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JEE Main Physics question

PhysicsWork, Energy and PowerJEE Main

The potential energy of a certain spring when stretched through a distance s is 10 J. The amount of work done (in joule) that must be done on this spring to stretch it through an additional distance s will be

Answer: (C)

Solution

Let PE(s) = ½ks² = 10 J. Then PE(2s) = ½k(2s)² = 4 × ½ks² = 4 × 10 = 40 J.

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Question ID VK-153538