Question

A vertically hung spring has a spring constant of 150 newtons per meter. A 2.00-kg mass is suspended from the spring and allowed to come to rest. Calculate the elongation of the spring produced by the suspended 2.00-kg mass. Write your answer in units of m.

PLEASE HELPPP!!

Answer 1

K= 150 N/M
M= 2Kg
Fs= Fg= 19.6N
PEs= Fs X = ½ (19.610) (13M)
= 121
PE = 1/2 K x 2
= .5 (150N/M) (1. 3m)^2
=1.27J