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Q.Explain the concept of elastic potential energy in a stretched wire and hence obtain the expression for it.

Andhra Pradesh BieapBIEAP Intermediate Board (1st Year) 2018Subjective· 4mImportance★★★★★
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Stretching a wire stores elastic PE equal to (1/2) x stress x strain x volume, or U = (1/2) x load x elongation.

Concept:

When an external force stretches a wire, work is done against the internal restoring (elastic) forces. Since the deformation is elastic, this work is not lost but is stored in the wire as elastic potential energy. When the force is removed, this stored energy is released and the wire regains its original length.

Derivation of the expression:

Let a wire of length L and area of cross-section A be stretched by a force. When the extension is x, the restoring force (within the elastic limit, obeying Hooke's law) is proportional to x:

F = (Y A / L) x

where Y is Young's modulus.

The force increases from 0 (at x = 0) to F (at extension l). To extend the wire by a further small amount dx, the work done is:

dW = F dx = (Y A / L) x dx

Total work done in stretching the wire to a final extension l:

W = integral from 0 to l of (Y A / L) x dx = (Y A / L) (l^2 / 2) = (1/2) (Y A l / L) l

But the final force F = (Y A / L) l, so:

W = (1/2) F l

This work is stored as elastic potential energy: …

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