Question

9. Derive an expression for the potential energy of an elastic stretched spring.​

Answer 1

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The Question is Give below

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Derive an expression for elastic potential energy in a stretched wire. Hooke's Law, F = -kx, where F is the force and x is the elongation. The energy stored can be found from integrating by substituting for force. The energy stored = kx2/2, where x is the final elongation.

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Answer 2

Answer:

hooke's law, F = -KX, where F is the force form integrating

by substituting for force, the energy stored=Kx square/2, where x is the final elongation