Elastic potential energy is given by:
[tex]E =\dfrac{1}{2}kx^2[/tex]
Where k is the spring constant in N/m and x is displacement from equilibrium position in m. Evaluating:
[tex]E =\dfrac{1}{2}kx^2 = \dfrac{1}{2}(45\;N/m)(1.4\;m)^2 = 44.1\;J[/tex]
A/ The potential energy is stored in the spring is 44.1 J.
