Hooke's Law (Cambridge International IGCSE Science)

Spec point — P1.5.9 · Define the spring constant as force per unit extension; recall and use the equation k = F / x "Define and use the term ‘limit of proportionality’ for a load–extension graph and identify this point on the graph (an understanding of the elastic limit is not required)"

Hooke's law

Extended Tier Only

• The relationship between the extension of an elastic object and the applied force is defined by Hooke's law
• Hooke's law states that:

The extension of an elastic object is directly proportional to the force applied, up to the limit of proportionality

• The force applied to the spring is sometimes referred to as the load

• Directly proportional means that as the force is increased, the extension increases

  • If the force is doubled, then the extension will double
  • If the force is halved, then the extension will also halve

• The limit of proportionality is the point beyond which the relationship between force and extension is no longer directly proportional

  • This limit varies according to the material

Exam Hint You may have met the term 'elastic limit' elsewhere. It is not on this syllabus, and using it in place of 'limit of proportionality' will not earn the mark.

Hooke's law states that a force applied to a spring will cause it to extend by an amount proportional to the force

• Hooke's law can be described by the following equation:

$F=kx$

• Where:

  • $F$ = force applied, measured in newtons (N)
  • $k$ = the spring constant, measured in newtons per metre (N/m)
  • $x$ = extension of spring, measured in metres (m)

Spring constant

• The spring constant is defined as:

The force per unit extension

• Therefore, the units are newtons per metre (N/m), though newtons per centimetre (N/cm) is equally acceptable and is often more convenient when extensions are measured in centimetres.

• The spring constant is a measure of how stiff the spring is

  • Stiff springs have a high spring constant
  • Stretchy springs have a low spring constant

• Any elastic object has a spring constant, not just springs

• The Hooke's law equation can be used to calculate the spring constant of a material

$k=\frac{F}{x}$

The load-extension graph

• Hooke's law is a linear relationship

  • This is represented by a straight line on a load-extension, or force-extension graph

Load-extension graph for a spring

Hooke's law is associated with the linear region of a load-extension graph. Beyond the limit of proportionality, Hooke's law no longer applies

Important features of the load-extension graph

• The linear portion of the graph

  • This represents the load or force under which the spring obeys Hooke's law
  • Load and extension are directly proportional
  • The gradient of the linear portion is equal to the spring constant for a load-extension graph
  • The gradient of the linear portion is equal to $\frac{1}{k}$ for an extension-load graph

• The limit of proportionality

  • This is the point at which the graph begins to curve
  • Beyond this point, load and extension are no longer proportional

• The curved portion of the graph

  • This is where the material does not obey Hooke's law
  • Load and extension are not proportional

Worked Example A force of 32 N pulls a spring from its orginal length of 8 cm to 18 cm. Calculate the spring constant of the spring in N/m.

[3]

Answer:

Step 1: List the known quantities

• Force, F = 32 N
• Original length = 8 cm
• Final length = 18 cm

Step 2: Write the relevant equation

$F=kx$

Step 3: Rearrange to make spring constant the subject

$k=\frac{F}{x}$ [1 mark]

Step 4: Calculate the extension, x

$x=\mathrm{final} \mathrm{length}-\mathrm{original} \mathrm{length}$

$x=18-8=10\mathrm{cm}$ [1 mark]

Step 5: Convert any units

• $x$ must be in metres (m) to get a spring constant in N/m

$x=\frac{10}{100}=0.1m$

Step 6: Substitute the values for force and extension

$k=\frac{32}{0.1}$

$k=320N/m$ [1 mark]

Worked Example The figure below shows the forces acting on a child who is balancing on a pogo stick. The child and pogo stick are not moving.

The spring constant of the spring on the pogo stick is 4900 N/m. The weight of the child causes the spring to compress elastically from a length of 40 cm to a new length of 33 cm.

Calculate the weight of the child.

[4]

Answer:

Step 1: List the known quantities

• Spring constant, k = 4900 N/m
• Original length = 40 cm
• Final length = 33 cm

Step 2: Write the relevant equation

$F=kx$ [1 mark]

Step 3: Calculate the extension, x

$x=\mathrm{final} \mathrm{length}-\mathrm{original} \mathrm{length}$

$x=33-40=-7\mathrm{cm}$

• A negative extension represents a compression of 7 cm [1 mark]

Step 4: Convert any units

• Since the spring constant is given in N/m, $x$ must be in metres (m)

$x=\frac{-7}{100}=-0.07m$ [1 mark]

Step 5: Substitute the values into the Hooke's law equation

$F=4900\times-0.07$

$F=-343N$

• The minus sign simply indicates the direction of the force, downwards in this case
• The child's weight is 343 N [1 mark]

Exam Hint Remember that the spring constant is calculated by dividing the load by the extension of the elastic object, not the total length of the elastic object.

In the exam, mark the limit of proportionality as a single precise point at the end of the straight section, and label it L. A mark placed on the curved part of the line, or a range drawn between two points, will not be accepted.