When a long spring is stretched by 2 cm, its potential energy is V. If the spring is stretched by 10 cm, the potential energy in it will be

Options

A. 10 V
B. 25 V
C. `V/5`
D. 5V

Explanation

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 (PE) stored in a spring is given by the formula:

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

where:
- \( k \) is the spring constant,
- \( x \) is the displacement from the equilibrium position (stretching of the spring).

### Step-by-Step Solution:

1. **Identify the initial conditions**:
When the spring is stretched by \( x_1 = 2 \) cm, the potential energy is given as \( V \).

\[ V = \frac{1}{2} k (2)^2 \]

2. **Calculate the potential energy for the second condition**:
Now, we need to find the potential energy when the spring is stretched by \( x_2 = 10 \) cm.

\[ PE_2 = \frac{1}{2} k (10)^2 \]

3. **Relate the two potential energies**:
Since we know that potential energy is proportional to the square of the displacement, we can set up a ratio:

\[ \frac{V}{PE_2} = \frac{x_1^2}{x_2^2} \]

Plugging in the values:

\[ \frac{V}{PE_2} = \frac{(2)^2}{(10)^2} = \frac{4}{100} = \frac{1}{25} \]

4. **Solve for \( PE_2 \)**:
Rearranging the ratio gives us:

\[ PE_2 = 25V \]

### Final Answer:
Thus, the potential energy in the spring when stretched by 10 cm is \( 25V \).

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