Question

How much spring potential energy (in Joules) can be held inside a spring with a spring constant of 335 N/m and a maximum elastic stretch of 23.7 cm?

Answer 1

The elastic potential energy of a spring is given by:

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

Then for this spring we have:

`U=(1)/(2)(335)(0.237)^2=9.41`

Therefore the potential energy is 9.41 J (rounded to two decimals)