Force, extension and Hooke’s law

Extension is a change in length

An object deforms when forces change its shape or size. A spring under tension becomes longer; under compression it becomes shorter. Extension x is loaded length L minus the original unloaded length L₀. It is a change in length, not the full length of the spring. Use a clearly marked reference point and the same point on the spring for every reading.

When known masses hang at rest from a vertical spring, the downward load is their total weight. Use F = mg, including the mass hanger when it is part of the load relative to the chosen zero. If the unloaded reference was measured with the hanger already attached, calculate additional force and additional extension consistently instead. Waiting until oscillations settle matters: during motion spring force need not equal the hanging weight.

Convert lengths before calculating stiffness in N m⁻¹: 1 cm = 0.01 m and 1 mm = 0.001 m. Convert grams to kilograms before using F = mg. These conversions often matter more than the final arithmetic. Record raw ruler readings as well as calculated extensions so that an incorrect reference value can be traced.

x = L − L₀ F = mg for a static hanging load

Symbols and units

• L is loaded spring length and L₀ is unloaded reference length, both in m; x = L − L₀ is extension in m.
• F is tensile load in N, m is total suspended mass in kg, and g is gravitational acceleration in m s⁻².

How the equation explains the idea

1. Measure both lengths from the same reference, then subtract before using Hooke’s law. A 0.18 m spring stretched to 0.22 m has x = 0.04 m, not 0.22 m.
2. For a stationary hanging load, upward spring force balances downward weight: F = mg. Include the hanger mass and avoid this balance while the mass accelerates.

Worked example

From ruler readings to extension

1. A spring is 12.0 cm long without its load and 15.5 cm long with a load.
2. Extension = 15.5 − 12.0 = 3.5 cm = 0.035 m.
3. A total suspended mass of 0.200 kg gives a load of 0.200 × 9.81 = 1.962 N.

The force acts over an extension of 0.035 m, not a length of 0.155 m.

Keep this idea. Measure the change in length, and be explicit about whether the mass hanger belongs to the load or the zero.

Hooke’s law describes a proportional region

Within the proportional region, increasing the deforming force causes a proportional increase in extension: F = kx when both are measured from the same unloaded zero. The stiffness k is force per unit extension, in N m⁻¹. A larger k means more force is needed for the same extension. It does not mean the object can necessarily support a larger force before breaking.

For two points within the same straight region, ΔF = kΔx. This difference form is useful with a preload or an offset in the readings: k is the gradient of force against extension. If force is plotted vertically and extension horizontally, calculate k = ΔF/Δx. If the axes are swapped, the gradient is 1/k. Always read the axis units: a gradient in N cm⁻¹ must be multiplied by 100 to express it in N m⁻¹.

The law is a model with a limit, not a universal rule for every deformation. In the proportional region the ratio F/x is constant when measured from a correct zero. Beyond that region a single constant k no longer predicts all points. The spring’s restoring force acts opposite to its extension; the equation F = kx here uses the positive magnitude of the external stretching force, not a signed restoring-force vector.

F = kx k = ΔF / Δx within the proportional region

Symbols and units

• F is tensile force in N; x is extension in m; k is spring stiffness in N m⁻¹.
• ΔF/Δx is the gradient of force vertically against extension horizontally.

How the equation explains the idea

1. Within the proportional region F = kx. Compare two points: F₂ − F₁ = k(x₂ − x₁), so k = ΔF/Δx.
2. A 200 N m⁻¹ spring needs 6 N for 0.03 m extension; doubling extension to 0.06 m doubles force to 12 N while the model remains valid.
3. If the axes are reversed, extension–force slope is 1/k. Stiffness describes resistance to extension, not the breaking load.

Worked example

Determine stiffness from two graph points

1. On the straight force–extension line, two points are (0.010 m, 1.5 N) and (0.030 m, 4.5 N).
2. k = (4.5 − 1.5)/(0.030 − 0.010) = 3.0/0.020 = 150 N m⁻¹.
3. A 6.0 N force would predict x = 6.0/150 = 0.040 m, only if that force remains within the proportional region.

Stiffness is 150 N m⁻¹.

Keep this idea. Stiffness is a gradient with stated axis units, and a constant stiffness applies only in the proportional region.

Proportionality and elasticity are different tests

The limit of proportionality is where the force–extension relationship stops being a straight line through the appropriate zero. Beyond it, Hooke’s law with a constant k no longer applies. A material can still return to its original length after unloading even when its loading curve was nonlinear. A curved graph alone does not establish permanent damage.

Elastic deformation is reversible: after the force is removed, the original dimensions are recovered. Plastic deformation leaves a permanent change in dimensions. The elastic limit is the boundary beyond which unloading leaves permanent deformation. It need not coincide exactly with the limit of proportionality. To investigate it, compare the unloaded length with the initial length after each loading cycle; a loading graph by itself may not show enough evidence.

Some materials show a yield region where extension increases greatly with only a small increase in force. The detailed shape depends on the material and specimen, so do not assume every spring has a distinct yield point. Compression graphs use the same ideas with compression as a positive deformation magnitude. Tensile stress, strain, Young modulus and breaking stress are covered in the materials lessons; stiffness here describes the object, not a geometry-independent material property.

Worked example

What does a curved graph prove?

1. A loading graph curves after 4 N, but the object returns to its initial length when unloaded from 5 N.
2. The changing gradient shows it has exceeded its limit of proportionality.
3. Returning to the original length shows that this deformation was still elastic at 5 N.

Non-Hookean behaviour does not by itself mean plastic deformation.

Keep this idea. Loading shape tests proportionality; recovery on unloading tests elasticity.

Measure a trustworthy force–extension graph

Clamp the spring securely, place a vertical ruler close to it and attach a light pointer perpendicular to the ruler. Record the unloaded pointer position. Add known loads in modest steps, wait for oscillations to settle and read at eye level to reduce parallax. Record total load and extension for each step. Keep clear of the space beneath the masses, use a tray to catch falling loads, and follow local laboratory supervision and load limits.

Take several readings across the intended range and repeat readings or loading cycles when it is safe. A wider usable extension range reduces the relative effect of a fixed reading uncertainty, but do not overload the spring merely to improve precision. Plot force vertically against extension horizontally, label units and draw a best-fit line through the linear region. Use widely separated points on that line for the gradient, rather than forcing a line through every noisy point or relying on one small extension.

Extension is obtained by subtracting two position readings. If each has an estimated uncertainty of ±1 mm, a conservative absolute uncertainty in their difference is ±2 mm. This is 10% of a 20 mm extension but 2% of a 100 mm extension. Repeats help assess and reduce random scatter; they do not automatically remove a systematic zero error or parallax. Recheck alignment and the unloaded reference when the line has an unexpected intercept.

Remove loads in steps and record the unloading values. A residual extension after complete unloading is evidence of plastic deformation, provided it exceeds the measurement uncertainty. A spring experiment illustrates force–extension methods; it is not the complete Pearson Young modulus core practical, which requires different measurements and analysis.

Worked example

Choose a useful measurement improvement

1. A short 10 mm extension is found from two readings, each with ±1 mm uncertainty.
2. Conservative extension uncertainty = ±2 mm, or 20%.
3. A longer safe extension of 50 mm has the same absolute uncertainty but 4% relative uncertainty. A close pointer and eye-level reading reduce parallax.

Use a suitable safe measurement range and improve the reading method; simply repeating a biased reading does not fix the bias.

Keep this idea. Good data require a consistent zero, settled loads, controlled reading geometry, an appropriate range and unloading checks.

Area under the graph is work done

Stretching an object requires work because force acts through a displacement. For a slowly applied load, the area under a force–extension graph gives the work done in deforming the object. The units are newton × metre = joule. Within the linear elastic region, the graph is a triangle from the origin, so the stored elastic energy is ½Fx = ½kx².

The factor ½ matters: the applied force rises from zero to F while the object is stretched, so its mean value is F/2 in this linear case. Using Fx assumes the force had its final value during the whole deformation and overestimates the work. If extension doubles while k stays constant and the spring remains in its linear region, stored energy becomes four times as large.

For a nonlinear force–extension curve, estimate its area using small trapezia or grid squares; do not apply the triangular formula automatically. Work done during loading is only fully recoverable as elastic energy when the process is reversible without dissipation. An unloading curve can enclose an area with the loading curve; that difference corresponds to energy not returned mechanically. This lesson introduces graphical energy. Continue with the materials lessons for stress–strain relationships and energy analysis.

Linear elastic region: E = ½Fx = ½kx²

Symbols and units

• E is elastic energy in J; F is final tensile force in N; x is extension in m.
• k is stiffness in N m⁻¹. The formula assumes linear, reversible loading from zero extension.

How the equation explains the idea

1. Work done stretching is the area under force against extension. A straight line from zero makes a triangle with base x and height F.
2. Thus E = ½Fx; substituting F = kx gives E = ½kx². Doubling x multiplies energy by four, although it only doubles force.
3. For nonlinear or dissipative loading, use the actual area and distinguish input work from recoverable elastic energy.

Worked example

Energy in a stretched spring

1. A spring of stiffness 200 N m⁻¹ is stretched by 0.030 m within its linear elastic region.
2. Final force F = kx = 200 × 0.030 = 6.0 N.
3. Energy = ½ × 6.0 × 0.030 = 0.090 J. This equals ½ × 200 × (0.030)².

The stored elastic energy is 0.090 J.

Keep this idea. Area gives deformation work; use ½kx² only for a linear elastic response from the unloaded reference.