The work done by external agent in stretching a spring of force constant k = 100 N/m from deformation x1 = 10 to deformation x2 = 20 cm.

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answer is 3.

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Detailed Solution

U2 = 12kx22U1 = 12kx21Wext = ∆U = U2-U1 = 12kx22-12kx12 = 12k(x2 2-x12) = 12×100×((20100)2-(10100)2) =150 J

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