Question

When a spring is stretched by 2 cm, it stores 100 J of energy. If it is stretched further by 2 cm, the stored energy will be increased by

• A. 100 J
• B. 200 J
• C. 300 J
• D. 400 J

Hint:

Doubling extension quadruples stored energy.

Answer 1

The Correct Option is C

To solve the given problem, we need to understand the behavior of a spring and how energy is stored in it when stretched. The potential energy stored in a spring when it is stretched or compressed is given by the formula:

\(E = \frac{1}{2} k x^2\)

where:

• \(E\) is the potential energy stored in the spring.
• \(k\) is the spring constant.
• \(x\) is the displacement from the equilibrium position (stretch or compression).

In the given scenario:

• When the spring is stretched by 2 cm, it stores 100 J of energy.

Using the energy formula:

\(100 = \frac{1}{2} k (0.02)^2\)

Solving for \(k\), we have:

\(100 = \frac{1}{2} k \times 0.0004\)

\(k = \frac{100 \times 2}{0.0004} = 500,000 \, \text{N/m}\)

Now, if the spring is stretched further by another 2 cm (total displacement = 4 cm or 0.04 m), the energy stored becomes:

\(E_{\text{total}} = \frac{1}{2} \times 500,000 \times (0.04)^2\)

Calculating the total energy:

\(E_{\text{total}} = \frac{1}{2} \times 500,000 \times 0.0016 = 400 \, \text{J}\)

The increase in stored energy due to the additional 2 cm stretch is:

\(E_{\text{increase}} = 400 - 100 = 300 \, \text{J}\)

Thus, the stored energy is increased by \(300 \, \text{J}\) when the spring is stretched further by 2 cm. Therefore, the correct answer is:

• 300 J