Difficulty: Introductory | Prerequisites: Basic algebra, understanding of force and Newton's laws.
Work and power sit at the heart of energy mechanics. This topic connects force and motion to the broader idea of energy transfer, bridging Newton's laws with conservation of energy. You will need a solid grasp of force, displacement, and basic trigonometry (cosine). If you can draw a free-body diagram and solve F = ma problems, you are ready for this material.
Work is the energy transferred when a force moves an object through a distance. Power is how quickly that work gets done. The work-energy theorem ties these together: the net work on an object equals its change in kinetic energy.
Work (W)
The transfer of energy to or from an object by means of a force acting through a displacement. Measured in joules (J). W = Fd cos θ.
Think of it as: the energy you actually deliver to something when you push or pull it along a path.
Power (P)
The rate at which work is done, or equivalently the rate of energy transfer. Measured in watts (W), where 1 W = 1 J/s. P = W/t.
In simple terms, this means: how fast you are spending or delivering energy.
Joule (J)
The SI unit of work and energy. One joule equals the work done when a force of one newton moves an object one metre in the direction of the force.
Think of it as: roughly the energy needed to lift a small apple one metre off the ground.
Watt (W)
The SI unit of power. One watt equals one joule per second.
In simple terms, this means: one unit of energy delivered every second.
Kinetic energy (KE)
The energy an object possesses due to its motion. KE = ½mv².
Think of it as: the energy a moving object would release if you brought it to a stop.
Gravitational potential energy (PE)
The energy an object possesses because of its position in a gravitational field. PE = mgh.
Think of it as: stored energy from height, waiting to convert back into motion.
Work-energy theorem
The net work done on an object equals its change in kinetic energy: W_net = ΔKE = KE₂ − KE₁.
In simple terms, this means: if you want to know how much an object sped up or slowed down, look at the total work done on it.
Work is done only when a force causes displacement. If you push a wall and it does not move, you have done zero work on the wall.
The equation is W = Fd cos θ, where:
F = magnitude of the applied force (in newtons)
d = displacement of the object (in metres)
θ = angle between the force vector and the displacement vector
When force and displacement point the same way (θ = 0°), cos θ = 1, so W = Fd. This is the maximum work for a given force and distance.
When force is perpendicular to displacement (θ = 90°), cos θ = 0, so W = 0. A centripetal force, for example, does no work.
Work is a scalar quantity. It can be positive (force in the direction of motion), negative (force opposing motion, e.g. friction), or zero.
Work scales linearly with both force and distance. Doubling the distance at the same force doubles the work. Tripling the height you lift a barbell triples the work required.
Power tells you how quickly work is being done: P = W / t.
Units: watts (W). 1 watt = 1 joule per second.
For the same amount of work, doing it in less time requires more power. 100 J in 0.50 s = 200 W; 100 J in 1.0 s = 100 W.
Power can also be expressed as P = Fv (force times velocity), which is useful when you know the speed rather than the time.
A higher-powered engine does not necessarily do more total work. It simply does the same work faster, or can sustain a higher rate of energy output.
The net work done on an object equals its change in kinetic energy: W_net = ΔKE = ½mv₂² − ½mv₁².
This is called the work-energy theorem. It connects the concepts of force, displacement, and speed in a single relationship.
Kinetic energy depends on velocity squared. If you double the speed, kinetic energy quadruples. This is why a car travelling at 80 mph skids four times as far as one at 40 mph (assuming the same braking force): doubling v means 4× the KE to dissipate, so 4× the stopping distance.
When you lift an object at constant speed, the work you do against gravity becomes gravitational potential energy: W = ΔPE = mgh.
Lifting a bucket with 100 J of work gives it 100 J of gravitational potential energy relative to its starting height. Lifting it twice as high doubles both the work and the stored PE to 200 J.
W = F d \cos\thetaP = \frac{W}{t}P = F vKE = \frac{1}{2} m v^2PE = m g hW_{\text{net}} = \Delta KE = KE_2 - KE_1Car braking distances depend on KE, which scales with v². This is why speed limits matter disproportionately: a small increase in speed produces a large increase in stopping distance.
Power ratings on appliances (a 1500 W kettle, a 60 W light bulb) tell you how fast they consume energy, not how much total energy they use. A 1500 W kettle run for two minutes uses the same energy as a 60 W bulb left on for 50 minutes.
Lifting heavy objects (cranes, elevators, gym equipment) is a direct application of W = Fd. The work done equals the weight times the height, regardless of the path taken (for gravity, a conservative force).
Students often think that holding a heavy object still requires work. It does not. If there is no displacement, W = 0, no matter how much force you exert.
Students frequently confuse work and power. More power does not mean more work. It means the same work done in less time.
A common error on stopping-distance problems is assuming that doubling speed doubles the skid distance. Because KE depends on v², doubling speed quadruples KE and therefore quadruples the stopping distance (assuming constant braking force).
Students sometimes say the work-energy theorem relates work to potential energy. The theorem specifically relates net work to the change in kinetic energy. Potential energy enters through the concept of conservative forces and total mechanical energy, which is a separate (though related) idea.
⚠️ Expect questions that change one variable (force, distance, speed, time) and ask how work, power, or stopping distance changes. Know which relationships are linear (W and d, W and F, P and W) and which are quadratic (KE and v).
⚠️ The cos θ factor is commonly tested. Be ready for problems where the force is at an angle to the displacement.
⚠️ Work-energy theorem problems often set the final KE to zero (object comes to rest) and ask you to find force or distance.
⚠️ Power problems frequently compare two scenarios with the same work but different times, or the same time but different forces.
True or false: A force applied perpendicular to an object's displacement does zero work. (True)
Fill in the blank: Power is measured in ______, which are equivalent to joules per second. (watts)
True or false: Doubling your speed doubles your kinetic energy. (False, it quadruples it)
Fill in the blank: The work-energy theorem states that net work equals the change in ______. (kinetic energy)
True or false: Lifting a box twice as high requires twice the work. (True)
Q: A person pushes a cart with a force of 10.0 N over a distance of 10.0 m. How much work is done on the cart?
A: W = Fd = 10.0 N × 10.0 m = 100 J.
Q: 100 J of work is done on a bicycle in 0.50 s. What is the power output?
A: P = W/t = 100 J / 0.50 s = 200 W.
Q: The same 100 J of work is done in 1.0 s instead. What is the power now?
A: P = W/t = 100 J / 1.0 s = 100 W. Doubling the time halves the power.
Q: You do 100 J of work to lift a bucket. What is its gravitational potential energy relative to its starting position?
A: 100 J. The work done against gravity is stored as gravitational PE.
Q: If the same bucket is lifted twice as high, what is its potential energy?
A: 200 J. PE = mgh, so doubling h doubles the PE (and requires twice the work).
Q: A car travelling at 40 mph skids 20 m when its brakes lock. How far will it skid at 80 mph?
A: 80 m. Doubling the speed quadruples the kinetic energy (since KE depends on v²). With the same braking force, the work-energy theorem gives W = Fd, so if KE quadruples and F stays constant, d must also quadruple: 4 × 20 m = 80 m.
Q: A barbell requires a certain amount of work to be lifted to a given height. How many times as much work is needed to lift it three times as high?
A: Three times as much. Work against gravity is W = mgh, so tripling h triples the work.
This material connects directly to conservation of energy, where kinetic and potential energy trade back and forth in systems like pendulums and roller coasters. The work-energy theorem is the bridge between Newton's second law (force and acceleration) and energy methods, so you will use it again whenever force-based and energy-based approaches overlap. Power reappears in electricity (P = IV) and thermodynamics (rate of heat transfer), so building intuition for it here pays off across the rest of the course.
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