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Sagot :
To find which scenario generates the most elastic potential energy, we will calculate the elastic potential energy for each scenario using the formula:
[tex]\[ U = \frac{1}{2} k x^2 \][/tex]
where:
- [tex]\( U \)[/tex] is the elastic potential energy,
- [tex]\( k \)[/tex] is the spring constant, and
- [tex]\( x \)[/tex] is the compression distance.
### Scenario A
For a spring with a spring constant [tex]\( k_a = 3 \frac{N}{m} \)[/tex] compressed a distance [tex]\( x_a = 1.0 \)[/tex] m:
[tex]\[ U_a = \frac{1}{2} k_a x_a^2 \][/tex]
[tex]\[ U_a = \frac{1}{2} \times 3 \times (1.0)^2 \][/tex]
[tex]\[ U_a = \frac{3}{2} \times 1 \][/tex]
[tex]\[ U_a = 1.5 \text{ Joules} \][/tex]
### Scenario B
For a spring with a spring constant [tex]\( k_b = 6 \frac{N}{m} \)[/tex] compressed a distance [tex]\( x_b = 0.8 \)[/tex] m:
[tex]\[ U_b = \frac{1}{2} k_b x_b^2 \][/tex]
[tex]\[ U_b = \frac{1}{2} \times 6 \times (0.8)^2 \][/tex]
[tex]\[ U_b = 3 \times 0.64 \][/tex]
[tex]\[ U_b = 1.92 \text{ Joules} \][/tex]
### Scenario C
For a spring with a spring constant [tex]\( k_c = 9 \frac{N}{m} \)[/tex] compressed a distance [tex]\( x_c = 0.6 \)[/tex] m:
[tex]\[ U_c = \frac{1}{2} k_c x_c^2 \][/tex]
[tex]\[ U_c = \frac{1}{2} \times 9 \times (0.6)^2 \][/tex]
[tex]\[ U_c = 4.5 \times 0.36 \][/tex]
[tex]\[ U_c = 1.62 \text{ Joules} \][/tex]
### Scenario D
For a spring with a spring constant [tex]\( k_d = 12 \frac{N}{m} \)[/tex] compressed a distance [tex]\( x_d = 0.4 \)[/tex] m:
[tex]\[ U_d = \frac{1}{2} k_d x_d^2 \][/tex]
[tex]\[ U_d = \frac{1}{2} \times 12 \times (0.4)^2 \][/tex]
[tex]\[ U_d = 6 \times 0.16 \][/tex]
[tex]\[ U_d = 0.96 \text{ Joules} \][/tex]
### Compare the Energies
Now that we have the elastic potential energies for each scenario:
- [tex]\( U_a = 1.5 \)[/tex] Joules
- [tex]\( U_b = 1.92 \)[/tex] Joules
- [tex]\( U_c = 1.62 \)[/tex] Joules
- [tex]\( U_d = 0.96 \)[/tex] Joules
The scenario with the most elastic potential energy is Scenario B with [tex]\( 1.92 \)[/tex] Joules.
Thus, Scenario B generates the most elastic potential energy.
[tex]\[ U = \frac{1}{2} k x^2 \][/tex]
where:
- [tex]\( U \)[/tex] is the elastic potential energy,
- [tex]\( k \)[/tex] is the spring constant, and
- [tex]\( x \)[/tex] is the compression distance.
### Scenario A
For a spring with a spring constant [tex]\( k_a = 3 \frac{N}{m} \)[/tex] compressed a distance [tex]\( x_a = 1.0 \)[/tex] m:
[tex]\[ U_a = \frac{1}{2} k_a x_a^2 \][/tex]
[tex]\[ U_a = \frac{1}{2} \times 3 \times (1.0)^2 \][/tex]
[tex]\[ U_a = \frac{3}{2} \times 1 \][/tex]
[tex]\[ U_a = 1.5 \text{ Joules} \][/tex]
### Scenario B
For a spring with a spring constant [tex]\( k_b = 6 \frac{N}{m} \)[/tex] compressed a distance [tex]\( x_b = 0.8 \)[/tex] m:
[tex]\[ U_b = \frac{1}{2} k_b x_b^2 \][/tex]
[tex]\[ U_b = \frac{1}{2} \times 6 \times (0.8)^2 \][/tex]
[tex]\[ U_b = 3 \times 0.64 \][/tex]
[tex]\[ U_b = 1.92 \text{ Joules} \][/tex]
### Scenario C
For a spring with a spring constant [tex]\( k_c = 9 \frac{N}{m} \)[/tex] compressed a distance [tex]\( x_c = 0.6 \)[/tex] m:
[tex]\[ U_c = \frac{1}{2} k_c x_c^2 \][/tex]
[tex]\[ U_c = \frac{1}{2} \times 9 \times (0.6)^2 \][/tex]
[tex]\[ U_c = 4.5 \times 0.36 \][/tex]
[tex]\[ U_c = 1.62 \text{ Joules} \][/tex]
### Scenario D
For a spring with a spring constant [tex]\( k_d = 12 \frac{N}{m} \)[/tex] compressed a distance [tex]\( x_d = 0.4 \)[/tex] m:
[tex]\[ U_d = \frac{1}{2} k_d x_d^2 \][/tex]
[tex]\[ U_d = \frac{1}{2} \times 12 \times (0.4)^2 \][/tex]
[tex]\[ U_d = 6 \times 0.16 \][/tex]
[tex]\[ U_d = 0.96 \text{ Joules} \][/tex]
### Compare the Energies
Now that we have the elastic potential energies for each scenario:
- [tex]\( U_a = 1.5 \)[/tex] Joules
- [tex]\( U_b = 1.92 \)[/tex] Joules
- [tex]\( U_c = 1.62 \)[/tex] Joules
- [tex]\( U_d = 0.96 \)[/tex] Joules
The scenario with the most elastic potential energy is Scenario B with [tex]\( 1.92 \)[/tex] Joules.
Thus, Scenario B generates the most elastic potential energy.
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