Question

1. 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:

Step 1: Find the spring constant \( k \)

Hooke's Law is \( F = kx \), where \( F \) is the force, \( k \) is the spring constant, and \( x \) is the displacement. We know \( F = 8.70 \, \text{N} \) and \( x = 0.28 \, \text{m} \). Solving for \( k \):
\( k=\frac{F}{x}=\frac{8.70}{0.28} \approx 31.07 \, \text{N/m} \)

Step 2: Determine the total displacement \( x_{\text{total}} \)

The initial stretch is \( 0.28 \, \text{m} \), and we add an additional \( 0.15 \, \text{m} \). So:
\( x_{\text{total}} = 0.28 + 0.15 = 0.43 \, \text{m} \)

Step 3: Calculate the elastic potential energy \( U \)

The formula for elastic potential energy is \( U=\frac{1}{2}kx^2 \). Substituting \( k \approx 31.07 \, \text{N/m} \) and \( x = 0.43 \, \text{m} \):
\( U=\frac{1}{2}(31.07)(0.43)^2 \approx \frac{1}{2}(31.07)(0.1849) \approx \frac{1}{2}(5.74) \approx 2.87 \, \text{J} \)

Answer:

The potential energy stored in the spring is approximately \( \boldsymbol{2.87 \, \text{J}} \) (or more precisely, follow the calculation steps for exactness).