Question

If it takes 5.00 J of work to stretch a Hooke’s-law spring 10.0 cm from its unstressed length, determine the extra work required to stretch it an additional 15.0 cm.

Answer 1

$$W=0.5kx^2\\5=0.5(0.1)^2k\\k=1000\frac{N}{m}\\W'=0.5kx'^2=0.5(1000)(0.15)^2=11.25\ J$$