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

Concept: The potential energy stored in a spring is given by the formula: \[ U = \frac{1}{2}kx^2 \] where \(k\) = spring constant and \(x\) = displacement from equilibrium.

Step 1: Substitute the given values. \[ k = 100\,\text{N/m}, \quad x = 0.2\,\text{m} \] \[ U = \frac{1}{2}\times100\times(0.2)^2 \]

Step 2: Compute the value. \[ (0.2)^2 = 0.04 \] \[ U = \frac{1}{2}\times100\times0.04 \] \[ U = 2\,\text{J} \] Thus, the potential energy stored in the spring is \[ \boxed{2\,\text{J}} \]