A long spring is stretched by 2 cm and the potential energy is V. IF the spring is stretched by 19 cm, its potential energy will be

Options

A. 361V/4
B. V/5
C. 5 V
D. 25 V

Answer

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

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

where:
- \( P \) is the potential energy,
- \( k \) is the spring constant,
- \( x \) is the displacement from the equilibrium position.

### Step 1: Calculate the potential energy when the spring is stretched by 2 cm.

Given:
- The spring is stretched by \( x_1 = 2 \) cm.
- The potential energy at this stretch is \( V \).

Using the formula:

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

This simplifies to:

\[ V = \frac{1}{2} k \cdot 4 = 2k \]

### Step 2: Find the potential energy when the spring is stretched by 19 cm.

Now, we need to calculate the potential energy when the spring is stretched by \( x_2 = 19 \) cm.

Using the formula again:

\[ P_2 = \frac{1}{2} k (19)^2 \]

This simplifies to:

\[ P_2 = \frac{1}{2} k \cdot 361 = \frac{361}{2} k \]

### Step 3: Substitute the value of \( k \) from the first case.

From the first case, we found that:

\[ k = \frac{V}{2} \]

Now, substituting this value of \( k \) into the equation for \( P_2 \):

\[ P_2 = \frac{361}{2} \left(\frac{V}{2}\right) \]

This simplifies to:

\[ P_2 = \frac{361V}{4} \]

### Final Answer

Thus, the potential energy when the spring is stretched by 19 cm is:

\[ P_2 = \frac{361V}{4} \]