Answers · Stretching springs

Spring constant and the limit of proportionality — answer key

Check meaning, not exact wording. These are teaching points, not official Cambridge marks.

1. SQ13 · Practice

Model answer. b. The spring constant is the force per unit extension: k = F / x, the force on the spring divided by its extension. With the force in newtons and the extension in metres, it is measured in N/m. A stiffer spring needs more force for each metre of extension, so it has a bigger spring constant.

A full answer includes

I chose b.
I wrote its unit: N/m.
I did not choose a, which is extension ÷ force, the other way round.

Also accept

b: the force per unit extension, k = F / x, in N/m.

Watch for

“a: the extension of the spring for each newton of load.” — That is the other way round, x / F, which is 1/k. The spring constant is force per unit extension, F / x.
“c: the largest load that the spring can safely hold.” — The spring constant says how much force each metre of stretch needs, not how much the spring can hold.
“d: the load at which the graph stops being a straight line.” — That is the limit of proportionality. The spring constant is the force per unit extension on the straight part.

2. SQ14 · Practice

Model answer. First change the extension to metres: 3.5 cm = 0.035 m. Then k = F / x = 4.2 N ÷ 0.035 m = 120 N/m. Dividing 4.2 by 3.5 gives 1.2, but that is in newtons per centimetre: 1.2 N/cm = 120 N/m.

Checked automatically

Spring constant: 120 N/m (within ±1 N/m)
- Common wrong answer 1.2 N/m: 1.2 is in newtons per centimetre. Change 3.5 cm to 0.035 m before you divide: 4.2 N ÷ 0.035 m.
- Common wrong answer 1.2 (no unit): 4.2 ÷ 3.5 = 1.2 is in N/cm. Change the extension to metres first, 0.035 m, and give the unit N/m.
- Common wrong answer 0.0083 : You divided the extension by the load. The spring constant is force ÷ extension: 4.2 N ÷ 0.035 m.
- Common wrong answer 14.7 : You multiplied the load by the extension. Divide instead: k = F ÷ x.
- Common wrong answer 0.833 : You divided 3.5 by 4.2, extension ÷ force. The spring constant is force ÷ extension, with the extension in metres.

A full answer includes

Method: x = 3.5 cm = 0.035 m.
Method: k = F / x = 4.2 N ÷ 0.035 m.
Answer: 120 N/m.
I did not write 1.2 N/m, which divides by the extension in centimetres.

Also accept

120 N/m (4.2 N ÷ 0.035 m).

Watch for

“k = 4.2 ÷ 3.5 = 1.2 N/m.” — 1.2 is in N/cm, because the extension was in centimetres. With the extension in metres, 4.2 N ÷ 0.035 m = 120 N/m.

3. SQ15 · Practice

Model answer. Rearrange k = F / x to x = F / k. Then x = 1.82 N ÷ 28 N/m = 0.065 m. There are 100 cm in a metre, so the extension is 0.065 × 100 = 6.5 cm.

Checked automatically

Extension: 6.5 cm (within ±0.05 cm)
- Common wrong answer 0.065 cm: 0.065 is the extension in metres. Change it to centimetres: × 100.
- Common wrong answer 0.065 (no unit): 0.065 is the extension in metres. Give it in centimetres, 0.065 × 100, with the unit cm.
- Common wrong answer 51 : You multiplied k by F. Rearrange k = F / x to x = F / k: 1.82 N ÷ 28 N/m.
- Common wrong answer 15.4 : You divided k by F. Use x = F / k: 1.82 N ÷ 28 N/m = 0.065 m.

A full answer includes

Method: x = F / k = 1.82 N ÷ 28 N/m.
Method: x = 0.065 m.
Answer: 6.5 cm.
I did not give 0.065 as the extension in centimetres.

Also accept

6.5 cm (0.065 m).

Watch for

“x = k × F = 51.” — Rearrange k = F / x by multiplying both sides by x and dividing by k: x = F / k.

4. SQ16 · Practice

Options as numbered on the pupil sheet: 1 — the force per unit extension of a spring; 2 — the point beyond which the extension is no longer proportional to the load; 3 — the increase in length of the spring; 4 — equal to the spring constant, on the straight part; 5 — equal to 1 ÷ the spring constant, on the straight part; 6 — the biggest load a spring can hold; 7 — the length of the spring with no load on it.

Model answer. The spring constant is the force per unit extension, k = F / x, in N/m. The limit of proportionality is the point beyond which the extension is no longer proportional to the load: where the line stops being straight. The extension is the increase in length of the spring. With the load up the side, the gradient of the straight part is load ÷ extension, F / x, which is the spring constant. With the extension up the side, the gradient is extension ÷ load, x / F, which is 1 ÷ k. Nothing here is the biggest load a spring can hold, or its length with no load.

A full answer includes

I matched spring constant to the force per unit extension.
I matched limit of proportionality to the point beyond which the extension is no longer proportional to the load.
I matched extension to the increase in length.
I matched the gradient with the load up the side to the spring constant.
I matched the gradient with the extension up the side to 1 ÷ the spring constant.
I did not match the limit of proportionality to the biggest load a spring can hold.

Also accept

spring constant – a; limit of proportionality – b; extension – c; gradient with the load up the side – d; gradient with the extension up the side – e.

Watch for

“limit of proportionality – the biggest load a spring can hold.” — Beyond the limit of proportionality the spring still stretches and still holds the load. It is only where the extension stops being proportional to the load.
“extension – the length of the spring with no load.” — That is its original length. The extension is how much longer it gets: the reading minus the zero reading.
“gradient with the extension up the side – equal to the spring constant.” — With the extension up the side, the gradient is extension ÷ load, x / F. That is 1 ÷ k, so on this graph a steeper line means a softer spring.

5. SQ17 · Practice

Model answer. (1) spring constant and (2) N/m: force ÷ extension is the force per unit extension, in newtons per metre. (3) half as big: the same force gives twice the extension, so the force per metre of extension is halved. (4) less: a smaller spring constant gives a less steep line when the force is up the side. (5) limit of proportionality: beyond 900 N the extension is no longer proportional to the force. (6) flattens: each newton now gives more extension than before, so the line bends over towards the extension axis. (7) beyond: k = F / x only describes the wire up to the limit of proportionality.

A full answer includes

(1) I wrote spring constant.
(2) I wrote N/m, or newtons per metre.
(3) I wrote half as big, halved, half or smaller.
(4) I wrote less or not as.
(5) I wrote limit of proportionality.
(6) I wrote flattens, or bends over towards the extension axis.
(7) I wrote beyond, above or more than.
I did not write that the longer wire has the bigger spring constant.

Also accept

spring constant; N/m (newtons per metre); half as big (halved, half, smaller); less (not as); limit of proportionality; flattens (bends over towards the extension axis); beyond (above, more than).

Watch for

“(3) twice as big.” — The longer wire stretches further for the same force, so it needs fewer newtons for each metre: its spring constant is smaller.
“(4) more.” — With the force up the side, a wire that stretches further for the same force has a less steep line.
“(5) breaking point.” — The wire does not break at 900 N; it keeps stretching. 900 N is where the extension stops being proportional to the force.
“(6) gets steeper.” — With the force up the side, beyond the limit each newton gives more extension than before, so the line bends over towards the extension axis: it flattens.

6. SQ18 · Practice

Model answer. Draw a big triangle under the line, from the origin to (10.0 cm, 4.0 N). The rise is 4.0 N and the run is 10.0 cm = 0.100 m. The gradient is 4.0 N ÷ 0.100 m = 40 N/m, and this is the spring constant. Without changing the run to metres you get 0.40, which is in N/cm.

Checked automatically

Spring constant: 40 N/m (within ±2 N/m)
- Common wrong answer 0.4 N/m: 0.40 is in newtons per centimetre. Change the run to metres before you divide: for example 4.0 N ÷ 0.100 m.
- Common wrong answer 0.4 (no unit): 0.40 is the gradient in N/cm. Change the run to metres, for example 4.0 N ÷ 0.100 m, and give the unit N/m.
- Common wrong answer 2.5 : You divided the extension by the load, 10 cm ÷ 4.0 N. The gradient is the rise, the load, divided by the run, the extension.
- Common wrong answer 0.025 : You divided the extension in metres by the load. Turn it over: k = load ÷ extension = 4.0 N ÷ 0.100 m.

A full answer includes

Method: I drew a big triangle on the line, for example from the origin to (10.0 cm, 4.0 N).
Method: run = 10.0 cm = 0.100 m.
Method: gradient = 4.0 N ÷ 0.100 m.
Answer: about 40 N/m.
I did not write 0.40 N/m, the gradient in N/cm.

Also accept

about 40 N/m (4.0 N ÷ 0.100 m).

Watch for

“k = 0.40 N/m.” — 0.40 is in N/cm, because the run was in centimetres. 0.40 N/cm = 40 N/m.

7. SQ24 · Practice

Model answer. Row B. On the straight parts, X has k = 2.7 N ÷ 0.030 m = 90 N/m and Y has k = 7.2 N ÷ 0.160 m = 45 N/m. X needs more force for each metre of extension, so X is stiffer, and its line is steeper. The limit of proportionality is where each straight part ends: 2.7 N for X and 7.2 N for Y, so Y has the larger load at its limit. Stiffness and the limit of proportionality are different things: a stiffer spring does not have to stay proportional to a bigger load.

A full answer includes

I worked out X’s spring constant as 90 N/m.
I worked out Y’s spring constant as 45 N/m.
I chose X as the stiffer spring.
I chose Y for the larger load at the limit: 7.2 N, not 2.7 N.
I chose row B.
I did not assume that the stiffer spring also has the larger limit of proportionality.

Also accept

Row B

Watch for

“Row A.” — X is stiffer, but its straight part ends at 2.7 N and Y’s at 7.2 N. A stiffer spring does not have to have the larger limit.
“Row C.” — Y stretches further, 16.0 cm, but for a bigger load. Per unit extension, X needs 90 N/m and Y only 45 N/m, so X is stiffer. And Y’s straight part goes up to the larger load.
“Row D.” — Y does have the larger limit, 7.2 N. But Y is the softer spring: it needs only 45 N/m, against X’s 90 N/m.

8. SQ19 · Practice

Model answer. Q. Up to Q, each 1.0 N adds 2.0 cm, and the points lie on a straight line through the origin. From Q to R, 1.0 N adds 3.5 cm, and from R to S it adds 4.5 cm: the extension is no longer proportional to the load, and the graph flattens. The limit of proportionality is where the straight line ends, at Q. The spring still stretches beyond it: S is not the biggest load it can hold.

A full answer includes

I chose Q.
I checked that the points up to Q lie on the straight line through the origin.
I checked that R and S fall below that line.
I did not choose R, the first point off the line.

Also accept

b: Q, the last point on the straight line.

Watch for

“P, because the line starts to change there.” — The points after P, up to Q, are still on the straight line through the origin. The limit is the last point on the straight part, not an earlier one.
“S, because it is the biggest load.” — The limit of proportionality is not the biggest load. It is where the line stops being straight, and the spring still stretches beyond it.
“R, the first point off the line.” — R is already beyond the limit. The limit is the end of the straight part, the last point on the line: Q.

9. SQ20 · Practice

Model answer. (a) The spring constant is in N/m, so change the extension to metres: 25 mm = 2.5 cm = 0.025 m. Rearrange k = F / x to F = k × x = 1200 N/m × 0.025 m = 30 N. (b) Weight = mass × g, so mass = weight ÷ g = 30 N ÷ 9.8 N/kg = 3.06 kg, which is 3.1 kg to 2 significant figures.

Checked automatically

(a) Weight of the bag: 30 N (within ±0.5 N)
- Common wrong answer 30000 : You used 25 mm as 25 m. 25 mm = 0.025 m.
- Common wrong answer 300 : You used 0.25 m. 25 mm = 2.5 cm = 0.025 m.
- Common wrong answer 3 : You used 0.0025 m. 25 mm = 2.5 cm = 0.025 m.
- Common wrong answer 48 : You divided k by the extension. Rearranged, k = F / x gives F = k × x.
(b) Mass of the bag, to 2 significant figures: 3.1 kg (2 significant figures)
- Common wrong answer 294 : You multiplied the weight by g. Mass = weight ÷ g.
- Common wrong answer 3.0 : You used g = 10 N/kg. Use 9.8 N/kg: 30 N ÷ 9.8 N/kg.

A full answer includes

(a) Method: x = 25 mm = 0.025 m.
(a) Method: F = k × x = 1200 N/m × 0.025 m.
(a) Answer: 30 N.
(b) Method: m = W ÷ g = 30 N ÷ 9.8 N/kg = 3.06 kg.
(b) Answer: 3.1 kg, to 2 significant figures.
I did not use 25 mm as 0.25 m, which gives 300 N.

Also accept

(a) 30 N; (b) 3.1 kg.

Watch for

“F = k ÷ x = 1200 ÷ 25 = 48 N.” — k = F / x rearranges to F = k × x: multiply, with the extension in metres.
“25 mm = 0.25 m.” — There are 1000 mm in a metre: 25 mm = 0.025 m.

10. SQ21 · Practice

Model answer. (a) The limit of proportionality is where the points leave the straight line. With 1.0 N steps, 3.0 N is the last point on the line and 4.0 N the first point off it, so the limit could be anywhere between 3.0 N and 4.0 N. (b) With the extra points, 3.5 N is the last point on the line and 3.75 N the first off it, so the limit lies between 3.5 N and 3.75 N. (c) The only loads that can narrow this range are between 3.5 N and 3.75 N, for example 3.6 N and 3.7 N. Loads above 3.75 N are already beyond the limit, and loads below 3.5 N are already known to be on the line.

A full answer includes

(a) I wrote between 3.0 N and 4.0 N.
(b) I wrote between 3.5 N and 3.75 N.
(c) I suggested two loads between 3.5 N and 3.75 N, such as 3.6 N and 3.7 N.
(c) I did not suggest loads above 3.75 N, which are already beyond the limit.

Also accept

(a) 3.0 N to 4.0 N; (b) 3.5 N to 3.75 N; (c) two loads between 3.5 N and 3.75 N, such as 3.6 N and 3.7 N.

Watch for

“(c) Loads above 4.0 N, to see the line bend more clearly.” — The bend is already known to start between 3.5 N and 3.75 N. Loads above that range are off the line and cannot narrow it down.
“(a) The limit is at 4.0 N, the first point off the line.” — 4.0 N is already beyond the limit. The limit lies somewhere between the last point on the line and the first point off it.

11. SQ22 · Practice

Model answer. (a) The points lie on a straight line through the origin up to 3.0 N, so the limit of proportionality is about 3.0 N. (b) Beyond 3.0 N the line curves upwards and gets steeper. (c) On this graph the gradient is extension ÷ load. A steeper line means more extension for each newton, so the spring is getting easier to stretch, not stiffer. With the load up the side, the same results would flatten. (d) On the straight part, 3.0 N gives 15.0 cm, so the gradient is 15.0 cm ÷ 3.0 N = 5.0 cm per newton. (e) This gradient is x / F, which is 1/k, not k. In metres it is 0.050 m/N, so k = 1 ÷ 0.050 m/N = 20 N/m. Check: 3.0 N ÷ 0.150 m = 20 N/m.

A full answer includes

(a) I wrote about 3.0 N.
(a) I chose it because the line stops being straight there.
(b) I wrote that the line curves upwards and gets steeper.
(c) I wrote that with the extension up the side, a steeper line means more extension for each newton.
(c) I wrote that the spring is getting easier to stretch, not stiffer.
(d) Method: gradient = 15.0 cm ÷ 3.0 N.
(d) Answer: 5.0 cm per newton (5.0 cm/N).
(e) I wrote that this gradient is extension ÷ load, which is 1/k.
(e) Method: 5.0 cm/N = 0.050 m/N.
(e) Method: k = 1 ÷ 0.050 m/N.
(e) Answer: 20 N/m.
I did not give 5, the gradient of this graph, as the spring constant.

Also accept

(a) about 3.0 N; (b) curves upwards; (c) steeper here means more extension per newton, easier to stretch; (d) 5.0 cm per newton (0.050 m/N); (e) 20 N/m, because the gradient is 1/k.

Watch for

“(e) k = 5.0 N/m, the gradient.” — That gradient is extension ÷ load, 1/k, in cm per newton. Turn it over, with metres: k = 1 ÷ 0.050 m/N = 20 N/m.
“(e) k = 1 ÷ 5.0 = 0.2 N/m.” — 0.2 is in newtons per centimetre. Change the gradient to metres per newton first, 0.050 m/N; then 1 ÷ 0.050 = 20 N/m.
“(a) The limit is 4.0 N, the last point.” — The limit of proportionality is where the line stops being straight, at about 3.0 N. The spring still stretches beyond it.

12. SQ23 · Practice

Model answer. (a) Up to the limit of proportionality, k = F / x holds, so x = F / k = 2.0 N ÷ 16 N/m = 0.125 m = 12.5 cm. (b) Her prediction is x = F / k = 2.4 N ÷ 16 N/m = 0.15 m = 15 cm. But 2.4 N is beyond the limit of proportionality, where the extension is no longer proportional to the load. Beyond the limit, each extra newton stretches the spring more than before, so the real extension will be bigger than 15 cm.

Checked automatically

(a) Extension at the limit of proportionality: 12.5 cm (within ±0.5 cm)
- Common wrong answer 0.125 cm: 0.125 m is right, but it is in metres. Change it to centimetres.
- Common wrong answer 8 : You divided k by F. Use x = F / k: 2.0 N ÷ 16 N/m.
- Common wrong answer 32 : You multiplied k by F. Use x = F / k: 2.0 N ÷ 16 N/m.
(b) Her prediction for 2.4 N: 15 cm (within ±0.5 cm)
- Common wrong answer 0.15 cm: 0.15 m is the prediction in metres. Give it in centimetres.
- Common wrong answer 38.4 : You multiplied k by F. Use x = F / k: 2.4 N ÷ 16 N/m.
- Common wrong answer 6.67 : You divided k by F. Use x = F / k: 2.4 N ÷ 16 N/m.

A full answer includes

(a) Method: x = F / k = 2.0 N ÷ 16 N/m = 0.125 m.
(a) Answer: 12.5 cm.
(b) Method: x = 2.4 N ÷ 16 N/m = 0.15 m.
(b) Answer: 15 cm.
I wrote that the real extension will be bigger than 15 cm.
I gave the reason: 2.4 N is beyond the limit, so the extension is no longer proportional to the load.
I did not give 0.125 or 0.15 as an extension in centimetres.

Also accept

(a) 12.5 cm; (b) 15 cm (15.0 cm); the real extension will be bigger, because 2.4 N is beyond the limit of proportionality.

Watch for

“(a) x = k ÷ F = 8.” — Rearrange k = F / x to x = F / k: 2.0 N ÷ 16 N/m.
“The real extension at 2.4 N will be 15 cm, as predicted.” — 2.4 N is beyond the limit, so the extension is no longer proportional to the load. The spring stretches further than F / k predicts.
“The real extension will be smaller, because the spring gets stiffer beyond the limit.” — With the load up the side, the graph flattens beyond the limit: each newton gives more extension than before, not less.

13. ST2 · Apply

Model answer. Row A. The other five points lie on one straight line through the origin: each 90 N adds 4.5 cm, so k = 90 N ÷ 0.045 m = 450 N ÷ 0.225 m = 2000 N/m. On that line, 270 N should give 3 × 4.5 cm = 13.5 cm, but she recorded 15.1 cm. Beyond a limit of proportionality the later points would stay off the line, but here 360 N and 450 N are back on it. So the 270 N reading is anomalous, probably misread, and it is left out when k is found.

A full answer includes

I checked that 90 N, 180 N, 360 N and 450 N all fit 4.5 cm for each 90 N.
I found that 270 N should give 13.5 cm, not 15.1 cm.
I noticed that the 360 N and 450 N points are back on the line, so 270 N is not the limit.
Method: k = 90 N ÷ 0.045 m.
Answer: 2000 N/m.
I chose row A.
I did not use the 270 N reading to find k.

Also accept

Row A

Watch for

“Row B.” — 20 is in newtons per centimetre: 90 N ÷ 4.5 cm. Change 4.5 cm to 0.045 m before you divide: 2000 N/m.
“Row C.” — Beyond a limit of proportionality, the later points stay off the line. Here 360 N and 450 N are back on it, so one reading was misread, not a limit.
“Row D.” — It is not a limit, and 270 N ÷ 0.151 m ≈ 1800 N/m uses the bad reading. Leave it out and use the points on the line.