A-Level Physics May/June 2025 Q5(b)(ii): The extension of the wire is initially 2.0 × 10⁻³m. Determine the work done to increase…

1 answer

• ▲0
  accepted ✓

  The work done on the wire is equal to the change in its elastic potential energy, $\Delta E_p$.

  The elastic potential energy $E_p$ stored in the wire is given by $E_p = \frac{1}{2}kx^2$, where $k$ is the spring constant and $x$ is the extension.

  The work done $W$ is therefore: $W = \Delta E_p = E_{p,final} - E_{p,initial}$ $W = \frac{1}{2}kx_2^2 - \frac{1}{2}kx_1^2 = \frac{1}{2}k(x_2^2 - x_1^2)$

  From part (b)(i), the spring constant $k = 1.25 \times 10^5 \text{ N m}^{-1}$.

  $W = \frac{1}{2} \times (1.25 \times 10^5) \times ((3.0 \times 10^{-3})^2 - (2.0 \times 10^{-3})^2)$ $W = \frac{1}{2} \times (1.25 \times 10^5) \times (9.0 \times 10^{-6} - 4.0 \times 10^{-6})$ $W = \frac{1}{2} \times (1.25 \times 10^5) \times (5.0 \times 10^{-6})$ $W = 0.3125 \text{ J}$

  Work done = $0.31 \text{ J}$ (to 2 s.f.)

  ---

  How the marks are awarded

  • C1 — Stating that work done is the change in elastic potential energy and using the correct formula, such as $E = \frac{1}{2}kx^2$ or stating $W = \frac{1}{2}k(x_2^2 - x_1^2)$.
  • C1 — Correctly substituting the initial and final extensions into the change in energy formula, using the difference of the squares. This mark allows for an error carried forward (ECF) for the value of 'k' from the previous question part.
  • A1 — Calculating the final correct answer of 0.31 J, correctly rounded to two significant figures.

  Common mistakes

  • Incorrectly calculating the work done using the change in extension squared, i.e., $W = \frac{1}{2}k(\Delta x)^2 = \frac{1}{2}k(x_2 - x_1)^2$. This is a very common conceptual error.
  • Calculating only the final elastic potential energy ($E_p = \frac{1}{2}kx_2^2$) and not subtracting the initial energy.
  • Using the area of a triangle ($W = \frac{1}{2}F_{final}x_{final}$) instead of the area of a trapezium, which represents the change in energy.
  • Making a calculation error when squaring the extensions, particularly with the powers of 10 (e.g., writing $(3.0 \times 10^{-3})^2$ as $9.0 \times 10^{-9}$).

  Examiner tip: Remember that the work done in changing extension is the change in stored energy ($"E_{final} - E_{initial}"$), which is mathematically different from the energy stored for the change in extension.

  ---

  AI-generated model answer, grounded in the official Cambridge mark scheme and reviewed by the MarkScheme team. Mark your own answer to this question →

  @markscheme_answers

Your answer

Sign in to answer this question.