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Q.Obtain an expression for the potential energy of a spring having spring constant k when it is extended through a length xx.

Assam AhsecAHSEC Higher Secondary (HS) 1st Year Examination 2019Subjective· 3mImportance★★★★★
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Work done against the spring's restoring force, integrated from 0 to x, gives U=12kx2U=\tfrac12kx^2.

By Hooke's law, when a spring of spring constant kk is stretched by a small displacement x′x' from its natural length, the restoring force it exerts is F=−kx′F=-kx' (opposing the extension). To stretch it further by dx′dx', an external agent must do work equal in magnitude to kx′ dx′kx'\,dx' against this restoring force.

Total work done in stretching the spring from 0 to xx: …

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