Cambridge International AS & A Level Mathematics 9709 (2026-2027) · Paper 4: Mechanics · official syllabus
ENERGY, WORK AND POWER LAB
Learn it by playing it
Four short missions on angled work, kinetic energy, a smooth slide and instantaneous engine power. Predict the rule, then calculate and test your answer. Keep forces, energy and power distinct.
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Mission 1 of 4
Work at an angle
A constant 20 N force acts at 60 degrees above a horizontal displacement of 5 m. Set the work done by this force. The diagram shows the force and its 10 N component along the displacement. The perpendicular component does no work for this horizontal displacement.
A force perpendicular to the displacement does...
White: the 20 N force. Purple: its horizontal component. The separate teal displacement arrow is in metres, on a different scale from the forces.
Your work is 100.0 J. Use the component along the horizontal displacement.
Study notes
Played the lab above? These notes are the reference version of what you just felt - use them to check exact formulas and the exam wording.
Work done by a constant force 4.5
Work done by a constant force is W=Fd cos theta, where theta is the angle between the force and the displacement of its point of application. Work and energy are measured in joules (J). Use the component of force along the displacement, not automatically the full force magnitude.
Worked example: a 20 N force at 60 degrees to a 5 m horizontal displacement does work 20(5)cos60=50 J. A force perpendicular to the displacement does no work. A force opposing the displacement does negative work.
A constant 6 N resistance over a 4 m path does work -24 J on the moving particle. Distinguish the signed work done by the resistance from the positive amount 24 J of mechanical energy transferred to other forms.
Kinetic and gravitational potential energy 4.5
Kinetic energy is (1/2)mv^2, using the magnitude of velocity. Gravitational potential energy relative to a chosen reference level is mgh. In these examples use g=10 m/s^2, and use the value supplied in an exam question.
A 2 kg particle at speed 6 m/s has kinetic energy 36 J. Raising it by a vertical height of 5 m increases gravitational potential energy by 100 J. The vertical height change matters, not the length of a sloping or curved path.
Only potential-energy differences affect the calculation; the zero reference can be chosen conveniently. Do not use a negative kinetic energy or equate a signed velocity directly to a positive speed without interpreting direction.
Choose one consistent work-energy equation 4.5
The net work done by all forces on a particle equals its change in kinetic energy. If gravity's work is counted directly, it is +mg times the vertical drop or -mg times the vertical rise.
Alternatively, include gravitational potential energy in the mechanical-energy total. Then the work done by the other forces equals the change in KE+GPE. Do not count the same gravitational effect both as gravity's work and as a potential-energy term on the same side of the balance.
Worked example: a 2 kg particle starts with speed 6 m/s and rises 1 m along a path of length 4 m against a constant 6 N resistance. Initial KE is 36 J, GPE gain is 20 J and resistance removes 24 J. The proposed journey to that endpoint is impossible under those conditions alone, because the calculation would require negative final KE. A negative result is a model warning, not an imaginary physical speed.
Smooth slides, curved paths and energy losses 4.5
With gravity and a smooth fixed guide alone, the normal reaction is perpendicular to motion and does no work. Mechanical energy is conserved. Energy can connect two points of a curved path without finding the acceleration at every point.
Worked example: a particle starts from rest and drops vertically by 5 m on a smooth slide, with g=10. From mg(5)=(1/2)mv^2, v=10 m/s. This gives speed at the lower point, not the direction of its velocity or the travel time. Assume the particle remains on the guide through the interval considered.
If that 2 kg particle also loses 20 J of mechanical energy through resistance, its final KE is 100-20=80 J and its speed is sqrt(80) m/s, approximately 8.94 m/s (2 d.p.). Do not claim mechanical-energy conservation when friction or resistance transfers energy out of that total.
Average and instantaneous power 4.5
Power is the rate of doing work, measured in watts (W), with 1 W=1 J/s. Average power is work done divided by elapsed time. Convert kW to W before combining power with newtons and metres per second.
For a force acting in the direction of motion, instantaneous power P=Fv. Thus F=P/v at a specified nonzero speed. This relates the driving force to the power at that instant; it is not the net force after subtracting resistance.
A motor doing 6000 J of work in 3 s has average power 2000 W=2 kW. At a particular instant, a 200 N driving force at speed 10 m/s has power 2000 W. Equal numerical power values do not imply the force or acceleration stayed constant throughout the motion.
A powered car on a hill 4.5
At an instant on an uphill journey, first use driving force D=P/v, then resolve along the hill: D-resistance-mg sin theta=ma. If the car moves downhill, choose a direction and redo the signs. State what the resistance opposes.
Worked example: a 1000 kg car moves uphill at 10 m/s with engine power 20 kW, resistance 500 N, sin theta=1/10 and g=10. Driving force is 20000/10=2000 N. The downhill weight component is 1000 N, so 2000-500-1000=1000a and instantaneous a=0.5 m/s^2 uphill.
At constant speed the resultant is zero, so this car would need power (500+1000)(10)=15000 W=15 kW at that same speed and slope. Constant engine power generally gives a changing driving force as speed changes, so do not apply constant-acceleration formulae to a whole interval without a valid reason.
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