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Work-energy problems with elastic strings

Edexcel IAL Mathematics · WME03 Mechanics 3

The work-energy principle from M2 now has a third kind of energy: elastic potential energy. Questions involve particles falling on elastic strings, springs launching objects and particles sliding on rough surfaces.

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The principle

Initial (KE + GPE + EPE) - work done against resistances = final (KE + GPE + EPE).

Write every term, then simplify. Measure heights from a level you choose and stick to it.

Falling on a string

A particle dropped from the fixed end O of a string of natural length l falls freely for l metres, then the string becomes taut. At the lowest point v = 0, so loss of GPE = EPE gained. The greatest speed is where the acceleration is zero, at the equilibrium position, where T = mg.

Examiner advice

Pearson's June 2025 report says candidates using "initial - final" often made sign errors, and many left out a kinetic energy term. Write a full equation such as ½mu2 + mgh + EPE_1 = ½mv2 + EPE_2 + work done, then substitute.

Worked example

A particle of mass 0.5 kg is attached to one end of an elastic string of natural length 1 m and modulus 24.5 N. The other end is fixed at O. The particle is released from rest at O. Find the greatest extension of the string.

  1. At the lowest point the particle has fallen (1 + x) m and v = 0.
  2. Loss of GPE = EPE: 0.5 × 9.8 × (1 + x) = 24.5x2/2.
  3. 12.25x2 - 4.9x - 4.9 = 0.

Answer: x = 0.863 m

Key idea: List every energy at the start and end, subtract work done against resistance, and set them equal.

Check yourself

  1. In the worked example, find the greatest extension, to 3 significant figures.
    Show answer

    0.863 m. Solving 12.25x2 - 4.9x - 4.9 = 0 gives x = 0.863 m.

  2. In the worked example, find the speed of the particle when the string first becomes taut, to 3 significant figures.
    Show answer

    4.43 m/s. It has fallen freely 1 m: v2 = 2 × 9.8 × 1, v = 4.43 m/s.

  3. In the worked example, find the greatest speed of the particle, to 3 significant figures.
    Show answer

    4.64 m/s. Greatest speed when T = mg: 24.5x/1 = 4.9, x = 0.2. Energy: ½(0.5)v2 = 0.5(9.8)(1.2) - 24.5(0.04)/2 = 5.39, v = 4.64 m/s.

  4. A particle of mass 0.4 kg on a smooth horizontal table is attached to a fixed point by an elastic string of natural length 0.5 m and modulus 20 N. It is held with the string stretched by 0.2 m and released. Find its speed when the string becomes slack.
    Show answer

    2 m/s. EPE = 20 × 0.04/1 = 0.8 J = ½(0.4)v2, so v = 2 m/s.

  5. Repeat the last question with the table rough, coefficient of friction 0.25. Find the speed when the string becomes slack, to 3 significant figures.
    Show answer

    1.74 m/s. Work against friction = 0.25 × 0.4 × 9.8 × 0.2 = 0.196 J. ½(0.4)v2 = 0.8 - 0.196, so v = 1.74 m/s.

  6. A spring of natural length 0.5 m and modulus 200 N is compressed by 0.1 m and used to fire a ball of mass 0.1 kg along a smooth horizontal surface. Find the speed of the ball as it leaves the spring, to 3 significant figures.
    Show answer

    6.32 m/s. EPE = 200 × 0.01/1 = 2 J = ½(0.1)v2, v = sqrt 40 = 6.32 m/s.

  7. A particle falls vertically on an elastic string and comes to rest at its lowest point. Which energy terms balance between release from rest and the lowest point (no air resistance)?
    • KE gained = EPE gained
    • GPE lost = EPE gained
    • GPE lost = KE gained
    • EPE lost = GPE gained
    Show answer

    GPE lost = EPE gained. It starts and ends at rest, so KE is zero at both ends.

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