Question

b. how much elastic potential energy is stored in the spring? 6. a spring stretches 0.28 m when a force of 8.70 n is applied. assuming hooke’s law applies to the spring, calculate the potential energy stored in the spring if the length is increased by an additional 0.15 m

Explanation:

Step1: Find the spring constant \( k \)

Using Hooke's Law \( F = kx \), where \( F = 8.70 \, \text{N} \) and \( x = 0.28 \, \text{m} \). Solve for \( k \):
\( k=\frac{F}{x}=\frac{8.70}{0.28}\approx31.07 \, \text{N/m} \)

Step2: Determine the total displacement \( x_{\text{total}} \)

Initial stretch: \( 0.28 \, \text{m} \), additional stretch: \( 0.15 \, \text{m} \).
\( x_{\text{total}} = 0.28 + 0.15 = 0.43 \, \text{m} \)

Step3: Calculate elastic potential energy \( U \)

Use the formula \( U=\frac{1}{2}kx^2 \). Substitute \( k \approx 31.07 \, \text{N/m} \) and \( x = 0.43 \, \text{m} \):
\( U=\frac{1}{2}(31.07)(0.43)^2 \approx\frac{1}{2}(31.07)(0.1849) \approx 2.86 \, \text{J} \)

Answer:

The elastic potential energy stored in the spring is approximately \( \boldsymbol{2.86 \, \text{J}} \) (or \( 2.9 \, \text{J} \) with rounding during steps).