Suppose that a 200g mass (0.20kg) is oscillating at the end of a spring upon a horizontal surface that is essentially friction-free. The spring can be both stretched and compressed and have a spring constant of 240N/m. It was originally stretched a distance of 12cm (0.12m) from its equilibrium (un-stretched) position prior to release What is its initial potential energy?
2 answers
PE=kA^2/2=240•(0.12)^2/2=1.728 J.
1.728J
1 answer