Question

A \(1.5\,\text{kg}\) block is attached to a spring with spring constant \(k=100\,\text{N m}^{-1}\) and displaced by \(0.2\,\text{m}\). Calculate the potential energy stored in the spring.

• A. \(1\,\text{J}\)
• B. \(2\,\text{J}\)
• C. \(3\,\text{J}\)
• D. \(4\,\text{J}\)

Hint:

Potential energy stored in a spring depends only on displacement and spring constant: \[ U=\frac{1}{2}kx^2 \] The mass of the block does \textbf{not affect} the elastic potential energy.

Answer 1

The Correct Option is B

Concept: The potential energy stored in a spring is given by the formula \[ U = \frac{1}{2}kx^2 \] where

• \(k\) = spring constant
• \(x\) = displacement from equilibrium position

Step 1: {Write the given values.} \[ k = 100\,\text{N m}^{-1}, \qquad x = 0.2\,\text{m} \] 
Step 2: {Substitute values into the formula.} \[ U = \frac{1}{2} \times 100 \times (0.2)^2 \] 
Step 3: {Calculate the value.} \[ (0.2)^2 = 0.04 \] \[ U = 50 \times 0.04 \] \[ U = 2\,\text{J} \] Thus, the potential energy stored in the spring is \(2\,\text{J}\). Hence, the correct option is (B).