A body of mass 100 gm is suspended from a massless spring of natural length 1 m. If the spring stretches through a vertical distance of 20 cm, the potential energy stored in the spring is (g = `10 ms^(-2)`)

Options

A. 1.2 J
B. 0.6 J
C. 0.1 J
D. 0.2 J

Answer

To find the potential energy stored in the spring when a mass is suspended from it, we can follow these steps:

### Step 1: Understand the given data
- Mass of the body (m) = 100 grams = 0.1 kg (since 1 gram = 0.001 kg)
- Natural length of the spring = 1 m
- Stretch in the spring (elongation) = 20 cm = 0.2 m (since 1 cm = 0.01 m)
- Acceleration due to gravity (g) = 10 m/s²

### Step 2: Calculate the force exerted by the mass
The force exerted by the mass due to gravity can be calculated using the formula:
\[ F = m \cdot g \]
Substituting the values:
\[ F = 0.1 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 1 \, \text{N} \]

### Step 3: Calculate the potential energy stored in the spring
The potential energy (PE) stored in the spring when it is stretched can be calculated using the formula for elastic potential energy:
\[ PE = \frac{1}{2} k x^2 \]
However, we can also use the work done against gravity to find the potential energy stored in the spring:
\[ PE = m \cdot g \cdot h \]
Where:
- \( h \) is the change in height, which is equal to the stretch in the spring (0.2 m).

Substituting the values:
\[ PE = 0.1 \, \text{kg} \cdot 10 \, \text{m/s}^2 \cdot 0.2 \, \text{m} \]
\[ PE = 0.1 \cdot 10 \cdot 0.2 = 0.2 \, \text{J} \]

### Step 4: Conclusion
The potential energy stored in the spring is:
\[ \text{Potential Energy} = 0.2 \, \text{Joules} \]