Answer to Question #108395 in Mechanics | Relativity for Tia

Question

What is the spring constant of a spring of negligible mass which gained 7.3 joules of potential energy as a result of being compressed 0.42 meters?

Answer 1

The potential energy of a spring

$$E_p=\frac{kx^2}{2}.$$

Hence, the spring constant

$$k=\frac{2E_p}{x^2}=\frac{2\times 7.3}{0.42^2}=83\:\rm N/m.$$