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Q.(a) State Hooke's law of elasticity.

(b) Derive an expression for potential energy stored in a stretched spring.
(c) The length of a steel wire increases by 0.5 cm when it is loaded with a mass of 5 kg. Calculate the work done in stretching the wire.
Kerala DhseKerala DHSE Plus One Board 2026Subjective· 5mImportance★★★★★
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Hooke's law: stress proportional to strain within the elastic limit; spring PE = (1/2) k x^2; work to stretch the wire = (1/2) x load x extension = 0.1225 J.

  1. Hooke's law of elasticity: Within the elastic limit, the stress developed in a body is directly proportional to the strain produced. That is, stress / strain = constant = modulus of elasticity (E).
  2. Potential energy stored in a stretched spring: When a spring is stretched by x, the restoring force at extension x' is F = k x' (Hooke's law). The work done in stretching it from 0 to x is U = integral from 0 to x of (k x') dx' = (1/2) k x^2. This work is stored as elastic potential energy: U = (1/2) k x^2.
  3. Work done in stretching the steel wire: …

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