The work-energy theorem, explained with examples
The work-energy theorem says that the net work done on an object equals its change in kinetic energy. It connects forces to motion without needing time or acceleration, and it is one of the most useful tools in AP Physics 1 and C.
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What work means in physics
In physics, work is done when a force acts on an object that moves in the direction of (a component of) the force. For a constant force F and a displacement d:
W = F d cosθ
where θ is the angle between the force and the displacement. Work is measured in joules: 1 J is the work done by a 1 N force acting over 1 m. Work is a scalar, so it has no direction, but it can be positive, negative or zero.
Positive, negative and zero work
- Positive work (θ < 90°): the force has a component along the motion, so it speeds the object up or transfers energy into it. Pushing a trolley forwards does positive work.
- Negative work (θ > 90°): the force has a component against the motion, so it takes energy out. Friction on a sliding box does negative work.
- Zero work (θ = 90°): the force is perpendicular to the motion. The normal force on a box sliding along a level floor, and the tension in a string keeping a ball in uniform circular motion, both do zero work.
Holding a heavy bag still feels tiring, but no work is done on the bag, because it does not move.
The work-energy theorem
Wnet = ΔK = ½mvf² − ½mvi²
The net work, the total work done by all the forces on an object (or equivalently the work done by the net force), equals the change in its kinetic energy. If the net work is positive, the object speeds up. If it is negative, the object slows down. If it is zero, the speed stays constant, although the direction may change, as in circular motion.
Why it is true
For a constant net force along the motion, Newton's second law gives F = ma, and kinematics gives vf² = vi² + 2ad. Multiply the kinematics equation by ½m: ½mvf² − ½mvi² = mad = Fd. So the net work Fd equals the change in kinetic energy. The theorem also holds for forces that vary, using integrals.
Worked examples
Example 1: work by a constant force
A 50 N force pulls a box 6.0 m along a floor. Find the work done if the force is horizontal, and if it acts at 30° above the horizontal.
- Horizontal: W = 50 × 6.0 = 300 J.
- At 30°: W = 50 × 6.0 × cos 30° ≈ 260 J. Only the horizontal component does work.
Example 2: finding a final speed
A 2.0 kg cart starts from rest and is pushed by a net force of 8.0 N over 4.0 m. Find its final speed.
- Wnet = 8.0 × 4.0 = 32 J = ½ × 2.0 × v².
- v² = 32, so v ≈ 5.7 m/s.
Example 3: several forces
A 10 kg box is pulled 5.0 m across a floor by a 60 N horizontal force against 20 N of friction. It starts at 1.0 m/s. Find its final speed.
- Net work = (60 − 20) × 5.0 = 200 J. Initial K = ½ × 10 × 1.0² = 5 J.
- Final K = 205 J, so v = √(2 × 205/10) ≈ 6.4 m/s.
Example 4: stopping distance
A 1200 kg car travelling at 20 m/s brakes with a constant 6000 N force. Find the stopping distance.
- Work done by brakes = −6000d = ΔK = 0 − ½ × 1200 × 20² = −240,000 J.
- d = 40 m.
- At 40 m/s, K is four times larger, so the stopping distance becomes 160 m: double the speed, four times the distance.
Example 5: work against gravity
A 3.0 kg box is lifted 2.0 m at constant speed. Find the work done by the lifting force and by gravity, and the net work.
- Constant speed, so the lifting force equals the weight, 29.4 N. Work by lift = 29.4 × 2.0 = 58.8 J.
- Gravity acts opposite to the motion: work = −58.8 J.
- Net work = 0, matching zero change in kinetic energy.
Work from a force-displacement graph
When the force changes with position, work is the area under a force-displacement graph. For a spring, F = kx gives a triangle, so the work to stretch it by x is ½kx². In AP Physics C, this area is written as an integral: W = ∫F dx.
Example 6: area under a graph
A force rises steadily from 0 to 12 N as an object moves from x = 0 to x = 3.0 m, then stays at 12 N until x = 5.0 m. Find the total work done.
- Triangle from 0 to 3 m: ½ × 3.0 × 12 = 18 J. Rectangle from 3 to 5 m: 2.0 × 12 = 24 J.
- Total work = 42 J.
Work, energy and power together
Power is the rate of doing work: P = W/t. For a force moving an object at constant velocity, P = Fv. A car engine that delivers 30 kW while driving at 25 m/s against resistance must be providing a driving force of 30,000/25 = 1200 N. Work, energy and power questions often combine all three ideas.
Common mistakes
- Using the full force instead of the component along the displacement.
- Forgetting that friction does negative work.
- Using the work of one force instead of the net work in the theorem.
- Thinking a force that does not move anything does work.
- Forgetting to square the speed in kinetic energy.
Practice questions
Try each one before opening the answer.
Q1. A 20 N force acts at 60° to a 3.0 m displacement. Find the work.
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Q2. Net work of −50 J is done on a moving object. What happens to its speed?
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Q3. A 0.5 kg ball speeds up from 2 m/s to 6 m/s. Find the net work done on it.
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Q4. How much work does gravity do on a satellite in a circular orbit?
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Frequently asked questions
Is the work-energy theorem on AP Physics 1?
Yes. Work, energy and the work-energy theorem form a unit of AP Physics 1, and AP Physics C: Mechanics covers them with integrals for variable forces.
How is it different from conservation of energy?
The work-energy theorem looks at one object and the work done on it by all forces. Conservation of energy looks at a system and tracks how energy moves between forms. Both give the same answers when used correctly.
Key takeaways
- W = Fd cosθ, in joules.
- Work can be positive, negative or zero.
- Net work equals the change in kinetic energy.
- Work is the area under a force-displacement graph.
- Double the speed, four times the stopping distance.