A long spring is stretched by x cm its P.E. is U. If the same spring is stretched by Nx cm the P.E. stored in it will become

Options

A. U/N
B. NU
C. `N^2U`
D. `U//N^3`

Answer

To solve the problem, we need to understand the relationship between the potential energy stored in a spring and the amount it is stretched. The potential energy (P.E.) stored in a spring can be calculated using Hooke's Law, which states that the force exerted by a spring is proportional to its displacement from the equilibrium position.

### Step-by-Step Solution:

1. **Understanding the Potential Energy Formula**:
The potential energy (U) stored in a spring when it is stretched by a distance \( x \) is given by the formula:
\[ U = \frac{1}{2} k x^2 \]
where \( k \) is the spring constant.

2. **Given Information**:
We know that when the spring is stretched by \( x \) cm, the potential energy is \( U \). Thus, we can write:
\[ U = \frac{1}{2} k x^2 \]

3. **Stretching the Spring by \( Nx \)**:
Now, if we stretch the same spring by \( Nx \) cm, we need to find the new potential energy \( U_1 \). According to the potential energy formula, we substitute \( Nx \) for \( x \):
\[ U_1 = \frac{1}{2} k (Nx)^2 \]

4. **Expanding the Expression**:
Expanding \( (Nx)^2 \):
\[ U_1 = \frac{1}{2} k N^2 x^2 \]

5. **Relating \( U_1 \) to \( U \)**:
We can relate \( U_1 \) to \( U \) by substituting the expression for \( U \):
\[ U_1 = N^2 \left(\frac{1}{2} k x^2\right) = N^2 U \]

6. **Final Result**:
Therefore, the potential energy stored in the spring when stretched by \( Nx \) cm is:
\[ U_1 = N^2 U \]

### Conclusion:
The potential energy stored in the spring when stretched by \( Nx \) cm becomes \( N^2 U \).