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

Step 1: Understanding the Question:
The question asks us to calculate the elastic potential energy stored in a spring when it is displaced from its equilibrium position.
We are provided with the spring constant (\(k\)), the displacement (\(x\)), and the mass of the block attached to it.
Step 2: Key Formula or Approach:
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 and \(x\) is the displacement from the equilibrium position. The mass of the block is not needed to calculate the potential energy stored in the spring itself.
Step 3: Detailed Explanation:
First, let's identify the given values from the problem statement:
- Spring constant, \(k = 100\,\text{N m}^{-1}\)
- Displacement, \(x = 0.2\,\text{m}\)
- Mass of the block, \(m = 1.5\,\text{kg}\) (This is extraneous information for this specific calculation).
Next, we substitute the relevant values into the potential energy formula:
\[ U = \frac{1}{2} \times (100\,\text{N m}^{-1}) \times (0.2\,\text{m})^2 \] Now, we perform the calculation step-by-step:
First, calculate the square of the displacement:
\[ (0.2)^2 = 0.04 \] Then, substitute this back into the equation:
\[ U = \frac{1}{2} \times 100 \times 0.04 \] \[ U = 50 \times 0.04 \] \[ U = 2\,\text{J} \] Thus, the potential energy stored in the spring is \(2\,\text{J}\).
Step 4: Final Answer:
The calculated potential energy is \(2\,\text{J}\), which matches option (B).