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Conservation of energy and the work-energy principle

Edexcel IAL Mathematics · WME02 Mechanics 2

The work-energy principle connects work done by forces to changes in energy. When only gravity does work, mechanical energy is conserved. Pearson questions often say which method to use, and using a different one scores nothing.

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Conservation of mechanical energy

If the only force doing work is gravity (normal reactions do no work), then

KE + PE stays constant.

This gives speeds at different heights without finding accelerations. It works for projectiles too: the speed at a given height does not depend on the angle of projection.

The work-energy principle

When other forces act:

work done by driving forces - work done against resistances = change in mechanical energy (KE + PE).

A neat way to write it: initial KE + initial PE + work done by driving forces = final KE + final PE + work done against resistances. Count gravity only once, either as PE or as work against the weight, never both (Pearson's examiners flag this error).

Exam technique

Worked example

A 2 kg particle is released from rest on a rough slope inclined at 30°. The coefficient of friction is 0.2. Find its speed after it has slid 5 m.

  1. PE lost = 2 × 9.8 × 5 sin 30° = 49 J.
  2. Friction = 0.2 × 2g cos 30° = 3.39 N; work against it = 17.0 J.
  3. KE gained = 49 - 17.0 = 32.0 J = (1/2)(2)v2.

Answer: v = 5.66 m/s

Key idea: Only gravity working: KE + PE constant. Otherwise: initial energy + work by driving forces = final energy + work against resistances. Count gravity once.

Check yourself

  1. A particle is released from rest and slides 4 m down a smooth plane inclined at 30°. Find its speed, to 3 significant figures.
    Show answer

    6.26 m/s. Height lost 2 m. (1/2)v2 = 9.8 × 2, v = 6.26 m/s.

  2. A ball is thrown at 8 m/s from a point 20 m above the ground. Use energy to find its speed when it hits the ground, to 3 significant figures.
    Show answer

    21.4 m/s. (1/2)v2 = (1/2)(64) + 9.8 × 20, so v2 = 456 and v = 21.4 m/s, whatever the direction of throw.

  3. A 0.5 kg ball falls from rest through 30 m and hits the ground at 20 m/s. Find the constant air resistance, to 3 significant figures.
    Show answer

    1.57 N. PE lost = 147 J, KE gained = 100 J, so work against resistance = 47 J = R × 30, R = 1.57 N.

  4. A 3 kg particle is projected at 10 m/s up a rough plane inclined at α, where sin α = 0.6. The coefficient of friction is 0.5. Find how far it travels up the plane before stopping, to 3 significant figures.
    Show answer

    5.10 m. KE lost 150 J = PE gained + work against friction = (3g × 0.6)d + (0.5 × 3g × 0.8)d = 29.4d, so d = 5.10 m.

  5. A 2 kg particle slides across a rough horizontal floor, slowing from 6 m/s to rest in 4.5 m. Find the coefficient of friction, to 3 significant figures.
    Show answer

    0.408. KE lost = 36 J = μ(2g)(4.5), so μ = 36/88.2 = 0.408.

  6. A particle is projected from ground level at 15 m/s. Find its speed when it is 5 m above the ground, to 3 significant figures.
    Show answer

    11.3 m/s. v2 = 225 - 2(9.8)(5) = 127, so v = 11.3 m/s.

  7. A question says "Use the work-energy principle to find the speed of P at B." A student uses suvat instead and gets the right answer. What happens?
    • Full marks, since the answer is correct
    • Half marks
    • The method earns no credit because a specified method was not used
    • Marks for accuracy only
    Show answer

    The method earns no credit because a specified method was not used. Pearson's examiners state that where a method is specified, other methods gain no credit.

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Last updated 8 November 2026. Spotted a mistake? Tell us.