The shape of the lesson
By the end of the lesson, learners can
- Area under the line: Stretch the spring and watch the shaded triangle under the force-extension line grow.
- Elastic versus plastic: The behaviour changes at the elastic limit.
- Reading the graph: The force-extension graph carries both the stiffness and the stored energy.
- Energy: Eₚ = ½Fx = ½kx², the triangle area.
- Graph: gradient is k, area is the stored energy.
- Plastic: beyond the elastic limit, energy is not all recovered.
One picture every idea returns to
The work done in stretching, and so the elastic potential energy stored, is the area under the force-extension graph. For a spring obeying Hooke law this is a triangle, Eₚ = ½Fx = ½kx². Elastic deformation is fully recovered when the load is removed; plastic deformation, beyond the elastic limit, is permanent.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Energy stored in a stretch.. Stretching a spring does work, and that work is stored as elastic potential energy, ready to be released. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Area under the line | Stretch the spring and watch the shaded triangle under the force-extension line grow. Project tab 3 of the lab (Elastic potential energy) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Elastic versus plastic | The behaviour changes at the elastic limit. | Pairs |
| 21 to 30 min | Teach and check: Reading the graph | The force-extension graph carries both the stiffness and the stored energy. | Cooperative Concept Mapping |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 15 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Reading the graph, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Cooperative Concept Mapping
Teams receive the key terms on cards and arrange them on a large sheet, drawing and labelling the links. Compare maps between teams before showing the model map.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| The stored energy is the triangle area ½Fx, not Fx; using the full rectangle doubles the answer. | Because Eₚ = ½kx² depends on x², doubling the extension stores four times the energy. And beyond the elastic limit not all the work done is recovered, because energy goes into permanent (plastic) deformation and heating. |
Elastic potential energy: worksheet
6 questions, 15 marks. Written in the style of the papers; no past paper question is reproduced.
State how the work done in stretching a spring can be found from its force-extension graph.
A spring of constant 300 N m⁻¹ is stretched by 0.080 m. Find the elastic potential energy stored.
A force-extension graph for a spring is a straight line reaching 20 N at an extension of 0.050 m. Find the energy stored at this extension.
A spring obeying Hooke law stores 4.5 J of energy when stretched by 0.30 m. Find its spring constant.
Explain why, when a wire is stretched beyond its elastic limit and the load removed, the energy recovered is less than the work done in stretching it.
A catapult's stretched elastic stores 8.0 J of elastic potential energy. It launches a 0.040 kg stone. Assuming all the stored energy becomes kinetic energy, find the launch speed.
Answer key and marking notes
- It is the area under the force-extension graph ✓
- For a Hooke-law spring this is the area of a triangle, ½Fx ✓
- Eₚ = ½kx² = ½ × 300 × 0.080² ✓
- Eₚ = 0.96 J ✓
- Energy = area of the triangle = ½ × force × extension ✓
- Eₚ = ½ × 20 × 0.050 ✓
- Eₚ = 0.50 J ✓
- Eₚ = ½kx², so 4.5 = ½ × k × 0.30² ✓
- 4.5 = 0.045k ✓
- k = 100 N m⁻¹ ✓
- Beyond the elastic limit the deformation is partly plastic (permanent) ✓
- Some of the work done goes into rearranging the structure and heating, and is not returned, so less energy is recovered ✓
- Energy conservation: ½mv² = 8.0 ✓
- v² = (2 × 8.0) / 0.040 = 400 ✓
- v = 20 m s⁻¹ ✓
18 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
The work done in stretching a spring equals the:
- Aarea under the force-extension graph
- Bgradient of the graph
- Cfinal force only
- Dextension only
The elastic potential energy in a Hooke-law spring is:
- Akx²
- Bkx
- C½kx²
- D½kx
A spring of constant 400 N m⁻¹ is stretched 0.050 m. The energy stored is:
- A1.0 J
- B0.50 J
- C10 J
- D20 J
A force of 15 N stretches a spring 0.040 m (Hooke law). The energy stored is:
- A0.60 J
- B0.15 J
- C375 J
- D0.30 J
Doubling the extension of a Hooke-law spring changes the stored energy to:
- Afour times
- Bdouble
- Chalf
- Dthe same
Within the elastic region, the loading and unloading lines on a force-extension graph:
- Aform a loop
- Bcross at the origin only
- Ccoincide
- Dare perpendicular
Beyond the elastic limit, the energy recovered on unloading is:
- Aequal to the work done
- Bless than the work done in stretching
- Cgreater than the work done
- Dzero
A spring stores 2.0 J when stretched 0.10 m. Its spring constant is:
- A200 N m⁻¹
- B20 N m⁻¹
- C40 N m⁻¹
- D400 N m⁻¹
A catapult stores 18 J and launches a 0.090 kg stone. The maximum speed is:
- A20 m s⁻¹
- B400 m s⁻¹
- C10 m s⁻¹
- D14 m s⁻¹
On a force-extension graph for a Hooke-law spring, the stored energy is the area of a:
- Arectangle
- Bcircle
- Ctriangle
- Dtrapezium that is not a triangle
A spring of constant 120 N m⁻¹ is compressed 0.10 m. The energy stored is:
- A1.2 J
- B0.60 J
- C6.0 J
- D12 J
Using E = ½Fx rather than Fx matters because the force:
- Ais constant
- Bacts at an angle
- Cis negative
- Dgrows from zero as the spring stretches
A spring stores 0.80 J at an extension of 0.080 m. The force at that extension is:
- A20 N
- B10 N
- C40 N
- D6.4 N
Elastic potential energy is stored when an object is:
- Amoving fast
- Bhigh above the ground
- Cdeformed within its elastic limit
- Dhot
Two springs of constant k and 3k are stretched by the same x. The ratio of energy stored is:
- A3 : 1
- B1 : 3
- C1 : 9
- D1 : 1
A bungee cord (k = 80 N m⁻¹) stretches 5.0 m. The elastic energy stored is:
- A400 J
- B200 J
- C2.0 kJ
- D1.0 kJ
When a stretched elastic band within its elastic limit is released, the stored energy is:
- Areturned as kinetic energy
- Blost as heat entirely
- Cdestroyed
- Dconverted to mass
A loading-unloading curve for rubber forms a loop (hysteresis). The area of the loop represents the:
- Atotal energy stored
- Bspring constant
- Cenergy transferred to heat in one cycle
- Dbreaking stress
Quiz key
Work = area under the F-x graph.
Eₚ = ½Fx = ½kx².
E = ½kx² = ½ × 400 × 0.050² = 0.50 J.
E = ½Fx = ½ × 15 × 0.040 = 0.30 J.
E ∝ x², so doubling x gives 4 times the energy.
Elastic behaviour means the spring retraces its path, so the lines coincide.
Some energy goes into permanent deformation, so less is recovered.
E = ½kx², so k = 2E/x² = 4.0/0.010 = 400 N m⁻¹.
½mv² = 18 gives v = √(36/0.090) = 20 m s⁻¹.
The straight line from the origin gives a triangular area, ½Fx.
E = ½kx² = ½ × 120 × 0.10² = 0.60 J.
The average force is half the final force, giving the factor of one half.
E = ½Fx, so F = 2E/x = 1.6/0.080 = 20 N.
Elastic PE is the energy stored in a stretched or compressed elastic object.
E = ½kx², so with equal x the energies are in the ratio of the spring constants, 1 : 3.
E = ½kx² = ½ × 80 × 25 = 1000 J = 1.0 kJ.
Elastic deformation returns its stored energy, mainly as kinetic energy.
The loop area is the energy converted to internal energy (heat) per cycle.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.