Question #100378

A helical spring is extended by a force which increase uniformally from 0 to 600 newtons. The corresponding extension of the spring is 200mm. What is the total work done?

Expert's answer

W=0.5FxW=0.5Fx

W=0.5(600)(0.2)=60 JW=0.5(600)(0.2)=60\ J


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