Energy: Synthesis and Exam Skills

You have now met all the key ideas of Topic P7: energy stores and transfers, conservation of energy, work done, power, kinetic energy, gravitational and elastic potential energy, efficiency, and thermal transfer. In the exam, the hardest marks come not from a single equation but from combining these ideas — for example, working out how high a rollercoaster will rise from its speed, or how much power a crane needs to lift a load in a given time. This final lesson of Topic P7 (Energy) of OCR Gateway Science A pulls the equations together, shows how to chain them in multi-step problems, and sets out the most common mistakes so you can avoid them under exam pressure.

By the end of this lesson you should be able to select and combine the P7 energy equations, solve multi-step problems involving gravitational, kinetic and elastic stores, calculate power and efficiency in context, and avoid the common P7 exam errors.

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The P7 Equations Gathered Together

Almost every P7 calculation uses one (or more) of these equations. Knowing them cold, and knowing the units, is the foundation of every multi-step problem.

Quantity	Equation	Units
Work done	$W = Fd$W=Fd	J (force in N, distance in m)
Power	$P = \dfrac{W}{t} = \dfrac{E}{t}$P=tW​=tE​	W (energy in J, time in s)
Kinetic energy	$E_k = \dfrac{1}{2}mv^2$Ek​=21​mv2	J (mass in kg, speed in m/s)
Gravitational PE	$E_p = mgh$Ep​=mgh	J ($g = 9.8\text{ N/kg}$g=9.8 N/kg, height in m)
Elastic PE	$E_e = \dfrac{1}{2}ke^2$Ee​=21​ke2	J ($k$k in N/m, extension in m)
Efficiency	$\dfrac{\text{useful output}}{\text{total input}}$total inputuseful output​	no units (×100 for %)

The link that ties them together is conservation of energy: when energy moves from one store to another with nothing dissipated, the energy lost from one store equals the energy gained by another. That is what lets you write expressions like $mgh = \dfrac{1}{2}mv^2$mgh=21​mv2 (gravitational → kinetic) or $\dfrac{1}{2}ke^2 = \dfrac{1}{2}mv^2$21​ke2=21​mv2 (elastic → kinetic).

Exam Tip: Before calculating, write down which store loses energy and which gains it, then set them equal (if nothing is dissipated). Picking the right pair of equations is half the battle — the arithmetic is the easy part.

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Gravitational ↔ Kinetic Transfers

The most common multi-step problem is energy moving between the gravitational potential and kinetic stores — a falling object speeding up, or a moving object rising and slowing down. Ignoring air resistance, set:

$mgh = \dfrac{1}{2}mv^2$mgh=21​mv2

Because the mass cancels, $v = \sqrt{2gh}$v=2gh​ (for a fall) and $h = \dfrac{v^2}{2g}$h=2gv2​ (for a rise).

Worked Example 1 — Falling then finding the height

A ball is thrown straight up and leaves the hand at $7 \text{ m/s}$7 m/s. Ignoring air resistance, how high does it rise? ($g = 9.8\text{ N/kg}$g=9.8 N/kg.)

Step 1 — the kinetic energy at the bottom becomes gravitational potential energy at the top: $\dfrac{1}{2}mv^2 = mgh$21​mv2=mgh, so $h = \dfrac{v^2}{2g}$h=2gv2​.

Step 2 — square the speed: $v^2 = 7^2 = 49$v2=72=49.

Step 3 — substitute: $h = \dfrac{49}{2 \times 9.8} = \dfrac{49}{19.6}$h=2×9.849​=19.649​.

Step 4 — calculate: $h = 2.5 \text{ m}$h=2.5 m.

Answer: the ball rises $2.5 \text{ m}$2.5 m. (Notice the mass was not needed — it cancels.)

Exam Tip: For a rise-and-fall problem, the mass cancels in $mgh = \dfrac{1}{2}mv^2$mgh=21​mv2, so you often don't need it. Use $v = \sqrt{2gh}$v=2gh​ to find a speed from a height, and $h = \dfrac{v^2}{2g}$h=2gv2​ to find a height from a speed.

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Combining Work, Power and Energy

Many questions ask you to find a power by first finding an energy (often from $W = Fd$W=Fd or $E_p = mgh$Ep​=mgh), then dividing by time. Always do the energy step first, then the power step.

Worked Example 2 — Power to lift a load

A crane lifts a $400 \text{ kg}$400 kg load to a height of $15 \text{ m}$15 m in $20 \text{ s}$20 s. Calculate the useful power output. ($g = 9.8\text{ N/kg}$g=9.8 N/kg.)

Step 1 — find the energy transferred to the gravitational potential store: $E_p = mgh = 400 \times 9.8 \times 15 = 58\,800 \text{ J}$Ep​=mgh=400×9.8×15=58800 J.

Step 2 — find the power: $P = \dfrac{E}{t} = \dfrac{58\,800}{20}$P=tE​=2058800​.

Step 3 — calculate: $P = 2940 \text{ W}$P=2940 W.

Answer: the useful power output is $2940 \text{ W}$2940 W (about $2.94 \text{ kW}$2.94 kW).

Worked Example 3 — Power and efficiency together

The crane in Worked Example 2 has a useful power output of $2940 \text{ W}$2940 W but draws $4200 \text{ W}$4200 W from its motor. Calculate its efficiency.

Step 1 — write the equation: efficiency $= \dfrac{\text{useful power output}}{\text{total power input}}$=total power inputuseful power output​.

Step 2 — substitute: efficiency $= \dfrac{2940}{4200}$=42002940​.

Step 3 — calculate: efficiency $= 0.7 = 70\%$=0.7=70%.

Answer: the crane is $70\%$70% efficient.

Exam Tip: Do the steps in order: energy first ($E_p = mgh$Ep​=mgh or $W = Fd$W=Fd), then power ($P = \dfrac{E}{t}$P=tE​), then efficiency if asked. Keep the time in seconds and the mass in kilograms throughout.

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A Full Multi-Step Worked Problem

Here is a problem that chains together gravitational potential energy, kinetic energy, conservation of energy and dissipation — exactly the kind of synoptic question that earns the top marks.

Worked Example 4 — A rollercoaster with friction

A rollercoaster car of mass $600 \text{ kg}$600 kg starts from rest at the top of a hill $25 \text{ m}$25 m high. ($g = 9.8\text{ N/kg}$g=9.8 N/kg.) (a) Calculate the gravitational potential energy of the car at the top, relative to the bottom. (b) Assuming no energy is dissipated, calculate the speed of the car at the bottom. (c) In reality the car reaches the bottom at $20 \text{ m/s}$20 m/s. Calculate its actual kinetic energy at the bottom, and hence the energy dissipated by friction and air resistance.

Part (a):

Step 1 — write the equation: $E_p = mgh$Ep​=mgh.

Step 2 — substitute: $E_p = 600 \times 9.8 \times 25$Ep​=600×9.8×25.

Step 3 — calculate: $E_p = 147\,000 \text{ J}$Ep​=147000 J.

Part (a) answer: $147\,000 \text{ J}$147000 J.

Part (b):

Step 1 — with no dissipation, all the gravitational potential energy becomes kinetic energy: $\dfrac{1}{2}mv^2 = 147\,000 \text{ J}$21​mv2=147000 J.

Step 2 — rearrange for speed: $v = \sqrt{\dfrac{2E_k}{m}} = \sqrt{\dfrac{2 \times 147\,000}{600}} = \sqrt{\dfrac{294\,000}{600}} = \sqrt{490}$v=m2Ek​​​=6002×147000​​=600294000​​=490​.

Step 3 — calculate: $v = 22.1 \text{ m/s}$v=22.1 m/s.

Part (b) answer: $22.1 \text{ m/s}$22.1 m/s (about $22 \text{ m/s}$22 m/s).

Part (c):

Step 1 — actual kinetic energy at the bottom: $E_k = \dfrac{1}{2}mv^2 = \dfrac{1}{2} \times 600 \times 20^2 = 0.5 \times 600 \times 400 = 120\,000 \text{ J}$Ek​=21​mv2=21​×600×202=0.5×600×400=120000 J.

Step 2 — energy dissipated = gravitational potential energy at the top − actual kinetic energy at the bottom = $147\,000 - 120\,000$147000−120000.

Step 3 — calculate: dissipated $= 27\,000 \text{ J}$=27000 J.

Part (c) answer: the actual kinetic energy is $120\,000 \text{ J}$120000 J, so $27\,000 \text{ J}$27000 J has been dissipated to the thermal store of the surroundings by friction and air resistance.