A spring 60 cm long is stretched 2 cm, when subjected to a force of 20 gram weight what would be the length, when a force of 500 gram weight is applied.?

Answer

To solve the problem, we need to determine the new length of the spring when a force of 500 grams is applied. We will use Hooke's Law, which states that the force exerted by a spring is proportional to the displacement (stretch) from its original length.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Original length of the spring (L₀) = 60 cm
- Stretch when 20 g weight is applied (y₁) = 2 cm
- Weight applied (m₁) = 20 g = 0.02 kg (since 1 g = 0.001 kg)
- Weight to be applied (m₂) = 500 g = 0.5 kg

2. **Convert Units:**
- Convert the stretch from cm to meters for consistency in SI units:
- y₁ = 2 cm = 0.02 m

3. **Calculate the Spring Constant (k):**
- According to Hooke's Law:
\[ F = k \cdot y \]
- The force (F₁) due to the weight of 20 g is:
\[ F₁ = m₁ \cdot g = 0.02 \, \text{kg} \cdot 9.81 \, \text{m/s²} = 0.1962 \, \text{N} \]
- Now, using Hooke's Law:
\[ k = \frac{F₁}{y₁} = \frac{0.1962 \, \text{N}}{0.02 \, \text{m}} = 9.81 \, \text{N/m} \]

4. **Calculate the Stretch (y₂) for 500 g Weight:**
- The force (F₂) due to the weight of 500 g is:
\[ F₂ = m₂ \cdot g = 0.5 \, \text{kg} \cdot 9.81 \, \text{m/s²} = 4.905 \, \text{N} \]
- Using Hooke's Law again:
\[ y₂ = \frac{F₂}{k} = \frac{4.905 \, \text{N}}{9.81 \, \text{N/m}} = 0.5 \, \text{m} = 50 \, \text{cm} \]

5. **Calculate the Total Length of the Spring:**
- The total length of the spring when the 500 g weight is applied:
\[ L = L₀ + y₂ = 60 \, \text{cm} + 50 \, \text{cm} = 110 \, \text{cm} \]

### Final Answer:
The length of the spring when a force of 500 grams is applied will be **110 cm**.

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