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CBSE Class 11-science Answered

Q.Derive an expression of elastic strain energy gained by a stretched wire. What happens to this energy when deforming force is withdrawn? State hooke's law and represent it graphically.
Asked by neerajprataphazarika | 24 Sep, 2018, 10:52: PM
answered-by-expert Expert Answer
Let us assume Force F is applied to stretch the wire of length so that elongation l is taking place.
 
Workdone W = Stored potential energy = F×l.........................(1)
 
Young's modulus Y of material of the wire is defined as,  begin mathsize 12px style Y space equals fraction numerator begin display style bevelled F over A end style over denominator begin display style bevelled l over L end style end fraction end style...................(2)
where A is area of cross section. Frome eqn.(2), we write that F = Y×A×(l/L) ..................(3)
 
From eqn.(1) and eqn.(3),  W = Y×(A/L)×l2 ......................(4)
 
(A/L) can be written in terms of mass of wire m and density ρ,  as    (A/L) = M/(ρL2) ....................(5)
 
hence using eqn.(4) and (5),  workdone W = Y×(M/ρ)×( l/L )2 
 
workdone by the force is stored as potential energy.
when the applied force is withdrawn, stored energy is utilised to restore the original condition of wire.
 
Hooks law :- Within the proportional limit or elstic limit, for small deformations, the stress and strain are proportional to each other.
 
Answered by Thiyagarajan K | 25 Sep, 2018, 11:06: AM
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