JEE Main Physics question
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: (A)
Solution
(Same as Q4.) PE(2s) = 4×10 = 40 J. Additional work = 40−10 = 30 J.