Calculate acceleration from the resultant force
Use F = ma on the signed total force rather than one convenient individual force.
Exam scope and assessment notes
Extended only (Supplement) · Physics 0972, 2026–2028, section 1.5.1. Use k = F/x within the proportional range, identify its limit on a load–extension graph, and apply F = ma to the resultant force on the chosen body. Circular motion is qualitative here: compare force, speed, radius and mass while preserving the specified fixed variables. The circular-force formula and an understanding of the elastic limit are not required by this section. Written practice is self-checked; cumulative assessment, supervised practical experience and human subject-teacher review remain pending. Cambridge states that IGCSE 0972 has the same subject content as IGCSE 0625, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- Calculate collinear resultants.
- Distinguish acceleration from existing velocity.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the resultant of 10 N right and 4 N left?
What you will learn
- Recall and use resultant force = mass × acceleration.
- State the acceleration direction and convert mass to kilograms.
The F in this equation is the resultant
For a constant-mass body in the ordinary laboratory inertial frame, F = ma uses its resultant force. Rearranging gives a = F/m. With right positive, a 3 kg trolley pulled by 10 N right against 4 N resistance left has F = 6 N and a = 2 m/s² right. Dividing the pull alone by mass would miss the resistance.
Use kg for mass, N for force and m/s² for acceleration. A 500 g mass is 0.500 kg; a 2.0 N resultant gives a = 2.0/0.500 = 4.0 m/s². Units are consistent because one newton is one kg·m/s².
Acceleration follows resultant direction
Force and acceleration are in the same direction for this positive-mass model. A 3 kg body with resultant 6 N left has acceleration 2 m/s² left. It may nevertheless be moving right at that instant and slowing down; its existing velocity does not have to match the acceleration direction.
If the resultant is zero, acceleration is zero and velocity does not change. Non-zero force over a specified interval can be combined with earlier motion methods when the acceleration model is constant. Do not infer a time or final velocity from F = ma alone when the required interval and starting velocity are absent.
Pause and explain
A 3 kg body has 6 N resultant left while moving right. What is its acceleration?
Worked example
A 3 kg trolley is pulled 10 N right against 4 N resistance left. Find acceleration.
Show the worked solution
- Rightward resultant = 10 − 4 = 6 N.
- a = F/m = 6/3 = 2 m/s².
- Acceleration points right with the resultant.
Answer 2 m/s² right.
Try a topic test
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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 3 kg trolley has 10 N pull right and 4 N resistance left. Find acceleration.
Give me a hint
First find the signed force total on the trolley.
Compare my reasoning
- F = 10 − 4 = 6 N right.
- a = 6/3 = 2 m/s².
- The acceleration is rightward.
2 m/s² right.
Look for these in your work
- Find the resultant.
- State acceleration direction.
A mass of 500 g has a 2.0 N rightward resultant. Find acceleration.
Give me a hint
Use kilograms in F = ma.
Compare my reasoning
- 500 g = 0.500 kg.
- a = 2.0/0.500 = 4.0 m/s².
- It points right with the resultant.
4.0 m/s² right.
Look for these in your work
- Convert mass.
- Use the resultant direction.
A 3 kg body currently moves right but has 6 N resultant left. Calculate acceleration and explain its initial effect.
Give me a hint
Acceleration describes a change in velocity, not necessarily the current direction.
Compare my reasoning
- a = 6/3 = 2 m/s² left.
- That opposes the present rightward velocity.
- The body initially slows while still moving right.
2 m/s² left; it initially slows while moving right.
Look for these in your work
- Calculate the acceleration.
- Separate acceleration from velocity.
Common mistakes
- Use resultant rather than driving force alone.
- Convert grams to kilograms.
- Acceleration direction need not equal current velocity direction.
What can you explain now?
What acceleration does 2 N resultant give a 500 g mass?
Compare with the explanation
4 m/s² in the resultant's direction.
Use mass 0.500 kg before dividing force by mass.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Write a force balance before F = ma and explain a case with opposite acceleration and velocity directions.
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.