When a 4 kg mass is hung vertically on a light string that obeys Hooke's law, the spring stretches by 2 cm. The work required to be done by an external agent in stretching this spring by 5 cm will be (`g=9.8(metrs)/(sec^2)`)

Options

A. 4.9 J
B. 2.4 J
C. 0.495 J
D. 0.24 J

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Determine the force exerted by the 4 kg mass
The force exerted by the mass can be calculated using the formula:
\[ F = m \cdot g \]
Where:
- \( m = 4 \, \text{kg} \)
- \( g = 9.8 \, \text{m/s}^2 \)

Calculating this gives:
\[ F = 4 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 39.2 \, \text{N} \]

### Step 2: Calculate the spring constant \( k \)
Using Hooke's law, the spring constant \( k \) can be calculated using the formula:
\[ k = \frac{F}{x} \]
Where:
- \( F = 39.2 \, \text{N} \)
- \( x = 2 \, \text{cm} = 0.02 \, \text{m} \)

Calculating \( k \):
\[ k = \frac{39.2 \, \text{N}}{0.02 \, \text{m}} = 1960 \, \text{N/m} \]

### Step 3: Calculate the work done in stretching the spring by 5 cm
The work done on the spring when it is stretched by a distance \( x \) is given by:
\[ W = \frac{1}{2} k x^2 \]
Where:
- \( k = 1960 \, \text{N/m} \)
- \( x = 5 \, \text{cm} = 0.05 \, \text{m} \)

Calculating the work done:
\[ W = \frac{1}{2} \cdot 1960 \, \text{N/m} \cdot (0.05 \, \text{m})^2 \]
\[ W = \frac{1}{2} \cdot 1960 \cdot 0.0025 \]
\[ W = 2.45 \, \text{J} \]

### Final Answer
The work required to be done by an external agent in stretching the spring by 5 cm is approximately **2.45 Joules**.

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