Question

Calculate the potential energy of a \(1.5\,\text{kg}\) block attached to a spring with \(k = 100\,\text{N/m}\) displaced by \(0.2\,\text{m}\).

• A. \(1\,\text{J}\)
• B. \(2\,\text{J}\)
• C. \(4\,\text{J}\)
• D. \(6\,\text{J}\)

Hint:

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

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The problem asks for the elastic potential energy stored in a spring when it is stretched or compressed from its equilibrium position. We are given the spring constant, the displacement, and the mass of the block attached.
Step 2: Key Formula or Approach:
The elastic potential energy (\(U\)) stored in a spring is given by Hooke's Law for potential energy:
\[ U = \frac{1}{2}kx^2 \] where \(k\) is the spring constant and \(x\) is the displacement from the equilibrium position. The mass of the block (\(m\)) is not needed to calculate the potential energy stored in the spring itself.
Step 3: Detailed Explanation:
We are given the following values:
Spring constant, \(k = 100\,\text{N/m}\)
Displacement, \(x = 0.2\,\text{m}\)
Mass of the block, \(m = 1.5\,\text{kg}\) (This is extra information)
Substitute the relevant values into the potential energy formula:
\[ U = \frac{1}{2} \times (100\,\text{N/m}) \times (0.2\,\text{m})^2 \] First, calculate the square of the displacement:
\[ (0.2)^2 = 0.04 \] Now, complete the calculation:
\[ U = \frac{1}{2} \times 100 \times 0.04 = 50 \times 0.04 = 2\,\text{J} \] Step 4: Final Answer:
The potential energy stored in the spring is \(2\,\text{J}\).