The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8 cm the potential energy stored in it is :

Options

A. `4U`
B. `8U`
C. `16U`
D. `(U)/(4)`

Answer

To solve the problem, we will use the formula for the potential energy stored in a spring, which is given by:

\[ U = \frac{1}{2} k x^2 \]

where:
- \( U \) is the potential energy,
- \( k \) is the spring constant,
- \( x \) is the displacement (or extension) of the spring from its equilibrium position.

### Step-by-Step Solution:

1. **Identify the potential energy for the first case:**
When the spring is stretched by 2 cm, the potential energy is given as \( U \).
\[ U = \frac{1}{2} k (2)^2 \]
\[ U = \frac{1}{2} k \cdot 4 = 2k \]

2. **Calculate the potential energy for the second case:**
Now, when the spring is stretched by 8 cm, we need to find the new potential energy \( U' \).
\[ U' = \frac{1}{2} k (8)^2 \]
\[ U' = \frac{1}{2} k \cdot 64 = 32k \]

3. **Relate the two potential energies:**
We have \( U = 2k \) and \( U' = 32k \). To find the relationship between \( U' \) and \( U \), we can express \( U' \) in terms of \( U \):
\[ U' = \frac{32k}{2k} \cdot U = 16U \]

4. **Conclusion:**
Therefore, the potential energy stored in the spring when stretched by 8 cm is:
\[ U' = 16U \]

### Final Answer:
The potential energy stored in the spring when stretched by 8 cm is \( 16U \).

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