1.5 Work and energy
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AS content in Unit 1 (Mechanics and Materials): specification points 17–20. Power and efficiency are on the next page; elastic strain energy is covered with Materials.
What this page covers
7 points
- Work done by a force, and the joule
- Work done by a force at an angle to the displacement
- Work done from the area under a force–displacement graph
- Kinetic energy, and braking distances from work done = change in kinetic energy
- Changes in gravitational potential energy near the Earth's surface
- The principle of conservation of energy, with and without resistive forces: ramps, roller coasters and pendulums
- Using energy to find the speed of a projectile
In one minute
6 points
- Energy gives a second way into mechanics: instead of following forces moment by moment, you compare energy stores at the start and the end, and this still works when the path curves.
- Work done = force × distance moved in the direction of the force: \(\Delta W = F\Delta s\). Work done = energy transferred. Unit: joule, 1 J = 1 N m.
- A force at angle \(\theta\) to the displacement does work \(F\Delta s\cos\theta\). A force at 90° to the motion does no work. Work done = area under a force–displacement graph.
- \(E_{\text{k}} = \tfrac12 mv^2\): double the speed, four times the kinetic energy — and four times the braking distance.
- \(\Delta E_{\text{grav}} = mg\Delta h\), where \(\Delta h\) is the vertical height change. The path does not matter.
- Energy is never created or destroyed. With resistive forces: loss of \(E_{\text{grav}}\) = gain in \(E_{\text{k}}\) + work done against resistive forces.
Contents
13 sections
- What is work done?
- What if the force is at an angle to the motion?
- How do you find work from a force–displacement graph?
- What is kinetic energy?
- What is gravitational potential energy?
- How do you use the principle of conservation of energy?
- How does energy give the speed of a projectile?
- Exam tips for 1.5
- Common misconceptions
- How this topic connects
- Key equations
- Check your understanding
- Specification checklist
What is work done?
In physics "work" has a precise meaning: a force does work when its point of application moves in the direction of the force. Work is how energy is transferred mechanically, so counting work is counting energy.
- Work done = energy transferred. When a force does 500 J of work, 500 J of energy is transferred from one store to another.
- The joule: 1 J is the work done when a force of 1 N moves its point of application 1 m in the direction of the force, so \(1\ \text{J} = 1\ \text{N m} = 1\ \text{kg m}^2\,\text{s}^{-2}\).
- Work is a scalar, even though force and displacement are vectors.
- No displacement, no work: holding a heavy bag still, or pushing against a wall that does not move, does no work on the object.
A cleaner pushes a floor-polishing machine 25 m along a straight corridor with a constant horizontal force of 64 N. Calculate the work done by the cleaner on the machine.
The force acts along the direction of motion, so use \(\Delta W = F\Delta s\) directly: \(\Delta W = 64 \times 25 = \) 1.6 × 10³ J (1600 J)
The weight of the machine acts vertically and the motion is horizontal, so the weight does no work here.
What if the force is at an angle to the motion?
Only the component of the force along the displacement does work. For a force \(F\) at angle \(\theta\) to the displacement, the component along the motion is \(F\cos\theta\), so
| Angle between force and displacement | Work done | Example |
|---|---|---|
| 0° | \(F\Delta s\) (the most possible) | a horizontal push along a level floor |
| between 0° and 90° | \(F\Delta s\cos\theta\), positive | pulling a suitcase by its handle |
| 90° | zero | weight and normal contact force on a body moving along level ground |
| 180° | negative: the force removes energy from the body | friction and air resistance, braking forces |
A traveller pulls a wheeled suitcase 120 m along a level airport floor. The handle makes an angle of 40° with the horizontal and the pull along the handle is 35 N. The suitcase moves at constant velocity. (a) Calculate the work done by the pull. (b) State the work done on the suitcase by its weight. (c) Determine the work done against resistive forces.
(a) Component of the pull along the displacement: \(F\cos\theta = 35\cos 40^\circ = 26.8\ \text{N}\). Work done \(= 26.81 \times 120 = 3217\ \text{J} \approx\) 3.2 × 10³ J
(b) The weight is perpendicular to the horizontal displacement: work done by the weight = 0
(c) Constant velocity, so the resultant horizontal force is zero and the resistive forces total 26.8 N backwards: work done against resistive forces = 3.2 × 10³ J — the same energy, transferred to the surroundings as thermal energy.
Using sin instead of cos, or multiplying the whole pull by the distance. Draw the angle between the force and the direction of motion; the component next to that angle (\(F\cos\theta\)) is the one that does work.
How do you find work from a force–displacement graph?
For a constant force, \(\Delta W = F\Delta s\) is the area of a rectangle under a horizontal line on a graph of \(F\) against \(s\). The same is true for any force: work done = area under the force–displacement graph.
- Straight-line sections: split the area into rectangles and triangles.
- Curved graphs: count squares (or estimate with thin strips), and multiply by the work represented by one square.
- The force plotted must be the component along the displacement.
The graph above shows how the force of a winch cable on a small boat varies with the distance the boat moves up a slipway. The cable is parallel to the slipway. Determine the total work done by the cable over the first 10.0 m.
Step 1 Triangle, 0 to 4.0 m: \(\tfrac12 \times 4.0 \times 600 = 1200\ \text{J}\)
Step 2 Rectangle, 4.0 m to 10.0 m: \(6.0 \times 600 = 3600\ \text{J}\)
Step 3 Add: \(\Delta W = 1200 + 3600 = \) 4.8 × 10³ J. Check the units of the area: N × m = J.
Before you use an area, check the axis units. If the displacement is in cm or the force in kN, convert first — or convert the area at the end — or the answer will be out by a power of ten.
What is kinetic energy?
Where does ½mv² come from?
Let a constant resultant force \(F\) accelerate a mass \(m\) from rest through a distance \(s\). The work done is \(W = Fs = mas\). From \(v^2 = u^2 + 2as\) with \(u = 0\), \(as = \tfrac12 v^2\). So \(W = \tfrac12 mv^2\): all the work done by the resultant force becomes kinetic energy. This is the work–energy idea: the work done by the resultant force on a body equals its change in kinetic energy.
- \(E_{\text{k}} \propto v^2\): doubling the speed multiplies the kinetic energy by four.
- Braking: a constant braking force \(F\) does negative work over the braking distance \(d\) until all the kinetic energy has gone: \(Fd = \tfrac12 mv^2\), so \(d = mv^2/2F\) — the braking distance is proportional to \(v^2\).
A car of mass 1350 kg travels at 22.0 m s⁻¹ on a level road. The driver brakes and the car stops in a straight line. The braking force is constant at 8.10 kN. (a) Calculate the kinetic energy of the car before braking. (b) Calculate the braking distance. (c) Without further use of the formula for kinetic energy, deduce the braking distance from 11.0 m s⁻¹ with the same braking force.
(a) \(E_{\text{k}} = \tfrac12 \times 1350 \times 22.0^2 = \) 3.27 × 10⁵ J
(b) Work done by the brakes = kinetic energy removed: \(Fd = E_{\text{k}}\), so \(d = 3.267 \times 10^5 \div 8100 = \) 40.3 m
(c) Half the speed → one quarter of the kinetic energy → one quarter of the distance: \(d = 40.33 \div 4 = \) 10.1 m
Squaring the mass or forgetting to square the speed. Write \(\tfrac12 \times m \times v^2\) with the numbers in that order, and square the speed before you multiply.
What is gravitational potential energy?
- It comes from work done against the weight: lifting a body steadily through \(\Delta h\) needs a force \(mg\) over a distance \(\Delta h\).
- Only changes in \(E_{\text{grav}}\) matter; you may choose any convenient zero level (the ground, the bottom of the slope).
- The path does not matter: up a ramp, up stairs or straight up, the gain is the same for the same \(\Delta h\).
A crate of mass 25 kg is moved onto a platform 1.20 m high, either by lifting it straight up or by pushing it up a ramp 3.50 m long. Resistive forces are negligible. (a) Calculate the gain in gravitational potential energy of the crate. (b) Calculate the smallest force, parallel to the ramp, that will push the crate up the ramp.
(a) Use the vertical height only: \(\Delta E_{\text{grav}} = 25 \times 9.81 \times 1.20 = \) 294 J
(b) Work done along the ramp = gain in \(E_{\text{grav}}\): \(F \times 3.50 = 294.3\), so \(F = \) 84.1 N
Lifting straight up needs at least \(mg = 245\ \text{N}\). The ramp reduces the force, not the energy.
Using the length along a slope as \(\Delta h\). In \(mg\Delta h\) the height is always vertical: for a slope of length \(L\) at angle \(\theta\) to the horizontal, \(\Delta h = L\sin\theta\).
How do you use the principle of conservation of energy?
No resistive forces
Loss of gravitational potential energy = gain in kinetic energy: \(mg\Delta h = \tfrac12 mv^2 - \tfrac12 mu^2\). The mass cancels, giving \(v^2 = u^2 + 2g\Delta h\), whatever the shape of the path.
With resistive forces
Friction and air resistance do negative work on the body. The energy is not destroyed — it is transferred to the surroundings, mostly as thermal energy (and some sound).
work done against resistive forces = average resistive force × distance moved along the path
This balance is not in the formulae list — derive it from the three equations that are.
A reliable method
- Mark the start and end points and choose the zero of height.
- Write the energy balance in words first ("loss of GPE = gain in KE + energy transferred to the surroundings").
- Substitute the stores at the start and the end; solve for the unknown.
A roller-coaster car and its riders have a total mass of 650 kg. The car passes point A, 32.0 m above the ground, at 2.0 m s⁻¹, then runs down to point B, 4.0 m above the ground. The track from A to B is 48 m long. (a) Calculate the speed at B if resistive forces are negligible. (b) The speed measured at B is 22.0 m s⁻¹. Calculate the energy transferred to the surroundings between A and B, and the average resistive force.
(a) Loss of GPE = gain in KE; the mass cancels: \(v^2 = 2.0^2 + 2 \times 9.81 \times (32.0 - 4.0) = 553.4\), so \(v = \) 23.5 m s⁻¹
(b) Loss of GPE from A to B: \(650 \times 9.81 \times 28.0 = 178\,542\ \text{J}\). Gain in KE: \(\tfrac12 \times 650 \times (22.0^2 - 2.0^2) = 156\,000\ \text{J}\). The difference has gone to the surroundings: \(178\,542 - 156\,000 = 22\,542\ \text{J} \approx\) 2.25 × 10⁴ J
Work done against resistive forces = average force × distance along the track: \(F = 22\,542 \div 48 = 469.6\ \text{N} \approx\) 470 N
Pendulums
- A pendulum bob swings on an arc; the string pulls at 90° to the motion, so the tension does no work.
- Only the vertical drop matters: released at angle \(\theta\) on a string of length \(L\), the bob falls \(h = L(1 - \cos\theta)\) to its lowest point.
- At the lowest point \(E_{\text{k}}\) is greatest; at the ends of the swing \(E_{\text{k}} = 0\) and \(E_{\text{grav}}\) is greatest.
- In a real pendulum the swings die away as energy is transferred to the surroundings by air resistance.
For a speed with no resistive forces you do not need the mass — it cancels. For the energy transferred to the surroundings, or a resistive force, you do. If a question gives you a mass, check which kind of question it is.
How does energy give the speed of a projectile?
With no air resistance, a projectile's speed at any height follows from energy alone: \(v^2 = u^2 + 2g\Delta h\), where \(\Delta h\) is the vertical drop. The launch angle does not matter for the speed: kinetic energy is a scalar and the loss of \(E_{\text{grav}}\) depends only on the drop. Energy gives the speed but not the direction or the time. For those, use the component method from 1.1.
A stone is thrown from the top of a cliff 18.0 m above the sea with a speed of 12.0 m s⁻¹. Air resistance is negligible. (a) Calculate the speed at which the stone hits the sea. (b) Show that a launch at 40° above the horizontal gives the same answer using velocity components.
(a) Loss of GPE = gain in KE: \(v^2 = 12.0^2 + 2 \times 9.81 \times 18.0 = 144 + 353.2 = 497.2\), so \(v = \) 22.3 m s⁻¹
(b) Components at launch: \(u_{\text{H}} = 12.0\cos 40^\circ = 9.19\ \text{m s}^{-1}\), \(u_{\text{V}} = 12.0\sin 40^\circ = 7.71\ \text{m s}^{-1}\). Vertical speed at the sea (18.0 m below), horizontal speed unchanged: \(v_{\text{V}}^2 = 7.71^2 + 2 \times 9.81 \times 18.0\), so \(v_{\text{V}} = 20.3\ \text{m s}^{-1}\). Then \(v = \sqrt{9.19^2 + 20.31^2} = \) 22.30 m s⁻¹ — the same speed
Exam tips for 1.5
For work done, sketch the force and the displacement and mark the angle between them. Work done = (component of force along the displacement) × displacement. Forces at 90° to the motion, such as the normal contact force on level ground, contribute nothing.
In \(mg\Delta h\), \(\Delta h\) is the vertical change in height. For a slope, \(\Delta h = L\sin\theta\); for a pendulum, \(\Delta h = L(1 - \cos\theta)\). The distance along the path goes into the work done against resistive forces instead.
"Loss of GPE = gain in KE + work done against friction" makes clear which store is which, and means a slip in one term need not cost the method. Then substitute.
Masses in kg (not g), speeds in m s⁻¹ (convert km h⁻¹ by dividing by 3.6), forces in N (not kN). Kinetic energy depends on \(v^2\), so a unit slip in the speed is squared too.
In a "show that", give your result to one more significant figure than the value quoted. In an "explain" about energy, name the stores and the transfer: "gravitational potential energy is transferred to kinetic energy, and some is transferred to the surroundings as thermal energy by work done against air resistance".
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| Every force on a moving body does work on it. | A force at 90° to the displacement has no component along it. | Only the component along the displacement does work: \(\Delta W = F\Delta s\cos\theta\). |
| Holding a heavy box still is hard work, so work is done on the box. | The box does not move, so \(\Delta s = 0\). | No work is done on the box; the energy your muscles use stays in your body as thermal energy. |
| Doubling the speed doubles the kinetic energy. | \(E_{\text{k}}\) depends on \(v^2\). | Double the speed gives four times the kinetic energy and four times the braking distance. |
| A longer, gentler ramp gives the load more potential energy. | \(\Delta E_{\text{grav}}\) depends only on the vertical height change. | Same \(\Delta h\), same \(\Delta E_{\text{grav}}\); the ramp only reduces the force needed. |
| Friction destroys energy. | Energy is conserved. | Friction transfers energy to the surroundings, mostly as thermal energy. |
| A heavier object reaches the bottom of a smooth slope faster. | Both \(mg\Delta h\) and \(\tfrac12 mv^2\) are proportional to \(m\). | With no resistive forces the mass cancels: \(v = \sqrt{2g\Delta h}\) from rest. |
How this topic connects
1.1 Motion, vectors and projectiles
\(v^2 = u^2 + 2as\) multiplied by \(\tfrac12 m\) is the work–energy statement: \(Fs = \tfrac12 mv^2 - \tfrac12 mu^2\). Energy gives a projectile's speed; components give its direction and time.
Power and efficiency (next page) and Materials (Unit 1)
Power is the rate of doing work, and efficiency compares the useful work done with the total energy supplied — both build directly on the work done calculated here. The area under a force–extension graph is the work done in stretching a material — the same "area under force against distance" idea.
Momentum and collisions (1.3 and Unit 4)
Momentum is conserved in every collision, but kinetic energy is conserved only in an elastic one; in the others some is transferred to the surroundings.
Key equations
| Equation | Use | In the formulae list? |
|---|---|---|
| \(\Delta W = F\Delta s\) | work done by a force along the displacement | Yes |
| \(\Delta W = F\Delta s\cos\theta\) | force at angle \(\theta\) to the displacement | No — learn it (resolve the force) |
| work done = area under \(F\)–\(s\) graph | varying force | No — learn it |
| \(E_{\text{k}} = \tfrac12 mv^2\) | kinetic energy | Yes |
| \(\Delta E_{\text{grav}} = mg\Delta h\) | change in gravitational potential energy near the Earth's surface | Yes |
| \(v^2 = u^2 + 2g\Delta h\) | speed after a vertical drop \(\Delta h\), no resistive forces, any path | No — derive it from \(mg\Delta h = \Delta(\tfrac12 mv^2)\) |
| \(mg\Delta h = \Delta E_{\text{k}} + F_{\text{r}}d\) | energy balance with an average resistive force \(F_{\text{r}}\) over a path length \(d\) | No — derive it |
| \(h = L(1 - \cos\theta)\) | vertical drop of a pendulum bob | No — derive it (trigonometry) |
Check your understanding
1. Define work done and give its SI unit in base units.
Work done is the force multiplied by the distance moved in the direction of the force. Unit: \(\text{J} = \text{N m} = \text{kg m}^2\,\text{s}^{-2}\).
2. A 2.0 kg bag of shopping is lifted 1.5 m onto a table. Calculate its gain in gravitational potential energy.
\(\Delta E_{\text{grav}} = 2.0 \times 9.81 \times 1.5 = \) 29 J.
3. Calculate the kinetic energy of a 0.16 kg cricket ball moving at 30 m s⁻¹.
\(E_{\text{k}} = \tfrac12 \times 0.16 \times 30^2 = \) 72 J.
4. A waiter carries a tray across a level room at constant speed and constant height. How much work does the waiter's upward force do on the tray?
None. The upward force is at 90° to the horizontal displacement, so it has no component along the motion.
5. A 12 kg sledge slides from rest down a snowy slope with a vertical drop of 6.0 m and reaches 9.0 m s⁻¹. How much energy is transferred to the surroundings?
Loss of GPE \(= 12 \times 9.81 \times 6.0 = 706\ \text{J}\); gain in KE \(= \tfrac12 \times 12 \times 9.0^2 = 486\ \text{J}\); energy transferred \(= 706 - 486 = \) 220 J.
Specification checklist
- (17) I can calculate the work done by a force from force × displacement, using the component of the force along the displacement when the force is at an angle to the motion.
- (17) I can find the work done from the area under a force–displacement graph.
- (18) I can calculate kinetic energy from mass and speed, and use work done = change in kinetic energy, for example for braking distances.
- (19) I can calculate a change in gravitational potential energy near the Earth's surface from mass, \(g\) and the vertical change in height.
- (20) I can apply conservation of energy to problems involving work done, kinetic energy and gravitational potential energy, including energy transferred to the surroundings by resistive forces.