To solve this problem, we need to understand the relationship between the spring constant and the elastic potential energy stored in a spring when it is stretched or compressed. The elastic potential energy ([tex]\(E_{PE}\)[/tex]) in a spring is given by the formula:
[tex]\[ E_{PE} = \frac{1}{2} k x^2 \][/tex]
where:
- [tex]\( k \)[/tex] is the spring constant (in N/m).
- [tex]\( x \)[/tex] is the displacement from the equilibrium position (in meters).
Given that all springs are stretched to the same distance ([tex]\(x\)[/tex]), the displacement ([tex]\( x \)[/tex]) remains constant for all springs. Therefore, the elastic potential energy is directly proportional to the spring constant [tex]\(k\)[/tex] since:
[tex]\[ E_{PE} \propto k \][/tex]
This implies that the greater the spring constant [tex]\( k \)[/tex], the greater the elastic potential energy stored in the spring.
We are given the following spring constants:
- [tex]\( k_W = 24 \, \text{N/m} \)[/tex]
- [tex]\( k_X = 35 \, \text{N/m} \)[/tex]
- [tex]\( k_Y = 22 \, \text{N/m} \)[/tex]
- [tex]\( k_Z = 15 \, \text{N/m} \)[/tex]
To determine the order of the springs based on the amount of elastic potential energy from greatest to least, we should rank them based on their spring constants in descending order.
1. [tex]\( k_X = 35 \, \text{N/m} \)[/tex]
2. [tex]\( k_W = 24 \, \text{N/m} \)[/tex]
3. [tex]\( k_Y = 22 \, \text{N/m} \)[/tex]
4. [tex]\( k_Z = 15 \, \text{N/m} \)[/tex]
So, the order of the springs based on the amount of elastic potential energy from greatest to least is:
[tex]\[ X, W, Y, Z \][/tex]
Therefore, the correct listing is:
[tex]\[ \text{\textbf{X, W, Y, Z}} \][/tex]
