Let's analyze each scenario step-by-step to determine the elastic potential energy stored in the springs and identify which scenario generates the most energy. The formula for elastic potential energy [tex]\( U \)[/tex] stored in a spring is given by:
[tex]\[
U = \frac{1}{2} k x^2
\][/tex]
where:
- [tex]\( k \)[/tex] is the spring constant,
- [tex]\( x \)[/tex] is the compression distance.
### Scenario A
- Spring constant, [tex]\( k_A = 3 \, \frac{N}{m} \)[/tex]
- Compression distance, [tex]\( x_A = 1.0 \, m \)[/tex]
Elastic potential energy:
[tex]\[
U_A = \frac{1}{2} \times 3 \, \frac{N}{m} \times (1.0 \, m)^2 = 1.5 \, J
\][/tex]
### Scenario B
- Spring constant, [tex]\( k_B = 6 \, \frac{N}{m} \)[/tex]
- Compression distance, [tex]\( x_B = 0.8 \, m \)[/tex]
Elastic potential energy:
[tex]\[
U_B = \frac{1}{2} \times 6 \, \frac{N}{m} \times (0.8 \, m)^2 = 1.92 \, J
\][/tex]
### Scenario C
- Spring constant, [tex]\( k_C = 9 \, \frac{N}{m} \)[/tex]
- Compression distance, [tex]\( x_C = 0.6 \, m \)[/tex]
Elastic potential energy:
[tex]\[
U_C = \frac{1}{2} \times 9 \, \frac{N}{m} \times (0.6 \, m)^2 = 1.62 \, J
\][/tex]
### Scenario D
- Spring constant, [tex]\( k_D = 12 \, \frac{N}{m} \)[/tex]
- Compression distance, [tex]\( x_D = 0.4 \, m \)[/tex]
Elastic potential energy:
[tex]\[
U_D = \frac{1}{2} \times 12 \, \frac{N}{m} \times (0.4 \, m)^2 = 0.96 \, J
\][/tex]
### Summary of Elastic Potential Energies
- [tex]\( U_A = 1.5 \, J \)[/tex]
- [tex]\( U_B = 1.92 \, J \)[/tex]
- [tex]\( U_C = 1.62 \, J \)[/tex]
- [tex]\( U_D = 0.96 \, J \)[/tex]
### Conclusion
The maximum elastic potential energy is [tex]\( 1.92 \, J \)[/tex]. Therefore, Scenario B (A [tex]\(6 \, \frac{N}{m}\)[/tex] spring compressed [tex]\(0.8 \, m\)[/tex]) generates the most elastic potential energy.
