Figure 1 shows a stretched spring - AQA - GCSE Physics Combined Science - Question 1 - 2021 - Paper 2

Measure the original length of the spring and the extended length of the spring (with the metre rule).

The extension is calculated as:

$\text{extension} = \text{extended length} - \text{original length}$

Given:

• Extension, $e = 0.080 \, \text{m}$
• Spring constant, $k = 40 \, \text{N/m}$

The formula for elastic potential energy is:

$E_e = \frac{1}{2} k e^2$

Substituting the given values:

$E_e = \frac{1}{2} \times 40 \, \text{N/m} \times (0.080 \, \text{m})^2$

Calculating:

$E_e = 0.128 \, \text{J}$

The equation is:

$F = k \times e$

or alternatively, $F = k e$

Given:

• Force, $F = 300 \, \text{N}$
• Extension, $e = 0.40 \, \text{m}$

Using the equation:

$F = k \times e$

We can rearrange it to find the spring constant:

$k = \frac{F}{e}$

Substituting the values:

$k = \frac{300 \, \text{N}}{0.40 \, \text{m}} = 750 \, \text{N/m}$