Figure 1 shows a boy bouncing on a trampoline - AQA - GCSE Physics - Question 1 - 2022 - Paper 1

Using the formula for elastic potential energy:

$E = 0.5 imes k imes x^2$

Where:

• $k = 120,000 \, \text{N/m}$ (spring constant)
• $x = 0.015 \, \text{m}$ (extension)

Substituting the values:

$E = 0.5 \times 120,000 \times (0.015)^2$ $E = 0.5 \times 120,000 \times 0.000225$ $E = 13.5 \, \text{J}$

Since each spring stores 13.5 J, for 40 springs, the total energy is:

$\text{Total energy} = 40 \times 13.5 \, \text{J} = 540 \, \text{J}$

The maximum kinetic energy after bouncing is 45% of the initial kinetic energy:

$\text{Max KE} = 0.45 \times 600 \, \text{J} = 270 \, \text{J}$

Energy is transferred to the surroundings.