Force and extension

1 Plan and risk assessment

Hooke’s law: up to the limit of proportionality, the extension of a spring is directly proportional to the force on it. The spring constant k is the force needed per metre of extension, so a stiffer spring has a bigger k.

F = k e

Prediction

2 Apparatus

• ✓Clamp stand with two bosses and clamps Holds the spring and the metre rule.
• ✓Spring The object being stretched.
• ✓Metre rule (resolution 1 mm) Measures the position of the pointer.
• ✓Pointer (splint or wire) Marks the bottom of the spring against the scale.
• ✓100 g mass hanger Carries the masses; counts as the first 100 g.
• ✓100 g slotted masses (×7) Each adds 0.98 N of weight.
• ✓G-clamp or heavy weight Stops the stand toppling.
• ✓Eye protection In case the spring snaps or springs off the hook.

3 Method: your choices

• Which version are you doing?: AQA required practical 6: force and extensionFind the spring constant and the limit of proportionality.
• Secure the clamp standSafety: DoneAbout 800 g will hang from the clamp arm. A secured stand cannot topple.
• Line up zero with the pointer: DoneEach reading is then the extension itself, with no zero error from a pointer resting between marks.
• Put on eye protectionSafety: DoneA spring can snap or jump off its hook when loaded.
• Read again while unloading: Read again while unloadingTwo readings at each force, so you can work out a mean, and you can see whether the spring went back to its own length.
• Read at eye level: Crouch so your eye is level with the pointerParallax is kept to a fraction of a millimetre, so each extension and k are close to the true values.
• Wait for the spring to stop bouncing: Wait until the pointer is stillEach reading is of the pointer at rest, so the points lie close to the line.

Anomaly: No anomalous reading.

6 Analysis

1. Extension

   extension = pointer reading − pointer reading with no masses

   Work out the extension of the spring when the force on it is 4.90 N.

   19.0 cm Right first time

2. Shape of the graph

   What does your graph show?

   A straight line through the origin for every load: extension is directly proportional to force. Right first time

3. Gradient (extension per newton)

   gradient = change in y ÷ change in x

   3.87 cm/N Right first time

4. Limit of proportionality

   Up to what force was the extension directly proportional to the force? Give the last point on the straight part.

   Every load: it never reached its limit of proportionality Right first time

5. Spring constant

   F = k e, so k = F ÷ e = 1 ÷ gradient (with e in m)

   Your gradient is 3.87 cm/N. Use it to calculate the spring constant k in N/m.

   25.9 N/m Right first time

7 Conclusion and evaluation

Conclusion

Sources of error

Improvement

Repeatable or reproducible

9 Exam questions on your data

On your own readings, self-marked against the mark scheme, with error carried forward where it applies. Total 33/33.

1. 01.1[2 marks]

   You investigated how the extension of a spring depends on the force applied to it.

   The spring hung from a clamp stand with a pointer fixed to its bottom end. You added 100 g masses one at a time and read the position of the pointer on a vertical metre rule.

   Give two control variables in this investigation.

   Your answer

   The original length of the spring. The diameter of the spring.

   2/2 self-marked

   • The (original) length of the spring
   • The diameter of the spring (or thickness of its wire)
   • The material the spring is made from
   • The spring itself / its spring constant (use the same spring)
   • The number of coils (turns) on the spring

   Self-marked: 2 of 2 marks. This is your own judgement against the mark scheme.

   Model answer: The original length of the spring. The diameter of the spring.

2. 01.2[1 mark]

   You fixed a pointer to the bottom of the spring and read its position on the metre rule each time.

   Give one reason why this improves the measurements of extension.

   Your answer

   The pointer marks the same point on the spring for every reading.

   1/1 self-marked

   • The same point on the spring is measured every time (a fixed reference point)
   • The pointer can be close to (touching) the scale, so parallax is reduced
   • A thin pointer is easier to read precisely against the millimetre marks than the end of the coil or hook

   Self-marked: 1 of 1 mark. This is your own judgement against the mark scheme.

   Model answer: The pointer marks the same point on the spring for every reading.

3. 01.3[1 mark]

   You used a set square to check that the metre rule was vertical.

   Explain why the metre rule needs to be vertical.

   Your answer

   The spring stretches vertically, so a rule at an angle would give inaccurate measurements of the extension.

   1/1 self-marked

   • So that the extension is measured accurately (a rule at an angle gives wrong readings of extension)

   Self-marked: 1 of 1 mark. This is your own judgement against the mark scheme.

   Model answer: The spring stretches vertically, so a rule at an angle would give inaccurate measurements of the extension.

4. 01.4[1 mark]

   You repeated your readings while taking the masses off the spring.

   Give one reason why it is good practice to repeat readings.

   Your answer

   To identify any anomalous results.

   1/1 self-marked

   • To identify (spot) anomalous results, or to reduce the effect of random error (allow: to calculate a mean spring constant; ignore more accurate or more reliable on their own)

   Self-marked: 1 of 1 mark. This is your own judgement against the mark scheme.

   Model answer: To identify any anomalous results.

5. 01.5[1 mark]

   Write down the equation that links extension (e), force (F) and spring constant (k).

   Your answer

   F = k e

   1/1 self-marked

   • F = k e or force = spring constant × extension (allow F = k x, and any correct rearrangement, in words or symbols)

   Self-marked: 1 of 1 mark. This is your own judgement against the mark scheme.

   Model answer: force = spring constant × extension F = k e

6. 02.1[4 marks]

   You plotted extension in cm on the y-axis against force in N on the x-axis.

   On the straight part of your graph, one of your readings was a force of 7.84 N with an extension of 30.3 cm.

   Calculate the spring constant of the spring.

   Give the unit.

   Your answer

   25.9 N/m

   Working: e = 30.3 cm = 0.303 m 7.84 = k × 0.303 k = 7.84 ÷ 0.303 k = 25.9 N/m

   4/4

   • Extension in metres: 30.3 cm = 0.303 m
   • Substitution: 7.84 = k × 0.303, or k = 7.84 ÷ 0.303
   • k = 25.9
   • Unit: N/m

   Full marks. That is exactly what the mark scheme is looking for.

   Model answer: e = 30.3 cm = 0.303 m 7.84 = k × 0.303 k = 7.84 ÷ 0.303 k = 25.9 N/m

7. 02.2[3 marks]

   elastic potential energy = 0.5 × spring constant × (extension)² Ee = ½ k e²

   Calculate the elastic potential energy stored in the spring when the extension is 15.2 cm.

   Use the spring constant you calculated earlier in this question and the Physics Equations Sheet.

   Give your answer to 2 significant figures.

   Your answer

   0.30

   Working: e = 15.2 cm = 0.152 m Ee = 0.5 × 25.9 × 0.152² Ee = 0.30 J

   3/3

   • Extension in metres: 15.2 cm = 0.152 m
   • Substitution: Ee = 0.5 × 25.9 × 0.152² (allow their value of k)
   • Ee = 0.30 J (allow error carried forward from the spring constant)

   Full marks. Well done.

   Model answer: e = 15.2 cm = 0.152 m Ee = 0.5 × 25.9 × 0.152² Ee = 0.30 J

8. 02.3[3 marks]

   A student stretches a spring with spring constant 26 N/m. It stores 0.1872 J of elastic potential energy.

   elastic potential energy = 0.5 × spring constant × (extension)² Ee = ½ k e²

   Calculate the extension of the spring.

   Use the Physics Equations Sheet.

   Give your answer in centimetres.

   Your answer

   12 cm

   Working: 0.1872 = 0.5 × 26 × e² e² = (2 × 0.1872) ÷ 26 = 0.0144 e = √0.0144 = 0.12 m e = 12 cm

   3/3

   • Rearrangement: e² = 2 × 0.1872 ÷ 26 (= 0.0144)
   • Square root: e = 0.12 m
   • e = 12 cm

   Full marks. Clear and precise.

   Model answer: 0.1872 = 0.5 × 26 × e² e² = (2 × 0.1872) ÷ 26 = 0.0144 e = √0.0144 = 0.12 m e = 12 cm

9. 03.1[2 marks]

   How does a graph of force against extension show that extension is directly proportional to force?

   Your answer

   The line is straight and it passes through the origin.

   2/2 self-marked

   • The line is straight (allow: constant gradient; the line does not curve)
   • and it passes through the origin (dependent on the first mark)

   Self-marked: 2 of 2 marks. This is your own judgement against the mark scheme.

   Model answer: The line is straight and it passes through the origin.

10. 03.2[1 mark]

    Your spring stayed proportional for every load you used. A classmate used a different spring and kept adding masses.

    Table 2 shows the classmate’s results.

    Which reading in Table 2 is the first one taken beyond the limit of proportionality?

    Select one row.

    Table 2

    Force in N	Extension in cm
    0.00	0.0
    0.98	4.1
    1.96	8.0
    2.94	11.9
    3.92	16.0
    4.90	20.0
    5.88	25.8
    6.86	33.5
    7.84	43.2

    Your answer

    the reading at 5.88 N

    1/1

    • The reading at 5.88 N: the first point that is off the straight line through the earlier points

    Full marks. Clear and precise. Up to 4.90 N each extra 0.98 N adds about the same extension. At 5.88 N the extension jumps by much more, so the graph curves away from the straight line.

    Model answer: The reading at 5.88 N. Up to 4.90 N the extension goes up by about the same amount for each mass; at 5.88 N it increases by much more.

11. 03.3[2 marks]

    Table 2

    Force in N	Extension in cm
    0.00	0.0
    0.98	4.1
    1.96	8.0
    2.94	11.9
    3.92	16.0
    4.90	20.0
    5.88	25.8
    6.86	33.5
    7.84	43.2

    Describe what a classmate’s results in Table 2 show about the spring for forces beyond the limit of proportionality.

    Your answer

    Beyond the limit of proportionality the extension is no longer directly proportional to the force, so the graph curves. Each extra mass increases the extension by more than before.

    2/2 self-marked

    • Extension is no longer (directly) proportional to force / the graph is no longer a straight line
    • Each extra newton (mass) gives a bigger increase in extension than before

    Self-marked: 2 of 2 marks. This is your own judgement against the mark scheme.

    Model answer: Beyond the limit of proportionality the extension is no longer directly proportional to the force, so the graph curves. Each extra mass increases the extension by more than before.

12. 03.4[2 marks]

    When you took all the masses off, your spring went back to its original length.

    A spring that has been stretched is elastically deformed.

    What does this mean?

    Your answer

    The spring returns to its original length when the force is removed.

    2/2 self-marked

    • It returns to its original length (shape)
    • when the force (load) is removed (dependent on the first mark)

    Self-marked: 2 of 2 marks. This is your own judgement against the mark scheme.

    Model answer: The spring returns to its original length when the force is removed.

13. 04[2 marks]

    You moved your eye level with the pointer for every reading.

    Explain why.

    Your answer

    To avoid parallax error: viewed from above or below, the pointer lines up with the wrong mark on the scale, so the readings of extension would be inaccurate.

    2/2 self-marked

    • Parallax error: viewed from above or below, the pointer appears against a different mark on the scale
    • So the readings (and extensions) would be inaccurate: too large if read from above
    • So the spring constant would be inaccurate (too small if the extensions are too large)

    Self-marked: 2 of 2 marks. This is your own judgement against the mark scheme.

    Model answer: To avoid parallax error: viewed from above or below, the pointer lines up with the wrong mark on the scale, so the readings of extension would be inaccurate.

14. 05[2 marks]

    Give two safety precautions for this investigation.

    Your answer

    Clamp the stand to the bench so it cannot topple over. Wear eye protection in case the spring snaps.

    2/2 self-marked

    • Clamp the stand to the bench or weight its base so it cannot topple
    • Keep feet clear of the masses / put a tray (or soft landing) underneath in case they fall
    • Wear eye protection in case the spring snaps or flies off
    • Make sure the masses hang over the base (foot) of the stand
    • Screw the boss on tightly
    • Do not overload the spring: limit the total mass / add masses gently

    Self-marked: 2 of 2 marks. This is your own judgement against the mark scheme.

    Model answer: Clamp the stand to the bench so it cannot topple over. Wear eye protection in case the spring snaps.

15. 06[6 marks]

    Apparatus: clamp stand, boss and clamp, spring, metre rule, pointer, 100 g mass hanger and 100 g slotted masses.

    Describe a method a student could use to investigate how the extension of a spring depends on the force applied to it.

    Your answer

    1. Clamp the stand to the bench and hang the spring from the clamp. 2. Clamp a metre rule vertically next to the spring and fix a pointer to the bottom of the spring. 3. With no masses on the spring, read the position of the pointer at eye level. This is the original length. 4. Hang a 100 g mass (0.98 N) on the spring and wait until the spring stops moving. 5. Read the new position of the pointer at eye level. 6. Calculate the extension: extension = new length − original length. 7. Repeat, adding one 100 g mass at a time, up to 800 g. 8. Plot a graph of extension (y-axis) against force (x-axis). The straight part shows extension is directly proportional to force; the spring constant is 1 ÷ its gradient. 9. Wear eye protection in case the spring snaps.

    6/6 self-marked

    • Hang the spring from a clamp on a clamp stand
    • Fix a metre rule vertically beside the spring
    • Measure the original (unstretched) length: the position of the pointer (the bottom of the spring) with no masses
    • Use a pointer on the bottom of the spring as the reference point
    • Add masses (weights) to the spring
    • Add them one at a time in equal steps (e.g. 100 g, about 1 N)
    • Measure the new length (position of the pointer) after each mass
    • Extension = new length − original length
    • Repeat for a range of masses (e.g. up to 800 g)
    • Read the scale at eye level, and wait for the spring to stop moving
    • Plot a graph of extension against force (straight line through the origin; spring constant = 1 ÷ gradient)
    • Safety: eye protection, secure the stand, keep feet clear of falling masses

    Self-marked: Level 3, 6 of 6 marks. You ticked 12 of 12 indicative points. This is your own judgement against the mark scheme.

    Model answer: 1. Clamp the stand to the bench and hang the spring from the clamp. 2. Clamp a metre rule vertically next to the spring and fix a pointer to the bottom of the spring. 3. With no masses on the spring, read the position of the pointer at eye level. This is the original length. 4. Hang a 100 g mass (0.98 N) on the spring and wait until the spring stops moving. 5. Read the new position of the pointer at eye level. 6. Calculate the extension: extension = new length − original length. 7. Repeat, adding one 100 g mass at a time, up to 800 g. 8. Plot a graph of extension (y-axis) against force (x-axis). The straight part shows extension is directly proportional to force; the spring constant is 1 ÷ its gradient. 9. Wear eye protection in case the spring snaps.

Unseen data: the same skills, new apparatus

1. 01[3 marks]

   A student hangs a bag of mass 600 g from a spring in a luggage scale. The spring stretches by 22 mm. The gravitational field strength is 9.8 N/kg.

   Calculate the spring constant of the spring.

   Your answer

   267

   Working: 0.6 * 9.8 = 5.88 22 / 1000 = 0.022 5.88 / 0.022

   3/3

   • W = m g = 0.6 × 9.8 = 5.88 N.
   • e = 22 mm = 0.022 m.
   • k = F ÷ e, in N/m.

   Full marks. Your answer meets all the points for this question.

   Model answer: W = m g = 0.6 × 9.8 = 5.88 N; e = 22 mm = 0.022 m. k = F ÷ e = 5.88 ÷ 0.022 = 267 N/m.

2. 02[2 marks]

   A designer tests three springs for a toy. Each spring has the same 2.0 N force on it, below its limit of proportionality.

   Extension with a 2.0 N force

   Spring	Extension in cm
   A	8.0
   B	4.0
   C	3.2

   Which TWO statements are correct? Tick (✓) two boxes.

   • ✓ Spring C has the greatest spring constant. (your choice)
   • ✓ For the same force, spring A stretches most because it has the smallest spring constant. (your choice)
   • Spring A has the greatest spring constant, because it stretched the most.
   • All three springs have the same spring constant, because the force was the same.

   Your answer

   Spring C has the greatest spring constant.; For the same force, spring A stretches most because it has the smallest spring constant.

   2/2

   • Spring C has the greatest spring constant.
   • For the same force, spring A stretches most because it has the smallest spring constant.

   Full marks. Well done.

   Model answer: k = F ÷ e with the same F, so the smallest extension means the greatest spring constant: spring C (63 N/m). Spring A stretched most: the smallest k (25 N/m).

3. 03[2 marks]

   Before using a newtonmeter, a student sees that it reads 0.2 N with nothing hanging from it. With a stone hanging from it, it reads 3.5 N.

   Which TWO statements are correct? Tick (✓) two boxes.

   • ✓ This is a zero error, a kind of systematic error: every reading is too high by the same amount. (your choice)
   • ✓ The weight of the stone is 3.3 N: take 0.2 N from each reading, or reset the newtonmeter to zero before using it. (your choice)
   • This is a random error: repeating the reading and taking a mean will remove it.
   • The weight of the stone is 3.7 N.

   Your answer

   This is a zero error, a kind of systematic error: every reading is too high by the same amount.; The weight of the stone is 3.3 N: take 0.2 N from each reading, or reset the newtonmeter to zero before using it.

   2/2

   • This is a zero error, a kind of systematic error: every reading is too high by the same amount.
   • The weight of the stone is 3.3 N: take 0.2 N from each reading, or reset the newtonmeter to zero before using it.

   Full marks. That is exactly what the mark scheme is looking for.

   Model answer: A newtonmeter that does not read zero with no load has a zero error: a systematic error that adds 0.2 N to every reading. Repeats would all be wrong by the same amount. Subtract it (3.5 − 0.2 = 3.3 N) or reset it to zero first.