Question
Doubling which quantity would cause the largest increase in a spring’s elastic potential energy?
displacement
spring constant
mass
gravitational acceleration
displacement
spring constant
mass
gravitational acceleration
Answers
GPT-4o mini
The elastic potential energy \( U \) stored in a spring is given by the formula:
\[
U = \frac{1}{2} k x^2
\]
where:
- \( U \) is the elastic potential energy,
- \( k \) is the spring constant, and
- \( x \) is the displacement from the spring's equilibrium position.
Let's assess the impact of doubling each quantity:
1. **Displacement (\( x \))**: If we double the displacement \( x \) from \( x \) to \( 2x \):
\[
U_{\text{new}} = \frac{1}{2} k (2x)^2 = \frac{1}{2} k (4x^2) = 2 U
\]
Doubling the displacement quadruples the elastic potential energy.
2. **Spring Constant (\( k \))**: If we double the spring constant \( k \) from \( k \) to \( 2k \):
\[
U_{\text{new}} = \frac{1}{2} (2k) x^2 = k x^2 = 2U
\]
Doubling the spring constant doubles the elastic potential energy.
3. **Mass**: The mass does not directly affect the elastic potential energy, as it is not part of the formula. Therefore, doubling the mass does not change the elastic potential energy \( U \).
4. **Gravitational Acceleration (\( g \))**: Similar to mass, gravitational acceleration does not appear in the spring elastic potential energy formula, so doubling \( g \) would also not affect \( U \).
From these evaluations, doubling the **displacement** causes the largest increase in the spring's elastic potential energy, increasing it by a factor of four.
\[
U = \frac{1}{2} k x^2
\]
where:
- \( U \) is the elastic potential energy,
- \( k \) is the spring constant, and
- \( x \) is the displacement from the spring's equilibrium position.
Let's assess the impact of doubling each quantity:
1. **Displacement (\( x \))**: If we double the displacement \( x \) from \( x \) to \( 2x \):
\[
U_{\text{new}} = \frac{1}{2} k (2x)^2 = \frac{1}{2} k (4x^2) = 2 U
\]
Doubling the displacement quadruples the elastic potential energy.
2. **Spring Constant (\( k \))**: If we double the spring constant \( k \) from \( k \) to \( 2k \):
\[
U_{\text{new}} = \frac{1}{2} (2k) x^2 = k x^2 = 2U
\]
Doubling the spring constant doubles the elastic potential energy.
3. **Mass**: The mass does not directly affect the elastic potential energy, as it is not part of the formula. Therefore, doubling the mass does not change the elastic potential energy \( U \).
4. **Gravitational Acceleration (\( g \))**: Similar to mass, gravitational acceleration does not appear in the spring elastic potential energy formula, so doubling \( g \) would also not affect \( U \).
From these evaluations, doubling the **displacement** causes the largest increase in the spring's elastic potential energy, increasing it by a factor of four.