Changes in energy — AQA GCSE Combined Science

Kinetic energy is the energy stored in an object because it is moving. It depends on the object's mass and on its speed, and it is measured in joules (J). To calculate it, use the equation Ek = 0.5 × m × v², where m is mass in kilograms and v is speed in metres per second. For example, a 2 kg ball moving at 3 m/s has kinetic energy 0.5 × 2 × 3² = 9 J. Doubling the speed has a much bigger effect than doubling the mass, because speed is squared in the equation. This links to energy transfers: when a moving object slows down, its kinetic energy decreases and is transferred to other stores, such as the thermal store of the brakes via mechanical work.

This equation lets you calculate kinetic energy in joules when mass is in kilograms and speed is in metres per second. The symbol 0.5 is the same as one half, and speed² means speed multiplied by itself. For example, a 1500 kg car travelling at 20 m/s has kinetic energy 0.5 × 1500 × 20² = 300 000 J, or 300 kJ. Because speed is squared, doubling speed multiplies kinetic energy by four for the same mass. You may also need to rearrange the equation: mass = kinetic energy ÷ (0.5 × speed²), and speed = √(kinetic energy ÷ (0.5 × mass)). Always convert units first, then substitute, then calculate, and give the unit J with your answer.

This equation calculates the kinetic energy stored by an object due to its motion. Eₖ is kinetic energy in joules (J), m is mass in kilograms (kg), and v is speed in metres per second (m/s). The speed is squared, so doubling speed quadruples kinetic energy. For example, a 2 kg ball moving at 3 m/s has Eₖ = ½ × 2 × 3² = 9 J. When using the equation, first identify the mass and speed, convert grams to kilograms and km/h to m/s if needed, square the speed, multiply by mass, then halve. Rearranging allows calculation of mass or speed from kinetic energy. This equation applies to any moving object in GCSE physics and is central to analysing energy transfers in collisions, braking, and sports.

Kinetic energy is the energy stored by a moving object. The symbol Eₖ represents kinetic energy, and its unit is the joule (J). One joule is the energy transferred when a force of one newton moves an object one metre. In calculations, kinetic energy is found using Eₖ = ½mv², where m is mass in kilograms and v is speed in metres per second. Because speed is squared, small changes in speed cause large changes in kinetic energy. For example, a 1000 kg car travelling at 10 m/s has Eₖ = ½ × 1000 × 10² = 50 000 J. Understanding the unit and symbol is essential for correct substitution and for interpreting energy values in contexts such as braking distances and safety design.

Mass measures the amount of matter in an object and is measured in kilograms (kg). In energy calculations, mass must be in kg before substitution into equations such as kinetic energy E_k = ½mv² or gravitational potential energy E_p = mgh. For example, a 2.5 kg ball moving at 4 m/s has E_k = ½ × 2.5 × 4² = 20 J. If a mass is given in grams, convert by dividing by 1000: 500 g = 0.5 kg. Mass is not the same as weight; weight is a force in newtons and depends on gravitational field strength. In the equation E_p = mgh, m is the mass in kg, g is gravitational field strength in N/kg, and h is height in m. Always check units before calculating energy changes.

Speed, v, is the rate of change of distance with time and is measured in metres per second (m/s). In energy calculations, speed must be in m/s before substitution into the kinetic energy equation E_k = ½mv². For example, a 1000 kg car travelling at 15 m/s has E_k = ½ × 1000 × 15² = 112500 J. If speed is given in km/h, convert to m/s by dividing by 3.6: 72 km/h = 20 m/s. Speed is a scalar quantity, while velocity is a vector. When using E_k = ½mv², the speed is squared, so doubling speed quadruples kinetic energy. Always check that speed is in m/s and mass in kg before calculating energy changes.

When a spring is stretched, work is done against the spring's stiffness and energy is transferred to the elastic potential store of the spring. The amount stored depends on how stiff the spring is and how far it has been stretched from its natural length. The equation that links these quantities is Ee = 0.5 × k × e², where Ee is elastic potential energy in joules (J), k is the spring constant in newtons per metre (N/m) and e is the extension in metres (m). Extension means the increase in length beyond the unstretched length, not the total length. For example, a spring with k = 200 N/m stretched by 0.10 m stores 0.5 × 200 × (0.10)² = 1.0 J. The equation applies provided the spring has not been stretched beyond its limit of proportionality, so the force–extension graph remains a straight line through the origin.

This equation gives the energy transferred to the elastic potential store of a spring when it is stretched. In symbols, Ee = 0.5 × k × e², where Ee is measured in joules (J), k is the spring constant in newtons per metre (N/m) and e is the extension in metres (m). The extension is the increase in length from the spring's natural length. The squared term means that doubling the extension quadruples the energy stored, because 2² = 4. For example, a spring with k = 100 N/m stretched by 0.20 m stores 0.5 × 100 × (0.20)² = 2.0 J. The equation is valid while the spring obeys Hooke's law, so the force–extension graph is a straight line through the origin and the spring has not passed its limit of proportionality.

This equation gives the elastic potential energy stored in a stretched or compressed spring, provided the spring obeys Hooke’s law. Eₑ is the energy in joules (J), k is the spring constant in newtons per metre (N/m), and e is the extension or compression in metres (m). The extension e is the change in length from the natural length, not the total length. Because e is squared, doubling the extension quadruples the stored energy. For example, a spring with k = 200 N/m stretched by e = 0.10 m stores Eₑ = ½ × 200 × (0.10)² = 1.0 J. The factor ½ arises because the force increases linearly from zero to ke, so the average force is ½ke. Always convert centimetres to metres and grams to kilograms where needed before substituting.

This condition means the equation Eₑ = ½ke² is valid only while the spring obeys Hooke’s law, where force is directly proportional to extension. The limit of proportionality is the point beyond which force is no longer proportional to extension; the spring may still stretch but the graph curves. If the limit is exceeded, the simple equation no longer gives the stored elastic potential energy accurately and must not be used. For example, a spring with k = 200 N/m stretched by 0.10 m stores 1.0 J, but if stretched beyond its limit, the linear relationship breaks down. In calculations, check that the extension is within the linear region before using Eₑ = ½ke².

Elastic potential energy is the energy stored in a stretched or compressed elastic object, such as a spring or elastic band, because work has been done to change its shape. When a force stretches a spring, energy is transferred from the store of the object doing the stretching to the elastic potential energy store of the spring. The amount stored depends on the spring constant and the extension. For a linear spring, E_e = ½ k x², where E_e is in joules, J, k is the spring constant in newtons per metre, N/m, and x is the extension in metres, m. Doubling the extension quadruples the energy because x is squared. When the spring is released, this stored energy is transferred to other stores, for example kinetic energy of a moving object or thermal energy through friction.

The spring constant, k, is a measure of the stiffness of a spring or elastic object. It is defined as the force needed to extend or compress the object by one metre, and its unit is newtons per metre, N/m. A larger k means a stiffer spring that needs a greater force for the same extension. For a linear spring that obeys Hooke's law, force is directly proportional to extension, so F = k x, where F is in newtons, N, and x is the extension in metres, m. Rearranging gives k = F ÷ x. The spring constant also appears in the elastic potential energy equation E_e = ½ k x². In an experiment, a student hangs masses on a spring, records the extension for each force, and plots a force-extension graph; the gradient of the straight-line region equals k.

Extension, e, is the increase in length of a spring or elastic object when a force stretches it, measured in metres (m). It is not the total stretched length: if a spring hangs at 12 cm and stretches to 20 cm, the extension is 8 cm, which must be converted to 0.08 m before substitution. Extension is a change in length, so it is found by subtracting the original length from the stretched length. In elasticity calculations, e is squared in the equation for elastic potential energy, Ee = ½ke², so the unit must be metres. A metre rule or tape measure can measure lengths, and the difference gives e. Extension increases linearly with force up to the limit of proportionality.

When an object is lifted, work is done against gravity and energy is transferred to the gravitational potential energy store. The gain is calculated using Ep = mgh, where m is mass in kilograms (kg), g is gravitational field strength in newtons per kilogram (N/kg), and h is height in metres (m). On Earth, g is 9.8 N/kg in AQA GCSE questions. For example, lifting a 2 kg mass by 3 m gives Ep = 2 × 9.8 × 3 = 58.8 J. The equation gives the gain in gravitational potential energy, so it assumes the object started at the reference level. If the object is lowered, the same equation gives the energy transferred away from the store.

Gravitational potential energy (g.p.e.) is the energy stored in an object because of its position in a gravitational field. You calculate it by multiplying three quantities: the object's mass in kilograms, the gravitational field strength in newtons per kilogram, and the vertical height in metres. For example, lifting a 2 kg book 1.5 m on Earth, where g = 9.8 N/kg, gives g.p.e. = 2 × 9.8 × 1.5 = 29.4 J. The height must be the vertical change, not the distance moved along a slope. Doubling any one factor doubles the energy stored. This relationship lets you compare energy transfers when objects are raised or lowered, and it links to kinetic energy and conservation of energy.

E_p = m g h is the symbol equation for gravitational potential energy. E_p is the energy stored in joules, m is the mass in kilograms, g is the gravitational field strength in newtons per kilogram, and h is the vertical height in metres. For example, a 0.5 kg ball raised 2 m on Earth stores E_p = 0.5 × 9.8 × 2 = 9.8 J. The equation shows that energy stored is directly proportional to mass, gravitational field strength and height. You can rearrange it to make m, g or h the subject. It is used alongside kinetic energy equations to solve problems about falling objects and energy transfers, assuming no energy is lost to friction or air resistance.

Gravitational potential energy, E_p, is the energy a mass stores because of its height in a gravitational field. Lifting an object means a force equal to its weight, W = m × g, moves it upward through a vertical distance, so the energy transferred to the gravitational store is E_p = m × g × h. E_p is measured in joules, J, the same unit as every other energy form, because one joule is one newton-metre. For example, lifting a 2 kg bag by 3 m with g = 9.8 N/kg gives E_p = 2 × 9.8 × 3 = 58.8 J. Only the vertical height change matters, not the path taken. When the object falls, that stored energy is transferred to kinetic and thermal stores, so E_p lost equals energy gained elsewhere if no energy is dissipated.

Gravitational field strength, g, describes the gravitational force acting on each kilogram of mass at a point. Its unit, N/kg, follows from weight = mass × gravitational field strength, W = m × g, so g is the weight per unit mass. Near the Earth's surface g is about 9.8 N/kg, often rounded to 10 N/kg, while the Moon's value is about 1.6 N/kg. In any calculation the value of g will be given, so read it from the question rather than assuming 9.8 or 10. For example, a 5 kg mass on a planet with g = 4 N/kg has weight 20 N, and lifting it 2 m stores E_p = 5 × 4 × 2 = 40 J. Because g is a field strength, it acts on every kilogram equally, so doubling the mass doubles the weight but leaves g unchanged.

In calculations about gravitational potential energy, height is the vertical distance an object is raised or lowered, measured in metres (m). It is not the distance along a slope and not the total height above sea level unless that is the reference chosen. For example, lifting a 2 kg box from the floor to a shelf 1.5 m above the floor gives h = 1.5 m, so the energy transferred to the gravitational store is about 2 × 9.8 × 1.5 = 29.4 J. If the box is moved sideways at constant height, h does not change, so no gravitational potential energy is added. Always identify the start and end heights, subtract to find the change, and convert centimetres or kilometres to metres before substituting into an equation.