Work, Energy and Power
Specification: OCR A H556 | Section: 3.3 | Teaching time: 5–6 hours
- Define work done by a force and use W = Fx cos θ to calculate it
- State and apply the principle of conservation of energy
- Recall and derive the equations for kinetic energy (Ek = ½mv²) and gravitational potential energy (Ep = mgh)
- Describe energy transfers between gravitational potential energy and kinetic energy
- Define power, derive P = Fv, and calculate the efficiency of a mechanical system
Work and Conservation of Energy
In physics, work is done when a force causes a displacement of an object in the direction of the force. If you push a box across a floor, you do work. If you push against a solid wall and nothing moves, no work is done — no matter how tired you feel.
Here F is the force (N), x is the displacement (m), and θ is the angle between the force and the displacement. The unit of work is the joule (J): 1 J = 1 N m.
When the force acts parallel to the displacement (θ = 0°), cos θ = 1 and W = Fx. When the force is perpendicular (θ = 90°), cos θ = 0 and no work is done. This is why a satellite in circular orbit experiences no work done by gravity — the force is always perpendicular to the velocity.
Work done = energy transferred. When a force does positive work on an object, energy is transferred to the object. Negative work means energy is transferred from it. This is the crucial link between force and energy.
The principle of conservation of energy states that energy cannot be created or destroyed — it can only be transferred from one form to another. The total energy of a closed system remains constant. Energy exists in many forms: kinetic, gravitational potential, thermal, elastic, nuclear, chemical, electromagnetic. It can be transferred between these forms, but the total always stays the same.
This principle constrains what is physically possible: no machine can output more energy than is put in, and no process can simply make energy vanish.
Knowledge Check — Work and Energy
A student pushes a trolley with a horizontal force of 45 N over a distance of 8.0 m. Calculate the work done.
- W = Fx = 45 × 8.0 = 360 J ✓
- Force is parallel to displacement so θ = 0°, cos 0° = 1
A force of 200 N is applied at 30° above the horizontal to move a crate 5.0 m across a floor. Calculate the work done by this force.
- W = Fx cos θ = 200 × 5.0 × cos 30° = 200 × 5.0 × 0.866 = 866 J ✓
A weightlifter holds a 100 kg barbell stationary above their head for 5 seconds. Explain why no work is done on the barbell during this time.
- Work requires displacement in the direction of the force ✓
- The barbell is stationary (x = 0), so W = 0 J, even though the weightlifter exerts an upward force ✓
Kinetic and Potential Energies
Kinetic energy (Ek) is the energy an object possesses because of its motion. The faster it moves and the more massive it is, the more kinetic energy it has.
Starting from v² = u² + 2as, with initial velocity u = 0:
v² = 2as → s = v² / 2a
Work done: W = Fs = mas = ma(v² / 2a) = ½mv²
Since work done = energy transferred, this is the kinetic energy.
Gravitational potential energy (Ep) is the energy an object possesses because of its position in a gravitational field. Lifting an object vertically does work against gravity and stores GPE.
Here m is mass (kg), g is gravitational field strength (9.81 N kg⁻¹ near Earth's surface), and h is the change in height (m).
To lift an object at constant velocity, the applied force equals its weight: F = mg.
Work done against gravity: W = Fs = mg × h = mgh
This work done is stored as gravitational potential energy.
When an object falls, gravitational potential energy is converted to kinetic energy. When it rises, kinetic energy is converted back to GPE. In the absence of air resistance, the total mechanical energy (Ek + Ep) is conserved: ½mv² + mgh = constant. This is a direct application of the conservation of energy.
A classic example: a roller coaster at the top of a loop has maximum Ep and minimum Ek. As it descends, Ep is converted to Ek and the coaster speeds up. At the bottom, Ek is maximum and Ep is minimum (assuming the reference level is the lowest point).
Knowledge Check — KE and GPE
Calculate the kinetic energy of a 1200 kg car travelling at 18 m s⁻¹.
- Ek = ½mv² = ½ × 1200 × 18² = ½ × 1200 × 324 = 1.94 × 10⁵ J ✓
A 0.50 kg ball is thrown vertically upward with initial speed 12 m s⁻¹. Using conservation of energy, calculate the maximum height it reaches. (Ignore air resistance.)
- At the bottom: Ek = ½ × 0.50 × 12² = 36 J; Ep = 0 ✓
- At max height: Ek = 0; Ep = mgh = 36 J (conservation of energy) ✓
- h = 36 / (0.50 × 9.81) = 7.3 m (2 s.f.) ✓
A pendulum bob of mass 0.20 kg is released from rest at a height of 0.15 m above the lowest point. Calculate its speed at the lowest point.
- Ep at top = mgh = 0.20 × 9.81 × 0.15 = 0.294 J ✓
- By conservation: ½mv² = 0.294 at the bottom ✓
- v = √(2 × 0.294 / 0.20) = √2.94 = 1.7 m s⁻¹ ✓
Power
Power is the rate at which work is done (or energy is transferred). A powerful machine doesn't necessarily do more work — it does the same work in less time.
The unit of power is the watt (W): 1 W = 1 J s⁻¹. A 60 W light bulb transfers 60 J of energy every second.
Starting from P = W/t and W = Fx:
P = Fx / t
Since x/t = v (average velocity): P = Fv
This means power equals force × velocity — very useful for transport problems.
The equation P = Fv is particularly useful for vehicles. A car engine must produce enough power to overcome resistive forces (drag, friction) at a given speed. At higher speeds, more power is needed even though the force might stay the same, because v is larger. Conversely, when accelerating from rest, the engine delivers power at low v but high F.
When solving power problems with vehicles, remember: at constant speed, the driving force equals the total resistive force. The engine's power output goes entirely into overcoming resistance. When accelerating, the driving force must exceed resistance — use F = ma alongside P = Fv.
Efficiency measures how much of the input energy is usefully transferred:
No real machine is 100% efficient — some energy is always dissipated, usually as thermal energy due to friction. A car engine might be 25–40% efficient; the rest of the chemical energy from the fuel is lost as heat. Improving efficiency is a major engineering challenge with real consequences for sustainability and cost.
Knowledge Check — Power and Efficiency
A motor lifts a 500 kg load vertically through 12 m in 8.0 s. Calculate the minimum power output of the motor.
- W = mgh = 500 × 9.81 × 12 = 58 860 J ✓
- P = W/t = 58 860 / 8.0 = 7.4 kW (2 s.f.) ✓
- (This is the minimum — actual power would be higher due to losses)
A car of mass 900 kg travels at a constant speed of 25 m s⁻¹ on a level road. The total resistive force is 620 N. Calculate the power developed by the engine.
- At constant speed, driving force = resistive force = 620 N ✓
- P = Fv = 620 × 25 = 15 500 W = 15.5 kW ✓
An electric motor has an input power of 2.4 kW and an efficiency of 78%. Calculate the useful output power and the power wasted.
- Useful output = 0.78 × 2400 = 1872 W ≈ 1.9 kW ✓
- Wasted power = 2400 − 1872 = 528 W ≈ 0.53 kW ✓
- (Wasted energy is dissipated as thermal energy) ✓
Energy Conservation Simulator
Exam-Style Questions
A crate of mass 25 kg is pulled at constant speed along a rough horizontal floor by a rope inclined at 35° to the horizontal. The tension in the rope is 150 N and the crate moves a distance of 9.0 m.
(a) Calculate the work done by the tension force. [2 marks]
(b) Explain why the work done by the tension is not equal to the increase in kinetic energy of the crate. [2 marks]
- (a) W = Fx cos θ = 150 × 9.0 × cos 35° = 150 × 9.0 × 0.819 = 1110 J ✓ (1.1 kJ) ✓
- (b) The crate moves at constant speed so there is no change in kinetic energy ✓
- The work done by the tension is balanced by the work done against friction — energy is transferred to thermal energy (internal energy of surfaces) rather than kinetic energy ✓
A ball of mass 0.15 kg is thrown vertically upwards with an initial kinetic energy of 12 J. Air resistance can be ignored.
(a) Calculate the initial speed of the ball. [2 marks]
(b) Calculate the maximum height the ball reaches. [2 marks]
(c) State and explain how the maximum height would change if air resistance were not negligible. [2 marks]
- (a) ½mv² = 12 → v² = 24/0.15 = 160 → v = 12.6 m s⁻¹ ✓
- (b) By conservation: mgh = 12 → h = 12/(0.15 × 9.81) = 8.2 m ✓
- (c) The maximum height would be lower ✓ because some initial kinetic energy would be transferred to thermal energy (due to work done against air resistance) rather than all being converted to GPE ✓
A car of mass 1500 kg accelerates from rest to a speed of 28 m s⁻¹ in 12 s on a level road. The total resistive force is constant at 450 N.
(a) Calculate the kinetic energy of the car at 28 m s⁻¹. [1 mark]
(b) Calculate the average power developed by the engine. [4 marks]
- (a) Ek = ½ × 1500 × 28² = ½ × 1500 × 784 = 588 000 J ✓
- (b) Average acceleration: a = (28 − 0)/12 = 2.33 m s⁻² ✓
- Driving force: F = ma + resistance = 1500 × 2.33 + 450 = 3500 + 450 = 3950 N ✓
- Distance: s = ½ × 2.33 × 12² = 168 m ✓
- Average power: P = W/t = Fx/t = 3950 × 168 / 12 = 55.3 kW ✓
A hydroelectric power station uses water falling from a reservoir 85 m above the turbines. Water flows through the turbines at a rate of 450 kg s⁻¹. The system has an efficiency of 72%.
(a) Calculate the gravitational potential energy lost per second by the water. [2 marks]
(b) Calculate the useful output power of the station. [2 marks]
(c) State one assumption made in this calculation. [1 mark]
- (a) GPE per second = mgh/t = 450 × 9.81 × 85 = 376 000 W ≈ 376 kW ✓ (mass flow rate × g × h)
- (b) Useful output = 0.72 × 376 000 = 271 kW ✓
- (c) Assumption: all of the GPE is transferred to KE of the water at the turbine (no energy lost to friction in the pipe / the water has negligible velocity at the reservoir surface) ✓
Topic Summary
Work Done
W = Fx cos θ. Work is done when a force causes displacement. 1 J = 1 N m. Work done = energy transferred.
Conservation of Energy
Energy cannot be created or destroyed — only transferred between forms. Total energy of a closed system is constant.
Kinetic Energy
Eₖ = ½mv². Derivable from suvat and Newton's second law. Doubled speed → four times the KE.
Gravitational PE
Eₚ = mgh. Valid for uniform gravitational fields (near Earth's surface). Derivable from W = Fs.
Power
P = W/t (watts). P = Fv — power equals force × velocity. Essential for vehicle and machine problems.
Efficiency
η = (useful output / total input) × 100%. No machine is 100% efficient — energy is always dissipated.