Question

A spring with a spring constant of 45 N/m is pulled 1.4 m away from its equilibrium position. How much potential energy is stored in the spring?

Answer 1

Elastic potential energy is given by:

$E =\dfrac{1}{2}kx^2$

Where k is the spring constant in N/m and x is displacement from equilibrium position in m. Evaluating:

$E =\dfrac{1}{2}kx^2 = \dfrac{1}{2}(45\;N/m)(1.4\;m)^2 = 44.1\;J$

A/ The potential energy is stored in the spring is 44.1 J.