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Elastic potential energy questions
A stretched string holds energy, and the amount is the area under the Hooke's law line. Adding that term to the energy equation extends the work-energy principle to every spring problem on the paper.
6 original questions · 23 marks · the elastic potential energy notes · Further Mechanics 1
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Write down the formula for the energy stored in a stretched elastic string, and explain where the factor of one half comes from.
Worked answer
EPE = λx²/(2l). The tension grows linearly from 0 to λx/l as the string stretches, so the work done stretching it is the area of the triangle under the Hooke's law graph, half the base x times the height λx/l. B1 for the formula, B1 for the triangle under the Hooke's law graph. Using the final tension throughout would double the answer.An elastic string of natural length 2 m and modulus 80 N is stretched by 0.5 m. Find the energy stored.
Worked answer
EPE = 80 × 0.5²/(2 × 2) = 80 × 0.25/4 = 5 J. M1 for substituting into λx²/(2l), A1 for 5 J. Checking by area: the tension at that extension is 20 N, and ½ × 0.5 × 20 = 5 J.A particle of mass 0.2 kg on a smooth horizontal table is attached to an elastic string of natural length 1 m and modulus 20 N. It is pulled out to an extension of 0.4 m and released. Find its speed when the string reaches its natural length.
Worked answer
EPE = 20 × 0.4²/(2 × 1) = 20 × 0.16/2 = 1.6 J. The table is smooth and horizontal, so all of it becomes kinetic energy: ½(0.2)v² = 1.6 gives v² = 16 and v = 4 m/s. M1 for the elastic energy, A1 for 1.6 J, M1 for equating it to kinetic energy, A1 for 4 m/s. Beyond the natural length the string goes slack and the particle keeps that speed.A particle of mass 0.5 kg hangs from an elastic string of natural length 1 m and modulus 39.2 N attached to a ceiling. It is pulled down until the string is 1.5 m long and released. Find its speed when the string returns to its natural length, taking g = 9.8 m/s².
Worked answer
Elastic energy released = 39.2 × 0.5²/(2 × 1) = 4.9 J. The particle rises 0.5 m, gaining 0.5(9.8)(0.5) = 2.45 J of potential energy. What remains is kinetic: ½(0.5)v² = 2.45, so v² = 9.8 and v = 3.13 m/s. M1 for the elastic energy, A1 for 4.9 J, B1 for the gain in potential energy, M1 for the energy equation, A1 for 3.13 m/s. Forgetting the height change would give 4.43 m/s.State the two commonest slips in setting up an energy equation with an elastic string, and how to avoid each.
Worked answer
First, using the total length in place of the extension. Write l and x down separately before substituting anything. Second, measuring heights from different levels at the two instants. Choose one zero level, mark it on the diagram and use it in both energy totals. B1 B1 for the two slips, B1 for the remedies. A third slip, less common, is forgetting that a string stores nothing once it goes slack.A particle of mass 1 kg lies against a spring of natural length 0.5 m and modulus 50 N, compressed by 0.1 m, on a rough horizontal surface with coefficient of friction 0.2. Find the speed of the particle as it leaves the spring and the total distance it travels before coming to rest. Take g = 9.8 m/s².
Worked answer
Energy stored = 50 × 0.1²/(2 × 0.5) = 0.5 J, and the frictional force is 0.2 × 1 × 9.8 = 1.96 N.
The particle leaves the spring at the natural length, after 0.1 m. By then friction has taken 1.96(0.1) = 0.196 J, leaving 0.5 − 0.196 = 0.304 J as kinetic energy, so ½(1)v² = 0.304 and v = 0.780 m/s.
For the total distance, all the elastic energy is eventually spent against friction. So 1.96d = 0.5 and d = 0.255 m, of which the last 0.155 m is travelled clear of the spring. M1 for the stored energy, A1 for 0.5 J, B1 for the frictional force, M1 for the energy equation over the 0.1 m of contact, A1 for 0.780 m/s, M1 for spending all the stored energy against friction, A1 for 0.255 m.
Friction acts over the whole 0.255 m, including the 0.1 m of contact, so it must not be applied only to the free part. A spring is used rather than a string because only a spring can push.
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