A trampoline is made from a sheet of material held in place by stretched springs - AQA - GCSE Physics Combined Science - Question 3 - 2022 - Paper 1

The springs undergo a change in elastic potential energy. Initially, at position A, they are stretched, storing elastic potential energy. As the child lands at position B, this elastic potential energy reaches a maximum, reflecting the springs' increased stretch at that position.

The energy dissipated in the surroundings typically manifests as thermal energy due to air resistance and vibrations transmitted through the ground. As the child lands, energy is lost to the surroundings, leading to an increase in the internal or thermal energy of the particles in the surrounding area.

To find the extension of each spring at position B, we use the formula for elastic potential energy:

$E_p = \frac{1}{2} k x^2$

Given that the elastic potential energy ( $E_p$) is 8.1 J, we already calculated the spring constant ( $k$) as 3125 N/m using the preliminary data from position A. Thus,

$8.1 = \frac{1}{2} \times 3125 \times x^2$

Solving for $x$ gives:

$x^2 = \frac{2 \times 8.1}{3125}$

$x^2 = 0.005184$

$x = \sqrt{0.005184} \approx 0.072 \text{ m}$

The extension of each spring when the child is at position B is approximately 0.072 m.

The work done by the child is equal to the total energy transferred by the child. Thus, the correct box to tick is:

• The total energy transferred by the child.