A long spring is stretched by 2 cm, its potential energy is U. IF the spring is stretched by 10 cm, the potential energy stored in it will be

Options

A. U/25
B. U/5
C. 5 U
D. 25 U

Explanation

To solve the problem of potential energy stored in a spring when stretched, we can follow these steps:

### Step 1: Understand the formula for potential energy in a spring
The potential energy (U) stored in a spring when it is stretched or compressed is given by the formula:
\[ U = \frac{1}{2} k x^2 \]
where \( k \) is the spring constant and \( x \) is the elongation (displacement from the equilibrium position).

### Step 2: Calculate the potential energy when the spring is stretched by 2 cm
Given that when the spring is stretched by 2 cm, the potential energy is \( U \):
- Convert 2 cm to meters: \( x_1 = 0.02 \, \text{m} \)
- Substitute into the formula:
\[ U = \frac{1}{2} k (0.02)^2 = \frac{1}{2} k (0.0004) = 0.0002 k \]
This gives us:
\[ U = 0.0002 k \]

### Step 3: Calculate the potential energy when the spring is stretched by 10 cm
Now, we need to find the potential energy when the spring is stretched by 10 cm:
- Convert 10 cm to meters: \( x_2 = 0.10 \, \text{m} \)
- Substitute into the formula:
\[ U' = \frac{1}{2} k (0.10)^2 = \frac{1}{2} k (0.01) = 0.005 k \]
This gives us:
\[ U' = 0.005 k \]

### Step 4: Relate the two potential energies
Now, we can relate \( U' \) to \( U \):
- From Step 2, we have \( U = 0.0002 k \)
- From Step 3, we have \( U' = 0.005 k \)

To find the ratio of \( U' \) to \( U \):
\[ \frac{U'}{U} = \frac{0.005 k}{0.0002 k} = \frac{0.005}{0.0002} = 25 \]

### Step 5: Express \( U' \) in terms of \( U \)
Thus, we can express \( U' \) as:
\[ U' = 25 U \]

### Conclusion
The potential energy stored in the spring when it is stretched by 10 cm is \( 25 U \).

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