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Work, energy and power

A-Level Physics · Topic 5

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5.1

Work, energy and power

Syllabus
  1. understand the concept of work, and recall and use $\text{work done} = \text{force} \times \text{displacement in the direction of the force}$
  2. recall and apply the principle of conservation of energy
  3. recall and understand that the efficiency of a system is the ratio of useful energy output from the system to the total energy input
  4. use the concept of efficiency to solve problems
  5. define power as work done per unit time
  6. solve problems using $P = W/t$
  7. derive $P = Fv$ and use it to solve problems

Source: Cambridge International syllabus

A row of wind turbines Wind turbines transfer the kinetic energy of the wind into electrical energy.

Work done by a force

Work is done when a force moves its point of contact along the line of the force. The work done by a constant force $F$ that causes a displacement 位移 $s$ is

$$W = F \cdot s \cdot \cos\theta,$$

where $\theta$ is the angle between the force and the displacement. Only the component 分量 of the force along the displacement does work.

A block on a surface pulled by a force F at angle theta to the horizontal displacement s; F is split into the component F cos theta along s and the perpendicular component F sin theta, and only F cos theta does work Only the component of the force along the displacement ($F\cos\theta$) does work

Worked example. A child pulls a sledge $5.0\ \text{m}$ across the snow with a rope, using a force of $20\ \text{N}$ at $60°$ to the ground. Find the work done by the rope.

$$W = Fs\cos\theta = 20 \times 5.0 \times \cos 60° = 20 \times 5.0 \times 0.50 = 50\ \text{J}.$$

Unit: $\text{J} = \text{N m}$. Work is a scalar 标量.

Special cases:

  • force in the same direction as the motion ($\theta = 0$): $W = Fs$, positive work, energy 能量 given to the object.
  • force at right angles to the motion ($\theta = 90°$): $W = 0$. The normal contact force 支持力 on a car on a flat road does no work.
  • force opposite to the motion ($\theta = 180°$): $W = -Fs$, negative work, energy taken from the object (for example friction 摩擦力).

Two diagrams: top, the force F points along the displacement s, giving positive work W = +Fs and energy to the object; bottom, the force F points opposite to the displacement s, giving negative work W = -Fs and taking energy from the object Positive work ($W = +Fs$): force along the motion (top). Negative work ($W = -Fs$): force opposite to the motion, e.g. friction (bottom)

For an object moving up a slope at angle $\alpha$ to the horizontal 水平, the work done against gravity in rising a height $h$ is $mgh$, while the work done by a horizontal push over the slope length $L$ uses $\cos\alpha$.

Conservation of energy

Energy is never made or destroyed — it only changes from one form to another, or moves from one object to another. In a closed system 封闭系统, the total energy stays constant. This is conservation of energy 能量守恒.

When you write an energy equation, list every form the energy starts as and ends as. Common forms in this syllabus: kinetic, gravitational potential, elastic potential energy 弹性势能, electrical 电能, thermal, sound, chemical energy 化学能.

A ball rolling down a frictionless 无摩擦 ramp 斜坡 turns gravitational potential energy 重力势能 into kinetic energy 动能: $mgh = \tfrac{1}{2} m v^{2}$, so $v = \sqrt{2gh}$. With friction, some of this energy becomes thermal energy 热能 of the ramp and the air.

Three energy bars for a ball descending a frictionless ramp: at the top all the energy is GPE, at the middle it is half GPE and half KE, at the bottom it is all KE — the total height stays the same On a frictionless ramp, GPE turns into KE while the total energy stays constant

Worked example. A ball is released from rest at the top of a smooth ramp $1.2\ \text{m}$ high. Find its speed at the bottom (take $g = 9.81\ \text{m s}^{-2}$).

All the gravitational potential energy becomes kinetic energy, so $v = \sqrt{2gh}$ (the mass cancels):

$$v = \sqrt{2 \times 9.81 \times 1.2} \approx 4.9\ \text{m s}^{-1}.$$

Efficiency

The efficiency 效率 of a system is

$$\text{efficiency} = \frac{\text{useful energy output}}{\text{total energy input}} \times 100\%.$$

The same idea with power:

$$\text{efficiency} = \frac{\text{useful power output}}{\text{total power input}} \times 100\%.$$

Efficiency is always less than 100% in a real system, because some input energy becomes "useless" forms — usually thermal energy.

Of the total energy input, a thick arrow shows the useful output and a thinner arrow the energy wasted as heat; efficiency is the useful output divided by the total input Efficiency: only part of the energy input leaves as useful output; the rest is wasted, usually as heat

For an electric motor lifting a load with efficiency $\eta$ at voltage 电压 $V$ and current 电流 $I$, the useful output power is $\eta V I$. From this you can find a force, a lifting speed, or a tension 张力.

Worked example. An electric motor lifts a $50\ \text{kg}$ load at a steady $0.40\ \text{m s}^{-1}$ while drawing $250\ \text{W}$ of electrical power. Find its efficiency (take $g = 9.81\ \text{m s}^{-2}$).

The useful output power is $P = mgv = 50 \times 9.81 \times 0.40 = 196\ \text{W}$, so

$$\text{efficiency} = \frac{196}{250} \times 100\% \approx 78\%.$$
Explore

Energy flow & efficiency

The input energy divides into useful work and wasted energy; efficiency = useful ÷ input, and useful + wasted always equals the input.

Explore

Work, energy & power

PE + KE = constant

Work transfers energy; as it falls, PE becomes KE with the total fixed.

Vocabulary Train
English Chinese Pinyin
work gōng
displacement 位移 wèi yí
component 分量 fèn liàng
scalar 标量 biāo liàng
energy 能量 néng liàng
normal contact force 支持力 zhī chí lì
friction 摩擦力 mó cā lì
horizontal 水平 shuǐ píng
closed system 封闭系统 fēng bì xì tǒng
conservation of energy 能量守恒 néng liàng shǒu héng
elastic potential energy 弹性势能 tán xìng shì néng
electrical energy 电能 diàn néng
chemical energy 化学能 huà xué néng
frictionless 无摩擦 wú mó cā
ramp 斜坡 xié pō
gravitational potential energy 重力势能 zhòng lì shì néng
kinetic energy 动能 dòng néng
thermal energy 热能 rè néng
efficiency 效率 xiào lǜ
voltage 电压 diàn yā
current 电流 diàn liú
tension 张力 zhāng lì
5.1

Power

Power 功率 is the rate of doing work, or the rate of transferring energy:

Power is work per second: the same work done in less time means more power The same work done in less time means more power

$$P = \frac{W}{t} = \frac{\Delta E}{\Delta t}.$$

Unit: $\text{W} = \text{J s}^{-1}$. Power is a scalar.

Power, force and velocity

For an object moving at velocity 速度 $v$ with a force $F$ along the direction of motion, in a short time $\Delta t$ the displacement is $v\,\Delta t$ and the work done is $F v\,\Delta t$. Dividing by $\Delta t$:

$$P = F v.$$

This is one of the most useful results in mechanics.

Worked example. A car travels at a steady $25\ \text{m s}^{-1}$ against a total resistive force of $600\ \text{N}$. Find the output power of its engine.

At constant speed the driving force equals the resistive force, so

$$P = Fv = 600 \times 25 = 15\,000\ \text{W} = 15\ \text{kW}.$$
  • For a car at constant velocity $v$ on a flat road, the engine power must balance the total resistive force: $P = F_{\text{resist}} \cdot v$. If the drag 阻力 grows with $v^{2}$, doubling the speed roughly quadruples the power needed.
  • For lifting a weight 重力 $mg$ straight up at constant speed $v$, the useful output power is $P = mg \cdot v$.
  • For an aircraft hovering at a fixed height, the lift force equals the weight, and a large power is needed because air must be pushed downwards all the time.
Vocabulary Train
English Chinese Pinyin
power 功率 gōng lǜ
velocity 速度 sù dù
drag 阻力 zǔ lì
weight 重力 zhòng lì
5.2

Gravitational potential energy

Syllabus
  1. derive, using $W = Fs$, the formula $\Delta E_{\text{P}} = mg\Delta h$ for gravitational potential energy changes in a uniform gravitational field
  2. recall and use the formula $\Delta E_{\text{P}} = mg\Delta h$ for gravitational potential energy changes in a uniform gravitational field
  3. derive, using the equations of motion, the formula for kinetic energy $E_{\text{K}} = \frac{1}{2}mv^2$
  4. recall and use $E_{\text{K}} = \frac{1}{2}mv^2$

Source: Cambridge International syllabus

Energy exchange in a pendulum

In a uniform gravitational field (close to a planet's surface), the change in gravitational potential energy of mass 质量 $m$ rising or falling through a height $\Delta h$ is

$$\Delta E_{\text{P}} = m g \Delta h.$$

Where it comes from

The work done against gravity to raise a mass $m$ slowly (no change in kinetic energy) through height $\Delta h$ equals the gravitational potential energy gained:

  • the gravitational force on the mass is $mg$ downwards,
  • the force needed to lift it slowly is $mg$ upwards,
  • the work done by this force is $W = F \cdot s = mg \cdot \Delta h$,
  • this work becomes $\Delta E_{\text{P}}$.

So $\Delta E_{\text{P}} = mg \Delta h$. To use it you need $m$ and $\Delta h$ (and $g$). You do not need speed or time.

A steep path and a gentle zig-zag path both rise to the same height h; the gain in gravitational PE is mgh either way, because only the height matters Gravitational PE depends only on the height risen, not the path taken

Vocabulary Train
English Chinese Pinyin
mass 质量 zhì liàng
Exercise sheet
5.2

Kinetic energy

The kinetic energy of an object of mass $m$ moving at speed $v$ is

$$E_{\text{K}} = \tfrac{1}{2} m v^{2}.$$

Where it comes from

Apply a resultant force $F$ to a mass $m$ that starts at rest. It speeds up evenly from $0$ to $v$ over a displacement $s$. From $v^{2} = u^{2} + 2as$ with $u = 0$,

$$s = \frac{v^{2}}{2a}.$$

The work done on the mass is

$$W = F \cdot s = m a \cdot \frac{v^{2}}{2a} = \tfrac{1}{2} m v^{2}.$$

All this work becomes kinetic energy, so $E_{\text{K}} = \tfrac{1}{2} m v^{2}$.

Kinetic energy and momentum

Combining $p = mv$ and $E_{\text{K}} = \tfrac{1}{2} m v^{2}$:

$$E_{\text{K}} = \frac{p^{2}}{2m}.$$

This is handy when the momentum 动量 is given but not the velocity. For a momentum change from $p_{1}$ to $p_{2}$ at constant mass, the change in kinetic energy is $(p_{2}^{2} - p_{1}^{2}) / (2m)$.

Explore

Kinetic & potential energy

PE + KE = constant

Potential energy turns into kinetic energy — the total never changes.

Vocabulary Train
English Chinese Pinyin
momentum 动量 dòng liàng
5.2

Using energy methods

A useful plan for problems that mix forces and energy:

  1. Find the start and end states. Write the kinetic and potential energies in each.
  2. List any work done by outside forces (friction, a push). Friction usually takes energy out; a push can add it.
  3. Conservation of energy: $E_{\text{start}} + W_{\text{in}} = E_{\text{end}} + W_{\text{lost as heat etc.}}$.

Examples:

  • A box pushed at constant velocity up a ramp of length $L$ rising by $h$: $E_{\text{K}}$ does not change, so the work done by the push goes into $\Delta E_{\text{P}}$ plus the work done against friction.
  • A block sliding into a spring 弹簧 with kinetic energy $E_{\text{K}}$ on a frictionless surface: at greatest compression 压缩 $x$, all the kinetic energy has become elastic potential energy $\tfrac{1}{2} k x^{2}$ (where $k$ is the spring constant 劲度系数).
  • A ball dropped from height $h_{1}$ that bounces to height $h_{2}$: the ratio $h_{2}/h_{1}$ is the fraction of mechanical energy kept, $h_{2}/h_{1} = (v_{\text{up}}/v_{\text{down}})^{2}$.
  • A projectile 抛体 thrown to the same height at different angles: the final speed is the same (only the height matters); use components to get its direction.

A block moving at speed v (kinetic energy one-half m v squared) slides into a spring; at greatest compression x all that energy has become elastic PE one-half k x squared A block's kinetic energy becomes elastic potential energy as it compresses the spring

Explore

Conservation of energy

Drop the mass and watch GPE turn into KE. With no friction the total energy stays the same — that's the energy method.

Vocabulary Train
English Chinese Pinyin
spring 弹簧 tán huáng
compression 压缩 yā suō
spring constant 劲度系数 jìn dù xì shù
projectile 抛体 pāo tǐ
5.2

Exam tips

  • Work $=$ force $\times$ distance moved in the direction of the force — use $Fs\cos\theta$ when they are at an angle.
  • Use conservation of energy: loss in GPE $=$ gain in KE ($+$ work done against resistance).
  • Power $=$ work $/$ time $= Fv$; efficiency $=$ useful output $\div$ total input.
  • GPE uses the vertical height gained, not the distance along a slope.

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