Physics · Energy

Elastic potential energy of a spring

Q: How are elastic energy and Hooke law related?

For $Fs = -Ds$, spring force is the negative derivative of potential energy: Fs = −dU/ds. Integration gives $U = \frac{1}{2}Ds^{2} + C$. The constant shifts energy zero without changing forces. A measured nonlinear force curve instead requires integrating that curve rather than using a fixed quadratic formula. Elastic does not automatically mean linear. For a conservative nonlinear spring, stored energy follows the actual force-curve integral. Outside the linear range, D cannot remain the same constant multiplier. Whether energy exceeds or falls below $\frac{1}{2}Ds^{2}$ depends on the curve; there is no universal direction of the deviation.

Q: Why does double displacement store four times the energy?

For constant D and energy zero at natural length, $\frac{U(2s)}{U}(s) = 4$. $\frac{A 400 N}{m}$ spring stores 2 J at 10 cm and 8 J at 20 cm, so the energy increase between those states is 6 J. The fourfold rule concerns absolute extension from the force-free state, not an arbitrary additional displacement. For a vertical spring let s be positive downward. Total potential is Utotal(s) = ½Ds² − mgs + C and static equilibrium is sEq = mg/D. With q = s − sEq, total potential becomes $\frac{1}{2}Dq^{2}$ plus a constant. Gravity cancels the spring energy’s linear term in this total; the term does not disappear from spring energy alone.

For an ideal linear spring, $U = \frac{1}{2}Ds^{2}$ is stored energy relative to its force-free natural state. Between pre-extended states use the difference of squared extensions, not the square of extra displacement.

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Formula

Calculate

E_{Sp ann} = \frac{1}{2} D s^{2}

E_Spann in J · D in N/m · s in m

Variables & units

Variables & units
Symbol	Meaning	Unit
$E_{\mathrm{Spann}}$	Spring energy relative to the force-free state	J
$D$	Constant linear spring stiffness	N/m
$s$	Length change relative to force-free natural length	m

LaTeX

E_{Spann} = \frac{1}{2} D s^2

QUANTA STUDYquanta-study.de

Diagram: a straight line through the origin shows the spring force D·s over the displacement s, the shaded triangle beneath it is the elastic energy E_Spann = ½·D·s². — Quasistatic external stretching force Ds grows linearly from force-free natural length. Triangular area gives stored energy U = ½Ds² with U(0) = 0.

Spring potential energy calculator

Find the stored energy, spring constant or extension of an ideal spring, including the energy gained between two extensions.

Calculation model

Find

Spring constant D

Spring constant, e.g. 200 N/m: then a force of 2 N stretches the spring by 1 cm.

Extension s from relaxed length

Extension or compression from the relaxed length, e.g. 10 cm; you may enter compression as negative.

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

Working · rearrange for E, substitute, evaluate

1. Rearrange for E
$E=\frac12Ds^2$
2. Substitute values (in SI units)
- s = 10 cm = 0.1 m
$E=\frac12\cdot200\cdot0.1^2$
3. Result
E = 1 J

Result: Energy E

E = 1 J

With D = 200 N/m and s = 10 cm, E = 1 J.

Quick tips

Twice the extension, four times the energy, because s is squared. Convert s to metres first: 10 cm is 0.1 m.
For a pre-stretched spring do not square the difference: ΔE = ½ · D · (s₂² − s₁²), not ½ · D · (s₂ − s₁)².

Assumptions and limits

Ideal conservative linear spring with D > 0 and zero energy at force-free natural length. s is signed length change. The inverse gives only |s|; ±|s| fit equal tension/compression behaviour. This is not total energy of a vertical load.

Derivation & background

For an ideal linear spring, quasistatic external force grows as Ds from the force-free state. Its integral gives triangular area $\frac{1}{2}Ds^{2}$ . Spring force on the body is −Ds and spring work is −ΔU. With preload s₀, extra displacement q adds energy $Ds_{0}q + \frac{1}{2}Dq^{2}$ . A lossless horizontal spring oscillator exchanges spring and kinetic energy; vertical arrangements must also account for gravitational energy.

Exam blueprint

Validity range

Valid for a conservative linear spring of constant stiffness D. s is measured from force-free natural length with $U = 0$ there. Elastic does not itself imply linear. Preload requires energy differences; vertically loaded equilibrium also involves gravity. Losses, hysteresis and plastic deformation require a different balance.

Derivation steps

Integrate the work of a linearly growing external force from the force-free state.

1Measure s from the force-free natural length and use constant linear stiffness $D > 0$ .
2For quasistatic lossless stretching Fext = Ds; spring force on the body is −Ds.
3External work is ∫₀ˢDσdσ, not final force times full displacement.
4The integral or triangular area gives $U = \frac{1}{2}Ds^{2}$ with $U(0) = 0$ .
5Between states $\Delta{}U = \frac{1}{2}D(s_{2}^{2} - s_{1}^{2})$ ; spring work is −ΔU.
6For preload s₀ and extra displacement q, retain $\Delta{}U = Ds_{0}q + \frac{1}{2}Dq^{2}$ ; do not confuse loaded equilibrium with natural length.

Rearrangements

Displacement magnitude

|s|=\sqrt{\frac{2U}{D}}

Energy gives magnitude, not direction; energy zero is at natural length.

Spring constant

D = \frac{2 E}{s^2}

$D =\frac{ 2E}{s^{2}}$ requires $s \ne{} 0$ and $E > 0$ for positive stiffness. If $E = s = 0$ , D is not uniquely determined.

Task variant

A spring $(D =\frac{ 100 N}{m})$ stores 8 J. How far is it displaced?

Approach: $|s|=\sqrt{\frac{2U}{D}}$ .
Substitution: $|s|=\sqrt{\frac{16}{100}}\,\mathrm m=0.4\,\mathrm m$ .

Check: $\frac12\cdot100\cdot0.4^2\,\mathrm J=8\,\mathrm J$ . Direction is unspecified.

An ideal spring gun with a massless spring (D = 500 N/m, s = 6 cm) accelerates an initially stationary 20 g ball horizontally without losses. Find its speed at natural spring length.

Approach: $\frac12Ds^2=\frac12mv^2$ .
Spring energy is $U=0.9\,\mathrm J$ .
Thus $|v|=\sqrt{\frac{2U}{m}}=\sqrt{\frac{1.8}{0.02}}\,\mathrm{m\,s^{-1}}\approx9.49\,\mathrm{m\,s^{-1}}$ .

Check: Substitution into kinetic energy returns 0.9 J.

Common mistakes

Dropping the factor $\frac{1}{2} (E = D\cdot{}s^{2})$ .

The force grows from zero; only the triangle area $\frac{1}{2}\cdot{}D\cdot{}s^{2}$ is stored.

Computing $E = F\cdot{}s$ with the final force.

That overestimates by a factor of 2; the mean force $\frac{1}{2}\cdot{}F_{\mathrm{max}}$ is correct.

Inserting the displacement in cm.

Convert s to metres; the error enters quadratically via s² (factor 10,000).

Exam context

Standard: energy conservation of spring versus kinetic or potential energy (spring gun, trampoline) and the energy balance of a spring pendulum between turning point and equilibrium.

Formula cluster

Forms of energy

The third mechanical energy form besides kinetic and potential energy.

F = k·x

Hooke's Law (Spring Force)

Ekin = ½mv²

Kinetic Energy

Epot = mgh

Potential Energy (Gravitational)

Worked example

A spring with $D =\frac{ 400 N}{m}$ is stretched by $s = 10 cm$ : $E = \frac{1}{2} \times{} 400 \times{} 0.1^{2} = 2 J$ . Twice the displacement (20 cm) stores four times as much due to $s^{2}$ : 8 J.

Applications

Spring pendulums and oscillation energy, shock absorbers, sports bows and springboards, mechanical clockworks, spring accumulators in brakes

How do you calculate with Elastic Energy of a Spring?

Here is how to work through a typical Elastic Energy of a Spring (E = ½·D·s²) task step by step:

1
Task
A spring $(D =\frac{ 100 N}{m})$ stores 8 J. How far is it displaced?
Solution path
1. Approach: $|s|=\sqrt{\frac{2U}{D}}$ .
2. Substitution: $|s|=\sqrt{\frac{16}{100}}\,\mathrm m=0.4\,\mathrm m$ .
Check: $\frac12\cdot100\cdot0.4^2\,\mathrm J=8\,\mathrm J$ . Direction is unspecified.
2
Task
An ideal spring gun with a massless spring (D = 500 N/m, s = 6 cm) accelerates an initially stationary 20 g ball horizontally without losses. Find its speed at natural spring length.
Solution path
1. Approach: $\frac12Ds^2=\frac12mv^2$ .
2. Spring energy is $U=0.9\,\mathrm J$ .
3. Thus $|v|=\sqrt{\frac{2U}{m}}=\sqrt{\frac{1.8}{0.02}}\,\mathrm{m\,s^{-1}}\approx9.49\,\mathrm{m\,s^{-1}}$ .
Check: Substitution into kinetic energy returns 0.9 J.

Understand and apply the equation

A deformed spring can accelerate an object when released. Its stored energy comes from work during deformation. Understand the factor of one half using the force–displacement area, then calculate the effects of prestretch and the chosen energy reference.

Which displacement belongs in the equation

The required holding force increases while you deform the spring. Work is therefore less than final force times total distance: the largest force acts only at the end. A force–displacement sketch shows this directly. For an initially relaxed linear spring, the area is triangular; with existing prestretch, an additional rectangular area contributes.[1]

D is constant stiffness in N/m, often written k. s is signed length change from the force-free natural length. An ideal spring with the same linear behaviour in tension and compression stores equal energy at +s and −s. Its force on the moving body is instead $F_s=-Ds$ and has a direction sign.[1]

U(s)=\frac12Ds^2+C,\qquad F_s=-\frac{\mathrm dU}{\mathrm ds}=-Ds

C fixes the energy zero; changing it does not affect forces or energy differences.

“Rest position” is ambiguous without a qualifier. A vertically loaded spring can be stationary while extended. Its spring energy is then nonzero if zero is set at natural length. Displacement from loaded equilibrium is a different coordinate requiring total potential energy, including gravity.[3]

Deriving one half from work

Stretch the ideal spring slowly from $s=0$ so kinetic energy stays unchanged and losses are absent. External force then balances spring force: $F_{\mathrm{ext}}=Ds$ . Input mechanical work is the integral of this varying force. Along a line through the origin, the displacement-average force is only half the final force.[2]

W_{\mathrm{ext}}=\int_0^s D\sigma\,\mathrm d\sigma=\frac12Ds^2

Geometrically this is a triangular area in the external-force–displacement plot.

\Delta U=\frac12D(s_2^2-s_1^2),\qquad W_{\mathrm{spring}}=-\Delta U

Work done by the spring has the opposite sign to its potential-energy change.

Relaxing from s₁ > 0 to zero lowers spring energy and makes spring work on the body positive. Further stretching raises its energy and makes spring work negative. Rapid stretching can also produce kinetic energy, so external work need not equal spring-energy increase alone.

Squares, rearrangements and units

[Ds^2]=\frac{\mathrm N}{\mathrm m}\mathrm m^2=\mathrm{N\,m}=\mathrm J,\quad |s|=\sqrt{\frac{2U}{D}},\quad D=\frac{2U}{s^2}

These rearrangements assume C = 0 and D > 0; solving for D requires s ≠ 0.

Energy determines |s|, not whether the spring is stretched or compressed. Convert 12 cm to 0.12 m before squaring. A stiffness of 2 N/mm equals 2000 N/m. The squared conversion factor matters: 10 cm is 0.1 m, not 10 m.

Original numerical check: with $D=250\,\mathrm{N\,m^{-1}}$ , $s=0.04\,\mathrm m$ stores 0.20 J and $s=0.12\,\mathrm m$ stores 1.80 J. The increase is 1.60 J. Squaring the displacement change gives $\frac12D(0.08\,\mathrm m)^2=0.80\,\mathrm J$ , incorrect for these two states.

With U = 0 and D > 0, s = 0. Conversely, U = 0 and s = 0 do not determine D: any positive stiffness fits. Positive energy at s = 0 contradicts the chosen zero reference. Negative U is excluded with this energy zero in the ideal linear model.[1]

Example: a preloaded spring

Additional energy beyond preload

Given: D = 300 N/m, initial extension s₀ = 0.05 m, additional extension q = 0.03 m. Ideal linear spring, quasistatic stretching.
Find: Find energy increase and external stretching work.

Approach: The spring is already extended initially. Additional quasistatic stretching work therefore equals final minus initial energy of the same linear spring.
$\begin{aligned}s_2&=s_0+q\\{}\Delta U&=\frac D2(s_2^2-s_0^2)\\{}W_{\mathrm{ext}}&=\Delta U\end{aligned}$
Initial and final extensions use the same natural length.
$\begin{aligned}s_1&=0.05\,\mathrm m\\{} s_2&=0.08\,\mathrm m\end{aligned}$
Calculate both state energies separately.
$\begin{aligned}U_1&=\tfrac12\cdot300\cdot0.05^2=0.375\,\mathrm J\\{} U_2&=0.960\,\mathrm J\end{aligned}$
Find the difference instead of squaring only the extra extension.
$\Delta U=0.960-0.375=0.585\,\mathrm J$

Result: Additional spring energy and quasistatic external work are +0.585 J; the spring performs −0.585 J of work.

Check: Force grows from 15 N to 24 N. Trapezoidal area: ½(15+24)·0.03 = 0.585 J. Incorrectly using ½Dq² gives only 0.135 J.

Added work is a trapezoid

QUANTA STUDYquanta-study.de

The spring is already extended by 5 cm before you stretch it to 8 cm. For D = 300 N/m, holding force rises from 15 to 24 N. Only the shaded area is added work: 0.45 J plus 0.135 J gives 0.585 J.[1][2]

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Example: stored energy becomes motion

Kinetic energy describes pure translation of the cart here. The following example therefore assumes a horizontal frictionless track and a massless spring.[3]

A horizontal cart without losses

Given: m = 0.50 kg, D = 200 N/m, release from rest at s = +0.10 m; massless ideal spring and frictionless horizontal track.
Find: Find speed on reaching natural length.

Approach: The cart starts at rest, and ideal spring energy is zero on reaching natural length. Without friction, initial elastic energy becomes kinetic energy completely.
$\frac12Ds^2=\frac12mv^2$
Calculate initial energy stored only in the stretched spring.
$U_0=\tfrac12\cdot200\cdot0.10^2=1.00\,\mathrm J$
At natural length U = 0, leaving translational kinetic energy.
$\begin{aligned}\tfrac12mv^2&=U_0\\{} v&=\sqrt{\frac{2U_0}{m}}\end{aligned}$
Insert the specified mass.
$v=\sqrt{\frac{2}{0.50}}=2.00\,\mathrm{m/s}$

Result: Speed is 2.00 m/s. With s positive rightward, the cart moves leftward at its first zero crossing.

Check: Kinetic check: ½·0.50·2² = 1 J. Energy fixes speed; direction follows from initial state and restoring force.

Distinguishing preload from loaded equilibrium

\Delta U=D s_0q+\frac12Dq^2

Additional extension q of a preloaded spring retains the linear term Ds₀q.

For a vertical spring let s be positive downward. Total potential is $U_{\mathrm{total}}(s)=\frac12Ds^2-mgs+C$ and static equilibrium is $s_{\mathrm{eq}}=\frac{mg}{D}$ . With $q=s-s_{\mathrm{eq}}$ , total potential becomes $\frac12Dq^2$ plus a constant. Gravity cancels the spring energy’s linear term in this total; the term does not disappear from spring energy alone.[3]

U_{\mathrm{total}}(s)=\frac12D(s-s_{\mathrm{eq}})^2+C_1,\qquad s_{\mathrm{eq}}=\frac{mg}{D}

The constant C₁ differs from the original energy zero.

A tension device with a constant initial force may have $F_{\mathrm{ext}}(q)=F_0+Dq$ . Input energy relative to $q=0$ is then $F_0q+\frac12Dq^2$ . Specify the actual force curve and permitted displacement range; “preloaded” does not justify applying the zero-force formula arbitrarily.

Linearity, recoverability and losses

Elastic does not automatically mean linear. For a conservative nonlinear spring, stored energy follows the actual force-curve integral. Outside the linear range, D cannot remain the same constant multiplier. Whether energy exceeds or falls below ½Ds² depends on the curve; there is no universal direction of the deviation.

Hysteresis, internal friction and plastic deformation may reduce recovered work relative to loading input. A damping force is not an extra part of ideal stored spring energy. Heavy springs, rotating attachments or friction also introduce other energy terms; $\frac12Ds^2=\frac12mv^2$ alone is then insufficient.[2][3]

Five exercises with solutions

Calculate · Foundation

D = 150 N/m and s = −0.08 m, with energy zero at natural length. Find U and Fs.

Show hint

The square makes energy positive; force stays signed.

Learning goal: Distinguish spring force and energy for negative displacement.

Show solution and calculation

Approach: Displacement is signed relative to natural length.
Energy uses its square, while spring force retains its sign with the restoring minus sign.
Result: 0.48 J and +12 N.

Check: At negative displacement the force points back in the positive direction; energy equals its value at +0.08 m.

\begin{aligned}U&=\frac12Ds^2=\frac12\cdot150\cdot(-0.08)^2\,\mathrm J=0.48\,\mathrm J\\F_s&=-Ds=-150\cdot(-0.08)\,\mathrm N=+12\,\mathrm N\end{aligned}

[1]

Calculate · Application

A linear spring of D = 100 N/m is stretched further from 0.10 m to 0.20 m. Find added energy.

Show hint

Use the difference of squares, not the square of the difference.

Learning goal: Find additional energy in a prestretched spring.

Show solution and calculation

Approach: Subtract initial energy from final energy.
Both displacements refer to natural length; existing prestretch must be retained.
Result: 1.50 J added energy.

Check: Holding force increases from 10 to 20 N. Its mean value of 15 N over an additional 0.10 m also gives 1.50 J of work.

\begin{aligned}\Delta U&=\frac12D(s_2^2-s_1^2)\\&=\frac12\cdot100\cdot(0.20^2-0.10^2)\,\mathrm J\\&=1.50\,\mathrm J\end{aligned}

[1]

Rearrange · Application

D = 80 N/m stores 0.64 J with zero at natural length. Find displacement magnitude.

Show hint

Take the square root; energy does not give direction.

Learning goal: Find displacement magnitude from energy.

Show solution and calculation

Approach: Solve the energy equation for displacement magnitude.
The positive square root gives magnitude, not direction.
Result: approximately 0.1265 m or 12.65 cm.

Check: Substitution without early rounding returns 0.64 J. Both positive and negative displacement fit the ideal model; whether a real spring can compress that far requires a separate check.

\begin{aligned}|s|&=\sqrt{\frac{2U}{D}}\\&=\sqrt{\frac{2\cdot0.64}{80}}\,\mathrm m\\&=\sqrt{0.016}\,\mathrm m\approx0.1265\,\mathrm m\end{aligned}

[1]

Interpret · Application

An ideal linear spring of D=200 N/m is either extended by +3 cm or compressed by −3 cm, with zero energy at natural length. Compare energies and signed spring forces; positive direction is extension.

Show hint

Energy is quadratic; spring force is linear with a minus sign.

Learning goal: Explain symmetric energy and antisymmetric force.

Show solution and calculation

Approach: Equal displacement magnitudes have equal squares, giving 0.09 J in each case.
Spring forces are −6 N in extension and +6 N in compression.
Both restore natural length.
The mirrored energy curve therefore has equal heights but opposite signs of its negative slope.

\begin{aligned}U(\pm0.03\,\mathrm m)&=\frac12D(0.03\,\mathrm m)^2=0.09\,\mathrm J\\{}F_s(+0.03\,\mathrm m)&=-6\,\mathrm N\\{}F_s(-0.03\,\mathrm m)&=+6\,\mathrm N\end{aligned}

[1]

Error analysis · Application

An initially relaxed ideal spring of D=100 N/m is extended quasistatically to 0.10 m. Someone multiplies final force 10 N by distance and calls 1 J stored energy. Correct this using the force-displacement area.

Show hint

Force starts at zero and rises linearly.

Learning goal: Correct endpoint-force times displacement for spring work.

Show solution and calculation

Approach: Area under applied force is triangular, not a rectangle holding final force across the whole distance.
Mean force 5 N across 0.10 m gives 0.5 J, agreeing with quadratic energy.
The wrong value is twice as large because final force was not present throughout the earlier displacement.

W_{\mathrm{ext}}=\int_0^sDx\,\mathrm dx=\frac12Ds^2=0.5\,\mathrm J

[1]

Key points

Spring energy grows with squared displacement. Additional stretching requires the difference between endpoint energies; do not square only the change in displacement. In a quasistatic lossless model, that increase equals external work.

Sources for this explanation

Quanta · Updated October 2, 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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Change history

October 2, 2026Clarified zero cases in energy inverses and linked mechanical work and kinetic energy in context; preserved existing calculations and diagram.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Revised the learning introduction and solution paths; independently checked model assumptions and calculations against primary sources.

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Frequently asked questions about Elastic Energy of a Spring

How do you calculate the elastic energy of a spring?+

For an ideal linear spring with energy zero at natural length, $U = \frac{1}{2}Ds^{2}$ . $D =\frac{ 400 N}{m}$ and $s = 0.10 m$ give 2 J. Convert centimetres to metres first and establish what s is measured from. A negative displacement also stores positive energy when a suitable linear compression model applies. “Rest position” is ambiguous without a qualifier. A vertically loaded spring can be stationary while extended. Its spring energy is then nonzero if zero is set at natural length. Displacement from loaded equilibrium is a different coordinate requiring total potential energy, including gravity.

Where does the factor ½ in the formula come from?+

During linear stretching from the force-free state, external force rises from zero to Ds. Its displacement-average value is $\frac{Ds}{2}$ , so the integral or triangular area is $\frac{1}{2}Ds^{2}$ . For $D =\frac{ 400 N}{m}$ and $s = 0.10 m$ , final force is 40 N, average force over displacement is 20 N and work is 2 J. Relaxing from $s_{1} > 0$ to zero lowers spring energy and makes spring work on the body positive. Further stretching raises its energy and makes spring work negative. Rapid stretching can also produce kinetic energy, so external work need not equal spring-energy increase alone.

How are elastic energy and Hooke law related?+

For $Fs = -Ds$ , spring force is the negative derivative of potential energy: Fs = −dU/ds. Integration gives $U = \frac{1}{2}Ds^{2} + C$ . The constant shifts energy zero without changing forces. A measured nonlinear force curve instead requires integrating that curve rather than using a fixed quadratic formula. Elastic does not automatically mean linear. For a conservative nonlinear spring, stored energy follows the actual force-curve integral. Outside the linear range, D cannot remain the same constant multiplier. Whether energy exceeds or falls below $\frac{1}{2}Ds^{2}$ depends on the curve; there is no universal direction of the deviation.

How do you convert spring energy into speed?+

For an initially stationary mass on a frictionless horizontal track with an ideal massless spring, reaching natural length gives $\frac{1}{2}mv^{2} = \frac{1}{2}Ds_{0}^{2}$ . Thus $|v| = |s_{0}|\sqrt{(\frac{D}{m})}$ . $D =\frac{ 200 N}{m}$ , $m = 0.50 kg$ and $s_{0} = 0.10 m$ give $\frac{2 m}{s}$ . Direction additionally follows from the motion history. Hysteresis, internal friction and plastic deformation may reduce recovered work relative to loading input. A damping force is not an extra part of ideal stored spring energy. Heavy springs, rotating attachments or friction also introduce other energy terms; $\frac{1}{2}Ds^{2} = \frac{1}{2}mv^{2}$ alone is then insufficient.

Why does double displacement store four times the energy?+

For constant D and energy zero at natural length, $\frac{U(2s)}{U}(s) = 4$ . $\frac{A 400 N}{m}$ spring stores 2 J at 10 cm and 8 J at 20 cm, so the energy increase between those states is 6 J. The fourfold rule concerns absolute extension from the force-free state, not an arbitrary additional displacement. For a vertical spring let s be positive downward. Total potential is Utotal(s) = ½Ds² − mgs + C and static equilibrium is sEq = mg/D. With q = s − sEq, total potential becomes $\frac{1}{2}Dq^{2}$ plus a constant. Gravity cancels the spring energy’s linear term in this total; the term does not disappear from spring energy alone.

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[1]OpenStax, University Physics Vol. 1, §15.2 Energy in Simple Harmonic Motion

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

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v = a \cdot t + v_0

Velocity-Time Law

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Elastic potential energy of a spring

For an ideal linear spring,

U = \frac{1}{2}Ds^{2}

is stored energy relative to its force-free natural state. Between pre-extended states use the difference of squared extensions, not the square of extra displacement.

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Symbol

Meaning

Unit

E_{\mathrm{Spann}}

Spring energy relative to the force-free state

D

Constant linear spring stiffness

N/m

s

Length change relative to force-free natural length

Derivation & background

Exam blueprint

Validity range

Derivation steps

Integrate the work of a linearly growing external force from the force-free state.

1Measure s from the force-free natural length and use constant linear stiffness $D > 0$ .
2For quasistatic lossless stretching Fext = Ds; spring force on the body is −Ds.
3External work is ∫₀ˢDσdσ, not final force times full displacement.
4The integral or triangular area gives $U = \frac{1}{2}Ds^{2}$ with $U(0) = 0$ .
5Between states $\Delta{}U = \frac{1}{2}D(s_{2}^{2} - s_{1}^{2})$ ; spring work is −ΔU.
6For preload s₀ and extra displacement q, retain $\Delta{}U = Ds_{0}q + \frac{1}{2}Dq^{2}$ ; do not confuse loaded equilibrium with natural length.

How do you calculate with Elastic Energy of a Spring?

Here is how to work through a typical Elastic Energy of a Spring (E = ½·D·s²) task step by step:

Task

A spring $(D =\frac{ 100 N}{m})$ stores 8 J. How far is it displaced?

Solution path

Approach: $|s|=\sqrt{\frac{2U}{D}}$ .
Substitution: $|s|=\sqrt{\frac{16}{100}}\,\mathrm m=0.4\,\mathrm m$ .

Check: $\frac12\cdot100\cdot0.4^2\,\mathrm J=8\,\mathrm J$ . Direction is unspecified.

Task

An ideal spring gun with a massless spring (D = 500 N/m, s = 6 cm) accelerates an initially stationary 20 g ball horizontally without losses. Find its speed at natural spring length.

Solution path

Approach: $\frac12Ds^2=\frac12mv^2$ .
Spring energy is $U=0.9\,\mathrm J$ .
Thus $|v|=\sqrt{\frac{2U}{m}}=\sqrt{\frac{1.8}{0.02}}\,\mathrm{m\,s^{-1}}\approx9.49\,\mathrm{m\,s^{-1}}$ .

Check: Substitution into kinetic energy returns 0.9 J.

Which displacement belongs in the equation

U(s)=\frac12Ds^2+C,\qquad F_s=-\frac{\mathrm dU}{\mathrm ds}=-Ds

C fixes the energy zero; changing it does not affect forces or energy differences.

Deriving one half from work

W_{\mathrm{ext}}=\int_0^s D\sigma\,\mathrm d\sigma=\frac12Ds^2

Geometrically this is a triangular area in the external-force–displacement plot.

\Delta U=\frac12D(s_2^2-s_1^2),\qquad W_{\mathrm{spring}}=-\Delta U

Work done by the spring has the opposite sign to its potential-energy change.

Squares, rearrangements and units

[Ds^2]=\frac{\mathrm N}{\mathrm m}\mathrm m^2=\mathrm{N\,m}=\mathrm J,\quad |s|=\sqrt{\frac{2U}{D}},\quad D=\frac{2U}{s^2}

These rearrangements assume C = 0 and D > 0; solving for D requires s ≠ 0.

Example: a preloaded spring

Additional energy beyond preload

Given: D = 300 N/m, initial extension s₀ = 0.05 m, additional extension q = 0.03 m. Ideal linear spring, quasistatic stretching.
Find: Find energy increase and external stretching work.

Approach: The spring is already extended initially. Additional quasistatic stretching work therefore equals final minus initial energy of the same linear spring.
$\begin{aligned}s_2&=s_0+q\\{}\Delta U&=\frac D2(s_2^2-s_0^2)\\{}W_{\mathrm{ext}}&=\Delta U\end{aligned}$
Initial and final extensions use the same natural length.
$\begin{aligned}s_1&=0.05\,\mathrm m\\{} s_2&=0.08\,\mathrm m\end{aligned}$
Calculate both state energies separately.
$\begin{aligned}U_1&=\tfrac12\cdot300\cdot0.05^2=0.375\,\mathrm J\\{} U_2&=0.960\,\mathrm J\end{aligned}$
Find the difference instead of squaring only the extra extension.
$\Delta U=0.960-0.375=0.585\,\mathrm J$

Result: Additional spring energy and quasistatic external work are +0.585 J; the spring performs −0.585 J of work.

Check: Force grows from 15 N to 24 N. Trapezoidal area: ½(15+24)·0.03 = 0.585 J. Incorrectly using ½Dq² gives only 0.135 J.

Added work is a trapezoid

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Example: stored energy becomes motion

Kinetic energy describes pure translation of the cart here. The following example therefore assumes a horizontal frictionless track and a massless spring.[3]

A horizontal cart without losses

Given: m = 0.50 kg, D = 200 N/m, release from rest at s = +0.10 m; massless ideal spring and frictionless horizontal track.
Find: Find speed on reaching natural length.

Approach: The cart starts at rest, and ideal spring energy is zero on reaching natural length. Without friction, initial elastic energy becomes kinetic energy completely.
$\frac12Ds^2=\frac12mv^2$
Calculate initial energy stored only in the stretched spring.
$U_0=\tfrac12\cdot200\cdot0.10^2=1.00\,\mathrm J$
At natural length U = 0, leaving translational kinetic energy.
$\begin{aligned}\tfrac12mv^2&=U_0\\{} v&=\sqrt{\frac{2U_0}{m}}\end{aligned}$
Insert the specified mass.
$v=\sqrt{\frac{2}{0.50}}=2.00\,\mathrm{m/s}$

Result: Speed is 2.00 m/s. With s positive rightward, the cart moves leftward at its first zero crossing.

Check: Kinetic check: ½·0.50·2² = 1 J. Energy fixes speed; direction follows from initial state and restoring force.

Distinguishing preload from loaded equilibrium

\Delta U=D s_0q+\frac12Dq^2

Additional extension q of a preloaded spring retains the linear term Ds₀q.

U_{\mathrm{total}}(s)=\frac12D(s-s_{\mathrm{eq}})^2+C_1,\qquad s_{\mathrm{eq}}=\frac{mg}{D}

The constant C₁ differs from the original energy zero.

Linearity, recoverability and losses

Five exercises with solutions

Calculate · Foundation

D = 150 N/m and s = −0.08 m, with energy zero at natural length. Find U and Fs.

Show hint

The square makes energy positive; force stays signed.

Learning goal: Distinguish spring force and energy for negative displacement.

Show solution and calculation

Approach: Displacement is signed relative to natural length.
Energy uses its square, while spring force retains its sign with the restoring minus sign.
Result: 0.48 J and +12 N.

Check: At negative displacement the force points back in the positive direction; energy equals its value at +0.08 m.

\begin{aligned}U&=\frac12Ds^2=\frac12\cdot150\cdot(-0.08)^2\,\mathrm J=0.48\,\mathrm J\\F_s&=-Ds=-150\cdot(-0.08)\,\mathrm N=+12\,\mathrm N\end{aligned}

[1]

Calculate · Application

A linear spring of D = 100 N/m is stretched further from 0.10 m to 0.20 m. Find added energy.

Show hint

Use the difference of squares, not the square of the difference.

Learning goal: Find additional energy in a prestretched spring.

Show solution and calculation

Approach: Subtract initial energy from final energy.
Both displacements refer to natural length; existing prestretch must be retained.
Result: 1.50 J added energy.

Check: Holding force increases from 10 to 20 N. Its mean value of 15 N over an additional 0.10 m also gives 1.50 J of work.

\begin{aligned}\Delta U&=\frac12D(s_2^2-s_1^2)\\&=\frac12\cdot100\cdot(0.20^2-0.10^2)\,\mathrm J\\&=1.50\,\mathrm J\end{aligned}

[1]

Rearrange · Application

D = 80 N/m stores 0.64 J with zero at natural length. Find displacement magnitude.

Show hint

Take the square root; energy does not give direction.

Learning goal: Find displacement magnitude from energy.

Show solution and calculation

Approach: Solve the energy equation for displacement magnitude.
The positive square root gives magnitude, not direction.
Result: approximately 0.1265 m or 12.65 cm.

Check: Substitution without early rounding returns 0.64 J. Both positive and negative displacement fit the ideal model; whether a real spring can compress that far requires a separate check.

\begin{aligned}|s|&=\sqrt{\frac{2U}{D}}\\&=\sqrt{\frac{2\cdot0.64}{80}}\,\mathrm m\\&=\sqrt{0.016}\,\mathrm m\approx0.1265\,\mathrm m\end{aligned}

[1]

Interpret · Application

An ideal linear spring of D=200 N/m is either extended by +3 cm or compressed by −3 cm, with zero energy at natural length. Compare energies and signed spring forces; positive direction is extension.

Show hint

Energy is quadratic; spring force is linear with a minus sign.

Learning goal: Explain symmetric energy and antisymmetric force.

Show solution and calculation

Approach: Equal displacement magnitudes have equal squares, giving 0.09 J in each case.
Spring forces are −6 N in extension and +6 N in compression.
Both restore natural length.
The mirrored energy curve therefore has equal heights but opposite signs of its negative slope.

\begin{aligned}U(\pm0.03\,\mathrm m)&=\frac12D(0.03\,\mathrm m)^2=0.09\,\mathrm J\\{}F_s(+0.03\,\mathrm m)&=-6\,\mathrm N\\{}F_s(-0.03\,\mathrm m)&=+6\,\mathrm N\end{aligned}

[1]

Error analysis · Application

An initially relaxed ideal spring of D=100 N/m is extended quasistatically to 0.10 m. Someone multiplies final force 10 N by distance and calls 1 J stored energy. Correct this using the force-displacement area.

Show hint

Force starts at zero and rises linearly.

Learning goal: Correct endpoint-force times displacement for spring work.

Show solution and calculation

Approach: Area under applied force is triangular, not a rectangle holding final force across the whole distance.
Mean force 5 N across 0.10 m gives 0.5 J, agreeing with quadratic energy.
The wrong value is twice as large because final force was not present throughout the earlier displacement.

W_{\mathrm{ext}}=\int_0^sDx\,\mathrm dx=\frac12Ds^2=0.5\,\mathrm J

[1]

Key points

Sources for this explanation

Quanta · Updated October 2, 2026

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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.

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Change history

October 2, 2026Clarified zero cases in energy inverses and linked mechanical work and kinetic energy in context; preserved existing calculations and diagram.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Revised the learning introduction and solution paths; independently checked model assumptions and calculations against primary sources.

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Quanta. Elastic Energy of a Spring. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/spannenergie-feder

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Frequently asked questions about Elastic Energy of a Spring

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