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GCSE Physics · GCSEChapter 15 · Work done, energy and springs
GCSE · Chapter 15

Work done, energy and springs

By the end of this chapter you will be able to

  • Calculate work done, and explain that it is the energy transferred
  • Use F = k e to find force, extension or spring constant
  • Read and explain a force-extension graph, including the limit of proportionality
  • [H] Calculate the elastic potential energy stored in a stretched spring

You should already know

  • Energy transfers and stores (Chapter 1)
  • Forces and their units (Chapter 13)

Where this leads

  • Energy resources and efficiency (Chapters 2 and 4)
  • Motion and acceleration (Chapter 16)

GCSE Physics. Content follows the GCSE (9-1) subject content for England (Edexcel and AQA). Content references: Calculate the work done by a force as W = F x d, and recognise it as energy transferred, Explain work done against friction and against gravity, Use Hooke's law F = k e within the limit of proportionality, and interpret force-extension graphs, Describe the required practical: investigate the extension of a spring when masses are added, [H] Calculate the elastic potential energy stored in a spring, E = 1/2 k e squared.

Exercise P  Do you remember?

1
FWhat is 20 x 3? Write the number.
2
FWhat is 0.04 x 300? Write the number.
3
FWhat is the unit of energy? Write the unit.
4
FWhat is 100 g in kg? Write the number.
5
FWhich is a non-contact force: friction or gravity? Write the word.
15.1

Work done and energy transfer

GCSE
Calculate the work done by a force as W = F x d, and recognise it as energy transferred, Explain work done against friction and against gravity
Video lesson: Work done and energy transferComing soon
Context. Mr Lee pushes a heavy cupboard 3 m across a room. He is tired and asks, “Have I done any work, or is it just tiring?” In physics, work is done whenever a force moves an object through a distance. The energy used is transferred to the cupboard or to the surroundings.

Work done by a force is the energy transferred when the force moves an object through a distance in the direction of the force:

work done W (J) = force F (N) x distance d (m)

One joule is the work done when a force of 1 N moves an object through 1 m. Work is done against friction when a force overcomes friction, and the energy is transferred to the thermal store of the surfaces. Work is done against gravity when an object is lifted, and the energy goes into its gravitational potential store.

Key ideas

  • W = F x d. Force in newtons, distance in metres, work in joules.
  • Work done = energy transferred.
  • 1 J = 1 N m.
  • Work against friction becomes thermal energy; work against gravity becomes gravitational potential energy.
Worked example 15.1.1
A force of 20 N moves a box 3 m in the direction of the force. Calculate the work done. Write the number of joules.

Method. Use W = F x d.

  1. Take the numbers. F = 20 N and d = 3 m.
    F = 20, d = 3
  2. Multiply. W = 20 x 3 = 60 J.
    20 x 3 = 60
  3. Write the answer.
    60 J

Check. 60 / 3 = 20 N. ✓

Worked example 15.1.2
A friction force of 5 N acts on a sledge while it moves 4 m. Calculate the work done against friction. Write the number of joules.

Method. Use W = F x d with the friction force.

  1. Take the numbers. F = 5 N and d = 4 m.
    F = 5, d = 4
  2. Multiply. W = 5 x 4 = 20 J of energy transferred to the thermal store.
    5 x 4 = 20
  3. Write the answer.
    20 J

Check. 20 / 4 = 5 N. ✓

Worked example 15.1.3
A 2 kg object is lifted through 3 m at a steady speed. Calculate the work done against gravity, using g = 9.8 N/kg. Write the number of joules.

Method. The lifting force equals the weight, so F = m g.

  1. Force. Weight = 2 x 9.8 = 19.6 N, which is the lifting force.
    19.6 N
  2. Work. W = 19.6 x 3 = 58.8 J.
    19.6 x 3 = 58.8
  3. Write the answer.
    58.8 J

Check. 58.8 / 3 = 19.6 N. ✓

Watch out

Watch out: the distance must be in the direction of the force. A force at right angles to the motion does no work in that direction.

Exam tip

Exam tip. Write W = F d, then the numbers with units (N and m), then the answer in joules.

Exercise 15A  Work done

1
FWhat is the unit of work done? Write the unit.
2
FA force of 6 N moves an object 5 m. Calculate the work done. Write the number of joules.
3
FA force of 12 N moves an object 2.5 m. Calculate the work done. Write the number of joules.
4
FA force does 50 J of work over a distance of 10 m. Find the force. Write the number of newtons.
5
FA force of 4 N does 24 J of work. Find the distance moved. Write the number of metres.
6
PSA 5 N friction force acts on a box as it slides 8 m. How much energy is transferred to the thermal store? Write the number of joules.
7
PSA 3 kg object is lifted at a steady speed through 2 m. Calculate the work done against gravity, with g = 9.8 N/kg. Write the number of joules.
8
RExplain in one sentence why the work done by a person pushing a wall that does not move is zero. Accept any sentence that says there is no distance moved in the direction of the force.
9
RA pupil says “work is done whenever you use energy.” Explain why this is not quite right, in one sentence. Accept any sentence that says work is done only when a force moves something through a distance.
10
PSA force of 15 N moves an object 4 m, but the force is at right angles to the motion. What is the work done in the direction of the motion? Write the number of joules.
11
PSA car engine does 60 000 J of work to move the car 400 m against a steady resistance. What is the average resistance force? Write the number of newtons.
12
PSA 0.5 kg book is lifted 1.2 m. Calculate the work done against gravity, using g = 9.8 N/kg. Write the number of joules.
13
RExplain why a lifted object gains gravitational potential energy when work is done on it, in one sentence. Accept any sentence that says the work done against gravity is transferred to the gravitational potential store.
14
FA 12 N force moves a box 0.5 m. Calculate the work done. Write the number of joules.
15
FA force does 90 J of work to move an object 15 m. Find the force. Write the number of newtons.
16
PSA 25 N force pushes a crate 3.2 m in the direction of the force. Calculate the work done. Write the number of joules.
15.2

Elastic deformation and Hooke's law

GCSE
Use Hooke's law F = k e within the limit of proportionality, and interpret force-extension graphs
Video lesson: Elastic deformation and Hooke's lawComing soon
Context. A spring is hung from a clamp. Each time a mass is added, the spring stretches a little more. When the mass is removed, the spring returns to its original length. This is elastic deformation: the change is temporary, and the spring returns to its shape.

Many springs obey Hooke's law: the extension is directly proportional to the force, as long as the force stays below the limit of proportionality.

F = k x e

where F is the force in newtons, k is the spring constant in N/m, and e is the extension in metres (the increase in length from the natural length). On a force-extension graph, the straight line through the origin shows Hooke’s law. Beyond the limit of proportionality the line curves, and the spring may be permanently stretched.

Force against extension (example spring)0123402468extension (cm)force (N)
Force against extension for a spring: a straight line through the origin, up to the limit of proportionality

Key ideas

  • Hooke's law: F = k e, within the limit of proportionality.
  • Extension e = length - natural length; use metres in the equation.
  • The spring constant k is the force per metre of extension (N/m).
  • Beyond the limit of proportionality the graph curves and Hooke’s law no longer applies.
Worked example 15.2.1
A spring has a spring constant of 300 N/m. A force of 6 N stretches it. Calculate the extension. Write the number of metres.

Method. Rearrange F = k e to e = F / k.

  1. Take the numbers. F = 6 N and k = 300 N/m.
    F = 6, k = 300
  2. Divide. e = 6 / 300 = 0.02 m.
    6 / 300 = 0.02
  3. Write the answer.
    0.02 m

Check. 300 x 0.02 = 6 N. ✓

Worked example 15.2.2
A spring stretches by 0.04 m under a force of 12 N. Calculate the spring constant. Write the number in N/m.

Method. Use k = F / e.

  1. Take the numbers. F = 12 N and e = 0.04 m.
    F = 12, e = 0.04
  2. Divide. k = 12 / 0.04 = 300 N/m.
    12 / 0.04 = 300
  3. Write the answer.
    300 N/m

Check. 300 x 0.04 = 12 N. ✓

Worked example 15.2.3
The spring constant is 49 N/m. What force gives an extension of 4 cm? Write the number of newtons.

Method. Convert the extension to metres first: 4 cm = 0.04 m.

  1. Convert. e = 4 cm = 0.04 m.
    0.04 m
  2. Multiply. F = 49 x 0.04 = 1.96 N.
    49 x 0.04 = 1.96
  3. Write the answer.
    1.96 N

Check. 1.96 / 0.04 = 49 N/m. ✓

Watch out

Watch out: convert the extension to metres before you use F = k e with k in N/m. 4 cm is 0.04 m, not 4 m.

Exam tip

Exam tip. On a graph, the gradient of the straight part is the spring constant k, in N/m.

Exercise 15B  Hooke's law and force-extension graphs

1
FWhat is the name of the law that says extension is proportional to force? Write the name.
2
FWhat is the unit of the spring constant? Write the unit.
3
FA spring with k = 100 N/m stretches by 0.05 m. Calculate the force. Write the number of newtons.
4
FA force of 8 N gives an extension of 0.2 m. Calculate the spring constant. Write the number in N/m.
5
PSA spring with k = 25 N/m is stretched by 4 cm. Calculate the force. Write the number of newtons.
6
PSA spring extends by 0.15 m under a force of 6 N. Calculate its spring constant. Write the number in N/m.
7
PSOn a force-extension graph, the straight line passes through the origin with gradient 49 N/m. What does the gradient tell you? Write the word: spring constant or extension.
8
RA spring is stretched beyond its limit of proportionality. Describe what happens to the graph, in one sentence. Accept any sentence that says the graph curves, and Hooke’s law no longer applies.
9
RExplain why a spring returns to its original length after a small force is removed, in one sentence. Accept any sentence that says it has elastic deformation, so it is within the limit and returns to its shape.
10
PSA spring with k = 80 N/m has a force of 4 N acting on it. What is the extension? Write the number of metres.
11
RA pupil says “doubling the force on a spring always doubles its extension, even beyond the limit.” Explain why this is wrong. Accept any answer that says doubling only holds within the limit of proportionality.
15.3

Required practical: extension of a spring

GCSE
Describe the required practical: investigate the extension of a spring when masses are added
Video lesson: Required practical: extension of a springComing soon
Context. The class hangs masses from a spring, measures the length each time, and works out the extension. They want to test whether the extension is proportional to the force.

Method (with a teacher). Clamp a spring to a stand and measure its natural length with a ruler, with the eye level with the pointer. Add a 50 g mass to the hanger, wait until the spring is still, and measure the new length. Repeat for each mass. Find the extension by subtracting the natural length. The weight of each mass is found from W = m g with g = 9.8 N/kg. Plot a graph of force against extension. The straight line through the origin shows Hooke’s law. Stop adding masses before the spring is damaged or the limit of proportionality is passed.

Safety. Wear eye protection. Keep the masses on the bench when not in use, and do not let a spring snap back. Use the masses as the teacher directs.

mass added (g)weight (N)extension (cm)
00.000.0
500.491.0
1000.982.0
1501.473.0
2001.964.0
2502.455.0
Example results: a spring with k = 49 N/m; masses in grams, weight from g = 9.8 N/kg

Key ideas

  • Measure the natural length first.
  • Extension = length with the load - natural length.
  • Weight = mass (kg) x 9.8 N/kg.
  • A straight line through the origin means Hooke’s law is obeyed.
Worked example 15.3.1
Using the example results, the mass added is 100 g. Calculate the weight of this mass. Write the number of newtons.

Method. Convert grams to kg, then W = m g.

  1. Convert. 100 g = 0.1 kg.
    0.1 kg
  2. Multiply. W = 0.1 x 9.8 = 0.98 N.
    0.1 x 9.8 = 0.98
  3. Write the answer.
    0.98 N

Check. ✓

Worked example 15.3.2
Using the example results, the extension for 100 g is 2.0 cm. Calculate the spring constant in N/m. Write the number in N/m.

Method. Use k = F / e, with e in metres.

  1. Convert. e = 2.0 cm = 0.02 m.
    0.02 m
  2. Divide. k = 0.98 / 0.02 = 49 N/m.
    0.98 / 0.02 = 49
  3. Write the answer.
    49 N/m

Check. ✓

Worked example 15.3.3
In the results, the extension for 250 g is 5.0 cm. Is the ratio of force to extension the same as for 100 g? Write yes or no.

Method. Compare F / e for both masses.

  1. 250 g. F = 2.45 N and e = 0.05 m, so F / e = 49 N/m.
    49
  2. 100 g. F = 0.98 N and e = 0.02 m, so F / e = 49 N/m.
    49
  3. Decide. The ratios are the same, so the extension is proportional to the force.
    yes
  4. Write the answer.
    yes

Check. ✓

Watch out

Watch out: measure the extension from the natural length, not from zero on the ruler. Read the scale at eye level to avoid parallax error.

Exam tip

Exam tip. For the practical, say how you would keep the variables controlled: the same spring, the same ruler and eye level, and waiting until the spring is still.

Exercise 15C  The practical and the results

1
FWhat must you measure first, before adding any mass to the spring? Write the word.
2
FWhat is the extension for a spring of natural length 10 cm, stretched to 14 cm? Write the number of cm.
3
PSIn the example results, what is the weight of the 200 g mass? Write the number of newtons.
4
FIn the example results, what is the extension for 200 g? Write the number of cm.
5
RWhy is the spring not loaded beyond its limit of proportionality in the practical? Write one reason in one sentence. Accept any sentence that says Hooke’s law no longer applies and the spring may be damaged.
6
FA student reads the scale from above the pointer. What error does this cause? Write the word: parallax or zero.
7
RWhy is the spring left to settle before the length is measured? Write one reason in one sentence. Accept any sentence that says the spring must be still for an accurate reading.
8
PSIn the results, the weight of the 150 g mass is 1.47 N. Calculate the extension using k = 49 N/m. Write the number of cm.
9
FA pupil plots force against extension and gets a straight line through the origin. What does this show? Write the words: Hooke’s law.
10
FName the variable that is changed on purpose in this practical. Write the word: mass or length.
11
FName one variable that must be kept the same in this practical. Write the word.
12
PSWhat is the extension when a 0.5 kg mass is hung from the spring in the example, using g = 9.8 N/kg and k = 49 N/m? Write the number of cm.
13
FA pupil records a natural length of 15.0 cm and a length of 19.0 cm with a load. What is the extension in cm? Write the number.
14
RSuggest one way to make the practical more accurate. Accept any answer that says repeat the readings and take a mean, or use a longer spring with a finer scale.
15.4

[H] Elastic potential energy

GCSE
[H] Calculate the elastic potential energy stored in a spring, E = 1/2 k e squared
Video lesson: [H] Elastic potential energyComing soon

[H] Higher tier only. The work done in stretching a spring is stored as elastic potential energy. For a spring that obeys Hooke’s law, the energy stored is the area under the force-extension graph: E = ½ k e2, where e is in metres and k is in N/m. Doubling the extension quadruples the stored energy.

Key ideas

  • [H] E = 1/2 k e squared; E in J, k in N/m, e in m.
  • [H] The energy stored equals the work done in stretching the spring.
  • [H] Doubling the extension quadruples the stored energy.
  • [H] Energy stored is the area under the force-extension graph up to the extension.
Worked example 15.4.1
[H] A spring of spring constant 200 N/m is stretched by 0.1 m. Calculate the energy stored. Write the number of joules.

Method. Use E = 1/2 k e squared.

  1. Square. e squared = 0.1 x 0.1 = 0.01.
    0.01
  2. Multiply. E = 0.5 x 200 x 0.01 = 1 J.
    0.5 x 200 x 0.01 = 1
  3. Write the answer.
    1 J

Check. ✓

Worked example 15.4.2
[H] A spring with k = 50 N/m is stretched by 0.2 m. Calculate the energy stored. Write the number of joules.

Method. Use E = 1/2 k e squared.

  1. Square. e squared = 0.04.
    0.04
  2. Multiply. E = 0.5 x 50 x 0.04 = 1 J.
    0.5 x 50 x 0.04 = 1
  3. Write the answer.
    1 J

Check. ✓

Worked example 15.4.3
[H] Using E = 1/2 k e squared, find the energy stored in a spring with k = 100 N/m stretched by 0.3 m. Write the number of joules.

Method. Use E = 1/2 k e squared.

  1. Square. e squared = 0.3 x 0.3 = 0.09.
    0.09
  2. Multiply. E = 0.5 x 100 x 0.09 = 4.5 J.
    0.5 x 100 x 0.09 = 4.5
  3. Write the answer.
    4.5 J

Check. ✓

Watch out

Watch out: [H] the extension is squared. Doubling the extension gives four times the energy, not twice.

Exam tip

Exam tip. [H] Write the equation, convert the extension to metres, square it, then multiply by 0.5 and by k.

Exercise 15D  [H] Elastic potential energy

1
F[H] Calculate the energy stored in a spring with k = 100 N/m and extension 0.2 m. Write the number of joules.
2
PS[H] A spring stores 4 J when stretched 0.4 m. Find its spring constant. Write the number in N/m.
3
PS[H] A spring of k = 20 N/m is stretched by 0.5 m. Calculate the energy stored. Write the number of joules.
4
PS[H] If the extension of a spring is doubled, by how many times is the stored energy increased? Write the number.
5
R[H] Explain why the energy stored in a stretched spring equals the work done in stretching it, in one sentence. Accept any sentence that says the work done against the spring force is stored as elastic potential energy.
6
PS[H] Calculate the energy stored in a spring of k = 400 N/m stretched by 0.05 m. Write the number of joules.
7
PS[H] A spring with k = 100 N/m stores 2 J of elastic potential energy. Find its extension. Write the number of metres.
8
F[H] Calculate the energy stored in a spring with k = 10 N/m stretched by 1 m. Write the number of joules.
9
F[H] In the equation E = 1/2 k e squared, which quantity is squared? Write the symbol.
10
PS[H] A spring stores 8 J when stretched by 0.4 m. Find the spring constant. Write the number in N/m.

Chapter summary

SkillMethod
Work doneW = F d (J); equal to the energy transferred
Hooke's lawF = k e within the limit of proportionality; k in N/m
GraphStraight line through origin: Hooke’s law; the gradient is k
PracticalNatural length, add masses, extension = length - natural length
[H] Elastic potential energyE = 1/2 k e squared

Can you…?

  • I can calculate work done using W = F d
  • I can use F = k e and find k from a graph
  • I can describe the spring and masses practical and its results
  • I can explain the limit of proportionality
  • [H] I can calculate elastic potential energy

Key terms

work done
energy transferred when a force moves an object through a distance
elastic deformation
a change in shape that goes away when the force is removed
Hooke's law
extension is proportional to force, within the limit of proportionality
spring constant (k)
force per unit extension, in N/m
limit of proportionality
the force beyond which Hooke’s law no longer applies
[H] elastic potential energy
energy stored in a stretched spring, E = 1/2 k e squared

Exercise 15Z  Check-up

1
A force of 9 N moves an object 4 m. Calculate the work done. Write the number of joules.
2
What is the spring constant if a force of 10 N gives an extension of 0.05 m? Write the number in N/m.
3
A spring with k = 60 N/m has an extension of 0.1 m. Calculate the force. Write the number of newtons.
4
Name the law that describes extension proportional to force within the limit. Write the name.
5
What is the work done against friction by a 6 N force moving a block 2 m? Write the number of joules.
6
A 4 kg object is lifted at a steady speed through 1.5 m, with g = 9.8 N/kg. Calculate the work done. Write the number of joules.
7
What is the unit of the spring constant? Write the unit.
8
A spring extends 3 cm under a force of 1.5 N. Calculate the spring constant in N/m. Write the number.
9
Explain why a spring returns to its original shape after a small force is removed, in one sentence. Accept any sentence that says the deformation is elastic and temporary.
10
[H] Calculate the elastic potential energy of a spring with k = 10 N/m stretched by 0.3 m. Write the number of joules.

Total 29 marks. Allow about 29 minutes.

Exercise 15X  Exam-style questions

1
FA force of 35 N moves a trolley 6 m. Calculate the work done. Write the number of joules.[2 marks]
2
PSA spring has a natural length of 12 cm and stretches to 16 cm under a 2 N load. Calculate the spring constant in N/m. Write the number.[3 marks]
3
PSA 0.5 kg object is lifted at a steady speed through 2 m, with g = 9.8 N/kg. Calculate the work done against gravity. Write the number of joules.[2 marks]
4
RDescribe how to carry out the required practical to investigate how the extension of a spring changes when masses are added. Accept any answer that names the natural length, adds masses in steps, measures the length each time, calculates the extension and plots force against extension.[5 marks]
5
RExplain why the work done in stretching a spring is stored, and what happens to this energy when the spring is released. Accept any answer that says the energy is stored as elastic potential energy, which becomes kinetic energy as the spring returns.[3 marks]
6
PSA student uses a spring with k = 40 N/m and a 4 N force. Explain, using the graph and the equation, why the extension is 0.1 m. Write the extension in metres.[3 marks]
7
PS[H] A spring with k = 80 N/m is stretched by 0.15 m. Calculate the elastic potential energy stored. Write the number of joules.[3 marks]
8
RExplain why a force-extension graph curves beyond the limit of proportionality. Accept any answer that says the spring no longer obeys Hooke’s law, and extension increases faster or more slowly than the force.[3 marks]
9
PSA 10 kg box is pushed 5 m across a floor against a constant friction force of 40 N. Calculate the work done against friction, and the energy transferred to the thermal store. Write the number of joules.[2 marks]
10
PSA spring stretches by 6 cm under a load of 3 N. The natural length is 20 cm. Calculate the length of the spring under the load, in cm. Write the number of cm.[3 marks]
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