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Density and Hooke's law
Two ideas start the materials story: density, which says how much matter a material packs into each cubic metre, and Hooke's law, which says how a material fights back when stretched, at least for a while. Where the law gives out is as important as where it holds.
IN THIS TOPIC
- Use ρ = m/V, including the conversion between g cm−3 and kg m−3.
- Apply F = kΔL up to the limit of proportionality, and distinguish that limit from the elastic limit.
- Find the energy stored from the area under a force-extension graph, and describe what changes past the elastic limit.
WHAT YOU PROBABLY THINK
Hooke's law holds right up until the spring snaps.
Density
The density of a material is its mass per unit volume:
measured in kg m−3. It is a property of the material, not of any particular lump of it: cut a brick in half and the mass and volume both halve, leaving the ratio untouched. Water sits at 1000 kg m−3, a benchmark worth carrying, and materials denser than the fluid they sit in sink while less dense ones float.
Handbooks often quote densities in g cm−3, and the conversion is a factor students misplace constantly: 1 g cm−3 = 1000 kg m−3, because a kilogram is a thousand grams while a cubic metre is a million cubic centimetres. Convert before substituting, never after.
Hooke's law, and where it stops
Load a spring or a wire and it extends. For modest loads the response is beautifully simple, Hooke's law: the force is proportional to the extension,
where k, the spring constant or stiffness, is measured in N m−1 and belongs to that particular object. Hooke's law applies to compression as well as stretching.
The law fails in two stages, and the exam separates them. The limit of proportionality is where the graph stops being straight. The elastic limit, usually a little beyond it, is where deformation stops being reversible: load past this point and the object no longer returns to its original length when released.
The energy stored
Stretching a spring transfers energy into it, the elastic strain energy, equal to the area under the force-extension graph. While the graph is straight that area is a triangle:
and substituting F = kΔL gives the equivalent form 12k(ΔL)2, which is worth learning: it says doubling the extension quadruples the stored energy. When the graph curves, the half-formula fails but the area does not, so count squares under the curve instead.
That stored energy is fully recoverable while the deformation is elastic, and energy conservation follows it around. Release a catapult and the strain energy becomes kinetic energy of the projectile; fire it upwards and the kinetic energy becomes gravitational potential energy in turn.
Past the elastic limit
Load beyond the elastic limit and the material deforms plastically: its internal structure rearranges permanently. Unload it and the contraction follows a line parallel to the original elastic section, but displaced, meeting the extension axis away from the origin.
That intercept is the permanent extension. The energy accounting changes too: the work done during loading exceeds the energy recovered during unloading, and the difference, the area between the two curves, has gone into rearranging the material and warming it. Energy put into plastic deformation does not come back.
THE EXAM BIT
- 1 g cm−3 = 1000 kg m−3. This conversion appears constantly and is missed constantly; a density answer of 2.7 when the data sheet world works in thousands should ring an alarm.
- Limit of proportionality and elastic limit are different points: the first is where the graph stops being straight, the second where deformation stops being reversible. Exam questions test the distinction directly.
- Strain energy is the area under the force-extension graph. Use 12FΔL only while the line is straight; on a curve, count squares.
- Read a permanent extension where the unloading line meets the extension axis, and expect it to run parallel to the original straight section.
- Volumes convert with the cube: 1 cm3 is 10−6 m3. Convert lengths first and cube afterwards.
CHECK YOURSELF
A spring with spring constant 25 N m−1 is stretched by 0.20 m, within its limit of proportionality. Find (a) the force applied and (b) the energy stored. (c) If the extension were doubled, still within the limit, what would the stored energy become?
Show a hint
Energy depends on the square of the extension.
Show the answer
(a) F = kΔL = 25 × 0.20 = 5.0 N.
(b) E = 12k(ΔL)2 = ½ × 25 × 0.202 = 0.50 J.
(c) Doubling ΔL quadruples the energy, because of the square: 2.0 J. The graph shows why: the triangle under the line doubles in both base and height.
Straight line: use ½FΔL.
Curved: the energy is still the area.
No animated video for this topic yet; these notes stand alone. InkPhysics on YouTube.