Forces and elasticity — AQA GCSE Combined Science

A single force would make a stationary object accelerate or move, not just change shape. To deform an object without moving it, the forces must balance overall, so at least two forces are needed. For a spring, one hand pulls one way and the other hand pulls the opposite way; the two forces are equal in size and opposite in direction, so the resultant force is zero and the spring stays in place while stretching. The same applies to compression, where two inward pushes balance, and to bending, where forces at different points balance. Because the object remains stationary, the forces must be balanced, and balanced forces require more than one force.

When a stretching force acts on an object, the object may change shape. In elastic deformation, the object returns to its original length and shape once the force is removed, because the particles are pulled slightly apart but spring back to their original arrangement. A rubber band or a spring stretched gently behaves elastically. In inelastic deformation, the object does not return to its original length or shape when the force is removed; it stays permanently stretched. This happens when the force is large enough to move particles into new positions so they cannot return. Stretching a plastic bag until it tears or over-stretching a spring produces inelastic deformation. The key difference is therefore whether the original shape is recovered after the stretching force is removed.

For an elastic object such as a spring, extension means the increase in length from its original length. If the force applied is doubled, the extension doubles, provided the force stays below the limit of proportionality. This straight-line relationship is written as F = k × e, where F is the force in newtons, e is the extension in metres and k is the spring constant in newtons per metre. The spring constant measures stiffness: a stiffer spring has a larger k and stretches less for the same force. On a force-extension graph, the directly proportional region is a straight line through the origin. Beyond the limit of proportionality the line curves and the relationship is no longer directly proportional, so F = k × e cannot be used. A spring stretched past this limit may also show inelastic deformation.

This relationship describes Hooke's law for a spring or elastic object that returns to its original shape when the stretching force is removed. The force is the load applied, measured in newtons (N). The spring constant k measures stiffness in newtons per metre (N/m): a large k means a stiff spring needing a large force for a small extension. Extension e is the increase in length beyond the natural length, in metres (m), so e = stretched length − original length. For example, a spring of natural length 0.20 m stretches to 0.26 m under a 3.0 N load, so e = 0.06 m and k = F ÷ e = 3.0 N ÷ 0.06 m = 50 N/m. The law holds only up to the limit of proportionality; beyond that the graph of force against extension curves and the equation no longer applies.

This is the symbolic form of Hooke's law: F is the force in newtons (N), k is the spring constant in newtons per metre (N/m) and e is the extension in metres (m). The equation says that, for a spring or elastic object obeying Hooke's law, force and extension are directly proportional, so doubling the force doubles the extension. The spring constant k is the force needed per metre of extension; a spring with k = 50 N/m needs 50 N to stretch it by 1 m, or 3.0 N to stretch it by 0.06 m. Rearranged, k = F ÷ e and e = F ÷ k. The relationship is linear only up to the limit of proportionality; beyond that the force–extension graph curves and the equation no longer predicts behaviour. On such a graph, the gradient of the straight section equals k.

Force is a vector quantity measured in newtons (N). In the context of elasticity, the force F usually refers to the load or tension applied to a spring or elastic object, or the restoring force it exerts. One newton is the force needed to give a 1 kg mass an acceleration of 1 m/s². When a spring is stretched, the extension depends on the applied force, and the relationship is described by Hooke's law: F = k × e, where k is the spring constant and e is the extension. For example, a spring with k = 20 N/m stretched by 0.15 m requires F = 20 × 0.15 = 3 N. Students should recognise force as a vector, use newtons correctly in calculations, and distinguish the applied force from the weight of a hanging mass, where W = m × g.

The spring constant k is a measure of the stiffness of a spring or elastic object. It is defined by Hooke's law, F = k × e, where F is the force in newtons and e is the extension in metres. Rearranging gives k = F ÷ e, so k is the force needed to produce unit extension. Its unit is newtons per metre (N/m). A stiffer spring has a larger k and stretches less for the same force. For example, if a force of 6 N produces an extension of 0.03 m, then k = 6 ÷ 0.03 = 200 N/m. Students should determine k from the gradient of the linear region of a force–extension graph, convert units correctly, and recognise that k is constant only up to the limit of proportionality.

Extension is the increase in length of an elastic object when a stretching force acts on it. It is measured in metres, symbol m, and given the symbol e in equations such as force = spring constant × extension. To find e, measure the stretched length and subtract the original unstretched length: e = L − L₀. For example, a spring of original length 0.12 m stretched to 0.20 m has e = 0.20 m − 0.12 m = 0.08 m. Extension is not the same as total length, and it must be expressed in metres before substitution into an equation. A negative value would indicate compression rather than extension.

The force–extension relationship, force = spring constant × extension, also describes compression of an elastic object. In compression, the object shortens, and e represents the compression, the decrease in length. Compression is calculated as original length minus compressed length: e = L₀ − L. For example, a spring of original length 0.15 m compressed to 0.11 m has e = 0.15 m − 0.11 m = 0.04 m. The spring constant has the same meaning, and the force is the compressive force. Compression is still measured in metres, and the same equation applies as long as the elastic limit is not exceeded.

When a force stretches or compresses a spring, the force moves the spring's ends, so work is done. That transferred energy is stored as elastic potential energy. For a spring obeying Hooke's law, the force grows steadily from zero to F, so the average force is F/2 and the work done is W = ½Fe, where e is the extension. Because energy is conserved, this work equals the elastic potential energy stored, Ee = ½ke², since F = ke. This equality holds only while the spring is not inelastically deformed: if the force exceeds the elastic limit, the spring is permanently stretched, some energy heats the material and is not recovered, so work done exceeds the elastic energy stored. Example: stretching a spring by 0.20 m with a final force of 10 N stores ½ × 10 × 0.20 = 1.0 J.

A force–extension graph shows force on the vertical axis and extension on the horizontal axis. A linear relationship is a straight line: equal increases in force produce equal increases in extension. If this straight line passes through the origin, force is directly proportional to extension, so the spring obeys Hooke's law, F = ke, and the gradient is the spring constant k. A non-linear relationship is a curved line: the gradient changes as extension increases. For a metal spring, the graph is linear up to the limit of proportionality, then curves. To describe the difference, state the shape, whether the line passes through the origin, whether the gradient is constant, and what that means for proportionality.

In the linear region of a force–extension graph, force and extension are directly proportional, so the spring constant k is the constant ratio F ÷ e. Rearranging F = k e gives k = F ÷ e. For example, if a 6.0 N force stretches a spring by 0.030 m, then k = 6.0 N ÷ 0.030 m = 200 N/m. The gradient of the straight-line section of a force–extension graph also equals k, so a steeper line means a stiffer spring. Convert extension to metres before dividing, and remember that extension is the increase in length, not the total length. The unit of k is newtons per metre (N/m). This calculation applies only while the graph is a straight line through the origin; beyond the limit of proportionality the relationship is non-linear and k is not constant.

In this investigation, a spring or wire is stretched by adding masses or weights, and the total length is measured for each force. Extension is the increase in length, found by subtracting the original length from each measured length. Plotting force on the y-axis against extension on the x-axis gives a straight line through the origin while the spring obeys Hooke's law, showing that force and extension are directly proportional. The gradient of this linear section equals the spring constant k. Beyond the limit of proportionality the line curves, so equal increases in force produce larger increases in extension. Anomalous points that do not fit the pattern should be identified and either rechecked or ignored when drawing the line of best fit.

When a force stretches or compresses a spring, energy is transferred to the elastic potential energy store of the spring. The work done by the force equals the energy transferred, provided the spring has not been stretched beyond its limit of proportionality. For a spring obeying Hooke's law, force is proportional to extension, so the force rises steadily from zero to F. The average force is F ÷ 2, and work done = average force × extension, giving W = 0.5 × k × e². For example, a spring with spring constant 200 N/m stretched by 0.10 m stores 0.5 × 200 × 0.10² = 1.0 J. Always convert centimetres to metres before squaring, and check that the extension used is within the straight-line region of the force–extension graph.

This equation gives the energy stored in a stretched or compressed spring when it obeys Hooke's law. The spring constant k measures stiffness in newtons per metre, and the extension e is the change in length in metres, not the total length. Because the force needed grows in proportion to the extension, the energy stored is the area under the force–extension graph, a triangle of base e and height k × e, so Ee = 0.5 × k × e². For example, a spring of spring constant 150 N/m extended by 0.20 m stores 0.5 × 150 × 0.20² = 3.0 J. The equation works for compression too, using the compression as e. It fails beyond the limit of proportionality, where the graph curves and the stored energy must be found from the area under the graph instead.

This equation gives the elastic potential energy stored in a stretched or compressed spring, provided the spring has not been stretched beyond its limit of proportionality. Eₑ is the energy in joules (J), k is the spring constant in newtons per metre (N/m), and e is the extension or compression in metres (m). The factor ½ arises because the force needed grows steadily from zero to ke as the spring stretches, so the average force is ½ke and the work done is ½ke × e. For example, a spring with k = 200 N/m stretched by e = 0.10 m stores Eₑ = ½ × 200 × 0.10² = 1.0 J. Always convert centimetres to metres and grams to kilograms before substituting, and square the extension before multiplying by k.

This statement requires you to use the elastic potential energy equation, Eₑ = ½ke², to find any missing quantity and to link the stored energy to transfers in a system. You may be asked to find the energy stored when a spring is stretched, the spring constant when energy and extension are known, or the extension when energy and k are known. Rearranging is often needed: k = 2Eₑ ÷ e² and e = √(2Eₑ ÷ k). For example, if 0.50 J is stored in a spring with k = 100 N/m, then e = √(2 × 0.50 ÷ 100) = 0.10 m. When the spring is released, the elastic potential energy store decreases and equal energy is transferred to kinetic, gravitational potential or thermal energy stores, depending on the situation.

This required practical explores how a spring stretches when a force is applied. You hang a spring from a clamp and add masses one at a time, recording the total weight as the force and the extension as the increase in length from the original. Plot force on the y-axis against extension on the x-axis. The graph is a straight line through the origin while the spring obeys Hooke's law, so force is directly proportional to extension. Beyond the limit of proportionality the line curves and the spring may not return to its original length. The gradient of the straight section equals the spring constant k, measured in N/m. Use F = kx to calculate k, and take repeat readings to reduce random error.