A-Level Mathematics May/June 2024 Q3: A train of mass 180000kg ascends a straight hill of length 1.5km, inclined at an angle…

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  We apply the work-energy principle, which states that the net work done on an object is equal to its change in mechanical energy. The work done by the engine less the work done against resistance equals the sum of the changes in kinetic and potential energy.

  Work Done by Engine (WD) - Work Done against Resistance = Change in KE + Change in PE

  $WD_{engine} - WD_{resistance} = \Delta KE + \Delta PE$

  Rearranging to find the work done by the engine:

  $WD_{engine} = \Delta PE + \Delta KE + WD_{resistance}$

  First, we identify the given values and convert units:

  • Mass, $m = 180000$ kg
  • Distance, $d = 1.5$ km $= 1500$ m
  • Angle, $\theta = 1.5^{\circ}$
  • Initial speed, $u = 45$ ms⁻¹
  • Final speed, $v = 40$ ms⁻¹
  • Work done against resistance, $WD_{resistance} = 12000$ kJ $= 12,000,000$ J

  Next, we calculate the change in potential energy (gain in PE):

  • Vertical height gained, $h = d \sin \theta = 1500 \sin(1.5^{\circ})$
  • $\Delta PE = mgh = 180000 \times 10 \times 1500 \sin(1.5^{\circ})$
  • $\Delta PE = 70,677,760.4...$ J

  Then, we calculate the change in kinetic energy (loss in KE):

  • Initial KE, $KE_{initial} = \frac{1}{2}mu^2 = \frac{1}{2} \times 180000 \times 45^2 = 182,250,000$ J
  • Final KE, $KE_{final} = \frac{1}{2}mv^2 = \frac{1}{2} \times 180000 \times 40^2 = 144,000,000$ J
  • $\Delta KE = KE_{final} - KE_{initial} = 144,000,000 - 182,250,000 = -38,250,000$ J

  Now, substitute these values into the work-energy equation:

  $WD_{engine} = (70,677,760.4...) + (-38,250,000) + (12,000,000)$

  $WD_{engine} = 44,427,760.4...$ J

  Finally, convert the answer to kJ as requested:

  $WD_{engine} = \frac{44,427,760.4...}{1000}$ kJ

  $WD_{engine} = 44427.76...$ kJ

  Rounding to 3 significant figures:

  $WD_{engine} = 44400$ kJ

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  How the marks are awarded

  • B1 — Correctly calculating the gain in Potential Energy using the formula $\Delta PE = mgh$ with $h = 1500 \sin(1.5^{\circ})$, leading to the value $70,677,760.4...$ J.
  • B1 — Correctly calculating either the initial or final Kinetic Energy. The model answer shows both: $KE_{initial} = 182,250,000$ J and $KE_{final} = 144,000,000$ J.
  • M1 — Setting up a complete work-energy equation with the correct four terms (Work Done by engine, Work Done against resistance, change in KE, change in PE), even if there are sign errors. This is shown in the line: $WD_{engine} = \Delta PE + \Delta KE + WD_{resistance}$.
  • A1 — Obtaining the correct final answer of 44400 kJ. The mark requires the answer to be in kilojoules and correctly calculated and rounded.

  Common mistakes

  • Unit conversion errors, such as using 1.5 km instead of 1500 m for distance, or 12000 J instead of 12,000,000 J for work done against resistance.
  • Using the wrong trigonometric function for height, i.e., calculating $h = 1500 \cos(1.5^{\circ})$ instead of $1500 \sin(1.5^{\circ})$.
  • Sign errors in the work-energy equation, particularly with the change in kinetic energy. Forgetting that a decrease in speed means a loss of KE, which should be subtracted from the energy gains (or added as a negative value).
  • Omitting one of the components from the work-energy equation, for instance, forgetting to include the work done against resistance or the change in potential energy.

  Examiner tip: This question rewards the systematic application of the work-energy principle, ensuring all energy changes (Kinetic and Potential) and work done by external forces (driving and resistive) are correctly accounted for in a single equation.

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