SP15 · Forces and matterTopic 15 — Forces and matter
Springs, extension and elastic energy
Revise the key ideas
Forces deform objects
Stretching, compressing or bending an object involves more than one force, such as a pull at one end and a support force at the other. A single unopposed force can instead accelerate the object as a whole.
Elastic deformation means an object returns to its original shape or length after the deforming forces are removed. Inelastic deformation leaves a permanent change.
A spring can stretch elastically for small enough loads and become permanently deformed under a sufficiently large load. Elasticity is not a guarantee for every applied force.
Different materials and shapes deform differently. A sheet may bend and spring back under a small force, but folding can leave a permanent crease. A rubber band can be elastic even though its force–extension graph is curved (non-linear).
Extension x is stretched length minus original unstretched length. Compression is the reduction in length; equations below use the magnitude of deformation for a linear spring.Always subtract the original length before applying Hooke’s law.
The elastic limit and limit of proportionality are different. Beyond the elastic limit, removing the force leaves a permanent change in shape or length. Beyond the limit of proportionality, the force–extension graph is no longer a straight line through the origin.
Hooke’s law and spring stiffness
For a spring in its linear range, force is proportional to extension: F = kx. This is Hooke’s law. F is in N, extension x in m and spring constant k in N/m.
A larger spring constant means a stiffer spring: it needs a larger force for the same extension. k is not measured in N alone or in N m.
Rearrange to k = F/x and x = F/k. Convert centimetres or millimetres to metres before finding k in N/m.
If 4 N extends a spring by 0.02 m, k = 4/0.02 = 200 N/m. In the same linear range, an 8 N load would give 0.04 m extension.Greater load gives proportional extension only within the linear range.
The spring's restoring force opposes deformation. A hanging load at rest is supported by a spring force equal to its weight; use F = mg to convert known mass to load force.
Beyond the limit of proportionality, F is no longer directly proportional to x, so a single constant k cannot describe the whole curve using F = kx.
Force–extension graphs
For a Hooke’s-law spring, a force-versus-extension graph is a straight line through the origin. Force belongs on the vertical axis and extension on the horizontal axis for the rules here.
Its gradient is the spring constant: k = ΔF/Δx. A steeper line means a stiffer spring when the axes use the same units and scales.
A curved (non-linear) graph means equal increases in force no longer give equal increases in extension. This does not necessarily mean the object is inelastic. Remove the force and check whether it returns to its original length.
The limit of proportionality is where the graph first departs from its proportional straight-line behaviour. The elastic limit is where unloading starts leaving permanent deformation.Non-linear behaviour need not mean permanent deformation.
If plotting extension against force instead, the gradient is 1/k, not k. Always read the axes and units before using a gradient.
A straight line that does not go through the origin may indicate a measurement offset or a different chosen length reference. Direct proportionality specifically requires the origin.
Elastic potential energy
Work done stretching or compressing a spring transfers energy into its elastic potential store. For an ideal elastic spring, energy can be returned as it relaxes.
For a linear spring, E = ½kx² in joules, with k in N/m and x in m. The extension is squared; doubling extension quadruples energy at fixed k.
A 200 N/m spring extended by 0.02 m stores ½ × 200 × 0.02² = 0.04 J.
The area under a force–extension graph represents work done. For a straight line through the origin, the triangular area is ½Fx = ½kx².Area = ½ × 4 × 0.02 = 0.04 J; final force × extension would be twice too large.
Do not use final force × extension for a spring loaded from zero: force rises during stretching, so the average force is half the final force in the linear range.
For a curved force–extension graph, find work done from the area under the graph if enough data are given. Do not use ½kx² outside the straight-line range. Some work may heat the surroundings rather than being stored as elastic potential energy that can be recovered.
Spring extension and work: core practical
Suspend a spring securely from a clamp stand beside a vertical ruler. Add a pointer if possible and record the unloaded length before adding masses.
Add known masses in small steps, allow oscillations to settle and record length at eye level. Convert load mass to force using F = mg and calculate extension by subtracting unloaded length.Wait for oscillations to settle before measuring length.
Keep the same spring, ruler position and attachment points. A fixed pointer and viewing at eye level reduce parallax; select a ruler resolution suitable for small extensions.
Remove masses in steps as well and check whether the original length returns. Stop before damaging the spring unless specifically investigating its limits under supervision.
Repeat readings and plot force against extension. Determine k from the linear gradient using two well-separated points on the best-fit line rather than a single noisy reading.
Find work done from the graph area or ½kx² within the linear range. Use metres rather than centimetres to get work in joules.
Secure the stand against toppling, keep feet clear of falling masses and use eye protection where appropriate. Avoid overstretching the spring or leaving it oscillating while reading.
Pressure in gases and liquids
Pressure = force acting at right angles to a surface (normal force) ÷ area. It is measured in pascals: 1 Pa = 1 N/m². For the same force, a smaller area gives greater pressure, as with a sharp blade.
Atmospheric pressure decreases with height because there is less air above and the air is generally less dense. Pressure arises from air-particle collisions and the weight of the atmosphere, not from a solid ceiling.
Fluid pressure pushes at right angles (normal) to any surface. In a liquid at rest, pressure acts in all directions, not just downwards. Absolute liquid pressure also includes the atmospheric pressure acting on the liquid’s surface.
(Higher tier) Pressure due to a liquid column increases with density and vertical depth: Δp = ρgh (Higher tier). Use kg/m³, N/kg and metres to obtain Pa; h is vertical depth, not the length of a sloping tube.
For water with density 1000 kg/m³ at depth 2 m and g = 10 N/kg, gauge pressure is 20000 Pa. Add atmospheric pressure only if absolute pressure is requested.
(Higher tier) At the same vertical depth in a connected stationary liquid, pressure is the same regardless of container shape. A denser liquid produces a greater pressure difference for the same depth.
Upthrust and floating (Higher tier)
(Higher tier) A submerged object experiences more pressure on its lower surface than its upper surface. The resulting upward force is upthrust; pressure forces from the sides balance in a symmetric simple model.The larger lower-surface pressure gives a net upward force; force arrows are schematic.
(Higher tier) Upthrust equals the weight of fluid displaced by the object. An object partly immersed in a liquid displaces only the submerged volume; a fully immersed object displaces its full external volume.
(Higher tier) An object floats in equilibrium when upthrust equals weight. If weight exceeds maximum available upthrust it sinks; if upthrust initially exceeds weight it rises until a new balance is reached.
(Higher tier) Average density matters: a hollow boat can float even if its material is denser than water because its total mass divided by its displaced external volume can be low enough.
(Higher tier) The same fully submerged object has greater upthrust in a denser fluid. In a liquid of uniform density, putting it deeper does not increase its upthrust: pressure rises on both its top and bottom, but the difference between them stays the same.
(Higher tier) Calculate upthrust with displaced-fluid mass × g, or ρVg when density and displaced volume are known. Do not substitute the object's mass for the displaced fluid’s mass.
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