Use the resultant force to find signed acceleration
Apply Newton’s first and second laws to a constant-mass particle, keeping the chosen direction and all actual forces in one balance.
Exam scope and assessment notes
Mechanics (Paper 4, AS & A Level) · Mathematics 9709, 2026–2027, section 4.4. Use g = 10 m s⁻² throughout these problems. Each particle has positive constant mass. Forces and acceleration are constant within each stated stage; linear motion only. Use the resultant force in F = ma. A light taut inextensible string over a smooth pulley has one tension and equal acceleration magnitudes; those constraints cease if the string becomes slack. A rope can pull but cannot push; a light rigid tow-bar may exert tension or thrust. Resistance other than the stated friction is neglected unless explicitly supplied. In our rough-contact model, sliding friction has magnitude μR and opposes motion; at rest, friction adjusts with magnitude at most μR. Check contact and motion assumptions before continuing a solution. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Resolve and add signed force components.
- Distinguish velocity from acceleration.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which quantity belongs on the force side of F = ma?
What you will learn
- Calculate acceleration from the resultant force rather than one selected force.
- Interpret zero resultant and acceleration opposite to current motion.
Resultant force determines acceleration
Choose an axis along the linear motion and list the actual forces on the particle. For constant mass, their signed sum is ma. A 4 kg particle acted on by 18 N to the right and 6 N to the left has resultant 12 N rightwards, so a = 12/4 = +3 m s⁻² with right positive. There is no additional force called ma to add to the diagram: ma describes the consequence of the resultant.
If perpendicular acceleration is zero, resolve separately perpendicular to the path. On a horizontal surface with only weight and normal reaction vertically, R − mg = 0. This vertical equation determines contact forces; it does not replace the horizontal equation. A force need not point in the direction of the current velocity: a leftward resultant can slow a rightward particle, then make it speed up leftwards if that force continues.
Zero resultant means constant velocity
Newton’s first law says that zero resultant gives zero acceleration in this model. The particle can be stationary or moving at a constant non-zero velocity; the forces need not be absent. A 12 N forward pull balanced by 12 N resistance maintains a stated constant velocity. Conversely, at an instant when velocity is zero, acceleration may be non-zero: an unsupported particle at its highest point still accelerates downwards.
Mass measures inertia and is expressed in kilograms; force is measured in newtons. For a known resultant, a = F/m, so doubling positive mass halves acceleration. Newton’s third-law partners act on different bodies and do not cancel within one particle’s diagram. They cancel only when both interacting bodies are included in a larger system, which is useful later for connected particles.
Pause and explain
A particle moves right but has negative acceleration with right positive. What follows immediately?
Worked example
A 4 kg particle has forces 18 N rightwards and 6 N leftwards. Find its acceleration, taking right as positive.
Show the worked solution
- Include both actual horizontal forces: F = 18 − 6 = 12 N.
- Use F = ma: 12 = 4a.
- Thus a = +3 m s⁻². Weight and normal reaction balance vertically if vertical acceleration is zero.
Answer Acceleration +3 m s⁻², rightwards.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A 4 kg particle has forces 18 N right and 6 N left. Find its signed acceleration with right positive.
Give me a hint
Subtract opposing forces before dividing by the positive mass.
Compare my reasoning
- The signed resultant is 18 − 6 = 12 N.
- Newton’s second law gives 4a = 12.
- Therefore a = +3 m s⁻², in the rightward direction.
a = +3 m s⁻².
Look for these in your work
- Include both forces in the resultant.
- Report acceleration with a direction and correct units.
A 5 kg particle moving right at 6 m s⁻¹ has a constant 8 N rightward drive and 18 N leftward resistance. Find acceleration and velocity after 2 s if these forces continue.
Give me a hint
Keep the negative acceleration in v = u + at.
Compare my reasoning
- The resultant is 8 − 18 = −10 N, so a = −10/5 = −2 m s⁻².
- After 2 s, v = 6 + (−2) × 2 = +2 m s⁻¹.
- The particle is still moving right but more slowly; acceleration is leftwards.
a = −2 m s⁻²; v(2) = +2 m s⁻¹.
Look for these in your work
- Use resistance with the opposing sign.
- Distinguish the direction of acceleration from the still-positive velocity.
A 3 kg particle moves right at constant velocity 5 m s⁻¹. A forward drive D balances a stated 12 N resistance. Find D and acceleration; explain whether the particle is stationary.
Give me a hint
Constant velocity means zero resultant, not zero forces or zero speed.
Compare my reasoning
- Because the velocity is constant, a = 0.
- Newton’s second law gives D − 12 = 3 × 0, so D = 12 N.
- The particle remains moving right at 5 m s⁻¹; it is not stationary.
D = 12 N; a = 0; velocity remains +5 m s⁻¹.
Look for these in your work
- Use the stated constant velocity to set acceleration to zero.
- Do not infer rest from balanced forces.
Common mistakes
- Use the resultant, not just the driving force.
- Do not add ma as another force.
- Zero acceleration does not require zero velocity.
What can you explain now?
Does zero resultant force imply that a moving particle stops?
Compare with the explanation
No; it retains its constant velocity in this model.
Stopping requires a change of velocity, which requires a non-zero resultant over a time interval.
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Make it stick
Choose a direction, draw actual forces and set their signed resultant equal to ma.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.