A.3.5—Work-energy change

Syllabus
First assessment 2025
Objective
Level
HL

Relate Resultant Work to Energy Change

Work–energy theorem

The net work done by the resultant force on a system equals its change in kinetic energy:

Wnet=ΔEkW_{net}=\Delta E_k

Use force–distance area

For a variable resultant force, the signed area under the force–distance graph gives the work and therefore the kinetic-energy change.

Include all resultant forces

Friction, applied forces and gravity may each do work. Add their signed contributions before relating the result to the final kinetic energy.

Worked example from local Question Bank row 31356

A constant net force of 100N100\,\mathrm{N} moves an object from rest through 2.0m2.0\,\mathrm{m} until its speed is 10ms110\,\mathrm{m\,s^{-1}}.

Wnet=Fs=(100)(2.0)=200JW_{net}=Fs=(100)(2.0)=200\,\mathrm{J}
200=ΔEk=12m(10)20200=\Delta E_k=\frac12m(10)^2-0
m=4.0kgm=4.0\,\mathrm{kg}

The positive net work is exactly the object's kinetic-energy gain.

Common trap

Do not use the work of one force as the net work unless all other force contributions are zero or already included.

A.3.5 Exam Analysis

Assessment in practice

2–4 marks
How it is assessed

The evidence asks for a stopping distance after applied force is removed and for maximum speed from a force–distance graph.

Command terms

Determine / Calculate

What earns marks

Use the signed work done by the resultant force to find the change in kinetic energy. For a force that varies with distance, calculate the relevant graph area and combine it with the initial kinetic energy.

Watch for

Using the area under only one force curve or treating negative work as a negative kinetic energy rather than a change.

Representative question

Question 1

[Maximum number: 3]

A force of 14.0 N acts on the box for 0.35 m as shown. The force is then removed and the box continues to move. The box comes to rest after a further displacement d.

Determine d.