They go in the boxes in this order:
density
2.meter
3.matter
4.hypothesis
5.control
6.kilogram
Answer:
The elastic potential energy of the spring change during this process is 21.6 J.
Explanation:
Given that,
Spring constant of the spring, 
It extends 6 cm away from its equilibrium position.
We need to find the elastic potential energy of the spring change during this process. The elastic potential energy of the spring is given by the formula as follows :

So, the elastic potential energy of the spring change during this process is 21.6 J.
Express the distance in meters:
d= 2 cm =0.02 m
V = Ed = 50 N/C * 0.02 m = 1 N/C m = 1 V = +1.0 V