Edexcel IGCSE Physics 4(c) Work and Power — Notes
Written for Pearson Edexcel International GCSE Physics (4PH1), Science Double Award (4SD0) and Science Single Award (4SS0), linear and modular.
Which parts do I need?
Everything on this page is for all three routes: Single Award, Double Award and Physics need every section. The units for Topic 4 (joule and watt) are on the 4(b) Energy transfers page.
What this page covers
| Spec | Modular no. | You need to be able to… | Route |
|---|---|---|---|
| 4.11 | 3.11 | recall and use work done = force × distance moved along the line of the force | Single · Double · Physics |
| 4.12 | 3.12 | recognise that the work done equals the energy transferred | Single · Double · Physics |
| 4.13 | 3.13 | recall and use GPE = m × g × h | Single · Double · Physics |
| 4.14 | 3.14 | recall and use KE = ½ × m × v² | Single · Double · Physics |
| 4.15 | 3.15 | link GPE, KE and work done using conservation of energy | Single · Double · Physics |
| 4.16 | 3.16 | describe power as how quickly energy is transferred or work is done | Single · Double · Physics |
| 4.17 | 3.17 | use power = work done ÷ time taken | Single · Double · Physics |
Modular candidates: Topic 4 is Topic 3 in the modular specifications, so 4.11–4.17 are statements 3.11–3.17 of 4XPH1 Unit 1 (4WPH1) and 4XSD1 Physics Unit 5 (4WSD5).
In one minute
- Work done = force × distance moved in the direction of the force, \(W=F\times d\), in joules (J).
- Work done = energy transferred. 1 J of work transfers 1 J of energy.
- Gravitational potential energy: \(GPE=m\times g\times h\). Kinetic energy: \(KE=\tfrac12\times m\times v^2\).
- For a falling object with no air resistance, GPE lost = KE gained. With friction, GPE lost = KE gained + work done against friction.
- Power is the rate of transfer of energy: \(P=\dfrac{W}{t}\), in watts (W). 1 W = 1 J/s.
What is work done and how do you calculate it? Single · Double · Physics
When a force moves an object, the force does work on it, and energy is transferred mechanically.
\[ \text{work done}=\text{force}\times\text{distance moved}\qquad W=F\times d \]- \(W\) = work done (J), \(F\) = force (N), \(d\) = distance moved in the direction of the force (m).
- 1 joule is the work done when a force of 1 N moves an object 1 m in the direction of the force.
- If the object does not move, no work is done, however hard you push. If the object moves at right angles to the force, that force does no work.
- Rearranged: \(F=\dfrac{W}{d}\) and \(d=\dfrac{W}{F}\).
Where does the energy go? Work done against friction or air resistance transfers energy to the thermal store of the object and its surroundings. Work done lifting an object transfers energy to its gravitational store. Work done speeding up an object transfers energy to its kinetic store.
A worker pushes a crate 6.0 m across a floor at constant speed with a horizontal force of 180 N. (a) Calculate the work done. (b) State the energy store that gains this energy.
(a) \(W=F\times d=180\times6.0=\) 1080 J
(b) The speed is constant, so the kinetic store does not change. The energy goes to the thermal store of the crate, the floor and the surroundings, because the work is done against friction.
How do you calculate gravitational potential energy? Single · Double · Physics
Lifting an object does work against its weight, and that work is stored in its gravitational store. The change in that store is:
\[ \text{gravitational potential energy}=\text{mass}\times\text{gravitational field strength}\times\text{height}\qquad GPE=m\times g\times h \]- \(GPE\) in J, \(m\) in kg, \(g\) in N/kg, \(h\) = change in height in m (measured vertically, not along a slope).
- Pearson papers tell you to assume \(g=10\ \text{N/kg}\) on Earth; an answer using 9.8 N/kg is also accepted. This page uses 10 N/kg throughout.
- \(m\times g\) is the weight of the object, so GPE = weight × height — the same as the work done lifting it (\(W=F\times d\) with \(F\) = weight).
A climber of mass 55 kg climbs to a ledge 320 m higher than her starting point. Calculate the increase in her gravitational potential energy. Use \(g=10\ \text{N/kg}\).
\(GPE=m\times g\times h=55\times10\times320=\) 176 000 J (176 kJ)
How do you calculate kinetic energy? Single · Double · Physics
\[ \text{kinetic energy}=\tfrac12\times\text{mass}\times\text{speed}^2\qquad KE=\tfrac12\times m\times v^2 \]- \(KE\) in J, \(m\) in kg, \(v\) in m/s.
- Only the speed is squared — not the mass and not the ½. Square the speed first.
- Because of the square, doubling the speed makes the kinetic energy four times bigger.
- To find the speed: \(v=\sqrt{\dfrac{2\times KE}{m}}\). Do not forget the square root.
A car of mass 1200 kg is travelling at 15 m/s. (a) Calculate its kinetic energy. (b) The car speeds up to 30 m/s. Calculate its new kinetic energy and compare it with your answer to (a).
(a) \(KE=\tfrac12\times1200\times15^2=\tfrac12\times1200\times225=\) 135 000 J
(b) \(KE=\tfrac12\times1200\times30^2=\) 540 000 J. This is 540 000 ÷ 135 000 = 4 times as much: the speed has doubled and KE depends on speed squared.
Common error: working out \((\tfrac12\times1200\times15)^2\) or forgetting to square the speed.
How does conservation of energy link GPE, KE and work? Single · Double · Physics
Energy cannot be created or destroyed, so energy leaving one store must appear in others. This gives three common patterns.
| Situation | Energy equation |
|---|---|
| object falls, air resistance negligible | GPE lost = KE gained, so \(mgh=\tfrac12mv^2\) |
| object thrown upwards, air resistance negligible | KE lost = GPE gained |
| object moves with friction or air resistance | GPE lost = KE gained + work done against friction |
| a vehicle brakes to a stop on a level road | KE lost = work done by the brakes = braking force × braking distance |
A ball of mass 0.50 kg is dropped from a height of 3.2 m. Air resistance is negligible. Use \(g=10\ \text{N/kg}\). (a) Calculate the GPE lost by the ball as it falls to the ground. (b) Calculate its speed just before it hits the ground.
(a) \(GPE=m\times g\times h=0.50\times10\times3.2=\) 16 J
(b) KE gained = GPE lost = 16 J, so \(\tfrac12\times0.50\times v^2=16\)
\(v^2=\dfrac{2\times16}{0.50}=64\), \(v=\sqrt{64}=\) 8.0 m/s
Halfway down, the ball has 8 J in its gravitational store and 8 J in its kinetic store — the total stays 16 J.
A skier of mass 70 kg starts from rest at A and slides down a slope to B. B is 12 m vertically below A and the slope is 60 m long. At B her speed is 14 m/s. Use \(g=10\ \text{N/kg}\).
(a) Calculate the GPE lost and the KE gained. (b) Calculate the work done against friction. (c) Calculate the average friction force on the skier.
(a) GPE lost = \(70\times10\times12=\) 8400 J (use the vertical height, 12 m, not the slope length).
KE gained = \(\tfrac12\times70\times14^2=\tfrac12\times70\times196=\) 6860 J
(b) By conservation of energy, work done against friction = 8400 − 6860 = 1540 J
(c) \(W=F\times d\Rightarrow F=\dfrac{1540}{60}=25.7=\) 26 N (friction acts along the slope, so use the 60 m slope length here).
What is power and how do you calculate it? Single · Double · Physics
- \(P\) in watts (W), \(W\) = work done or energy transferred (J), \(t\) in s.
- 1 watt = 1 joule per second. 1 kW = 1000 W.
- Two machines can do the same work; the one that does it in less time is more powerful.
- Rearranged: \(W=P\times t\) and \(t=\dfrac{W}{P}\). Convert minutes to seconds first.
- Take care with symbols: \(W\) is used for work done (in J) and W is also the unit watt.
(a) A student of mass 60 kg runs up a flight of stairs with a vertical height of 4.5 m in 5.0 s. Calculate her useful power. Use \(g=10\ \text{N/kg}\).
(b) A kettle has a power of 2.5 kW. Calculate the energy it transfers in 2.0 minutes.
(a) Work done against her weight = \(m\times g\times h=60\times10\times4.5=2700\ \text{J}\) (her weight is 600 N)
\(P=\dfrac{W}{t}=\dfrac{2700}{5.0}=\) 540 W
(b) P = 2500 W, t = 2.0 × 60 = 120 s. Energy = \(P\times t=2500\times120=\) 300 000 J
Exam tips for 4(c)
Questions often begin “State the formula linking…” for one mark. Write the word equation or standard symbols, for example KE = ½ × m × v². The formula linking kinetic energy, mass and speed is one that is often misquoted.
For GPE, \(h\) is always the change in vertical height. A slope length is only used with \(W=F\times d\) for a force acting along the slope.
In longer questions, decide first which energy is useful and which store it ends up in, then calculate each energy separately and label it (“GPE lost = …”, “KE gained = …”). Clearly labelled steps earn method marks even if the final answer is wrong.
In \(KE=\tfrac12mv^2\), square the speed on its own first. When finding the speed, remember the square root at the end; an answer like 64 m/s in worked example 4 shows the root was missed.
Power in watts needs energy in joules and time in seconds. Convert minutes (× 60) and kJ (× 1000) before substituting.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| Pushing hard on a wall that does not move does work. | The distance moved is zero. | Work done = F × d = 0. |
| Doubling the speed doubles the kinetic energy. | KE depends on speed squared. | Doubling the speed gives four times the KE. |
| Heavier objects fall faster because they have more GPE. | In \(mgh=\tfrac12mv^2\) the mass cancels. | Without air resistance, all objects dropped from the same height land at the same speed. |
| Power and energy are the same thing. | Power includes time. | Power is energy transferred per second. |
| Use the slope length for GPE. | Only vertical height changes the gravitational store. | Use the vertical height in \(GPE=mgh\). |
Key equations
| Equation | Use it to find | Route | On the Equation List? |
|---|---|---|---|
| work done = force × distance moved, \(W=F\times d\) | energy transferred by a force | All | yes — all papers |
| \(GPE=m\times g\times h\) | change in gravitational store | All | yes — all papers |
| \(KE=\tfrac12\times m\times v^2\) | energy in the kinetic store | All | yes — all papers |
| \(P=\dfrac{W}{t}\) | power, or energy transferred in a time | All | yes — all papers |
| GPE lost = KE gained (+ work against friction) | speeds and heights in falling or sliding problems | All | — (conservation of energy) |
Pearson includes the Equation List with the question papers, but the first mark in many questions is for writing the relationship yourself — learn them anyway.
Check your understanding
1. A 2.0 kg book is lifted from the floor onto a shelf 1.8 m high. Calculate the gain in GPE. (g = 10 N/kg) All
\(GPE=2.0\times10\times1.8=\) 36 J
2. A runner of mass 70 kg has 2835 J of kinetic energy. Calculate her speed. All
\(v^2=\dfrac{2\times2835}{70}=81\), \(v=\) 9.0 m/s
3. A stone is dropped from 0.45 m above the ground. Ignoring air resistance, calculate its speed as it lands. (g = 10 N/kg) All
\(mgh=\tfrac12mv^2\Rightarrow v^2=2gh=2\times10\times0.45=9\), \(v=\) 3.0 m/s. The mass cancels, so it is not needed.
4. A car of mass 1500 kg travelling at 20 m/s brakes and stops in 40 m. Calculate the average braking force. All
KE = ½ × 1500 × 20² = 300 000 J. Work done by brakes = KE, so \(F=\dfrac{300\,000}{40}=\) 7500 N.
5. A lift motor raises a lift of mass 800 kg through 24 m in 16 s. Calculate the useful power output. (g = 10 N/kg) All
Work done = 800 × 10 × 24 = 192 000 J; \(P=\dfrac{192\,000}{16}=\) 12 000 W (12 kW).
6. A 2000 W heater is switched on for 3.0 minutes. Calculate the energy transferred. All
t = 3.0 × 60 = 180 s; \(W=P\times t=2000\times180=\) 360 000 J
7. Two students of equal mass run up the same stairs. One takes less time. Explain which develops more power. All
Both do the same work (same weight, same height). The one who takes less time transfers the energy at a greater rate, so develops more power (\(P=W/t\)).
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