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1} When a load of 10 kg is hung from the wire, then extension of 2 m is produced. Then work done by restoring force is_____________?
F = -kx is the restoring force k is the spring constant => load of 10 kg weighs 10g N => 10(9.81) = k(2) => k = 49.05 N/m therefore work done = forceₐᵥₑ x displacement in direction of force here forceₐᵥₑ(kx/2) and distance moved are oppositely directed W =[ -49.05 + 0] *2*2/2 = - 98.1 J this work done is stored in the wire as elastic potential energy(½kx²)
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