Kinematics and Newton's laws
| English | 中文 | Pinyin |
|---|---|---|
| resultant force/rɪˈzʌltənt fɔːs/ | 合力 | hé lì |
Which force accelerates the trolley?
- A trolley accelerates while friction opposes its motion. The engine's force alone does not equal mass times acceleration.
- This lesson studies resultant force 合力: The vector sum of all external forces acting on the modelled object.
Choose the mathematical structure
- Choose a positive direction and draw a force diagram. Use resultant F=ma. For constant acceleration, v=u+at and s=ut+at²/2. A particle model ignores size; a smooth surface ignores friction.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
Which description correctly defines resultant force?
The vector sum of all external forces acting on the modelled object.
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
A 5 kg trolley has driving force 18 N and resistance 3 N. The resultant is 15 N, so a=15/5=3 m/s². Starting from rest for 4 s gives v=0+3×4=12 m/s and s=0×4+3×4²/2=24 m.
Kinematics and Newton's laws
Choose a positive direction and draw a force diagram
Compare the model with the worked case and explain one change.
Find acceleration for mass 5 kg and resultant force 15 N.
Use the resultant: a=F/m=15/5=3.
Test a tempting shortcut
- The normal reaction need not equal weight on a slope. Connected-particle models need one equation per particle and a consistent acceleration relation. The constant-acceleration formulae cannot be used for arbitrary variable acceleration.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
The constant-acceleration formulae apply unchanged when acceleration varies with time. This claim is false. Explain which definition or assumption it violates.
Starting from rest at a=3 m/s², find speed after 4 s.
v=u+at=0+3×4=12.
The constant-acceleration formulae apply unchanged when acceleration varies with time.
The normal reaction need not equal weight on a slope. Connected-particle models need one equation per particle and a consistent acceleration relation. The constant-acceleration formulae cannot be used for arbitrary variable acceleration.
Interpret a new situation
- Explain what each idealisation permits and what it leaves out. For a velocity-time graph, gradient represents acceleration and signed area represents displacement.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
For that motion, find displacement after 4 s.
s=ut+at²/2=0+3×16/2=24.
Match each part of a complete solution to its purpose.
An assumption justifies the model; a check tests the result; interpretation connects it to the question.
Use this in your course
- 7357 · A-level · P. Match the target tier and specification before assigning extensions.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
The vector sum of all external forces acting on the modelled object. Choose the relationship, show the method, check its assumptions and interpret the result.