Physics · Mechanical Properties Of Solids · NEET
When you stretch a wire, the internal restoring force is NOT constant. At the start the wire is unstretched, so the force needed is almost zero. As it stretches more, the force grows steadily up to the final value F. Work = force × distance only works when force is constant. Here the force changes from 0 to F linearly, so we must use the AVERAGE force = (0 + F)/2 = ½F. That is exactly where the ½ comes from. Energy per volume u = ½ × stress × strain.
If you plot stress on the y-axis and strain on the x-axis (a stress-strain graph in the elastic region), you get a straight line through the origin. The energy density equals the AREA under this line. That area is a triangle, and the area of a triangle is ½ × base × height = ½ × strain × stress. A triangle is half of the rectangle stress × strain, which is why the answer is halved.
They are the same idea written two ways. The total energy stored in the whole wire is U = ½ × F × l (½ × load × extension). If you divide this by the wire's volume (A × L), you get the energy PER UNIT VOLUME: u = ½ × (F/A) × (l/L) = ½ × stress × strain. So U = ½ F l is the total energy, and u = ½ stress × strain is the density. Do not mix them up in NEET numerical problems — check whether the question asks for total energy (joules) or energy density (joules per cubic metre).
Yes, and this is the most common NEET form. Since stress = Y × strain (Young's modulus Y), you can substitute: u = ½ × stress × strain = ½ × Y × strain² = ½ × stress²/Y. All three forms are correct — pick whichever matches the data given in the question. The NEET 2023 problem gives Young's modulus and strain, so u = ½ Y × strain² is fastest there.
The amount of elastic potential energy per unit volume (in SI unit) of a steel wire of length 100 cm stretched by 1 mm is (Young's modulus of the wire = 2.0 × 10¹¹ N m⁻²)
When a block of mass M is suspended by a long wire of length L, the length of the wire becomes (L + l). The elastic potential energy stored in the extended wire is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Elastic energy density (energy per unit volume) is u = ½ × stress × strain. Using Young's modulus Y it also equals ½ Y × strain² or ½ × stress²/Y. All three are equivalent.
Hooke's law (stress = Y × strain) is about the FORCE-deformation relationship at a single instant. The energy formula sums up work done over the whole stretch, and because force grows from 0 to F, the total work uses the average force, adding the ½.
Total energy U = ½ F l (in joules). Energy PER UNIT VOLUME u = ½ stress × strain (in J/m³). Multiply the density by volume (A × L) to get the total, or divide the total by volume to get the density.
Joule per cubic metre (J m⁻³), which is the same as pascal (Pa) or N m⁻², because stress × strain has units of pressure (strain is unitless).
If a weight Mg is attached suddenly, gravity loses PE = Mgl, but only ½ Mgl is stored elastically. The other ½ Mgl is lost as heat, sound and vibrations. This is why the NEET 2019 answer is ½ Mgl, not Mgl.
