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 =(1)/(2)kx^2`

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

`E =(1)/(2)kx^2 = (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.