Question

A mass of \( 0.5 \, \text{kg} \) is attached to a spring with a spring constant \( k = 200 \, \text{N/m} \). The mass is displaced by \( 0.1 \, \text{m} \) from its equilibrium position. What is the potential energy stored in the spring?

• A. \( 1 \, \text{J} \)
• B. \( 0.5 \, \text{J} \)
• C. \( 2 \, \text{J} \)
• D. \( 0.25 \, \text{J} \)

Hint:

For potential energy stored in a spring, use the formula \( U = \frac{1}{2} k x^2 \), where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position.

Answer 1

The Correct Option is A

We are given a mass of \( m = 0.5 \, \text{kg} \) attached to a spring with a spring constant \( k = 200 \, \text{N/m} \). The mass is displaced by \( x = 0.1 \, \text{m} \) from its equilibrium position. We need to calculate the potential energy stored in the spring. Step 1: Recall the formula for potential energy stored in a spring The potential energy \( U \) stored in a spring is given by Hooke's Law: \[ U = \frac{1}{2} k x^2 \] Where: - \( k \) is the spring constant, - \( x \) is the displacement from the equilibrium position. Step 2: Substitute the given values into the formula Substitute \( k = 200 \, \text{N/m} \) and \( x = 0.1 \, \text{m} \) into the formula: \[ U = \frac{1}{2} \times 200 \times (0.1)^2 \] \[ U = \frac{1}{2} \times 200 \times 0.01 = 1 \, \text{J} \] Answer: The potential energy stored in the spring is \( 1 \, \text{J} \), so the correct answer is option (1).