ExaminerPrep › Question bank › IAL Physics › 1.8 Hooke's law and elastic strain energy
Edexcel International A Level Physics · Unit 1 · Mechanics and Materials
40 free, high-quality past-paper-style questions on Hooke's law and elastic strain energy for Edexcel IAL Physics, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Physics notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.
The stiffness k of a spring is defined by ΔF = kΔx.
Which of the following is the unit of k expressed in SI base units?
| Answer | Mark | Notes |
|---|---|---|
| k = ΔF/Δx, so its unit is N m−1. A newton is kg m s−2; dividing by a metre gives kg s−2. | ✓ 1 | Answer A |
Answer: A
Exam tip: Write the unit of each quantity in the defining equation, then cancel. The newton in base units, kg m s−2, comes from F = ma.
A spring is 120 mm long when no force acts on it. A force of 6.0 N stretches it to a length of 168 mm, within its limit of proportionality.
What is the stiffness of the spring?
| Answer | Mark | Notes |
|---|---|---|
| Extension = 168 − 120 = 48 mm = 0.048 m. k = F/Δx = 6.0 ÷ 0.048 = 125 N m−1. | ✓ 1 | Answer C |
Answer: C
Common error: Using a length instead of the extension. Hooke's law uses the change in length: subtract the original length first, then convert to metres.
A metal wire is stretched by gradually increasing the force on it.
Which statement describes the elastic limit of the wire?
| Answer | Mark | Notes |
|---|---|---|
| The elastic limit marks the end of elastic behaviour: up to it, the wire returns to its original length when the force is removed. Beyond it some extension is permanent. D. | ✓ 1 | Answer D |
Answer: D
Exam tip: Keep three points apart: the limit of proportionality (graph stops being straight), the elastic limit (deformation stops being reversible) and the yield point (large plastic flow begins).
A rubber band is stretched and then allowed to contract to its original length. The force–extension graph for unloading lies below the graph for loading, forming a loop.
What does the area enclosed by the loop represent?
| Answer | Mark | Notes |
|---|---|---|
| Area under loading curve = work done on the band; area under unloading curve = energy returned. The difference, the loop, is energy not returned: it is dissipated as thermal energy in the rubber (hysteresis). C. | ✓ 1 | Answer C |
Answer: C
Common error: Calling the loop “plastic deformation”. The band returns to its original length, so there is no permanent deformation; the loop is energy dissipated as thermal energy.
The seat of a child's ride-on toy rests on a single vertical compression spring. A force of 150 N compresses the spring by 14 mm. The spring obeys Hooke's law over the range used. Ignore the weight of the seat.
State Hooke's law.
Calculate the stiffness of the spring.
A child of weight 240 N sits on the seat. Calculate the compression of the spring.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the extension (or compression) of an object is (directly) proportional to the force applied | ✓ 1 | |
| provided the limit of proportionality is not exceeded | ✓ 1 | |
| Part (b) | ||
| k = ΔF/Δx = 150 ÷ 0.014 | ✓ 1 | |
| = 1.1 × 104 N m−1 (10 700 N m−1) | ✓ 1 | |
| Part (c) | ||
| 240 ÷ 10 714 = 0.0224 m = 22 mm (allow error carried forward) | ✓ 1 | |
Exam tip: A statement of Hooke's law without its condition is incomplete: always add “provided the limit of proportionality is not exceeded”.
A student tests a copper wire. Its force–extension graph is a straight line from the origin to an extension of 1.6 mm at a force of 48 N; beyond this the graph curves and its gradient decreases.
When the wire is loaded to an extension of 1.8 mm and the force is removed, the wire returns to its original length. When a second, identical wire is loaded to an extension of 6.0 mm and the force is removed, it is left with a permanent extension of 3.9 mm.
State the force at the limit of proportionality.
Calculate the stiffness of the wire on the straight section of the graph.
Explain how the results show that the elastic limit is beyond the limit of proportionality.
State what is meant by plastic deformation, and identify the evidence for it in these results.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 48 N | ✓ 1 | |
| Part (b) | ||
| 48 ÷ (1.6 × 10−3) = 3.0 × 104 N m−1 | ✓ 1 | |
| Part (c) | ||
| at 1.8 mm the wire is beyond the limit of proportionality / on the curved part of the graph | ✓ 1 | |
| but it returns to its original length, so the deformation is still elastic | ✓ 1 | |
| Part (d) | ||
| a permanent change of shape that remains when the force is removed, shown by the 3.9 mm permanent extension of the second wire | ✓ 1 | |
Common error: Treating the end of the straight line as the elastic limit. Elastic means “returns to its original length”, which can still happen on the curved part of the graph.
A student hangs loads on a spring and plots the extension of the spring against the force applied. The graph shows her results. The spring obeys Hooke's law over this range.
What is the stiffness of the spring?
| Answer | Mark | Notes |
|---|---|---|
| Force is on the horizontal axis, so the gradient of this graph is Δx/ΔF = 1/k. From the line, 6.0 N gives an extension of 4.0 cm = 0.040 m. k = ΔF/Δx = 6.0 ÷ 0.040 = 150 N m−1. | ✓ 1 | Answer C |
Answer: C
Common error: Taking the gradient as the stiffness without looking at the axes. With extension on the vertical axis, the gradient is 1/k, not k.
The graph shows how the force on a metal wire varies with its extension as the wire is loaded until it breaks at M. Point J is the limit of proportionality and point K is the elastic limit.
The wire is loaded to point L and the force is then gradually removed. Which statement describes what happens?
| Answer | Mark | Notes |
|---|---|---|
| L lies beyond the elastic limit K, so the wire has been deformed plastically. As the force is removed, only the elastic part of the extension is recovered, with the original stiffness: the unloading graph is a straight line parallel to OJ. It reaches zero force at a non-zero extension, the permanent extension. D. | ✓ 1 | Answer D |
Answer: D
Exam tip: To find a permanent extension from a graph, draw a line from the unloading point parallel to the straight section. Where it meets the extension axis is the permanent extension.
A student measures the extension of a spring from a toy trampoline as she adds weights to it. The graph shows her results and her best-fit line.
State how the graph shows that the spring obeys Hooke's law over this range.
Identify the anomalous result.
Use the best-fit line to determine the stiffness of the spring.
Calculate the elastic strain energy stored in the spring when the force on it is 5.0 N. Use the best-fit line.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the best-fit line is straight and passes through the origin, so the extension is proportional to the force | ✓ 1 | |
| Part (b) | ||
| the reading at 4.0 N (extension about 9.4 cm), which lies well off the line | ✓ 1 | |
| Part (c) | ||
| gradient = ΔF/Δx from a large triangle, e.g. 6.0 N ÷ 0.120 m | ✓ 1 | |
| = 50 N m−1 (allow 48–52 N m−1) | ✓ 1 | |
| Part (d) | ||
| Δx = 5.0 ÷ 50 = 0.10 m, so ΔEel = ½ × 5.0 × 0.10 = 0.25 J (allow error carried forward) | ✓ 1 | |
Exam tip: Draw the gradient triangle over more than half of the line, with its corners on gridline intersections, and leave the anomalous point out of the best-fit line.
The spring that closes a garden gate is tested by hanging weights from it and measuring its total length L. The graph shows the results and a best-fit line. g = 9.81 N kg−1.
State the length of the spring when no force acts on it.
Explain why the graph does not pass through the origin.
Calculate the stiffness of the spring.
An object of mass 0.35 kg is hung from the spring. Calculate the length of the spring.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 15.0 cm (the intercept on the length axis) | ✓ 1 | |
| Part (b) | ||
| L is the total length, which includes the natural length; only the extension (L − 15.0 cm) is proportional to the force | ✓ 1 | |
| Part (c) | ||
| extension at 6.0 N = 30.0 − 15.0 = 15.0 cm = 0.150 m | ✓ 1 | |
| k = 6.0 ÷ 0.150 = 40 N m−1 | ✓ 1 | |
| Part (d) | ||
| Δx = 0.35 × 9.81 ÷ 40 = 0.0858 m, so L = 15.0 + 8.6 = 23.6 cm (allow error carried forward) | ✓ 1 | |
Common error: Dividing a force by a length read straight off this graph. Subtract the natural length first: Hooke's law uses the extension, not the length.
Figure 1 shows force–extension graphs for three samples, X, Y and Z, each loaded and then unloaded. Loading is shown by a solid line and unloading by a dashed line. One sample is a steel spring, one is a rubber band and one is a polythene strip.
State which of the samples deformed only elastically. Give a reason based on the graphs.
Identify the rubber band, and state what the area between its loading and unloading curves represents.
State the permanent extension of sample Z.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| X and Y | ✓ 1 | |
| their extension returns to zero when the force is removed / the unloading graph ends at the origin | ✓ 1 | |
| Part (b) | ||
| Y: its unloading curve lies below its loading curve (a hysteresis loop) and it returns to its original length | ✓ 1 | |
| the energy dissipated as thermal energy in the rubber during the cycle | ✓ 1 | |
| Part (c) | ||
| 30 mm (allow 29–31 mm) | ✓ 1 | |
Common error: Calling every curved graph plastic. Rubber gives a curved graph with a loop, but it returns to its original length, so its deformation is elastic.
Figure 1 shows apparatus used to measure the stiffness of a spring. The magnified views show the metre rule next to the pointer with no load and with a load of mass 400 g. The scale increases downwards. g = 9.81 N kg−1.
Record the two pointer readings and calculate the extension of the spring.
Calculate the stiffness of the spring.
Describe how the student could check that the spring had not been stretched beyond its elastic limit.
Suggest one safety precaution for this experiment.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 14.6 cm and 22.8 cm (both needed; allow ± 0.1 cm) | ✓ 1 | |
| extension = 22.8 − 14.6 = 8.2 cm | ✓ 1 | |
| Part (b) | ||
| weight = 0.400 × 9.81 = 3.92 N | ✓ 1 | |
| k = 3.92 ÷ 0.082 = 48 N m−1 (47.9 N m−1) (allow error carried forward) | ✓ 1 | |
| Part (c) | ||
| remove the load and check that the pointer returns to its original reading of 14.6 cm | ✓ 1 | |
| Part (d) | ||
| clamp the stand to the bench or use a counterweight so it cannot topple / put a tray of sand or a soft mat below the masses / wear eye protection in case the spring snaps | ✓ 1 | |
Exam tip: Read the pointer with your eye level with it, at right angles to the rule, to avoid parallax, and record each reading to the resolution of the rule (0.1 cm).
Two springs are joined end to end and hung vertically. The upper spring has stiffness 40 N m−1 and the lower spring has stiffness 60 N m−1. A load of weight 6.0 N hangs from the lower spring. Both springs obey Hooke's law and have negligible mass.
What is the total extension of the two springs?
| Answer | Mark | Notes |
|---|---|---|
| In series each spring carries the whole 6.0 N. Extensions: 6.0 ÷ 40 = 0.15 m and 6.0 ÷ 60 = 0.10 m. Total = 0.25 m. (Equivalently 1/k = 1/40 + 1/60, k = 24 N m−1, 6.0 ÷ 24 = 0.25 m.) | ✓ 1 | Answer D |
Answer: D
Exam tip: In series every spring feels the full load, so the extensions add. The combined stiffness is always smaller than the smallest individual stiffness.
Two identical springs, each of stiffness 150 N m−1, hang side by side and are joined at the bottom by a light horizontal bar. A load of weight 45 N hangs from the centre of the bar. The springs obey Hooke's law.
What is the total elastic strain energy stored in the two springs?
| Answer | Mark | Notes |
|---|---|---|
| In parallel the combined stiffness is 300 N m−1, so the extension is 45 ÷ 300 = 0.15 m. Total energy = ½FΔx = ½ × 45 × 0.15 = 3.4 J (3.375 J). | ✓ 1 | Answer B |
Answer: B
Common error: Finding the energy in one spring and stopping. Each spring carries 22.5 N and stores 1.7 J; the question asks for the total.
A metal wire has stiffness 1.2 × 104 N m−1 on the straight section of its force–extension graph. It is loaded beyond its elastic limit to an extension of 9.0 mm, where the force is 64 N. The force is then removed, and the unloading line is parallel to the straight section.
What is the permanent extension of the wire?
| Answer | Mark | Notes |
|---|---|---|
| Extension recovered on unloading = F/k = 64 ÷ (1.2 × 104) = 5.3 × 10−3 m = 5.3 mm. Permanent extension = 9.0 − 5.3 = 3.7 mm. | ✓ 1 | Answer A |
Answer: A
Exam tip: On unloading from beyond the elastic limit, only the elastic part of the extension is recovered, along a line with the original stiffness. What is left is the permanent extension.
A force of 18 N extends a spring by 40 mm. The spring obeys Hooke's law.
What is the elastic strain energy stored in the spring?
| Answer | Mark | Notes |
|---|---|---|
| ΔEel = ½FΔx = ½ × 18 × 0.040 = 0.36 J. | ✓ 1 | Answer A |
Answer: A
Common error: Leaving the extension in millimetres. N × mm gives millijoules, so the answer comes out 1000 times too large.
A student hangs a spring from a clamp and reads the position of a pointer on its lower end against a vertical metre rule. She adds weights and records the length of the spring.
Force / N: 0, 1.0, 2.0, 3.0, 4.0, 5.0
Length / mm: 240, 263, 287, 310, 334, 371
Use the readings at 0 and 4.0 N to calculate the stiffness of the spring.
State the evidence in the results that the limit of proportionality has been exceeded at 5.0 N.
The uncertainty in the extension at 4.0 N is ± 1 mm. Calculate the percentage uncertainty in this extension.
Suggest one technique the student should use to make each length reading accurate.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| extension = 334 − 240 = 94 mm | ✓ 1 | |
| k = F/Δx = 4.0 ÷ 0.094 | ✓ 1 | |
| = 43 N m−1 (42.6 N m−1) | ✓ 1 | |
| Part (b) | ||
| the last 1.0 N increased the length by 37 mm, compared with 23–24 mm for each earlier 1.0 N | ✓ 1 | |
| Part (c) | ||
| 1 ÷ 94 × 100 = 1.1% | ✓ 1 | |
| Part (d) | ||
| read the scale at eye level, level with the pointer, to avoid parallax / check the rule is vertical with a set square / wait until the spring has stopped oscillating before reading | ✓ 1 | |
Exam tip: Find the limit of proportionality from the increments: equal steps in force should give equal steps in extension. The first step that breaks the pattern shows the limit has been passed.
The body of a small car is supported by four identical suspension springs, one near each wheel. Each spring has stiffness 3.0 × 104 N m−1. The body has mass 1100 kg and its weight is shared equally between the springs. The springs obey Hooke's law. g = 9.81 N kg−1.
Show that the combined stiffness of the four springs is 1.2 × 105 N m−1.
Calculate the compression of the springs due to the weight of the body.
Passengers of total mass 300 kg get into the car. Calculate the extra compression.
Calculate the total elastic strain energy now stored in the four springs.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the springs act in parallel, so stiffnesses add: 4 × 3.0 × 104 = 1.2 × 105 N m−1 | ✓ 1 | |
| Part (b) | ||
| weight = 1100 × 9.81 = 10 800 N | ✓ 1 | |
| compression = 10 791 ÷ 1.2 × 105 = 0.090 m (0.0899 m) | ✓ 1 | |
| Part (c) | ||
| 300 × 9.81 ÷ 1.2 × 105 = 0.025 m (0.0245 m) | ✓ 1 | |
| Part (d) | ||
| total compression = 0.0899 + 0.0245 = 0.114 m | ✓ 1 | |
| ΔEel = ½k(Δx)2 = ½ × 1.2 × 105 × 0.1142 = 790 J (786 J) (allow error carried forward) | ✓ 1 | |
Common error: Finding the extra energy from ½k × (extra compression)2. The springs were already compressed, so the energy depends on the total compression.
A hanging plant pot of weight 18 N is supported by two springs joined end to end. The upper spring P has stiffness 90 N m−1; the lower spring Q has stiffness 180 N m−1. Both obey Hooke's law and have negligible mass.
Show that the combined stiffness k of two springs in series, of stiffnesses k1 and k2, is given by 1/k = 1/k1 + 1/k2.
Calculate the total extension of the two springs.
Calculate the elastic strain energy stored in each spring, and state which spring stores more.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| each spring carries the same (whole) force F | ✓ 1 | |
| the extensions add: F/k = F/k1 + F/k2, and dividing by F gives the result | ✓ 1 | |
| Part (b) | ||
| 1/k = 1/90 + 1/180, so k = 60 N m−1 | ✓ 1 | |
| Δx = 18 ÷ 60 = 0.30 m | ✓ 1 | |
| Part (c) | ||
| P: ½ × 18 × 0.20 = 1.8 J and Q: ½ × 18 × 0.10 = 0.90 J | ✓ 1 | |
| the less stiff spring, P, stores more | ✓ 1 | |
Exam tip: With the same force on each spring, ½FΔx shows that the spring that stretches more — the less stiff one — stores more energy.
In a pinball machine, a ball of mass 0.080 kg rests against a plunger. The plunger spring, of stiffness 420 N m−1, is compressed by 0.065 m and then released, launching the ball along a horizontal channel. The spring obeys Hooke's law.
Calculate the elastic strain energy stored in the compressed spring.
Calculate the speed of the ball as it leaves the plunger, assuming all the stored energy becomes kinetic energy of the ball.
Suggest why the actual speed of the ball is less than this.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| F = 420 × 0.065 = 27.3 N, or use ½k(Δx)2 | ✓ 1 | |
| ΔEel = ½ × 27.3 × 0.065 = 0.89 J (0.887 J) | ✓ 1 | |
| Part (b) | ||
| ½mv2 = 0.887 | ✓ 1 | |
| v = √(2 × 0.887 ÷ 0.080) = 4.7 m s−1 (allow error carried forward) | ✓ 1 | |
| Part (c) | ||
| some of the energy becomes kinetic energy of the plunger and spring / is dissipated by friction as thermal energy | ✓ 1 | |
Exam tip: Energy questions with springs are conservation questions: write “elastic strain energy lost = kinetic energy gained” before you substitute.
A length of nylon fishing line is tested. Its force–extension graph is a straight line from the origin to the limit of proportionality at 24 N and 40 mm, which is also the elastic limit. The line is then loaded further, to an extension of 75 mm at a force of 30 N, and unloaded. The unloading graph is a straight line parallel to the first straight section.
Calculate the stiffness of the line on the straight section.
Calculate the elastic strain energy stored at the limit of proportionality.
Calculate the permanent extension of the line after unloading.
Calculate the energy released by the line as it is unloaded.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 24 ÷ 0.040 = 6.0 × 102 N m−1 (600 N m−1) | ✓ 1 | |
| Part (b) | ||
| ΔEel = ½FΔx = ½ × 24 × 0.040 | ✓ 1 | |
| = 0.48 J | ✓ 1 | |
| Part (c) | ||
| extension recovered = 30 ÷ 600 = 0.050 m | ✓ 1 | |
| permanent extension = 0.075 − 0.050 = 0.025 m (25 mm) | ✓ 1 | |
| Part (d) | ||
| area under the unloading line = ½ × 30 × 0.050 = 0.75 J | ✓ 1 | |
Model sketch
Exam tip: After plastic deformation, the energy given back is the triangle under the unloading line only; the rest of the work done stays in the line as permanent deformation and thermal energy.
Figure 1 shows three springs that obey Hooke's law and have negligible mass. Two identical springs hang side by side from a fixed support and are joined by a light bar. A third spring hangs from the centre of the bar and supports the load. The load is attached to the unstretched arrangement and lowered gently until it hangs at rest.
How far has the load moved down?
| Answer | Mark | Notes |
|---|---|---|
| The upper pair are in parallel: k = 60 + 60 = 120 N m−1, extension 12 ÷ 120 = 0.10 m. The lower spring carries the whole 12 N: extension 12 ÷ 40 = 0.30 m. Total = 0.10 + 0.30 = 0.40 m. | ✓ 1 | Answer B |
Answer: B
Exam tip: Work through a combination one stage at a time: find the force on each stage, then its extension, then add the extensions of stages that are in series.
The graph shows force against extension for two springs, P and Q. Each spring is stretched by a force of 6.0 N.
What is the value of the ratio
(elastic strain energy stored in Q) ÷ (elastic strain energy stored in P)?
| Answer | Mark | Notes |
|---|---|---|
| From the graph, at 6.0 N the extension of P is 2.0 cm and that of Q is 6.0 cm. With the same force, ΔEel = ½FΔx ∝ Δx, so the ratio = 6.0 ÷ 2.0 = 3.0. | ✓ 1 | Answer C |
Answer: C
Exam tip: With the same force, use ½FΔx: the less stiff spring stretches further and stores more energy. With the same extension, the stiffer spring stores more.
The elastic cord of a toy glider launcher becomes stiffer as it is stretched. The graph shows its force–extension graph; the marked points lie on gridline intersections.
Using strips of width 0.10 m, what is the best estimate of the elastic strain energy stored in the cord at an extension of 0.40 m?
| Answer | Mark | Notes |
|---|---|---|
| Area under the graph by trapezia: 0.10 × [(0 + 10)/2 + (10 + 30)/2 + (30 + 60)/2 + (60 + 100)/2] = 0.10 × (5 + 20 + 45 + 80) = 15 J. | ✓ 1 | Answer A |
Answer: A
Common error: Using ½FΔx for a curved graph. Here the curve lies below the straight line from the origin to the end point, so ½FΔx overestimates the energy.
Figure 1 shows a block held against a compressed spring on a smooth track. The block is released, leaves the spring and slides up the curved part of the track. The spring obeys Hooke's law; ignore air resistance. g = 9.81 N kg−1.
What is the greatest height h reached by the block above the horizontal part of the track?
| Answer | Mark | Notes |
|---|---|---|
| ΔEel = ½k(Δx)2 = ½ × 800 × 0.0502 = 1.0 J. All of it becomes gravitational potential energy: mgh = 1.0, so h = 1.0 ÷ (0.20 × 9.81) = 0.51 m. | ✓ 1 | Answer A |
Answer: A
Exam tip: Convert the compression to metres before squaring it: 5.0 cm = 0.050 m, and 0.0502 = 2.5 × 10−3 m2.
The graph shows force against extension for a soft metal wire. The wire is loaded to an extension of 6.0 mm and then unloaded along the dashed line.
How much energy is released by the wire as it is unloaded?
| Answer | Mark | Notes |
|---|---|---|
| The unloading line runs from (6.0 mm, 60 N) to (4.8 mm, 0), so the extension recovered is 1.2 mm. Energy released = area under the unloading line = ½ × 60 × 1.2 × 10−3 = 0.036 J. | ✓ 1 | Answer A |
Answer: A
Exam tip: Only the triangle under the unloading line is returned. The rest of the work done in loading stays in the wire as permanent deformation and thermal energy.
The graph shows force against extension for a spring. The spring is stretched from an extension of 20 mm to an extension of 60 mm.
What is the work done on the spring during this stretch, shown by the shaded area?
| Answer | Mark | Notes |
|---|---|---|
| The shaded area is a trapezium: ½ × (3.0 + 9.0) × 0.040 = 0.24 J. (Equivalently ½k(Δx22 − Δx12) = ½ × 150 × (0.0602 − 0.0202) = 0.24 J.) | ✓ 1 | Answer C |
Answer: C
Common error: Calculating ½k × (40 mm)2. That is only the triangle at the top of the shaded area; the spring already carried 3.0 N at the start of the stretch.
The graph shows force against extension for an aluminium wire that is loaded until it breaks. P is the limit of proportionality; the elastic limit is very close to P. Three readings are marked on the curve.
Calculate the stiffness of the wire for extensions up to P.
A second, identical wire is loaded to an extension of 3.0 mm and the force is then removed. Explain how the graph can be used to find the permanent extension of this wire, and determine its value.
Estimate the work done in stretching the wire to an extension of 3.0 mm.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 60 ÷ (0.50 × 10−3) = 1.2 × 105 N m−1 | ✓ 1 | |
| Part (b) | ||
| the unloading graph is a straight line from the point at 3.0 mm, parallel to OP (same gradient as the straight section) | ✓ 1 | |
| extension recovered = 90 ÷ 1.2 × 105 = 7.5 × 10−4 m = 0.75 mm | ✓ 1 | |
| permanent extension = 3.0 − 0.75 = 2.25 mm ≈ 2.3 mm (allow 2.2–2.3 mm) | ✓ 1 | |
| Part (c) | ||
| area under the graph: triangle to P plus trapezia between the marked readings: ½ × 60 × 0.50 + 0.50 × (60 + 83)/2 + 1.0 × (83 + 87)/2 + 1.0 × (87 + 90)/2 = 15 + 35.75 + 85 + 88.5 = 224 N mm | ✓ 1 | |
| = 0.224 J ≈ 0.22 J (allow 0.21–0.23 J) | ✓ 1 | |
Exam tip: After plastic deformation only the energy under the unloading line comes back — here ½ × 90 N × 0.75 mm ≈ 0.034 J, a small part of the 0.22 J of work done.
Figure 1 shows the force–extension graph for a spring S, and a baby bouncer that uses four springs identical to S. The seat and baby have a total mass of 10 kg. The springs obey Hooke's law and have negligible mass. g = 9.81 N kg−1.
Use the graph to show that the stiffness of S is 1.5 × 103 N m−1.
Show that the combined stiffness of the four springs in the bouncer is equal to the stiffness of one spring S.
Calculate the total extension of the arrangement due to the seat and baby.
Calculate the elastic strain energy stored in each spring.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 60 N ÷ 0.040 m = 1.5 × 103 N m−1 | ✓ 1 | |
| Part (b) | ||
| each strand is two springs in series: 1/k = 1/kS + 1/kS, so k = kS/2 | ✓ 1 | |
| the two strands are in parallel, so their stiffnesses add: 2 × kS/2 = kS | ✓ 1 | |
| Part (c) | ||
| 10 × 9.81 ÷ 1500 = 0.065 m (0.0654 m) | ✓ 1 | |
| Part (d) | ||
| each strand carries half the weight, 49.05 N, so each spring extends 49.05 ÷ 1500 = 0.0327 m | ✓ 1 | |
| ΔEel = ½ × 49.05 × 0.0327 = 0.80 J (allow error carried forward) | ✓ 1 | |
Common error: Giving every spring the full weight. Strands in parallel share the load; springs in series within a strand each carry that strand's share.
A spring that obeys Hooke's law is stretched from an extension of 2.0 cm to an extension of 6.0 cm.
The extra elastic strain energy stored is how many times the energy stored at 2.0 cm?
| Answer | Mark | Notes |
|---|---|---|
| ΔEel = ½k(Δx)2 ∝ (Δx)2. Energy at 6.0 cm = 32 = 9 times the energy at 2.0 cm, so the extra energy is 9 − 1 = 8 times it. | ✓ 1 | Answer C |
Answer: C
Exam tip: Read “extra” carefully: the ratio of the energies is 9, but the energy added is the difference, 9 − 1.
A strip of polymer is stretched and remains elastic. Its force–extension graph is a smooth curve whose gradient decreases. Readings from the curve are: extension 0, 0.020 m, 0.040 m and 0.060 m; force 0, 30 N, 48 N and 56 N.
Using strips of width 0.020 m, what is the best estimate of the elastic strain energy stored at an extension of 0.060 m?
| Answer | Mark | Notes |
|---|---|---|
| Area under the graph by trapezia: 0.020 × [(0 + 30)/2 + (30 + 48)/2 + (48 + 56)/2] = 0.020 × (15 + 39 + 52) = 2.1 J (2.12 J). | ✓ 1 | Answer B |
Answer: B
Exam tip: ½FΔx is only for a straight line through the origin. For a curve, add the areas of strips (or count squares).
A copper wire and a rubber band are each loaded to a large extension and then unloaded.
Compare the behaviour of the two samples. Your answer should refer to the shapes of the loading and unloading force–extension graphs, the final length of each sample and what happens to the energy.
Compare the behaviour of the copper wire and the rubber band.
| Answer | Mark | Notes |
|---|---|---|
| Indicative content: | — | |
| copper: the graph is a straight line at first (obeys Hooke's law), then curves beyond the limit of proportionality and extends a lot for little extra force after the yield point | ✓ 1 | |
| copper: on unloading the line is straight and parallel to the original straight section | ✓ 1 | |
| copper: it is left with a permanent extension — plastic deformation because the elastic limit was exceeded | ✓ 1 | |
| rubber: the loading graph is curved from the start (does not obey Hooke's law) and becomes steeper at large extension | ✓ 1 | |
| rubber: the unloading curve lies below the loading curve, but the band returns to (almost) its original length — the deformation is elastic | ✓ 1 | |
| energy: the area between the loading and unloading graphs is energy not returned — for rubber dissipated as thermal energy (hysteresis), for copper the work done in permanently deforming the wire | ✓ 1 |
Exam tip: In an extended comparison, deal with each point for both samples — shape when loading, shape when unloading, final length, energy — rather than describing one sample and leaving the comparison unstated.
A bungee cord of unstretched length 15 m is tied to a platform. A jumper of mass 65 kg is attached to its free end and steps off the platform from rest. Readings from the force–extension graph of the cord are:
Extension / m: 0, 5.0, 10.0, 15.0, 20.0
Force / kN: 0, 0.90, 2.0, 3.4, 5.2
Treat the jumper as a point mass and ignore air resistance and the mass of the cord. g = 9.81 N kg−1.
State how the readings show that the cord does not obey Hooke's law.
Estimate the elastic strain energy stored in the cord at an extension of 15.0 m.
Calculate the gravitational potential energy lost by the jumper in falling to the point where the extension of the cord is 15.0 m.
Deduce whether the jumper reaches an extension of 15.0 m.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| equal increases in extension give increasing increases in force (0.90, 1.1, 1.4, 1.8 kN) / force is not proportional to extension | ✓ 1 | |
| Part (b) | ||
| energy = area under the force–extension graph, using strips / trapezia | ✓ 1 | |
| 5.0 × (0 + 0.90)/2 + 5.0 × (0.90 + 2.0)/2 + 5.0 × (2.0 + 3.4)/2 = 2.25 + 7.25 + 13.5 kJ | ✓ 1 | |
| = 23 kJ (23.0 kJ) | ✓ 1 | |
| Part (c) | ||
| distance fallen = 15 + 15.0 = 30 m | ✓ 1 | |
| 65 × 9.81 × 30 = 1.9 × 104 J (19 100 J) | ✓ 1 | |
| Part (d) | ||
| no: the energy needed to stretch the cord to 15.0 m (23 kJ) is more than the gravitational potential energy available (19 kJ), so the jumper stops (at a smaller extension) first (allow error carried forward) | ✓ 1 | |
Exam tip: For a curved graph, find the area with strips and say that you are doing so. Then compare energies: an object can only reach a point if the energy available is at least the energy that must be stored there.
A railway wagon of mass 2.4 × 104 kg rolls at 0.80 m s−1 into the end of a siding, where it is stopped by two identical spring buffers acting side by side. Each buffer has stiffness 6.0 × 105 N m−1 and obeys Hooke's law. Ignore friction.
Calculate the kinetic energy of the wagon.
Calculate the maximum compression of the buffers.
Calculate the maximum deceleration of the wagon.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| ½ × 2.4 × 104 × 0.802 = 7.7 × 103 J (7680 J) | ✓ 1 | |
| Part (b) | ||
| the buffers are in parallel: k = 2 × 6.0 × 105 = 1.2 × 106 N m−1 | ✓ 1 | |
| ½k(Δx)2 = 7680 | ✓ 1 | |
| Δx = √(2 × 7680 ÷ 1.2 × 106) = 0.11 m (0.113 m) | ✓ 1 | |
| Part (c) | ||
| maximum force = kΔx = 1.2 × 106 × 0.113 = 1.36 × 105 N | ✓ 1 | |
| a = F/m = 1.36 × 105 ÷ 2.4 × 104 = 5.7 m s−2 (allow error carried forward) | ✓ 1 | |
Exam tip: The force from a spring is greatest at the greatest compression, so the maximum deceleration occurs when the wagon is momentarily at rest.
The graph shows how the elastic strain energy stored in a spring varies with its extension. The spring obeys Hooke's law throughout.
What is the stiffness of the spring?
| Answer | Mark | Notes |
|---|---|---|
| Read a clear point: at 4.0 cm = 0.040 m, E = 0.20 J. From E = ½k(Δx)2, k = 2E/(Δx)2 = 2 × 0.20 ÷ 0.0402 = 250 N m−1. | ✓ 1 | Answer D |
Answer: D
Exam tip: The energy graph is a curve because E ∝ (Δx)2, so its gradient is not the stiffness. Read one clear point and rearrange E = ½k(Δx)2.
The graph shows the loading and unloading curves for a rubber cord that is stretched to 8.0 cm and then released. The marked points are readings taken from the curves.
Using the readings and strips of width 2.0 cm, what percentage of the work done in stretching the cord is dissipated as thermal energy during the cycle?
| Answer | Mark | Notes |
|---|---|---|
| Work done in loading = 0.020 × [(0 + 4)/2 + (4 + 6)/2 + (6 + 8)/2 + (8 + 14)/2] = 0.020 × 25 = 0.50 J. Energy returned = 0.020 × [(0 + 1)/2 + (1 + 2)/2 + (2 + 5)/2 + (5 + 14)/2] = 0.020 × 15 = 0.30 J. Dissipated = 0.20 J, which is 0.20 ÷ 0.50 = 40% of the work done. | ✓ 1 | Answer B |
Answer: B
Exam tip: The area under the loading curve is the work done on the cord; the area under the unloading curve is the energy it gives back. The loop between them is the energy dissipated.
An archer draws back the string of a bow. The graph shows how the force needed to hold the string varies with the draw distance x; readings are marked on the curve. The arrow has a mass of 0.025 kg.
Explain why the energy stored in the bow at a draw distance of 0.50 m cannot be calculated using ½FΔx.
Estimate the elastic strain energy stored in the bow at a draw distance of 0.50 m.
When the string is released, 80% of the stored energy is transferred to kinetic energy of the arrow. Calculate the speed of the arrow.
Suggest what happens to the remaining 20% of the energy.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| ½FΔx applies only when force is proportional to extension (a straight line through the origin); this graph is curved | ✓ 1 | |
| Part (b) | ||
| energy = area under the graph, using five strips (trapezia) of width 0.10 m | ✓ 1 | |
| 0.10 × (22.5 + 65 + 100 + 127.5 + 147.5) | ✓ 1 | |
| = 46.25 J ≈ 46 J | ✓ 1 | |
| Part (c) | ||
| ½ × 0.025 × v2 = 0.80 × 46.25 = 37.0 J | ✓ 1 | |
| v = √(2 × 37.0 ÷ 0.025) = 54 m s−1 (allow error carried forward) | ✓ 1 | |
| Part (d) | ||
| it becomes kinetic energy of the bow limbs and string, or is dissipated as thermal energy and sound | ✓ 1 | |
Exam tip: Name the method for the area (trapezia or counting squares) and state the strip width. For a curve that bends towards the extension axis, as here, trapezia slightly underestimate the area.
A student investigates a spring launcher. A trolley of mass 0.50 kg is pushed against a spring, compressing it by Δx, and released on a horizontal track. A light gate measures the speed v of the trolley after it leaves the spring. The graph shows v2 plotted against (Δx)2, with a best-fit line.
Show that, if all the elastic strain energy becomes kinetic energy of the trolley, v2 = (k/m)(Δx)2, where k is the stiffness of the spring and m the mass of the trolley.
Determine the gradient of the best-fit line. Include its unit.
Calculate the stiffness of the spring.
Suggest why the value of k found in this way may be smaller than the true stiffness of the spring.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| ½k(Δx)2 = ½mv2 | ✓ 1 | |
| cancel the ½ and divide by m: v2 = (k/m)(Δx)2 | ✓ 1 | |
| Part (b) | ||
| large triangle, e.g. 1.40 ÷ (35 × 10−4) | ✓ 1 | |
| = 4.0 × 102 s−2 (400 s−2; allow 390–410 s−2; unit required) | ✓ 1 | |
| Part (c) | ||
| k = gradient × m = 400 × 0.50 = 2.0 × 102 N m−1 (allow error carried forward) | ✓ 1 | |
| Part (d) | ||
| not all the elastic strain energy becomes kinetic energy of the trolley — some becomes kinetic energy of the spring or is dissipated by friction — so each v2 is smaller than predicted | ✓ 1 | |
Exam tip: Arrange the equation in the form y = mx before you read the graph: here y = v2, x = (Δx)2 and the gradient is k/m. Remember the 10−4 in the axis label.
Figure 1 shows a buffer at the end of a toy-car track, drawn to scale. Both springs are fixed to the wall and obey Hooke's law. A toy car of mass 0.50 kg hits the plate and compresses the outer spring; once the plate reaches the end of the inner spring, both springs are compressed together. The greatest compression of the outer spring is 6.0 cm. Ignore friction and the mass of the plate.
Use Figure 1 to find the compression at which the plate first touches the inner spring, and calculate the force on the plate at that compression.
Calculate the force on the plate at the greatest compression of 6.0 cm.
Calculate the total elastic strain energy stored at the greatest compression.
Calculate the speed of the toy car just before it hits the plate.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 12.0 − 8.0 = 4.0 cm | ✓ 1 | |
| F = 800 × 0.040 = 32 N | ✓ 1 | |
| Part (b) | ||
| beyond 4.0 cm the springs act in parallel: k = 800 + 2400 = 3200 N m−1 | ✓ 1 | |
| F = 32 + 3200 × 0.020 = 96 N | ✓ 1 | |
| Part (c) | ||
| outer: ½ × 800 × 0.0602 = 1.44 J; inner: ½ × 2400 × 0.0202 = 0.48 J | ✓ 1 | |
| total = 1.92 J ≈ 1.9 J | ✓ 1 | |
| Part (d) | ||
| ½ × 0.50 × v2 = 1.92, so v = 2.8 m s−1 (allow error carried forward) | ✓ 1 | |
Exam tip: Find each spring's energy from its own compression: the inner spring is compressed only by the distance the plate moves after reaching it, here 2.0 cm.
A load of weight 6.0 N is attached to the lower end of a vertical spring that obeys Hooke's law. The load is held so that the spring is unstretched and is then released. The graph shows two energies plotted against the extension Δx of the spring as the load falls: the gravitational potential energy lost by the load and the elastic strain energy stored in the spring. Ignore air resistance and the mass of the spring.
State which graph, A or B, shows the gravitational potential energy lost by the load, and explain why it is a straight line.
Use the graph to determine the stiffness of the spring.
Explain why the load is momentarily at rest when the extension is 0.10 m.
Use the graph to find the extension at which the kinetic energy of the load is greatest, and the value of this kinetic energy.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| A, because the energy lost = weight × distance fallen = 6.0 × Δx, which is proportional to Δx | ✓ 1 | |
| Part (b) | ||
| a point on B, e.g. 0.60 J at 0.10 m, with ΔEel = ½k(Δx)2 | ✓ 1 | |
| k = 2 × 0.60 ÷ 0.102 = 120 N m−1 | ✓ 1 | |
| Part (c) | ||
| the graphs meet there: all the gravitational potential energy lost has been stored as elastic strain energy, so the kinetic energy is zero | ✓ 1 | |
| Part (d) | ||
| kinetic energy = vertical gap between A and B, which is greatest at Δx = 0.050 m (where kΔx = weight) | ✓ 1 | |
| kinetic energy = 0.30 − 0.15 = 0.15 J | ✓ 1 | |
Exam tip: With energy graphs, the kinetic energy is the gap between energy lost and energy stored. The gap is greatest where the spring force equals the weight — the equilibrium position.
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