The potential energy of a spring when stretched through a distance x is 10 J. What is the amount of work done on the same spring stretch it through an additional distance T?

Potential energy of the spring when stretched through a distance x.U=12kx2=10JWhen x becomes 2x, the potential energy will beU′=12k(2x)2=4×12kx2=4×10 = 40 J∴ Work done U′−U=40−10=30J