Work, Energy and Power

Energy is one of the most fundamental concepts in physics. Unlike forces which act in specific directions and have complex interactions, energy gives us a powerful scalar approach to solving problems. The work-energy method often provides simpler solutions than Newton’s second law, especially for problems involving changes in height or speed.

Work Done

Work is the transfer of energy by a force. When a force moves an object through a displacement, it does work on that object.

Work done = Force x displacement x cos(theta)

W = Fs cos(theta)

where F is the force in newtons, s is the displacement in metres, and theta is the angle between the force and the displacement. Work is measured in joules (J).

Important cases:

Angle	cos(theta)	Work Done	Physical Meaning
0°	1	W = Fs	Force in direction of motion — maximum work
0° < theta < 90°	0 < cos(theta) < 1	W = Fs cos(theta)	Partial work done
90°	0	W = 0	Force perpendicular to motion — no work
90° < theta < 180°	-1 < cos(theta) < 0	W < 0	Force partially opposes motion
180°	-1	W = -Fs	Force opposes motion — negative work (e.g. friction)

Worked Example 1

A person pushes a box 4.0 m along a floor with a force of 50 N at 30° below the horizontal. What is the work done by the pushing force?

W = Fs cos(theta) = 50 x 4.0 x cos(30°) = 200 x 0.866 = 173 J

Worked Example 2

A 10 kg box is dragged 5.0 m along a rough horizontal floor by a rope at 25° above horizontal with tension 80 N. The coefficient of friction is 0.30. Find the work done by each force.

Work by tension: W_T = 80 x 5.0 x cos(25°) = 400 x 0.906 = 362 J Work by weight: W_mg = mg x 5.0 x cos(90°) = 0 J (weight is perpendicular to horizontal displacement) Work by normal: W_N = N x 5.0 x cos(90°) = 0 J (normal is perpendicular to motion)

For friction, first find N: Vertical equilibrium: N + 80 sin(25°) = mg N = 10 x 9.81 - 80 x 0.423 = 98.1 - 33.8 = 64.3 N Friction = mu N = 0.30 x 64.3 = 19.3 N

Work by friction: W_f = 19.3 x 5.0 x cos(180°) = -96.5 J (negative because friction opposes motion)

Net work = 362 + 0 + 0 - 96.5 = 265.5 J

Kinetic Energy

Kinetic energy (KE) is the energy an object has due to its motion.

KE = (1/2)mv²

where m is the mass in kg and v is the speed in m/s. Kinetic energy is measured in joules.

The work-energy theorem states that the net work done on an object equals the change in its kinetic energy:

W_net = delta KE = (1/2)mv² - (1/2)mu²

Worked Example 3

A 1500 kg car accelerates from 10 m/s to 30 m/s. What is the change in kinetic energy?

delta KE = (1/2)(1500)(30²) - (1/2)(1500)(10²) = (1/2)(1500)(900 - 100) = (1/2)(1500)(800) = 600 000 J = 600 kJ

Why Braking Distance Depends on Speed Squared

If a car doubles its speed from v to 2v, its kinetic energy increases by a factor of 4 (since KE is proportional to v²). Since the braking force is roughly constant, the braking distance is also roughly 4 times longer. This is why:

Speed	KE Factor	Braking Distance Factor
v	1	1
2v	4	4
3v	9	9

At 60 mph you need 4 times the braking distance of 30 mph. This is a crucial road safety concept.

Gravitational Potential Energy

Gravitational potential energy (GPE) is the energy stored in an object due to its position in a gravitational field.

GPE = mgh

where m is mass in kg, g = 9.81 m/s², and h is the height above a chosen reference level in metres. GPE is measured in joules.

The choice of reference level (where h = 0) is arbitrary — only changes in GPE matter in calculations.

Worked Example 4

A 70 kg hiker climbs a 500 m mountain. What is the gain in gravitational potential energy?

delta GPE = mgh = 70 x 9.81 x 500 = 343 000 J = 343 kJ

Conservation of Energy

The principle of conservation of energy states:

Energy cannot be created or destroyed — it can only be transferred from one form to another.

In a system with no energy dissipation (no friction or air resistance), the total mechanical energy (KE + GPE) remains constant:

(1/2)mv₁² + mgh₁ = (1/2)mv₂² + mgh₂

flowchart LR
    A["Start: KE₁ + GPE₁"] --> B{"Energy dissipated\nby friction/drag?"}
    B -->|"No (ideal)"| C["KE₁ + GPE₁ = KE₂ + GPE₂"]
    B -->|"Yes (real)"| D["KE₁ + GPE₁ = KE₂ + GPE₂\n+ Energy dissipated"]

Worked Example 5

A ball is dropped from a height of 10 m. What is its speed just before hitting the ground? (Ignore air resistance)

Using conservation of energy: mgh = (1/2)mv² The mass cancels: gh = (1/2)v² v² = 2gh = 2 x 9.81 x 10 = 196.2 v = 14.0 m/s

Worked Example 6

A roller coaster car (mass 500 kg) starts from rest at a height of 40 m and descends to a height of 15 m. What is its speed at the lower point? (Ignore friction)

Energy conservation: mgh₁ = (1/2)mv² + mgh₂ mg(h₁ - h₂) = (1/2)mv² v² = 2g(h₁ - h₂) = 2 x 9.81 x (40 - 15) = 2 x 9.81 x 25 = 490.5 v = 22.1 m/s

Note: the mass cancels and the shape of the track does not matter — only the height difference determines the speed.

Worked Example 7: With Friction

A 5.0 kg block slides 3.0 m down a rough slope inclined at 30°. The coefficient of friction is 0.20. Find the speed at the bottom.

Height dropped: h = 3.0 sin(30°) = 1.5 m GPE lost: mgh = 5.0 x 9.81 x 1.5 = 73.6 J

Normal force: R = mg cos(30°) = 5.0 x 9.81 x 0.866 = 42.5 N Friction: F = 0.20 x 42.5 = 8.50 N Work against friction: 8.50 x 3.0 = 25.5 J

KE gained = 73.6 - 25.5 = 48.1 J (1/2)(5.0)v² = 48.1 v² = 19.2 v = 4.4 m/s

Dissipation and Efficiency

In real systems, energy is always dissipated — transferred to thermal energy (heat) by friction, air resistance, or other resistive forces. The total energy is still conserved, but useful mechanical energy decreases.

Work done against friction = friction force x distance = energy transferred to thermal energy.

Efficiency measures how much of the input energy is usefully transferred:

Efficiency = (useful energy output / total energy input) x 100%

Or equivalently:

Efficiency = (useful power output / total power input) x 100%

Worked Example 8

A motor lifts a 200 kg crate through 5.0 m. The motor uses 15 000 J of electrical energy. What is the efficiency?

Useful energy output (GPE gained) = mgh = 200 x 9.81 x 5.0 = 9810 J Efficiency = (9810 / 15000) x 100% = 65.4%

Energy dissipated = 15000 - 9810 = 5190 J (lost as heat in motor, friction in pulleys, etc.)

Power

Power is the rate of doing work or the rate of energy transfer.

P = W / t = E / t

where P is power in watts (W), W is work done in joules, and t is time in seconds.

One watt equals one joule per second: 1 W = 1 J/s.

There is also a very useful alternative formula. Since W = Fs:

P = Fs / t = F x (s/t) = Fv

P = Fv

where F is the force in the direction of motion and v is the velocity. This is particularly useful for vehicles moving at constant velocity or instantaneous power calculations.

Reference: Common Power Values

System	Typical Power
Human walking	~70 W
Human sprinting	~500 W
Cyclist (sustained)	75–200 W
Car engine	50–200 kW
Electric kettle	2–3 kW
Wind turbine (large)	2–5 MW
Power station	500–2000 MW

Worked Example 9

A car travels at a constant velocity of 30 m/s against a total resistive force of 800 N. What is the power output of the engine?

At constant velocity, the driving force equals the resistive force (Newton’s first law): F = 800 N. P = Fv = 800 x 30 = 24 000 W = 24 kW

Worked Example 10

A car engine produces a maximum power of 60 kW. If the total resistive force at high speed is 1200 N, what is the maximum speed?

P = Fv, so v = P/F = 60000/1200 = 50 m/s (about 112 mph)

Worked Example 11: Power on a Hill

A 1400 kg car drives at 20 m/s up a 6° slope against resistive forces of 500 N. What power must the engine deliver?

Forces opposing motion: Component of weight down slope = mg sin(6°) = 1400 x 9.81 x 0.1045 = 1436 N Resistive forces = 500 N Total opposing force = 1436 + 500 = 1936 N

At constant velocity: driving force = 1936 N P = Fv = 1936 x 20 = 38 720 W = 38.7 kW

Common Mistakes and Misconceptions

Confusing force and energy: Force is not energy. A force does work (transfers energy) only when it acts over a displacement.
Forgetting that friction does negative work: Friction removes kinetic energy from the system. The work done by friction is always negative.
Ignoring the angle in work calculations: W = Fs only when the force is parallel to the displacement. In general, W = Fs cos(theta).
Using the wrong height in GPE: GPE depends on vertical height, not the distance along a slope. For a slope of length L at angle theta, h = L sin(theta).
Thinking efficiency can exceed 100%: This is physically impossible. If your calculation gives efficiency > 100%, you have made an error.
Confusing instantaneous power with average power: P = Fv gives instantaneous power at velocity v. Average power = total work / total time.

Exam Tips for Edexcel Physics

Energy conservation problems are often quicker to solve than using F = ma and SUVAT. If a question involves changes in height and speed, try the energy approach first.
Always account for energy dissipated by friction: GPE lost = KE gained + work against friction.
When using P = Fv, make sure F is the force in the direction of motion, not just any force.
For "show that" questions involving efficiency, clearly show the useful output and total input before calculating the ratio.
Convert kW to W before substituting into equations (1 kW = 1000 W).
If a question asks for power "at a particular instant," use P = Fv with the velocity at that instant.

Key Points

Work = Fs cos(theta). Only the component of force in the direction of displacement does work.
KE = (1/2)mv². GPE = mgh.
Conservation of energy: total energy is constant; in the absence of dissipation, KE + GPE is constant.
With friction: GPE lost = KE gained + work done against friction.
Efficiency = useful output / total input x 100%.
Power = energy/time = Fv.
At constant velocity, driving force = resistive force, and P = Fv gives the power needed.
Braking distance is proportional to v² because KE is proportional to v².

A-Level Deep Dive: Work, Energy and Power

Spec mapping

Edexcel 9PH0 specification, Topic 2 — Mechanics, sub-topic 2.5 (Energy) requires candidates to define work done as the product of force and displacement in the direction of the force, calculate work done with W = Fs cos(theta) when force and displacement are not aligned, distinguish kinetic and gravitational potential energy, apply the principle of conservation of energy, and define and calculate power as the rate of doing work — including the instantaneous form P = Fv (refer to the official Pearson Edexcel specification document for exact wording). Energy methods then thread through Topic 4 (Materials) as elastic potential energy stored in stretched springs and wires, Topic 5 (Waves) as the energy carried per cycle by an oscillation, Topic 6 (Further mechanics) when work-energy and momentum methods are compared on collision problems, Topic 7 (Electric and magnetic fields) where work done in moving a charge against a field defines potential, and Topic 8 (Nuclear and particle physics) where mass-energy equivalence $E = mc^2$ extends the conservation principle. The Edexcel data, formulae and relationships booklet supplies $E_k = \tfrac{1}{2}mv^2$ , $\Delta E_{\mathrm{grav}} = mg\Delta h$ and $P = E/t$ , but does not state $P = Fv$ explicitly — the instantaneous-power form must be derived in working from $P = W/t$ and $W = Fs$ .

Worked example with full mark scheme

Question (8 marks):

An electric motor of input power 1.50 kW is used to lift a load of mass 240 kg vertically through a height of 12.0 m. The lift takes 28.0 s at constant speed. Take $g = 9.81$ m s⁻².

(a) Calculate the useful work done on the load. (2)

(b) Calculate the useful output power of the motor. (2)

(d) Suggest, with reasoning, two physical processes that account for the energy not transferred to gravitational potential energy of the load. (2)

Solution with mark scheme:

(a) Step 1 — identify the relevant energy transfer. At constant speed the kinetic energy of the load does not change, so the useful work done equals the gain in gravitational potential energy.

$W_{\mathrm{useful}} = \Delta E_{\mathrm{grav}} = mg\Delta h = 240 \times 9.81 \times 12.0$

M1 — correct substitution into $\Delta E_{\mathrm{grav}} = mg\Delta h$ (or equivalent statement $W = Fs$ with $F = mg$ ).

$W_{\mathrm{useful}} = 28\,253 \text{ J} \approx 2.83 \times 10^4 \text{ J}$

A1 — value to 3 sf with units (28.3 kJ acceptable).

(b) Step 1 — apply the definition of average power.

$P_{\mathrm{useful}} = \dfrac{W_{\mathrm{useful}}}{t} = \dfrac{28\,253}{28.0}$

M1 — correct use of $P = W/t$ with the value from (a) and the stated time.

$P_{\mathrm{useful}} = 1009 \text{ W} \approx 1.01 \text{ kW}$

A1 — answer to 3 sf with units.

$\eta = \dfrac{P_{\mathrm{useful}}}{P_{\mathrm{input}}} = \dfrac{1009}{1500}$

M1 — correct ratio set up; using the energy form $\eta = W_{\mathrm{useful}}/(P_{\mathrm{input}} \times t) = 28\,253/42\,000$ is equally credited.

$\eta = 0.673 = 67.3\%$

A1 — value as a percentage to 3 sf. Stating only "0.67" without indicating units or percentage is treated as ambiguous and may forfeit the A1 in tighter mark schemes.

(d) Two processes from:

Resistive heating in the motor windings ( $I^2 R$ losses dissipating to the surroundings).
Friction in the cable, pulleys and bearings, transferring energy to thermal energy and a small amount of sound.
Air resistance on the moving load and counterweight.
Energy stored briefly as elastic strain in the cable, partly recovered but partly dissipated.

B1 — first plausible mechanism named with a brief justification (e.g. "heating in the motor coils because of resistance to current").

B1 — second distinct mechanism (mechanical friction must be distinguished from resistive heating to count as a second mark).

Total: 8 marks (M3 A3 B2).

Specimen question modelled on the Edexcel 9PH0 Paper 1 format

Question (6 marks): A small car of mass 950 kg accelerates from rest along a horizontal road. After travelling 85 m it reaches a speed of 18 m s⁻¹. The total resistive force on the car is constant at 320 N throughout.

(a) Calculate the gain in kinetic energy of the car. (1)

(b) Calculate the work done against the resistive forces over the 85 m. (1)

(d) Calculate the instantaneous output power of the engine when the car is travelling at 18 m s⁻¹, assuming the driving force is the same as the average value found in (c). (2)

Mark scheme decomposition by AO:

Part	AO weighting	Earned by
(a)	AO1.1b	$\Delta E_k = \tfrac{1}{2}(950)(18)^2 - 0 = 1.54 \times 10^5$ J.
(b)	AO1.1b	$W_{\mathrm{res}} = F_{\mathrm{res}} s = 320 \times 85 = 2.72 \times 10^4$ J.
(c) M1	AO2.1	Work-energy statement: $W_{\mathrm{drive}} = \Delta E_k + W_{\mathrm{res}} = 1.54 \times 10^5 + 2.72 \times 10^4$ .
(c) A1	AO1.1b	Driving force $F_{\mathrm{drive}} = W_{\mathrm{drive}}/s = 1.81 \times 10^5 / 85 = 2.13 \times 10^3$ N (3 sf).
(d) M1	AO2.1	Instantaneous power $P = Fv$ , identified as the appropriate form because force and velocity are needed at one moment.
(d) A1	AO1.1b	$P = 2130 \times 18 = 3.83 \times 10^4$ W (38.3 kW).

Total: 6 marks split AO1 = 4, AO2 = 2. This is a classic mixed-AO energy item: AO1 marks for substitution into $E_k$ , $W = Fs$ and $P = Fv$ , AO2 marks for choosing the work-energy bookkeeping in (c) and the instantaneous form in (d).

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