Field, contact and elastic forces

Key idea: Forces describe interactions. Identify what is interacting with the chosen body before deciding the force type or direction.

GCE A-Level H1 Physics 2027

H1 Physics 8867 · Lesson 1 of 3

Check your understanding

By the end of this lesson, you should be able to

Describe gravitational, electric and magnetic field forces on the appropriate object.
Recognise normal, frictional and viscous contact forces without using out-of-scope coefficients.
Apply Hooke’s law within the proportional region.

Learn the idea

Big question: Which interaction produces each force, and what determines that force’s direction?

Choose the body before naming forces

A force diagram answers one question: what forces act on this chosen body? Weight is exerted by a gravitational field on mass; electric force by an electric field on charge; magnetic force by a magnetic field on a current-carrying conductor. Contact forces arise only where the body touches a surface, spring or fluid.

Naming the agent prevents invented forces. A flying ball has weight and perhaps drag, but it does not retain a forward force from the hand after contact ends.

Check your understanding: Why is a normal force not automatically equal to weight?

Its size is set by the complete force balance perpendicular to the surface; other forces or acceleration can make it different from weight.

Separate motion from force direction

Drag and friction oppose relative motion between contacting bodies or fluid layers. A spring force instead opposes deformation: a stretched spring pulls and a compressed spring pushes towards its natural length. Neither rule says that force must oppose the object’s instantaneous velocity.

Check your understanding: A compressed spring launches a block forward. Can spring force and velocity point the same way?

Yes. While the spring expands, its restoring force can accelerate the already forward-moving block in the same direction.

Scroll diagram horizontally to read all labels.

The proportional limit P ends the straight Hooke’s-law region. The elastic limit E is a different condition: deformation remains reversible only up to E.

Key ideas

Name the agent exerting each force.
Draw only forces acting on the chosen body.
The spring constant k is the gradient of a straight F–x graph.

Relationships to know

weight W = mg
electric force F = QE
spring force magnitude F = kx

Follow the reasoning

Worked example

Use Hooke’s law inside an equilibrium argument

Question: A 0.60 kg block rests on a rough horizontal table. A spring of force constant 80 N m⁻¹ attached on the right is stretched by 0.050 m. Find the friction force and describe all other forces.

Step 1: Calculate the spring interaction
Why: The extension is within the stated proportional regime, so Hooke’s law applies.
Working: Fspring = kx = 80(0.050) = 4.0 N to the right.
Step 2: Apply horizontal equilibrium
Why: The resting block has zero resultant horizontal force.
Working: Ffriction + 4.0 = 0, so friction is 4.0 N to the left.
Step 3: Complete the vertical model
Why: A free-body account must include every external force, even those not needed numerically.
Working: Weight = 0.60g downward and normal contact = 0.60g upward.

Answer: Static friction is 4.0 N left. Weight and normal contact are equal and opposite vertically; spring force and friction are equal and opposite horizontally.

Check: Both component resultants are zero, consistent with the block remaining at rest.

Now try it with support

Practise with support

A 2.0 N force extends a spring by 4.0 cm. Find k and predict the extension for 3.0 N if the spring remains proportional.

Hints

Convert 4.0 cm to metres.
Use the same k only because proportional behaviour is stated.

View the guided answer

k = 2.0/0.040 = 50 N m⁻¹. For 3.0 N, x = 3.0/50 = 0.060 m = 6.0 cm.

Your turn

Practise independently

A 0.60 kg block rests on a rough horizontal table and is attached to a spring of force constant 80 N m⁻¹ stretched by 0.050 m. Draw the forces and find the friction needed for equilibrium.

Check your answer

The forces are weight 0.60g downward, normal contact upward, spring force kx = 80(0.050) = 4.0 N towards the spring, and static friction 4.0 N opposite the spring force. Vertical forces balance and the horizontal resultant is zero.

Common mistakes and exam guidance

Watch out for

Drawing a continuing ‘push’ after an object has left the hand.
Applying F = kx after the force–extension graph has stopped being linear.

In an exam

A force label should include its type and, where useful, the agent.
If the question asks for a free-body diagram, do not add velocity or acceleration arrows as forces.

Put the ideas together

Exam-style practice [6 marks]

For each situation, name the force and state its direction: (i) a positive charge in an eastward electric field; (ii) a current-carrying wire that experiences an upward magnetic force; (iii) a sphere falling through oil below terminal speed. For the sphere, explain how the forces change as speed rises.

Plan before you answer

Identify the object responding to each field or contact interaction.
Use the field definition or relative-motion rule for direction.
For the sphere, connect rising speed to viscous force and resultant force.

View the marking points and model answer

Marking points

Positive charge: electric force east.
Wire: magnetic force upward as stated, acting on the current-carrying conductor.
Sphere: weight downward.
Sphere: viscous drag upward, opposing motion.
Drag increases as speed rises while weight remains approximately constant.
Downward resultant and acceleration decrease towards zero.

Model answer

The positive charge experiences electric force east, along the electric field. The wire experiences magnetic force upward. The falling sphere has weight downward and viscous drag upward. As its speed increases, drag increases while weight is nearly constant, so the downward resultant and acceleration decrease. At terminal speed the two forces become equal.

Finish from memory

Three-question recap

What interaction produces a normal force?
Check
Contact between a body and a surface.
When is F = kx valid?
Check
Within the region where force is proportional to extension or compression.
Does friction always oppose velocity?
Check
It opposes relative slipping or the tendency to slip at the contact, not necessarily the body’s overall velocity.

Try this next

Draw force diagrams for a falling object, a charged particle and a block on a surface before attempting calculations.

Continue to Moments, couples and centre of gravity Return to Forces and Moments

Continue with the next resource in this course.

Course and syllabus information

Course: GCE A-Level H1 Physics
Edition: GCE A-Level H1 Physics 2027