P4.5 Work Done & Forces & Springs — Physics with Kate
AQA GCSE Physics — 4.5.2 & 4.5.3
⚡ Work Done & Forces & Springs 🌪
From the work done by a bolt of lightning to the elastic snap of a spring — equations, energy transfers, Hooke's Law, and Required Practical 6.
⚡ stormy physics • work & energy • springs under pressure 🌪
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Work Done (4.5.2)
⚡ What Is Work Done?
When a force causes an object to move through a distance, work is done on the object.
Work done means energy is transferred — they are the same thing!
Work done = energy transferred (measured in joules, J).
1 joule of work is done when a force of 1 newton moves an object through a distance of 1 metre.
So 1 J = 1 N·m (one newton-metre).
🌪 Work Done — Diagram
Work done = Force × Distance moved in direction of force
🌪 The Work Done Equation
W = F × s
W = work done (joules, J)
F = force applied (newtons, N)
s = distance moved along the line of action of the force (metres, m)
The distance must be in the same direction as the force — this is a key exam detail.
☔ Worked Example
A 50 N force pushes a box 3 m along the floor. Calculate the work done.
W = F × s = 50 × 3 = 150 J
So 150 joules of energy have been transferred.
🌀 Work Done Against Friction
When work is done against friction, the energy is transferred to the thermal energy stores of the objects involved.
This causes the temperature to rise.
For example, rubbing your hands together — work done against friction heats them up.
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Forces & Elasticity (4.5.3)
⚡ Changing the Shape of an Object
To change the shape of an object (by stretching, bending, or compressing it), more than one force must be applied.
If only one force acted, the object would simply move (accelerate) rather than change shape.
🌪 Examples of Deformation
Stretching a spring: pull forces at both ends in opposite directions.
Compressing a sponge: push forces on both sides, squashing it together.
Bending a ruler: push down in the middle while supporting both ends.
☔ Stretching a Spring — Diagram
A spring at natural length vs stretched with an applied force
☔ Elastic vs Inelastic Deformation
Elastic deformation: the object returns to its original shape and length when the force is removed. Example: a spring being gently stretched.
Inelastic deformation: the object does NOT return to its original shape — it is permanently deformed. Example: bending a paperclip.
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Hooke's Law
⚡ The Relationship Between Force & Extension
The extension of an elastic object (like a spring) is directly proportional to the force applied, provided the limit of proportionality is not exceeded.
This is known as Hooke's Law.
F = k × e
F = force applied (newtons, N)
k = spring constant (newtons per metre, N/m)
e = extension (metres, m)
🌪 What Is the Spring Constant?
The spring constant k tells you how stiff the spring is.
A higher value of k = a stiffer spring (harder to stretch).
A lower value of k = a more flexible spring (easier to stretch).
Units: N/m (newtons per metre).
🌀 The Limit of Proportionality
The limit of proportionality is the point beyond which force and extension are no longer directly proportional.
Beyond this point, the spring begins to deform and the graph starts to curve.
Hooke's Law (F = ke) only applies up to the limit of proportionality.
☔ The Force-Extension Graph
Force-Extension graph showing Hooke's Law and the limit of proportionality
Plot force (N) on the y-axis and extension (m) on the x-axis.
Up to the limit of proportionality: a straight line through the origin — this shows a linear (proportional) relationship.
The gradient of this straight section equals the spring constant (k).
Beyond the limit of proportionality: the line curves (non-linear) — the spring no longer obeys Hooke's Law.
The point where the line starts to curve is the limit of proportionality.
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Elastic Potential Energy
⚡ Energy Stored in a Stretched Spring
A force that stretches or compresses a spring does work on the spring.
This work is stored as elastic potential energy in the spring.
If the spring has not been inelastically deformed, then the work done on the spring equals the elastic potential energy stored.
🌪 The Elastic Potential Energy Equation
Ee = ½ × k × e²
Ee = elastic potential energy (joules, J)
k = spring constant (N/m)
e = extension (metres, m)
This equation is only valid up to the limit of proportionality (i.e. while the spring obeys Hooke's Law).
☔ Worked Example
A spring has a spring constant of 25 N/m and is extended by 0.3 m. Calculate the elastic potential energy stored.
Ee = ½ × k × e² = 0.5 × 25 × 0.3² = 0.5 × 25 × 0.09 = 1.125 J
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Investigating Springs Required Practical 6
⚡ Aim & Method
Aim: investigate the relationship between force and extension for a spring.
Set up a clamp stand with a spring hanging from it and a ruler alongside.
Attach a pointer or marker to the bottom of the spring to help read the ruler accurately.
Measure and record the original length of the spring with no load.
Add masses one at a time (e.g. 100 g each = 1 N weight per mass).
For each mass, measure the new length and calculate the extension (new length − original length).
Repeat, adding more masses each time.
🌪 Apparatus Setup — Diagram
Required Practical 6 — Apparatus for investigating force and extension
🌪 Variables
Variable Type
Variable
Independent (what you change)
Force applied (weight of masses added)
Dependent (what you measure)
Extension of the spring
Control (what you keep the same)
Same spring, same starting position, same ruler
🌀 Analysing the Results
Plot a force-extension graph (force on y-axis, extension on x-axis).
A straight line through the origin shows the spring obeys Hooke's Law (force is directly proportional to extension).
The gradient of the straight-line section = the spring constant (k).
If the line curves at higher forces, the spring has passed the limit of proportionality.
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⚡ Exam Equations — What You Get & What You Must Know
The equations W = Fs, F = ke, and Ee = ½ke² are all on the equation sheet.
But you MUST be able to rearrange them to solve for any variable.
For W = Fs: rearrange for F = W/s or s = W/F.
For F = ke: rearrange for k = F/e or e = F/k.
For Ee = ½ke²: rearrange for k = 2Ee/e² or e = √(2Ee/k).
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Question 01 [2 marks]
A worker pushes a crate with a force of 200 N across a warehouse floor for a distance of 8 m. Calculate the work done.
⚡ Answer
W = F × s (1 mark). W = 200 × 8 = 1600 J (1 mark).
Question 02 [2 marks]
State the equation linking work done, force, and distance. Give the units for each quantity.
⚡ Answer
W = F × s (1 mark). Work done in joules (J), force in newtons (N), distance in metres (m) (1 mark).
Question 03 [2 marks]
A box is pushed along a rough floor. Explain what happens to the energy transferred when work is done against friction.
⚡ Answer
When work is done against friction, energy is transferred to the thermal energy stores of the surfaces in contact (1 mark). This causes the temperature to increase (1 mark).
Question 04 [2 marks]
Explain the difference between elastic deformation and inelastic deformation.
⚡ Answer
Elastic deformation: the object returns to its original shape and length when the force is removed (1 mark). Inelastic deformation: the object does not return to its original shape — it is permanently deformed (1 mark).
Question 05 [3 marks]
A spring has a spring constant of 40 N/m. A force of 6 N is applied. Calculate the extension of the spring in cm.
⚡ Answer
F = k × e, so e = F / k (1 mark). e = 6 / 40 = 0.15 m (1 mark). Convert: 0.15 × 100 = 15 cm (1 mark).
Question 06 [3 marks]
A spring is extended by 0.2 m and stores 1.0 J of elastic potential energy. Calculate the spring constant of the spring.
⚡ Answer
Ee = ½ × k × e² (1 mark). Rearrange: k = 2Ee / e² (1 mark). k = (2 × 1.0) / 0.2² = 2.0 / 0.04 = 50 N/m (1 mark).
Question 07 [2 marks]
What is meant by the ‘limit of proportionality’ for a spring?
⚡ Answer
The limit of proportionality is the point beyond which force and extension are no longer directly proportional (1 mark). Beyond this point, the force-extension graph is no longer a straight line / the line starts to curve (1 mark).
Question 08 [3 marks]
A student obtains a straight line through the origin on a force-extension graph for a spring. At higher forces, the line begins to curve. Explain what these two parts of the graph show about the spring's behaviour.
⚡ Answer
The straight line through the origin shows that force is directly proportional to extension — the spring obeys Hooke's Law (1 mark). The curve at higher forces shows the spring has exceeded the limit of proportionality (1 mark). Beyond this point, the extension increases by more for each additional unit of force — the relationship is no longer linear (1 mark).
Question 09 [4 marks]
Describe the method for Required Practical 6 — investigating the relationship between force and extension for a spring. Include the variables.
⚡ Answer
Set up a clamp stand with a spring hanging from it and a ruler alongside (1 mark). Measure the original length of the spring, then add masses one at a time, measuring the new length each time to calculate extension (new length − original length) (1 mark). The independent variable is the force (weight added) and the dependent variable is the extension (1 mark). Control variables: use the same spring, same starting position, and same ruler (1 mark).
Question 10 [3 marks]
A spring with a spring constant of 60 N/m is extended by 0.25 m. Calculate the elastic potential energy stored in the spring.
⚡ Answer
Ee = ½ × k × e² (1 mark). Ee = 0.5 × 60 × 0.25² (1 mark). Ee = 0.5 × 60 × 0.0625 = 1.875 J (1 mark).
Question 11 [2 marks]
Explain why more than one force is needed to stretch a spring.
⚡ Answer
If only one force were applied, the spring would simply accelerate / move in the direction of the force rather than changing shape (1 mark). To stretch (deform) the spring, forces must be applied in opposite directions at each end (1 mark).
Question 12 [6 marks]
A student investigates how the extension of a spring changes with force. They plot a force-extension graph that shows a straight line through the origin up to a force of 5 N, after which the line curves. Explain what the graph tells us about the spring, calculate the spring constant if the extension at 5 N is 0.20 m, and calculate the elastic potential energy stored at this point. Evaluate whether the elastic potential energy equation is valid at forces above 5 N.
⚡ Answer
The straight line through the origin shows the spring obeys Hooke's Law up to 5 N — force is directly proportional to extension (1 mark). The curve beyond 5 N shows the spring has exceeded the limit of proportionality and no longer obeys Hooke's Law (1 mark). Spring constant: k = F / e = 5 / 0.20 = 25 N/m (1 mark). Elastic PE: Ee = ½ × k × e² = 0.5 × 25 × 0.20² = 0.5 × 25 × 0.04 = 0.5 J (1 mark). The equation Ee = ½ke² is only valid up to the limit of proportionality (1 mark). Above 5 N, the spring has exceeded this limit, so the equation would not give an accurate value for the elastic potential energy stored (1 mark).