Find the force or mass needed for a specified acceleration
Rearrange F = ma and include resistance when an individual driving force is requested.
Exam scope and assessment notes
Extended only (Supplement) · Physics 0625, 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.
Before you begin
- Rearrange a product equation.
- Use signed constant-acceleration velocity changes.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which equation finds mass from resultant F and acceleration a?
What you will learn
- Find mass or resultant from acceleration information.
- Distinguish required resultant from driving and braking forces.
Solve for the requested quantity
To find mass, use m = F/a with a matched resultant and its acceleration. A 6 N resultant producing 3 m/s² gives mass 2 kg. To find a resultant, multiply mass by acceleration. A 1200 kg car requiring 0.80 m/s² eastward acceleration needs 960 N eastward resultant.
If that car also has 240 N westward resistance, its driving force must satisfy driving − 240 = 960 N. The required driving force is 1200 N east, not the 960 N resultant alone. List other forces acting on the body before reporting an individual pull, thrust or brake force.
Connect the force balance to an interval carefully
For a 900 kg car initially at 12 m/s east, a constant horizontal resultant 1800 N west gives acceleration 2 m/s² west. Taking east positive, a = −2 m/s² and after four seconds v = 12 − 2 × 4 = 4 m/s east. A backward resultant can slow forward motion without instantly reversing it.
That interval calculation assumes the stated resultant and mass remain constant over the four seconds. A changing resistance or braking force needs its own model. An acceleration value alone is insufficient for a final velocity without the initial value and time; avoid inventing missing conditions.
Pause and explain
A 1200 kg car needs a 960 N resultant against 240 N resistance. What forward drive is needed?
Worked example
A 1200 kg car needs acceleration 0.80 m/s² east against 240 N westward resistance. Find required driving force.
Show the worked solution
- Required resultant = 1200 × 0.80 = 960 N east.
- Driving − 240 = 960.
- Driving force = 1200 N east.
Answer 1200 N east; required resultant 960 N east.
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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.
For 1200 kg at 0.80 m/s² east against 240 N west resistance, find drive.
Give me a hint
Find the required resultant, then add the opposing resistance to obtain drive.
Compare my reasoning
- Fresultant = 1200 × 0.80 = 960 N east.
- Drive − 240 = 960 N.
- Drive = 1200 N east.
1200 N east.
Look for these in your work
- Calculate the required resultant.
- Include the resistance.
A 3 kg trolley needs acceleration 2 m/s² east against 4 N westward resistance. Find its required pull, then find the mass that a 6 N resultant accelerates at 3 m/s².
Give me a hint
The first requested force is an individual pull; the second supplied force is already resultant.
Compare my reasoning
- First resultant = 3 × 2 = 6 N east.
- Pull = 6 + 4 = 10 N east.
- Second mass = 6/3 = 2 kg.
10 N east pull; mass 2 kg.
Look for these in your work
- Distinguish pull and resultant.
- Rearrange for mass separately.
A 900 kg car initially moves at 12 m/s east. Its constant horizontal resultant is 1800 N west for 4 s. Find acceleration and final velocity.
Give me a hint
Choose east positive and keep the acceleration sign in the velocity update.
Compare my reasoning
- a = −1800/900 = −2 m/s².
- v = 12 + (−2) × 4 = 4 m/s.
- The positive velocity is still eastward, despite westward acceleration.
2 m/s² west; final velocity 4 m/s east.
Look for these in your work
- Use a signed acceleration.
- Check the final velocity direction.
Common mistakes
- Driving force and resultant differ when resistance acts.
- Use resultant F when rearranging for mass.
- An interval result requires a maintained acceleration model.
What can you explain now?
A 6 N resultant gives 3 m/s² acceleration. What is mass?
Compare with the explanation
2 kg.
Divide resultant force by its acceleration.
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Make it stick
Explain why the force needed to drive a resistive vehicle exceeds its required resultant.
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.