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JEE Main Physics question

PhysicsWork, Energy and PowerJEE Main

The PE of a certain spring when stretched from its natural length through a distance 0.3 m is 5.6 J. Find the amount of work (in joule) that must be done on this spring to stretch it through an additional distance 0.15 m.

Answer: 7

Solution

5.6 = ½k(0.3)² ⟹ k = 11.2/0.09 = 124.44 N/m. Additional work done to stretch from 0.3 m

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Question ID VK-153836