When a 20 g mass hangs attached to one end of a light spring of length 10 cm, the spring stretches by 2 cm. The mass is pulled down until the total length of the spring is 14 cm. The elastic energy, (in Joule) stored in the spring is :

Options

A. `4xx10^(-2)`
B. `4xx10^(-3)`
C. `8xx10^(-2)`
D. `8xx10^(-3)`

Answer

To solve the problem, we need to find the elastic energy stored in the spring when it is stretched. We will follow these steps:

### Step 1: Understand the problem
We have a spring of initial length 10 cm, which stretches by 2 cm when a mass of 20 g is attached to it. The spring is then pulled down to a total length of 14 cm.

### Step 2: Convert the mass to kilograms
The mass given is 20 g. We need to convert this to kilograms for our calculations.
\[ \text{Mass} = 20 \, \text{g} = 20 \times 10^{-3} \, \text{kg} = 0.02 \, \text{kg} \]

### Step 3: Calculate the spring constant (k)
When the mass is hanging at rest, the force due to the weight of the mass is equal to the spring force. The weight of the mass is given by:
\[ F = mg = 0.02 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 0.1962 \, \text{N} \]
The spring stretches by 2 cm (0.02 m), so we can use Hooke's law:
\[ F = kx \implies k = \frac{F}{x} = \frac{0.1962 \, \text{N}}{0.02 \, \text{m}} = 9.81 \, \text{N/m} \]

### Step 4: Determine the extension of the spring when pulled down to 14 cm
The spring's total length is now 14 cm, which means it has been stretched by:
\[ \text{Total extension} = 14 \, \text{cm} - 10 \, \text{cm} = 4 \, \text{cm} = 0.04 \, \text{m} \]

### Step 5: Calculate the elastic potential energy stored in the spring
The elastic potential energy (EPE) stored in a spring is given by the formula:
\[ EPE = \frac{1}{2} k x^2 \]
Substituting the values we have:
\[ EPE = \frac{1}{2} \times 9.81 \, \text{N/m} \times (0.04 \, \text{m})^2 \]
\[ EPE = \frac{1}{2} \times 9.81 \times 0.0016 \]
\[ EPE = \frac{1}{2} \times 0.015696 = 0.007848 \, \text{J} \]

### Final Answer
The elastic energy stored in the spring is approximately:
\[ EPE \approx 0.00785 \, \text{J} \]

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