Energy Stores, Transfers and Power
Energy Stores
Energy cannot be created or destroyed — it can only be transferred between different energy stores. You need to know eight energy stores.
The eight energy stores are: kinetic, thermal, chemical, elastic potential, gravitational potential, electrostatic, magnetic, and nuclear.
Kinetic energy depends on the mass and speed of an object. The equation is:
$$E_k = \frac{1}{2} m v^2$$
Where \(E_k\) is kinetic energy in joules (J), \(m\) is mass in kilograms (kg), and \(v\) is speed in metres per second (m/s).
Gravitational potential energy depends on the mass, height, and gravitational field strength:
$$E_p = m g h$$
Where \(E_p\) is gravitational potential energy in joules, \(m\) is mass in kg, \(g\) is gravitational field strength (about 9.8 N/kg on Earth), and \(h\) is height in metres. The exam always gives you the value of \(g\), and the worked examples below round it to 10 N/kg to keep the numbers simple.
Elastic potential energy is stored when an object is stretched or compressed:
$$E_e = \frac{1}{2} k e^2$$
Where \(E_e\) is elastic potential energy in J, \(k\) is the spring constant in N/m, and \(e\) is the extension in metres. This only applies up to the limit of proportionality.
Energy Transfers
Energy moves between stores by four transfer pathways: mechanically (by forces), electrically (by currents), by heating, and by radiation (light, sound, or other waves).
When a ball is thrown upwards, energy transfers from the kinetic store to the gravitational potential store. At the top, the ball momentarily stops — all kinetic energy has become gravitational potential energy. As it falls, the transfer reverses.
The law of conservation of energy states that energy can be transferred usefully, stored, or dissipated — but cannot be created or destroyed. The total energy of a closed system always stays the same.
A 2 kg ball is dropped from a height of 5 m. Calculate the speed just before it hits the ground (assume no air resistance).
Step 1: \(E_p = mgh = 2 \times 10 \times 5 = 100\) J
Step 2: Set \(E_k = E_p\), so \(\frac{1}{2}mv^2 = 100\)
Step 3: \(v^2 = \frac{2 \times 100}{2} = 100\), so \(v = 10\) m/s
Work Done and Power
When a force moves an object, work done equals the energy transferred:
$$W = F s$$
Where \(W\) is work done in joules, \(F\) is force in newtons (N), and \(s\) is distance along the line of action of the force in metres.
Power is the rate of energy transfer:
$$P = \frac{E}{t} \quad \text{and} \quad P = \frac{W}{t}$$
Where \(P\) is power in watts (W), \(E\) is energy transferred in joules, and \(t\) is time in seconds. One watt equals one joule per second.
If an object moves at constant speed, you can combine the equations: \(P = Fv\), where \(v\) is speed in m/s. This is useful for calculating the power output of engines or motors.