Energy, Work Done and Power
Work Done
When a force causes an object to move, energy is transferred from one store to another. The amount of energy transferred in this way is called work done. The equation is:
$$W = F \times d$$
Where \(W\) is work done in joules (J), \(F\) is the force in newtons (N), and \(d\) is the distance moved in the direction of the force in metres (m). One joule is the work done when a force of one newton moves an object one metre.
Work done equals energy transferred: \(W = Fd\). If the force is perpendicular to the direction of motion, no work is done — for example, the normal contact force on a horizontally sliding object does no work because it acts vertically.
A common exam application is work done against friction. When an object slides across a rough surface, friction transfers energy from the kinetic energy store to the thermal energy store of the surface and the object — both warm up slightly.
A crate is dragged 12 m across a warehouse floor against a constant friction force of 40 N. How much work is done against friction?
\(W = F \times d = 40 \times 12 = 480\) J
All 480 J is transferred to the thermal energy store of the crate and floor — they become slightly warmer. This energy is dissipated and cannot easily be recovered.
Kinetic Energy
Kinetic energy (KE) is stored in any moving object. The equation is:
$$E_k = \tfrac{1}{2}mv^2$$
Where \(E_k\) is kinetic energy in joules, \(m\) is mass in kilograms, and \(v\) is speed in metres per second. Notice that doubling the speed quadruples the kinetic energy — this is why high-speed impacts are so much more destructive than low-speed ones.
Gravitational Potential Energy
Gravitational potential energy (GPE) depends on the mass, height, and gravitational field strength of the location:
$$E_p = mgh$$
Where \(E_p\) is GPE in joules, \(m\) is mass in kg, \(g\) is gravitational field strength (10 N/kg at Earth's surface), and \(h\) is height above a chosen reference point in metres. Note that \(g\) is different on the Moon, Mars, or other planets — always check the value given in a question.
On Earth, gravitational field strength \(g = 10\) N/kg. As an object falls, GPE decreases and, if there is no air resistance, KE increases by exactly the same amount — total mechanical energy is conserved.
Conservation of Energy
The conservation of energy principle states that the total energy in a closed system remains constant. In practice, some energy is always dissipated as thermal energy due to friction or air resistance — but the total is never lost, only spread out.
A classic example is a falling object. At the top of its path, the object has maximum GPE and zero KE. As it falls, GPE converts to KE. Just before hitting the ground, nearly all GPE has become KE (ignoring air resistance). This gives us the useful shortcut: \(mgh = \tfrac{1}{2}mv^2\), which can be used to find the speed of a falling object without knowing the time of fall.
A cyclist and bike have a combined mass of 80 kg. They start from rest at the top of a smooth hill 5 m high. What is their speed at the bottom? (g = 10 N/kg)
Loss in GPE = gain in KE (smooth ramp, so no energy lost to friction):
\(mgh = \tfrac{1}{2}mv^2\)
\(80 \times 10 \times 5 = \tfrac{1}{2} \times 80 \times v^2\)
\(4000 = 40v^2\)
\(v^2 = 100\)
\(v = 10\) m/s
Note that mass cancels — the speed at the bottom is independent of how heavy the cyclist is.
Efficiency
Efficiency measures how well a device converts energy into a useful form. No real device is 100% efficient — some energy is always wasted, usually as thermal energy or sound.
$$\text{efficiency} = \frac{\text{useful energy output}}{\text{total energy input}}$$
Multiply by 100 for a percentage. An electric motor with efficiency 0.75 converts 75% of electrical energy into kinetic energy; the remaining 25% is dissipated as heat in the windings and bearings.
Power
Power is the rate at which work is done or energy is transferred:
$$P = \frac{E}{t} \quad \text{or equivalently} \quad P = \frac{W}{t}$$
Where \(P\) is power in watts (W), \(E\) (or \(W\)) is energy in joules, and \(t\) is time in seconds. One watt equals one joule per second. A machine with a power of 500 W transfers 500 J every second.
There is a second power equation useful when a steady force moves at constant velocity: \(P = Fv\), where \(F\) is force in newtons and \(v\) is speed in m/s. This is derived by combining \(P = W/t\) with \(W = Fd\): since \(d/t = v\), it follows that \(P = Fv\). Edexcel does not list this equation, so you never have to recall it — but it saves time when a question gives a steady force and a steady speed.
A car engine provides a constant driving force of 1800 N while the car travels at a steady speed of 25 m/s. Calculate the power output of the engine.
\(P = Fv = 1800 \times 25 = 45\,000\) W = 45 kW
At steady speed, the driving force equals the resistive forces (air resistance + friction), so this 45 kW is entirely used to overcome resistance.