A spring of force constant `800N//m` has an extension of 5cm. The work done in extending it from 5cm to 15cm is

Options

A. 16 J
B. 8 J
C. 32 J
D. 24 J

Explanation

To find the work done in extending a spring from an extension of 5 cm to 15 cm, we can use the formula for the work done on a spring, which is given by:

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

where:
- \( W \) is the work done,
- \( k \) is the spring constant,
- \( x \) is the extension of the spring from its natural length.

### Step 1: Convert the extensions from cm to meters
First, we need to convert the extensions from centimeters to meters:
- Initial extension \( x_1 = 5 \, \text{cm} = 0.05 \, \text{m} \)
- Final extension \( x_2 = 15 \, \text{cm} = 0.15 \, \text{m} \)

### Step 2: Calculate the work done at the final extension (15 cm)
Using the formula for work done:
\[ W_2 = \frac{1}{2} k x_2^2 \]
Substituting the values:
\[ W_2 = \frac{1}{2} \times 800 \, \text{N/m} \times (0.15 \, \text{m})^2 \]
\[ W_2 = \frac{1}{2} \times 800 \times 0.0225 \]
\[ W_2 = 400 \times 0.0225 \]
\[ W_2 = 9 \, \text{J} \]

### Step 3: Calculate the work done at the initial extension (5 cm)
Now, calculate the work done at the initial extension:
\[ W_1 = \frac{1}{2} k x_1^2 \]
Substituting the values:
\[ W_1 = \frac{1}{2} \times 800 \, \text{N/m} \times (0.05 \, \text{m})^2 \]
\[ W_1 = \frac{1}{2} \times 800 \times 0.0025 \]
\[ W_1 = 400 \times 0.0025 \]
\[ W_1 = 1 \, \text{J} \]

### Step 4: Calculate the work done in extending from 5 cm to 15 cm
The work done in extending the spring from 5 cm to 15 cm is given by:
\[ W = W_2 - W_1 \]
Substituting the values:
\[ W = 9 \, \text{J} - 1 \, \text{J} \]
\[ W = 8 \, \text{J} \]

### Final Answer
The work done in extending the spring from 5 cm to 15 cm is **8 joules**.

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