Forces, Extension & Hooke's Law | CAIE 9702 Paper 1 Notes

Chapters

Paper 1

Physics9702

Forces, Extension & Hooke's Law

A solid deforms when forces change its length or shape.

A tensile pair pulls outward and stretches the solid. A compressive pair pushes inward and shortens it.

Extension is the increase in length, not the final length.

Two hand-drawn metal bars. The left bar is longer than its dashed original outline, while the right bar is shorter than its dashed original outline.

tensile forces pull apartcompressive forces push together

A tensile pair pulls the ends apart and increases length. A compressive pair pushes the ends together and decreases length. The dashed outlines show the original shapes.

Measure the change in length first

Let the original length be $L_0$ and the stretched length be $L$ . The extension is the amount added to the original length.

x=L-L_0

Symbols

$x$ = extension (m)
$L$ = stretched length (m)
$L_0$ = original unstretched length (m)

Worked example

Do not use the final length as the extension

A spring changes from 12.0 cm to 17.0 cm.

Subtract the original length from the new length.
$x=17.0-12.0=5.0\text{ cm}=0.050\text{ m}$ .

Answer

x=0.050\text{ m}

A straight-line response has a simple rule

Add small loads one at a time. At first, doubling the applied force doubles the extension. Force and extension are then directly proportional.

Definition

1 mark

What is meant by Hooke's law?

Model answer: The extension of an object is proportional to the applied force, provided that the limit of proportionality is not exceeded.

F=kx

Symbols

$F$ = applied force, or load (N)
$k$ = spring constant, which measures stiffness (N m⁻¹)
$x$ = extension (m)

A larger $k$ means a stiffer specimen: more force is needed for the same extension.
On a graph of force against extension, $k$ is the gradient of the straight section.
The limit of proportionality is where the graph first stops being a straight line. Beyond it, $F=kx$ is no longer valid.

Change extension

k = 200 N m⁻¹ · x = 0.040 m · F = 8.0 N

Your turn9702/23/M/J/24 Q3(a), (c)(i)

State Hooke's law. A straight force-extension graph reaches 5.4 N at an extension of 60 mm. Calculate the spring constant. (9702/23/M/J/24 Q3(a), (c)(i))

Show worked answer

State that extension is proportional to applied force, provided the limit of proportionality is not exceeded.
Convert 60 mm to 0.060 m.
$k=F/x=5.4/0.060=90\text{ N m}^{-1}$ .

Answer

90\text{ N m}^{-1}

Common mistake

Do not write “extension equals force”. Hooke's law requires the word proportional and its condition. Also convert mm or cm to metres before using a spring constant in N m^-1.

Forces, Extension & Hooke's Law

A solid deforms when forces change its length or shape.

A tensile pair pulls outward and stretches the solid. A compressive pair pushes inward and shortens it.

Extension is the increase in length, not the final length.

tensile forces pull apartcompressive forces push together

A tensile pair pulls the ends apart and increases length. A compressive pair pushes the ends together and decreases length. The dashed outlines show the original shapes.

Measure the change in length first

Let the original length be $L_0$ and the stretched length be $L$ . The extension is the amount added to the original length.

x=L-L_0

Symbols

$x$ = extension (m)
$L$ = stretched length (m)
$L_0$ = original unstretched length (m)

Worked example

Do not use the final length as the extension

A spring changes from 12.0 cm to 17.0 cm.

Subtract the original length from the new length.
$x=17.0-12.0=5.0\text{ cm}=0.050\text{ m}$ .

Answer

x=0.050\text{ m}

A straight-line response has a simple rule

Add small loads one at a time. At first, doubling the applied force doubles the extension. Force and extension are then directly proportional.

Definition

1 mark

What is meant by Hooke's law?

Model answer: The extension of an object is proportional to the applied force, provided that the limit of proportionality is not exceeded.

F=kx

Symbols

$F$ = applied force, or load (N)
$k$ = spring constant, which measures stiffness (N m⁻¹)
$x$ = extension (m)

A larger $k$ means a stiffer specimen: more force is needed for the same extension.
On a graph of force against extension, $k$ is the gradient of the straight section.
The limit of proportionality is where the graph first stops being a straight line. Beyond it, $F=kx$ is no longer valid.

Change extension

k = 200 N m⁻¹ · x = 0.040 m · F = 8.0 N

Your turn9702/23/M/J/24 Q3(a), (c)(i)

State Hooke's law. A straight force-extension graph reaches 5.4 N at an extension of 60 mm. Calculate the spring constant. (9702/23/M/J/24 Q3(a), (c)(i))

Show worked answer

State that extension is proportional to applied force, provided the limit of proportionality is not exceeded.
Convert 60 mm to 0.060 m.
$k=F/x=5.4/0.060=90\text{ N m}^{-1}$ .

Answer

90\text{ N m}^{-1}

Common mistake

Do not write “extension equals force”. Hooke's law requires the word proportional and its condition. Also convert mm or cm to metres before using a spring constant in N m^-1.