JEE Main Physics question
The potential energy of a certain spring when stretched through a distance 'S' is 10 joule. The amount of work that must be done on this spring to stretch it through an additional distance 'S' will be
Answer: (A)
Solution
½kS² = 10 J (given). Additional work to stretch from S to 2S: W = ½k[(2S)² − S²] = 3 × (½kS²) = 3 × 10 = 30 J.