Practise Forces and elasticity. 19 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Stretching, squashing or bending an object needs more than one force. You need F = k e, the difference between elastic and inelastic deformation, the limit of proportionality, force–extension graphs and the required practical on a spring. The equation for elastic potential energy, Ee = ½ k e2, is given on the equations sheet.
Grade by grade
What you need to be able to do, from the first marks up to the top grade.
3
Explain why more than one force is neededTo stretch, bend or compress an object, forces must act on it in different directions.
4
Distinguish elastic and inelastic deformationAn elastically deformed object returns to its original shape when the forces are removed; an inelastically deformed one does not.
5
Calculate force or extension using F = k eFor example, k = 40 N/m and e = 0.15 m give F = 6.0 N.
5
Describe the spring extension required practicalHang known weights on a spring, measure its length each time, calculate the extension and plot force against extension.
6
Interpret linear and non-linear force–extension graphsA straight line through the origin shows F ∝ e with gradient k; the line curves beyond the limit of proportionality.
7
Calculate elastic potential energy storedUse Ee = 0.5 × k × e2, up to the limit of proportionality.
7
Link work done to energy storedIf the spring is not inelastically deformed, the work done stretching it equals the elastic potential energy stored.
Notes
Changing shape
To stretch, bend or compress an object you need more than one force. To stretch a spring, you pull on both ends, or hang a weight from it while a clamp holds the top.
Elastic deformation: the object returns to its original length and shape when the forces are removed.
Inelastic deformation: the object does not return to its original length and shape when the forces are removed.
Force and extension: F = k e
force = spring constant × extension: F = k e
F in newtons (N), k in newtons per metre (N/m), e in metres (m).
Extension is the increase in length: extension = stretched length − original length. The same equation works for compression.
The extension of an elastic object is directly proportional to the force applied, provided the limit of proportionality is not exceeded.
A stiffer spring has a larger spring constant: it needs a bigger force for each metre of extension.
Force–extension graphs
With force on the y-axis and extension on the x-axis, the graph is a straight line through the origin (linear) up to the limit of proportionality.
The gradient of this straight section is the spring constant, k.
Beyond the limit of proportionality the line curves (non-linear): extension is no longer proportional to force.
Elastic potential energy grade 7+
A force that stretches or compresses a spring does work, and elastic potential energy is stored in the spring.
Provided the spring is not inelastically deformed, the work done on the spring equals the elastic potential energy stored.
Ee = ½ k e2 (given on the equations sheet). Only use it up to the limit of proportionality.
Required practical: force and extension of a spring
Clamp the spring next to a vertical metre rule and measure its original length.
Hang weights in equal steps. Each time, measure the new length and calculate the extension. Read the ruler at eye level, using a pointer on the bottom of the spring.
Plot force (the weight added) against extension.
Cheatsheet
F = k e (force = spring constant × extension)
Units: F in N, k in N/m, e in m
Extension = stretched length − original length
Elastic deformation: returns to original shape; inelastic deformation: does not
F ∝ e up to the limit of proportionality
Gradient of a force–extension graph (force on the y-axis) = k
Ee = ½ k e2 (on the equations sheet) grade 7+
Work done stretching = elastic potential energy stored (if not inelastically deformed) grade 7+
How to answer each type of question
Calculate using F = k e
2 to 3 marks5
Work out the extension (new length − original length).
Convert it to metres.
Substitute into F = k e (or rearrange for k or e).
Example. A spring has a spring constant of 25 N/m. Its original length is 12.0 cm. A force stretches it to a length of 18.0 cm. The limit of proportionality is not exceeded. Calculate the force.
Show the model answer
extension = 18.0 − 12.0 = 6.0 cm = 0.060 m (1) F = 25 × 0.060 (1) F = 1.5 N (1)
Interpret a force–extension graph
2 to 3 marks6
Check which quantity is on each axis.
Use a point on the straight section to find k = F ÷ e.
A curve shows that the limit of proportionality has been exceeded.
Example. A student's force–extension graph for a spring is a straight line through the origin up to a force of 4.0 N, when the extension is 0.16 m. After that, the line curves. (a) Calculate the spring constant. (b) What does the curved part of the graph show?
Show the model answer
(a) k = 4.0 ÷ 0.16 (1) k = 25 N/m (1) (b) The limit of proportionality has been exceeded, so the extension is no longer directly proportional to the force (1).
Calculate elastic potential energy
2 to 3 marks7
Make sure the extension is in metres.
Square the extension first, then multiply by k and by 0.5.
Give the answer in J.
Example. A spring with a spring constant of 80 N/m is stretched by 0.25 m. It does not pass its limit of proportionality. Calculate the elastic potential energy stored in the spring.
Show the model answer
Ee = 0.5 × 80 × 0.252 (1) Ee = 2.5 J (1)
Describe the required practical
4 to 6 marks6
Say how the spring and ruler are set up and what you measure first.
Say how you change the force and what you measure each time.
Say how you calculate the extension and make readings accurate.
Say what you plot.
Example. Describe how a student could investigate the relationship between the force on a spring and its extension.
Show the model answer
Clamp the spring vertically next to a metre rule and measure its original length (1). Hang a known weight (e.g. a 1.0 N mass hanger) on the spring and measure the new length (1). Read the ruler at eye level / use a pointer attached to the bottom of the spring (1). Calculate the extension = new length − original length (1). Repeat for at least five different weights (1). Plot a graph of force against extension (1).
Shortcuts and memory tricks
Extension, not length: always subtract the original length first.
cm to m: divide by 100, so 6.0 cm = 0.060 m.
Stiff spring = steep force–extension line = big k.
For ½ k e2, square the extension first, then multiply by k, then halve.
Where marks are lost
Using the stretched length instead of the extension.
Forgetting to convert cm or mm to metres.
Using Ee = ½ k e2 beyond the limit of proportionality.
Plotting extension on the y-axis and calling the gradient k. With extension on the y-axis, the gradient is 1 ÷ k.
Squaring k × e instead of squaring only the extension.
Exam technique
F = k e must be recalled; Ee = ½ k e2 is on the equations sheet.
Before using a graph, check which quantity is on each axis.
In method questions, name the instrument (metre rule), what you measure (length), what you calculate (extension) and what you plot.
Cover the answers and test yourself. The app has these as flashcards that come back just before you'd forget them.
A spring has a spring constant of 40 N/m. Calculate the force needed to stretch the spring by 0.15 m. Use the equation: force = spring constant × extension
6.0 N
Complete the sentence. Work done in stretching a spring is stored as ............ energy, as long as the spring is not permanently stretched.
Elastic potential.
Name the type of deformation in which the spring stays permanently stretched.
Inelastic deformation.
Sample questions
Written for this site in the style of AQA exam questions. They are not taken from real past papers.
Question 1Easy6 marks
(a) A spring hangs from a clamp. A weight is hung from the bottom of the spring and the spring stretches. Why must more than one force act on the spring to stretch it? Tick (✓) one box.[1]
A single force would make the spring move instead of changing its shape.
A single force is always too small to stretch a spring.
Two forces are needed to double the extension.
The weight of the spring must be balanced first.
(b) Describe the difference between elastic deformation and inelastic deformation.[2]
(c) A spring has a spring constant of 40 N/m. Calculate the force needed to stretch the spring by 0.15 m. Use the equation: force = spring constant × extension[2]
(d) A diver stands on the end of a diving board and the board changes shape. Name the type of change of shape that happens to the diving board.[1]
Show the answer and mark scheme
(a)Answer: A single force would make the spring move instead of changing its shape.
(b)Answer: Elastic: returns to its original shape when the force is removed. Inelastic: does not return to its original shape.
elastic deformation: the object returns to its original shape / length when the force is removed
inelastic deformation: the object does not return to its original shape / length when the force is removed
(c)Answer: 6.0 N
F = 40 × 0.15
6.0 (N)
(d)Answer: Bending.
bending
Question 2Medium4 marks
An archer pulls back a bowstring, storing elastic potential energy in the bow.
(a) Describe the energy transfer that occurs from the moment the archer releases the bowstring to the moment just after the arrow leaves the bow.[2]
(b) Explain why the archer must apply a bigger force to pull the bowstring back further.[2]
Show the answer and mark scheme
(a)Answer: From the bow’s elastic potential energy store mainly to the arrow’s kinetic energy store; some is dissipated to the surroundings by heating and as sound.
energy is transferred from the elastic potential energy store of the bow mainly to the kinetic energy store of the arrow
some energy is dissipated to the surroundings: to thermal energy stores (because of friction and vibrations) and as sound
(b)Answer: A bigger extension of the bow needs a bigger force (like a spring, F = k e).
pulling the string back further gives the bow a greater extension (deformation)
the force needed increases with extension (like a spring, F = k e, if the bow stays within its limit of proportionality)
Question 3Hard10 marks
A student is given a spring and a set of 100 g slotted masses on a hanger.
(a) Plan an investigation to determine the spring constant of the spring and to find its limit of proportionality.[6]
(b) The student’s graph was a straight line through the origin up to a force of 6.0 N, which gave an extension of 0.080 m. Calculate the spring constant of the spring.[2]
(c) Calculate the elastic potential energy stored in the spring when the extension is 0.080 m. Use the Physics Equations Sheet.[2]
Show the answer and mark scheme
(a)
hang the spring from a clamp on a clamp stand with a metre rule clamped vertically beside it
measure the original length of the spring, using a pointer fixed to the bottom of the spring
add the masses one at a time (100 g has a weight of about 1 N) and record the new length each time
calculate each extension as new length − original length; read the ruler at eye level to avoid parallax
continue adding masses until the extensions stop increasing in equal steps
calculate the force on the spring for each load using W = m g
plot a graph of force against extension
the spring constant is the gradient of the straight-line section of the graph
the limit of proportionality is the point where the graph stops being a straight line
safety: wear eye protection and place a soft surface under the masses
Marked with levels of response: the full level descriptors are in the app.