Physics · Mechanical Properties Of Solids · NEET
Because the restoring force is not constant. When the wire is un-stretched the internal force is zero, and it grows in a straight line up to the maximum value F as the wire reaches full extension l. So the work done is the area under the force-vs-extension graph, which is a triangle: area = ½ × base × height = ½ × l × F. That is why U = ½ F l, not F l. Use the AVERAGE force (F/2), not the final force.
Half of it. U = ½ × F × l. A very common mistake is to write U = F × l because that is how work is calculated for a constant force. Here the force builds up gradually from 0 to F, so the correct work (and stored energy) is ½ F l. Only the AVERAGE force (F/2) acts over the full extension l.
The load Mg does work Mgl (it falls by l), but only HALF of that, ½Mgl, is stored as elastic potential energy in the wire. The other half is lost as heat and vibration when the mass settles. This is a classic NEET trap: the potential energy LOST by the falling mass (Mgl) is not the same as the elastic energy STORED in the wire (½Mgl).
For a small extra stretch dl, the force is F = YAl/L, so small work dW = F dl = (YA/L) l dl. Integrate from 0 to l: W = (YA/L) × l²/2 = ½ (YA/L) l². Since F = YAl/L, this equals ½ F l. This is the exact NCERT derivation in section 8.5.5.
Total stored energy is U = ½ F l (in joules) for the whole wire. Energy per unit volume (energy density) is u = U / (A×L) = ½ × stress × strain (in J/m³). Divide the total energy by the wire's volume (A×L) to get the density. The density form is used when the question gives Young's modulus and strain but not the actual force.
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.
U = ½ × F × l = ½ × (YA/L) × l², where F is the stretching force, l the extension, Y the Young's modulus, A the area of cross-section and L the original length. It can also be written as U = ½ × stress × strain × volume.
It comes from the work you do against the inter-atomic forces while stretching the wire. This work is stored inside the wire and is released when the force is removed and the wire springs back to its original length.
Yes, as long as the wire obeys Hooke's law (the force is proportional to extension). Then the force-extension graph is a straight line and the stored energy equals the triangular area ½ F l.
The joule (J), because it is an energy. Energy per unit volume (energy density) has the unit joule per cubic metre (J/m³).
