8.1–8.3, 8.8–8.9Energy stores, and the two equations that go with them
Whenever a system changes, energy moves between stores. Naming the stores at the start and at the end, and the pathway between them, is the structure of nearly every energy answer on this paper.
The stores to name are kinetic, gravitational potential, elastic potential, thermal, chemical, magnetic, electrostatic and nuclear. The pathways — the ways energy gets from one to another — are mechanical work, electrical work, heating, and radiation.
Two equations carry most of the calculations, and both come up in the context of work done by a force.
The gravitational potential energy gained when something is lifted:
change in GPE (J) = mass (kg) × gravitational field strength (N/kg) × change in vertical height (m)
The kinetic energy of something moving:
kinetic energy (J) = ½ × mass (kg) × (speed)²
The square in the second one is where marks are lost: double the speed and the kinetic energy goes up four times, not two. It is also why a small increase in the speed of a car makes such a large difference to how far it takes to stop.
Worked example
A 0.50 kg ball is dropped from a height of 1.8 m. Assuming no energy is transferred to the surroundings, calculate its speed just before it lands. (g = 10 N/kg)
- Change in GPE = 0.50 × 10 × 1.8 = 9.0 J.
- All of it becomes kinetic energy, so KE = 9.0 J.
- 9.0 = ½ × 0.50 × v², so v² = 9.0 ÷ 0.25 = 36.
- v = √36.
Answer: 6.0 m/s.
In a closed system — one where nothing enters or leaves — the total energy never changes. Energy is not used up and is not created; it only moves between stores, and some of it always ends up somewhere useless.
Common mistake
The height in the GPE equation is the VERTICAL height gained, not the distance moved along a slope. A box pushed 5 m up a ramp that rises 2 m has gained the GPE for 2 m.
8.4–8.6Work done
When a force causes an object to move through a distance, work is done on the object. So a force does work on an object when the force causes a displacement of the object.
The work done by a force on an object can be calculated using the equation:
work done = force × distance moved along the line of action of the force W = F s
work done, W, in joules, J; force, F, in newtons, N; distance, s, in metres.
You should be able to recall and apply this equation.
The phrase "along the line of action of the force" is doing real work in that definition. Only movement in the direction the force acts counts. Carrying a bag horizontally does no work against its weight, because the weight acts downwards and the movement is sideways.
Worked example
A force of 250 N pushes a box 6.0 m across a floor in the direction of the force. Calculate the work done.
- Write the equation: W = F s.
- The distance moved is along the line of action of the force, so it can be used directly.
- Substitute: W = 250 × 6.0.
- = 1500.
- Give the unit: joules.
Answer: 1500 J
One joule of work is done when a force of one newton causes a displacement of one metre.
1 joule = 1 newton-metre
You should be able to convert between newton-metres and joules — which, given that equation, means recognising that they are the same size of unit. A value of 350 Nm of work is 350 J.
You should also be able to describe the energy transfer involved when work is done. Doing work on an object transfers energy to it: pushing a box along the floor transfers energy from the chemical store of the person to the thermal store of the box and floor through friction, while lifting it transfers energy to its gravitational potential store.
Common mistake
Newton-metres appear in two different places at GCSE — as a unit of work done, where 1 Nm = 1 J, and as the unit of the moment of a force, where it is not a joule. Read which quantity the question is about.
- Work done
- the energy transferred when a force causes an object to move through a distance, calculated as force × distance moved along the line of action of the force
8.7Work done against friction
Work done against the frictional forces acting on an object causes a rise in the temperature of the object.
This is the same conservation argument as for energy transfers in a system, applied to a particular case. Energy is transferred by the force doing work, but if the object ends up moving no faster and no higher, that energy has not gone into its kinetic or gravitational potential store. It has been dissipated to the thermal stores of the object and its surroundings, so the temperature rises.
Worked example
A cyclist brakes and comes to a stop. Describe the energy transfers, and explain why the brakes become hot.
- Before braking, energy is in the kinetic store of the bicycle and rider.
- The brake blocks press against the wheel, so there is a large frictional force.
- The wheel moves against that frictional force, so work is done against friction.
- Work done against frictional forces causes a rise in temperature.
- The energy from the kinetic store is therefore transferred to the thermal store of the brakes, the wheel and the surroundings — which is why the brakes get hot.
Answer: Energy is transferred from the kinetic store of the bicycle to the thermal store of the brakes and surroundings, because work is done against friction — and work done against friction raises the temperature.
Common mistake
Friction does not destroy the kinetic energy. The energy is transferred to thermal stores, which is measurable as a temperature rise — the total energy is unchanged.
Before you practise
- work done = force × distance moved in the direction of the force, in joules.
- One joule is one newton metre — the work done when a force of 1 N moves an object 1 m.
- Work done by a force equals the energy transferred by it.
- Only the distance moved IN THE DIRECTION of the force counts.
- change in GPE = mass × gravitational field strength × change in vertical height.
- kinetic energy = ½ × mass × (speed)², so doubling the speed quadruples the energy.
- Work done against friction raises the temperature of the surfaces and is then dissipated to the surroundings.