Answer to Question #177195 in Physics for Pushpa

Question

Determine the extension of the spring when it has a spring constant k= 60 N/m with 25 J of potential energy.

Answer 1

$$PE=\dfrac{1}{2}kx^2,$$ $$x=\sqrt{\dfrac{2PE}{k}}=\sqrt{\dfrac{2\cdot25\ J}{60\ \dfrac{N}{m}}}=0.91\ m.$$