Question

A force of 2 Newton stretches an elastic material by 30mm. What additional force will stretch the material 40mm assuming that the elastic limit is not exceeded

Answer 1

To solve this problem, we can use Hooke's Law, which states that the force \( F \) applied to a spring (or elastic material) is directly proportional to the extension \( x \), as long as the elastic limit is not exceeded. The mathematical representation of Hooke's Law is:

\[ F = kx \]

where:
- \( F \) is the force applied,
- \( k \) is the spring constant, and
- \( x \) is the extension.

From the information given, a force of 2 Newtons stretches the elastic material by 30mm (0.03m). We can use this to find the spring constant \( k \).

### Step 1: Calculate the spring constant \( k \)

Using the values:
- \( F = 2 \, \text{N} \)
- \( x = 30 \, \text{mm} = 0.03 \, \text{m} \)

We can plug these values into Hooke's law:

\[ 2 = k \cdot 0.03 \]

To find \( k \), we rearrange the equation:

\[ k = \frac{2}{0.03} \approx 66.67 \, \text{N/m} \]

### Step 2: Find the force required to stretch the material by 40mm

Now, we want to find the force required to stretch the material by 40mm (0.04 m).

Using Hooke's law again:

\[ F = kx \]

Where \( x = 40 \, \text{mm} = 0.04 \, \text{m} \):

\[ F = 66.67 \cdot 0.04 \]

Calculating this gives:

\[ F \approx 2.67 \, \text{N} \]

### Step 3: Find the additional force required

Since we previously applied a force of 2 N to stretch it to 30 mm, we need to find the additional force required to stretch it to 40 mm:

\[ \text{Additional force} = F - \text{previous force} \]

\[ \text{Additional force} = 2.67 \, \text{N} - 2 \, \text{N} \]

\[ \text{Additional force} = 0.67 \, \text{N} \]

### Conclusion

The additional force required to stretch the elastic material by 40mm is approximately **0.67 Newtons**.