Hooke's law and elastic strain energy flashcards

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Hooke's law
The extension of an object is directly proportional to the force applied, provided the limit of proportionality is not exceeded.
Stiffness k
The force needed per unit extension, k = ΔF/Δx. Unit N m⁻¹ (= kg s⁻²). A property of the object, not just the material.
Extension (and compression)
Change in length: stretched length − original length (or original length − squashed length).
Limit of proportionality
The point beyond which force and extension are no longer proportional — the end of the straight section of the graph.
Elastic limit
The greatest force (or extension) for which the object returns to its original length when the force is removed.
Yield point
The point beyond which the material extends a great deal for little or no increase in force — large plastic flow begins.
Elastic deformation
A change of shape that is completely reversed when the force is removed.
Plastic deformation
A permanent change of shape that remains when the force is removed.
Elastic strain energy
The energy stored in an object because it has been deformed elastically; recoverable when the force is removed.
Hysteresis (rubber)
The unloading curve lies below the loading curve; the loop area is energy dissipated as thermal energy each cycle.
Hooke's law equation
ΔF = kΔx (Yes)
Elastic strain energy (linear region)
ΔE_el = ½FΔx (Yes)
Elastic strain energy in terms of k
ΔE_el = ½k(Δx)² (substitute F = kΔx into ½FΔx). (No — derive it)
Springs in series
1/k = 1/k₁ + 1/k₂ — same force in each, extensions add. (No — derive it)
Springs in parallel
k = k₁ + k₂ — same extension for each, forces add. (No — derive it)
Gradient of a force–extension graph (force on the y-axis)
The stiffness k, on the straight section. (With extension on the y-axis the gradient is 1/k.)
Area under a force–extension graph
The work done in stretching = elastic strain energy stored (if the deformation is elastic).
Unloading a metal wire from beyond the elastic limit
A straight line parallel to the original straight section; where it meets the extension axis gives the permanent extension.
Force–extension graph of rubber
Curved from the start (no Hooke's law region), stiffer at large extension; unloading curve below loading curve; returns to original length.
Force–compression graph of a compression spring
Straight line (Hooke's law) until the coils touch; then the force rises very steeply.
Area under a curved force–extension graph
Count squares (energy per square × number) or add trapezium strips ½(F₁ + F₂)Δx.
Why does a longer spring of the same wire have a smaller k?
Each coil stretches by the same amount for a given force, and there are more coils, so the total extension is larger.
Why are springs in series less stiff than either spring alone?
Each carries the full load, so both stretch and the extensions add.
How do you show a spring has not passed its elastic limit?
Remove the loads: the pointer returns to its original reading.
Why does a rubber band get warm when stretched and released repeatedly?
Hysteresis: less energy is returned on unloading than was put in; the difference (loop area) is dissipated as thermal energy.
Doubling the extension of a Hooke's-law spring changes the stored energy by what factor?
×4: the force also doubles, and E = ½FΔx ∝ (Δx)².
A 6.0 N force extends a spring by 24 mm. Stiffness?
k = 6.0 ÷ 0.024 = 250 N m⁻¹.
Springs of 200 N m⁻¹ and 300 N m⁻¹: combined stiffness in series and in parallel?
Series: 1/k = 1/200 + 1/300 → 120 N m⁻¹. Parallel: 500 N m⁻¹.
A spring extends 50 mm under 30 N (within the proportional region). Energy stored?
½ × 30 × 0.050 = 0.75 J.
Energy stored in a 2000 N m⁻¹ spring compressed by 4.0 cm?
½k(Δx)² = ½ × 2000 × 0.040² = 1.6 J.
A wire (k = 2.5 × 10⁴ N m⁻¹ on its straight section) is loaded to 6.0 mm at 50 N, past its elastic limit, then unloaded. Permanent extension?
Recovered = 50 ÷ 2.5 × 10⁴ = 2.0 mm; permanent = 6.0 − 2.0 = 4.0 mm.
A 500 N m⁻¹ spring compressed 2.0 cm launches a 10 g ball. Launch speed (no losses)?
E = ½ × 500 × 0.020² = 0.10 J; v = √(2 × 0.10 ÷ 0.010) = 4.5 m s⁻¹.
Units before using ½FΔx
Convert mm or cm to m first. N × mm gives mJ — an answer 1000 times too big.
Using ½FΔx on a curved graph
Wrong — ½FΔx is a triangle. For a curve, find the area by squares or strips.
Energy added between two extensions
Area of the trapezium ½(F₁ + F₂)(x₂ − x₁), or ½k(x₂² − x₁²) — not ½k(x₂ − x₁)².
Placing P, E and Y on a sketch
P at the end of the straight line; E at or just beyond P; Y further along, where the curve flattens. Never put Y on the straight part.

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