A.3.5—Work-energy change
- Syllabus
- First assessment 2025
- Objective
- —
- Level
- HL
Work–energy theorem
The net work done by the resultant force on a system equals its change in kinetic energy:
Wnet=ΔEk
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 100N moves an object from rest through 2.0m until its speed is 10ms−1.
Wnet=Fs=(100)(2.0)=200J
200=ΔEk=21m(10)2−0
m=4.0kg
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.
The evidence asks for a stopping distance after applied force is removed and for maximum speed from a force–distance graph.
Determine / Calculate
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.
Using the area under only one force curve or treating negative work as a negative kinetic energy rather than a change.
Representative question
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.
ALT 1
Ff=0.28×1.2×9.8=3.29 «N»
W done over 0.35 m=(14−3.29)×0.35=3.75<J/> d = «3.75 J / 3.29 N = » 1.14 «m»
ALT 2
a=(14−0.28×1.2×9.8)/1.2=8.92⟨ m s−2⟩v=(2)(8.92)(0.35)=2.50⟨ m s−1⟩d=⟨2.52/(2×0.28×9.8)=−1.14⟨ m∥
Allow ECF from MP1
Only award marks from one ALT.