Figure 3 shows a child on a playground toy - AQA - GCSE Physics - Question 2 - 2021 - Paper 1

To obtain results as shown in Figure 4, the student can follow these steps:

1. Set up a clamp stand: Fix the clamp stand securely to a bench to prevent it from falling.
2. Attach the spring: Connect the spring to the clamp.
3. Attach a ruler: Use a ruler alongside the spring to measure the changes in length as forces are applied.
4. Apply a force: Hang a known weight (e.g., 1 N) from the spring to observe its extension.
5. Record the measurements: Measure the new position of the bottom of the spring with the weight attached and calculate the extension.
6. Repeat: Change the weights and repeat the measurements for accurate data collection.

Risk Assessment: The clamp stand could fall and cause injury. To minimize risk, ensure the clamp is tightly fastened to the bench and that masses are securely attached, and always be aware of your surroundings while conducting the experiment.

The correct equation that links extension, force, and spring constant is:

Force = spring constant × extension.

From Figure 4, if we assume that the force applied to the spring is known (for instance, 5 N) and the extension measured is 0.125 m, we can calculate the spring constant (k) using Hooke's Law:

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

Substituting the values gives:

$k = \frac{5 \, N}{0.125 \, m} = 40 \, N/m$

Thus, the spring constant is 40 N/m.

Figure 4 shows a straight line graph indicating a linear relationship between the force applied to the spring and the extension produced. This supports the student's conclusion that the extension of the spring is directly proportional to the force applied, as a straight line through the origin demonstrates consistent proportionality. A constant gradient on the graph also suggests a constant spring constant.

The elastic potential energy (Eₑ) stored in a spring can be calculated using the formula:

$Eₑ = \frac{1}{2} k e^2$

Where:

• k = 13 N/m (spring constant)
• e = 0.20 m (20 cm converted to meters)

Substituting in the values:

$Eₑ = \frac{1}{2} \times 13 \, N/m \times (0.20 \, m)^2$

Calculating gives:

$Eₑ = \frac{1}{2} \times 13 \, N/m \times 0.04 \, m^2 = 0.26 \, J$

Thus, the elastic potential energy is 0.26 Joules.