Question

A 2 $\text{kg}$ mass is attached to a spring with spring constant $ k = 200, \text{N/m} $. If the mass is displaced by $ 0.1, \text{m} $, what is the potential energy stored in the spring?

• A. \( 1 \, \text{J} \)
• B. \( 0.5 \, \text{J} \)
• C. \( 2 \, \text{J} \)
• D. \( 0.2 \, \text{J} \)

Hint:

Remember: The potential energy stored in a spring is proportional to the square of the displacement.

Answer 1

The Correct Option is A

Step 1: Use the formula for potential energy stored in a spring
The potential energy \( U \) stored in a spring is given by the formula: \[ U = \frac{1}{2} k x^2 \] where: - \( k \) is the spring constant, - \( x \) is the displacement from the equilibrium position.
Step 2: Substitute the given values
Given: - Spring constant \( k = 200 \, \text{N/m} \) - Displacement \( x = 0.1 \, \text{m} \) \[ U = \frac{1}{2} \times 200 \times (0.1)^2 = \frac{1}{2} \times 200 \times 0.01 = 1 \, \text{J} \]
Answer:
Therefore, the potential energy stored in the spring is \( 1 \, \text{J} \). So, the correct answer is option (1).