The key idea
Begin with a diagram of the forces acting on one object. Find their resultant, then explain the resulting change in velocity or shape. This approach connects acceleration, resistance, springs and circular motion.
Understand it · 01
Choose one object and draw its interactions
A force diagram starts with a clear choice of object. Show weight, contact forces and any other interactions acting on that object as labelled arrows. The arrow direction represents the force direction; its length can show relative magnitude. Do not add a forward force merely because the object is moving forwards. Motion is the state you are trying to explain, not a separate force.
For forces along one line, choose a positive direction and add signed values. Forces in the same direction add; opposing forces subtract. On a horizontal surface with no vertical acceleration, the upward support force balances weight if no other vertical forces act. This vertical balance can coexist with an unbalanced horizontal force, so calculate each direction separately rather than adding perpendicular arrows as ordinary numbers.
Understand it · 02
Connect the resultant to changing velocity
Newton's second law uses the resultant force: F = ma for constant mass. A larger resultant produces a larger acceleration on the same mass. The same force produces a smaller acceleration on a larger mass. Acceleration has direction, so a force opposite the current velocity slows an object; the same force can speed it up after the object stops and reverses.
With zero resultant force, velocity stays constant. That includes an object remaining at rest, but also an object continuing in a straight line at steady speed. Everyday vehicles require a driving force at steady speed because drag and friction also act. The driving force balances these resistive forces; it does not imply a forward resultant. State which forces balance and what this means for acceleration.
Understand it · 03
Distinguish extension, proportionality and recovery
Measure a spring's original length before adding a load. Extension is the increase in length, so subtract the original length from every loaded length. In the proportional region, doubling the force doubles the extension and F = kx applies. On a force-versus-extension graph the gradient is k; if the axes are swapped, the gradient is 1/k. Read the axis labels before using a gradient.
The limit of proportionality is where the straight-line relationship ends. Elastic behaviour means the material returns to its original shape after unloading. These are related ideas but not identical: a material can behave elastically over a region that is not proportional. To investigate, add loads in steps, allow the spring to settle, measure extensions and then unload to check recovery. Avoid assuming every curved section proves permanent deformation.
Understand it · 04
Changing direction also needs a resultant force
An object travelling around a circle at constant speed continually changes its velocity because its direction changes. The resultant force points towards the circle's centre. It is supplied by an actual interaction, such as tension, friction or gravity; do not draw a second inward force simply labelled centripetal in addition to the real inward force. If that interaction disappears, the object initially continues along the tangent.
Action–reaction force pairs act on different objects. If a hand pushes a trolley, the trolley pushes the hand with an equal force in the opposite direction. These forces do not cancel on the trolley's own force diagram because only one acts on the trolley. Weight and the ground's support force on a resting trolley act on the same object, so they are not a Newton's third-law pair.
Key definitions
- Resultant force
- The single force that has the same effect as all the forces acting on an object combined.
- Hooke's law
- The extension of a spring is directly proportional to the load applied, provided the limit of proportionality is not exceeded.
- Friction
- A contact force that opposes relative motion, or the tendency for relative motion, between surfaces. Sliding friction transfers energy to the internal energy of the surfaces and surroundings.
- Limit of proportionality
- The point beyond which extension is no longer proportional to load, and the load-extension graph stops being a straight line.
Equations, units & conditions
Newton's second law
F = m × a
Units: N
Hooke's law
F = k × x
Units: N
k = spring constant (N/m), x = extension
Worked examples, step by step
Example 1
A 1200 kg car experiences a driving force of 3600 N and total resistive forces of 900 N. Calculate its acceleration.
Follow the method
- Resultant force = 3600 − 900 = 2700 N (forces oppose, so subtract).
- a = F ÷ m = 2700 ÷ 1200.
a = 2.25 m/s²
Practice questions with worked solutions
Try each question on paper before opening the solution. These are original academy questions; the suggested marks are a self-checking guide, not an official exam-board mark scheme. Difficulty labels describe this practice set, not predicted grades.
Build · 3 suggested marks
1. Find the resistive force
A 6.0 kg trolley is pulled horizontally with 25 N and accelerates at 3.0 m/s². Find the total horizontal resistive force.
Show solution and self-check: question 1
- Resultant force = ma = 6.0 × 3.0 = 18 N.
- Pull − resistance = resultant, so resistance = 25 − 18 = 7.0 N.
Resistive force = 7.0 N opposite the pull.
Self-check · one mark per point
- Uses resultant F = ma.
- Finds resultant 18 N.
- Subtracts to obtain resistance 7.0 N opposite the pull.
Build · 4 suggested marks
2. Find a loaded spring length
A spring of original length 8.0 cm extends by 2.0 cm under a 3.0 N force. Calculate its spring constant and its total length under a 4.5 N force, assuming it remains within its proportional region.
Show solution and self-check: question 2
- Initial extension = 0.020 m, so k = F/x = 3.0/0.020 = 150 N/m.
- New extension = 4.5/150 = 0.030 m = 3.0 cm.
- Total length = original length + extension = 8.0 + 3.0 = 11.0 cm.
k = 150 N/m; total loaded length = 11.0 cm.
Self-check · one mark per point
- Converts the first extension to metres.
- Finds k = 150 N/m.
- Finds new extension 3.0 cm.
- Adds the original length to obtain 11.0 cm.
Common mistakes & how to avoid them
- Using the total length of the spring instead of the extension in Hooke's law. Extension = stretched length minus original length — every time.
- Saying a resultant force is needed to keep an object moving. It is not: with zero resultant force an object continues at constant velocity. A resultant force is needed to change velocity.
- In circular motion, saying the speed changes. At constant speed in a circle the velocity changes (direction), so there is an acceleration towards the centre — caused by a resultant force towards the centre.
- Adding forces that act in opposite directions instead of subtracting. Draw the arrows before you calculate.
Test yourself
10 questions · instant marking
Question 1 of 10
A resultant force of 20 N acts on a 4 kg mass. The acceleration is:
Question 2 of 10
An object moves at constant velocity. The resultant force on it is:
Question 3 of 10
A spring is 12 cm long unstretched and 17 cm with a load. Its extension is:
Question 4 of 10
Hooke's law is valid:
Question 5 of 10
Forces of 12 N left and 8 N right act on an object. The resultant is:
Question 6 of 10
An object moves in a circle at constant speed. Its velocity is:
Question 7 of 10
The weight of a 5 kg object on Earth (g = 9.8 N/kg) is:
Question 8 of 10
A spring with spring constant 200 N/m is stretched by 0.03 m. The force is:
Question 9 of 10
Friction acting on a sliding box transfers kinetic energy mainly to:
Question 10 of 10
A car of mass 900 kg accelerates at 2 m/s² with resistive forces of 300 N. The driving force is:
Understand the why
Questions, explained
Start with the short answer, then follow the reasoning. Each explanation has a permanent link you can share with a classmate or return to when revising.
What does a resultant force do to an object?
A resultant force changes an object's velocity: its speed, its direction, or both. For constant mass, the acceleration is in the resultant-force direction and has magnitude F/m. The resultant is the combined effect of all forces on that object, not just the largest force.
Separate the object from its surroundings before adding forces. A trolley's horizontal pull and resistance act on the trolley, while the force the trolley exerts on the pulling hand belongs on a different object's diagram. Adding every interaction in the scene together would obscure the particular acceleration you are calculating.
A resultant opposite to the velocity initially reduces speed, whereas a resultant perpendicular to the velocity can change direction. With zero resultant force, there is no acceleration and the existing velocity continues. A force can also deform a real object, but a calculation using F = ma describes the acceleration of its centre of mass under the stated model.
Find the combined horizontal effect
A 5.0 kg trolley experiences 18 N rightwards and 8.0 N leftwards. Calculate its horizontal acceleration.
- Choose right as positive: resultant = 18 − 8.0 = 10 N.
- Acceleration = 10/5.0 = 2.0 m/s² to the right.
Acceleration = 2.0 m/s² rightwards.
How do you answer Hooke's law graph questions?
Read which quantity is on each axis, identify the straight proportional region and use extension rather than total spring length. A force-against-extension gradient gives k; an extension-against-force gradient gives 1/k. Do not extend the proportional model beyond the point where the measured relationship changes.
A spring graph can use centimetres even when the requested spring constant is in N/m. Either convert the extensions before finding the gradient or convert the resulting N/cm value afterwards. Choose well-separated points on the straight line, rather than taking a tiny triangle that magnifies reading uncertainty.
A graph that does not pass through the origin needs investigation. The horizontal quantity may be total length, there may be an initial tension, or there may be a measurement offset. Do not automatically call every straight line through the displayed points Hooke's law without checking what the measured variables and zero positions represent.
Use a large gradient triangle
On the proportional part of a force–extension graph, force rises from 1.0 N to 5.0 N while extension rises from 0.50 cm to 2.50 cm. Find k in N/m.
- Force change = 4.0 N; extension change = 2.00 cm = 0.0200 m.
- k = gradient = 4.0/0.0200 = 200 N/m.
Spring constant = 200 N/m.
Why does an object moving in a circle accelerate at constant speed?
Velocity includes direction, and direction changes continuously around a circle. That means velocity changes even if its magnitude, speed, stays constant. The acceleration and resultant force point towards the centre in uniform circular motion; an outward balancing force would remove the required inward resultant.
Compare velocity arrows at two nearby points on the circle. They can have equal lengths but different directions, so they are not the same vector. The inward resultant continually bends the path. It may be supplied by tension in a string, friction on a road or gravitational attraction, depending on the situation.
Centripetal names the inward role of the resultant rather than an extra interaction to add to a force diagram. If a string supplying the force breaks, the object initially follows the tangent at the release point, not a straight line directly away from the centre. Cambridge's qualitative circular-motion treatment belongs to Supplement content.
Why don't Newton's third-law forces cancel each other?
The two forces in a Newton's third-law pair act on different objects. They are equal and opposite, but only one belongs on each object's force diagram. Forces cancel when you calculate a resultant on the same chosen object or when treating both objects as one system.
When a hand pushes a trolley, the hand exerts a force on the trolley and the trolley exerts an opposite force on the hand. The trolley's acceleration depends on the force on the trolley together with its other forces, such as resistance. It does not depend on adding the reaction on the hand to that individual diagram.
For the combined hand-and-trolley system, the contact pair is internal and cancels in the system's total force balance. External forces then determine the combined momentum change. Also distinguish this pair from weight and support acting on one resting object: those may balance, but they are separate interactions and are not each other's third-law partners.
Are the elastic limit and the limit of proportionality the same?
They describe different conditions. The limit of proportionality marks the end of a proportional force–extension relationship. The elastic limit concerns whether the material returns to its original shape after unloading. A material can be elastic over a non-linear region, so a curved graph alone does not prove permanent deformation.
Hooke's law predicts a straight force–extension graph through the origin over the proportional region. Elastic behaviour instead requires a loading-and-unloading comparison. If you remove the force and the original length is recovered, the deformation was reversible even if doubling the force did not exactly double the extension.
To check recovery experimentally, record the unloaded length, add loads in controlled steps and then remove them. A remaining extension after the force has been removed is evidence of permanent deformation. Report what the data demonstrate rather than identifying every first visible curve as both limits at once; the detailed material terminology required varies between courses.
Why does a car need a driving force if it is travelling at constant speed?
A real car needs a driving force to balance resistive forces such as drag and rolling resistance. At constant speed along a straight, level road, the horizontal resultant is zero even though the individual forces are not. The driving force maintains that balance rather than producing forward acceleration.
If the engine stops supplying the required driving force while resistance continues, the forces no longer balance and the car slows. The need for a powered engine is therefore compatible with Newton's first law. Energy is still transferred to the surroundings through resistance during the steady-speed journey.
Compare this with an ideal frictionless surface: an object could keep moving at constant velocity without a continuing push. Also check the route. A car rounding a bend at constant speed requires a sideways resultant because its direction changes; constant speed on its own is insufficient to conclude that every component of resultant force is zero.
Separate driving force from resultant force
A car travels steadily in a straight line against 450 N of air resistance and 150 N of other resistance. Find the required driving force and horizontal resultant.
- Total resistance = 450 + 150 = 600 N.
- Steady straight-line velocity requires 600 N forwards, balancing resistance.
- Resultant = 600 − 600 = 0 N.
Driving force = 600 N forwards; horizontal resultant = 0 N.
Your learning checklist
Use these goals to check your understanding. Requirements vary by specification, tier and exam year; the official references below define your full course.
- Find the resultant of forces acting along a line
- Know that a resultant force changes the velocity of an object (speed and/or direction)
- Use F = ma for calculations
- Describe the effects of friction and air resistance
- Use Hooke's law (F = kx), the limit of proportionality, and load-extension graphs
- Describe circular motion: a perpendicular force causes a change in direction at constant speed (Extended)
Sources & how to cite these notes
Original explanations, illustrations and practice by Dr Desouky Physics Academy. Official specifications guide the course scope; questions and suggested marks on this page are our own revision material.
- Cambridge IGCSE Physics 0625 — 2026–2028 syllabus
- Pearson Edexcel International GCSE Physics 4PH1 — specification
Reference this page
Dr Desouky Physics Academy. IGCSE Physics: Forces. Updated 2026-09-25. https://www.drdesouky.com/revision/igcse-physics/forces
Use the arrow beside an explanation heading to link straight to that section. Spotted a problem? Contact the academy with the topic and the step you want us to check.