OxfordAQA International GCSE Physics: Work, kinetic, gravitational and elastic energy, and power (9203)

This guide follows section 3.2.1, Forces and energy (statements a to f), of the OxfordAQA International GCSE Physics (9203) specification, for exams from May/June 2018. The qualification is untiered, so all of it is for every candidate. Each statement has a worked example. For short recall notes use the revision notes, and for questions with answers use the practice set.

Related pages: the OxfordAQA International GCSE Physics hub, the printable checklist, the wider Energy overview, the Hooke’s law guide and the free diagnostics if you want to find your weak spots first.

What this unit covers

Spec	What you must be able to do
3.2.1 a	Define work done and use W = F × d
3.2.1 b	Explain that work against friction transfers energy by heating; discuss shuttle re-entry, meteorites and vehicle brakes
3.2.1 c	Calculate the elastic potential energy of a stretched spring, Ee = ½ × k × e², below the limit of proportionality
3.2.1 d	Calculate gravitational potential energy, Ep = m × g × h
3.2.1 e	Calculate kinetic energy, Ek = ½ × m × v²; compare doubling mass with doubling speed; apply this to road safety
3.2.1 f	Use power P = E / t and P = W / t

Energy is measured in joules (J) and power in watts (W). Use the value of g given in a question. The worked examples below use g = 10 N/kg.

Work done (3.2.1 a)

Work is done when a force makes an object move through a distance. The equation is:

work done (J) = force (N) × distance moved in the direction of the force (m)
W = F × d

A force of 1 N that moves an object through 1 m along its own line of action does 1 J of work. Only movement in the direction of the force counts. If you hold a heavy bag still, you feel tired but you do no work on the bag, because it does not move. A weightlifter holding a bar above the head does no work on the bar while it is still.

Energy is transferred when work is done, so the work done equals the energy transferred. A force of 10 N moving a trolley 3 m transfers 30 J.

Worked example 1. A porter pushes a luggage trolley with a steady force of 85 N for 24 m.

W = F × d = 85 × 24 = 2040 J

Worked example 2 (rearranging). A crate is dragged 12 m along a floor. The work done is 1560 J. Find the force.

F = W / d = 1560 / 12 = 130 N

Take care with units. If a distance is in centimetres, convert it to metres first.

Work against friction (3.2.1 b)

When you push an object across a rough surface, the friction force acts against the motion. Work is done against it. That energy does not disappear and is not stored as something useful. It is transferred by heating: the surfaces and the surrounding air get warmer.

Vehicle brakes

A braking system presses pads or shoes against a disc or drum that turns with the wheel. Friction between them does work. The kinetic energy of the vehicle is transferred to thermal energy in the brake parts, so the brakes become hot. Heavy or fast vehicles have more kinetic energy to remove, so their brakes get hotter.

Worked example 3. A car of mass 1100 kg is travelling at 20 m/s. The brakes bring it to rest with a constant braking force of 5500 N. Find the braking distance, ignoring air resistance.

Ek = ½ × m × v² = ½ × 1100 × 20² = 220 000 J
Work done by the brakes = Ek, so F × d = 220 000
d = 220 000 / 5500 = 40 m

The 220 000 J is now thermal energy in the brakes, tyres and road.

Shuttle re-entry and meteorites

A spacecraft returning from orbit and a meteoroid entering the atmosphere both move very fast. The air in front is compressed, and air resistance does work against the motion. The kinetic energy of the object is transferred by heating to the object’s surface and the air around it. A shuttle’s underside carries heat-resistant tiles to survive this. A small meteoroid has no such protection, so it gets hot enough to burn up and glow as a meteor.

Worked example 4. A meteoroid of mass 0.80 kg enters the atmosphere at 12 000 m/s.

Ek = ½ × 0.80 × 12 000² = 5.76 × 10⁷ J

That is 57.6 million joules, which is why the heating is so great.

Elastic potential energy (3.2.1 c)

Work is done to stretch a spring. That energy is stored as elastic potential energy. Provided the limit of proportionality has not been exceeded:

elastic potential energy (J) = ½ × spring constant (N/m) × (extension (m))²
Ee = ½ × k × e²

Extension is the new length minus the original length. The spring constant k is the force needed per metre of extension (from F = k × e, covered with Hooke’s law). On a force-extension graph, the area under the straight line is ½ × F × e, which is the same energy.

Worked example 5. A spring with k = 60 N/m is stretched by 0.20 m.

Ee = ½ × 60 × 0.20² = ½ × 60 × 0.04 = 1.2 J

Worked example 6 (rearranging). A spring with k = 300 N/m stores 0.96 J. Find the extension.

e² = 2 × Ee / k = 2 × 0.96 / 300 = 0.0064
e = √0.0064 = 0.08 m

Extension is squared, so doubling the extension makes the stored energy four times bigger. Past the limit of proportionality the spring no longer obeys Hooke’s law and this equation does not apply.

Gravitational potential energy (3.2.1 d)

Lifting an object vertically means doing work against its weight. The object gains gravitational potential energy:

Ep in J = m in kg × g in N/kg × h in m
Ep = m × g × h

Here h is how far the object rises vertically, not how far it travels up a ramp. Because m × g is the weight, this is the same as W = F × d with F equal to the weight.

Worked example 7. A climber of mass 65 kg climbs a vertical height of 340 m.

Ep = 65 × 10 × 340 = 221 000 J

Worked example 8 (rearranging). A 2.4 kg object gains 72 J of gravitational potential energy when lifted. Find the height.

h = Ep / (m × g) = 72 / (2.4 × 10) = 3.0 m

On a planet or moon with a smaller g, the same lift gives a smaller Ep.

Kinetic energy (3.2.1 e)

A moving object has kinetic energy that depends on its mass and its speed:

kinetic energy (J) = ½ × mass (kg) × (speed (m/s))²
Ek = ½ × m × v²

Worked example 9. A cyclist and bike have a total mass of 72 kg and travel at 8.0 m/s.

Ek = ½ × 72 × 8.0² = 2304 J

Speeds are often given in km/h. To change to m/s, divide by 3.6: 54 km/h is 15 m/s.

Doubling mass and doubling speed

Change	Effect on Ek	Cyclist example
Mass doubled, same speed	Twice as much	144 kg at 8.0 m/s: 4608 J
Speed doubled, same mass	Four times as much	72 kg at 16 m/s: 9216 J

Ek is proportional to m, but to v². Doubling v gives 2² = 4 times the energy.

Road safety

To stop, the brakes must remove all the kinetic energy by doing work: braking force × braking distance = Ek. With the same braking force, four times the energy needs four times the distance.

Worked example 10. A van of mass 1200 kg brakes with a constant force of 6000 N. Compare stopping from 10 m/s and from 20 m/s.

10 m/s: Ek = ½ × 1200 × 10² = 60 000 J, d = 60 000 / 6000 = 10 m
20 m/s: Ek = ½ × 1200 × 20² = 240 000 J, d = 240 000 / 6000 = 40 m

Doubling the speed made the braking distance four times as long. This is why a small increase in speed in a built-up area raises the risk so much: the vehicle takes much longer to stop, and if it hits something it has far more energy to transfer. A vehicle with twice the mass at the same speed has twice the energy to remove.

Power (3.2.1 f)

Power measures how quickly energy is transferred, or how quickly work is done. The unit is the watt (W), which is one joule per second.

power (W) = energy transferred (J) / time (s)       P = E / t
power (W) = work done (J) / time (s)                P = W / t

Worked example 11. A motor lifts a 120 kg load through 9.0 m in 15 s.

Work done = Ep gained = 120 × 10 × 9.0 = 10 800 J
P = W / t = 10 800 / 15 = 720 W

Worked example 12 (rearranging). A heater has a power of 2200 W. How much energy does it transfer in 2.5 minutes?

t = 2.5 × 60 = 150 s
E = P × t = 2200 × 150 = 330 000 J

Two machines can do the same work with different power. The machine that takes the shorter time has the larger power.

Putting it together

Worked example 13. A 0.50 kg ball is dropped from a height of 3.2 m. Ignore air resistance and find its speed just before it lands.

Ep lost = m × g × h = 0.50 × 10 × 3.2 = 16 J
This becomes kinetic energy: ½ × 0.50 × v² = 16
v² = 16 / 0.25 = 64
v = 8.0 m/s

Common errors

• Putting the whole length of a spring into Ee = ½ × k × e² instead of its extension.
• Forgetting to square the extension or the speed, or squaring after multiplying by one half.
• Doubling Ek when the speed doubles. The energy is four times as much.
• Measuring height along a slope rather than straight up in Ep = m × g × h.
• Leaving time in minutes in P = E / t. Use seconds.
• Saying energy is “used up” by friction. It is transferred by heating and spread into the surroundings.
• Writing kilojoules or km/h figures into an equation without converting.

Official syllabus

This guide is aligned to section 3.2.1 of the OxfordAQA International GCSE Physics (9203) specification, for exams May/June 2018 onwards, published by OxfordAQA.