Question

A 1.5 kg block is placed on a frictionless surface and attached to a spring with a spring constant of 100 N/m. If the block is displaced by 0.2 m from its equilibrium position, what is the potential energy stored in the spring?

• A. \( 2.0 \, \text{J} \)
• B. \( 1.0 \, \text{J} \)
• C. \( 0.5 \, \text{J} \)
• D. \( 3.0 \, \text{J} \)

Hint:

The potential energy in a spring depends on the square of the displacement from the equilibrium position. Always use the correct spring constant and displacement in your calculations.

Answer 1

The Correct Option is A

Given:

• Mass of the block: \( m = 1.5 \, \text{kg} \) 
• Spring constant: \( k = 100 \, \text{N/m} \)
• Displacement from equilibrium: \( x = 0.2 \, \text{m} \)

Step 1: Use the formula for the potential energy stored in a spring

The potential energy stored in a spring is given by Hooke's law: \[ U = \frac{1}{2} k x^2 \] where: - \( U \) is the potential energy stored in the spring, - \( k \) is the spring constant, - \( x \) is the displacement from the equilibrium position.

Step 2: Substitute the given values into the formula

\[ U = \frac{1}{2} \times 100 \, \text{N/m} \times (0.2 \, \text{m})^2 \] \[ U = \frac{1}{2} \times 100 \times 0.04 = 2 \, \text{J} \]

✅ Final Answer:

The potential energy stored in the spring is \( \boxed{2.0 \, \text{J}} \).