}$
Set zero at the endpoint and calculate initial energy.
$U_1=2.5\cdot9.81\cdot3.2=78.48\,\mathrm J$
Balance energy changes.
$\tfrac12mv^2=mgh$
Cancel mass and take the positive root.
$v=\sqrt{2gh}=\sqrt{2\cdot9.81\cdot3.2}\approx7.92\,\mathrm{m/s}$

Result: Potential energy decreases by 78.48 J and kinetic energy increases equally. Speed is about 7.92 m/s.

Check: Mass cancels from speed but not from energy. Motion is downward; the positive result gives speed only.

Does a ramp’s length or height difference determine potential-energy change?

Mechanical energy and energy transfers

E_{\mathrm{kin},1}+U_1+W_{\mathrm{other}}=E_{\mathrm{kin},2}+U_2

For a particle in the chosen fixed gravitational field, Wother is work by other forces.[3]

When is m·g·h no longer sufficient?

\Delta U=GMm\left(\frac1{r_1}-\frac1{r_2}\right)

For r₂ > r₁, energy increases. Small height changes relative to radius recover approximately mgΔh.

Five problems with solutions

Calculate · Foundation

An 8 kg mass rises 1.5 m with g = 9.81 m/s². Find its potential-energy increase.

Show hint

Use vertical height change.

Learning goal: Calculate potential-energy change from vertical lift.

Show solution and calculation

Approach: the required quantity is an increase, so use $\Delta U=mg\Delta h$ .
Multiply mass, g and the vertical height gain.
For 8 kg and a rise of 1.5 m, the increase is 117.72 J.

\begin{aligned}\Delta U&=mg\Delta h\\&=(8\,\mathrm{kg})(9.81\,\mathrm{m/s^2})(1.5\,\mathrm m)\\&=117.72\,\mathrm J\end{aligned}

[1][2]

Rearrange · Foundation

A 12 kg body gains 588.6 J potential energy. Find Δh for g = 9.81 m/s².

Show hint

Rearrange for height change.

Learning goal: Infer height change from energy gain.

Show solution and calculation

Approach: a given energy change determines height gain through Δh = ΔU/(mg).
Weight is 117.72 N, so 588.6 J corresponds to a 5 m height gain.

Check: weight times that height returns the specified energy. This calculation determines height difference, not automatically absolute height above the ground.

\begin{aligned}\Delta h&=\frac{\Delta U}{mg}\\&=\frac{588.6\,\mathrm J}{(12\,\mathrm{kg})(9.81\,\mathrm{m/s^2})}\\&=5\,\mathrm m\end{aligned}

[1][2]

Calculate · Application

A 3 kg mass descends from h = 4 m to h = 1 m. Find ΔU and gravitational work.

Show hint

Height change is negative; gravitational work has the opposite sign.

Learning goal: Compare signs for descent and gravitational work.

Show solution and calculation

Approach: subtract initial height from final height.
Height change is −3 m, so potential energy decreases by 88.29 J.
Gravity performs work with the opposite sign, +88.29 J.

Check: during downward motion weight and displacement point the same way, so gravitational work is positive.

\begin{aligned}\Delta h&=1\,\mathrm m-4\,\mathrm m=-3\,\mathrm m\\\Delta U&=mg\Delta h=(3\,\mathrm{kg})(9.81\,\mathrm{m/s^2})(-3\,\mathrm m)\\&=-88.29\,\mathrm J\\W_g&=-\Delta U=+88.29\,\mathrm J\end{aligned}

[1][2]

Interpret · Application

A 2 kg mass is moved from rest to rest to a point 3 m higher, with constant g=9.81 m/s². Path A is frictionless; path B has additional friction work −10 J. Compare potential-energy change and external work for both paths.

Show hint

ΔU depends only on endpoint heights; ΔK is zero.

Learning goal: Separate path-independent gravitational work from friction work.

Show solution and calculation

Approach: both paths have $\Delta U=mg\Delta h=58.86\,\mathrm J$ .
From $\Delta K=W_{\mathrm{ext}}+W_g+W_{\mathrm{friction}}=0$ , external work is 58.86 J for A and 68.86 J for B.

\begin{aligned}\Delta U&=2\cdot9.81\cdot3=58.86\,\mathrm J\\W_{\mathrm{ext}}&=\Delta U-W_{\mathrm{friction}}\end{aligned}

[2][3]

Error analysis · Foundation

The potential-energy zero is 10 m above a floor. A 1 kg mass falls frictionlessly from rest 2 m above the floor to it. A solution refuses calculation because both U values are negative. Find ΔU and kinetic energy just before contact, g=9.81 m/s².

Show hint

The fall balance requires only the energy difference.

Learning goal: Interpret negative potential energy as a reference choice rather than a motion prohibition.

Show solution and calculation

Approach: heights relative to the zero are −8 and −10 m, so initial U=−78.48 J and final U=−98.10 J; ΔU=−19.62 J.
The loss becomes 19.62 J kinetic energy.

Check: choosing the floor as zero gives the same change. Negative U is allowed and does not forbid motion; impact itself requires a separate contact model.

\begin{aligned}\Delta U&=mg(-10\,\mathrm m+8\,\mathrm m)=-19.62\,\mathrm J\\K_f&=-\Delta U=19.62\,\mathrm J\end{aligned}

[1][3]

What to take away

Potential energy belongs to the body–gravitational-source system. In a constant field, changes depend only on vertical height difference.
The zero level is arbitrary but must remain consistent within a calculation. Negative heights and energies below it are normal.
Lifting work equals potential-energy gain only under the appropriate energy balance. Account separately for acceleration, friction and efficiency where needed.

Sources for this explanation

Quanta · Updated October 1, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Change history

October 1, 2026Made zero-level choice, ramp height and the limit of the m·g·h approximation accessible through direct answers and headings; added contextual links to mechanical work, efficiency and central-field energy.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Added prerequisites and an intuitive introduction; clarified the derivation, worked-example approaches and exercise solutions, and added a summary.

Cite this reference page

Quanta. Potential Energy (Gravitational). Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/energie-potentiell

When citing a specific scientific statement, also cite the specialist source linked to that statement.

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Frequently asked questions about Potential Energy (Gravitational)

How do you calculate potential energy in a gravitational field?+

Why can you freely choose the zero level of height?+

How are potential and kinetic energy linked in free fall?+

When does

E_{\mathrm{pot}} = m\cdot{}g\cdot{}h

no longer apply?+

How do you rearrange

E_{\mathrm{pot}} = m\cdot{}g\cdot{}h

for the height?+

Potential energy in a uniform gravitational field

Variables & units

Potential energy calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Mechanical energy

Worked example

Applications

How do you calculate with Potential Energy (Gravitational)?

Getting an intuitive picture

Which zero level applies to gravitational potential energy?

Different zero, same energy change

From weight to energy

Lifting without a speed change

Units and useful rearrangements

Two examples with an energy balance

Lifting a crate with losses

Falling from rest without drag

Does a ramp’s length or height difference determine potential-energy change?

Mechanical energy and energy transfers

When is m·g·h no longer sufficient?

Five problems with solutions

An 8 kg mass rises 1.5 m with g = 9.81 m/s2. Find its potential-energy increase.

A 12 kg body gains 588.6 J potential energy. Find Δh for g = 9.81 m/s2.

A 3 kg mass descends from h = 4 m to h = 1 m. Find ΔU and gravitational work.

A 2 kg mass is moved from rest to rest to a point 3 m higher, with constant g=9.81 m/s2. Path A is frictionless; path B has additional friction work −10 J. Compare potential-energy change and external work for both paths.

The potential-energy zero is 10 m above a floor. A 1 kg mass falls frictionlessly from rest 2 m above the floor to it. A solution refuses calculation because both U values are negative. Find ΔU and kinetic energy just before contact, g=9.81 m/s2.

What to take away

Sources for this explanation

Frequently asked questions about Potential Energy (Gravitational)

Subject references and further reading

More formulas: Mechanics

Retain Potential Energy (Gravitational) for exams

Potential energy in a uniform gravitational field

Variables & units

Potential energy calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Mechanical energy

Worked example

Applications

How do you calculate with Potential Energy (Gravitational)?

Getting an intuitive picture

Which zero level applies to gravitational potential energy?

Different zero, same energy change

From weight to energy

Lifting without a speed change

Units and useful rearrangements

Two examples with an energy balance

Lifting a crate with losses

Falling from rest without drag

Does a ramp’s length or height difference determine potential-energy change?

Mechanical energy and energy transfers

When is m·g·h no longer sufficient?

Five problems with solutions

An 8 kg mass rises 1.5 m with g = 9.81 m/s2. Find its potential-energy increase.

A 12 kg body gains 588.6 J potential energy. Find Δh for g = 9.81 m/s2.

A 3 kg mass descends from h = 4 m to h = 1 m. Find ΔU and gravitational work.

A 2 kg mass is moved from rest to rest to a point 3 m higher, with constant g=9.81 m/s2. Path A is frictionless; path B has additional friction work −10 J. Compare potential-energy change and external work for both paths.

The potential-energy zero is 10 m above a floor. A 1 kg mass falls frictionlessly from rest 2 m above the floor to it. A solution refuses calculation because both U values are negative. Find ΔU and kinetic energy just before contact, g=9.81 m/s2.

What to take away

Sources for this explanation

Frequently asked questions about Potential Energy (Gravitational)

Subject references and further reading

More formulas: Mechanics

Retain Potential Energy (Gravitational) for exams

An 8 kg mass rises 1.5 m with g = 9.81 m/s². Find its potential-energy increase.

A 12 kg body gains 588.6 J potential energy. Find Δh for g = 9.81 m/s².

A 2 kg mass is moved from rest to rest to a point 3 m higher, with constant g=9.81 m/s². Path A is frictionless; path B has additional friction work −10 J. Compare potential-energy change and external work for both paths.

The potential-energy zero is 10 m above a floor. A 1 kg mass falls frictionlessly from rest 2 m above the floor to it. A solution refuses calculation because both U values are negative. Find ΔU and kinetic energy just before contact, g=9.81 m/s².

An 8 kg mass rises 1.5 m with g = 9.81 m/s². Find its potential-energy increase.

A 12 kg body gains 588.6 J potential energy. Find Δh for g = 9.81 m/s².

A 2 kg mass is moved from rest to rest to a point 3 m higher, with constant g=9.81 m/s². Path A is frictionless; path B has additional friction work −10 J. Compare potential-energy change and external work for both paths.

The potential-energy zero is 10 m above a floor. A 1 kg mass falls frictionlessly from rest 2 m above the floor to it. A solution refuses calculation because both U values are negative. Find ΔU and kinetic energy just before contact, g=9.81 m/s².