A-Level Physics · ⁨A-Level Физика⁩

Centre of gravity

The weight 重力 of a large object can be treated as acting at one single point, called the centre of gravity 重心 (the same as the centre of mass 质心 in a uniform gravitational field). For a uniform, regular shape — a rectangle, a sphere, a uniform rod — the centre of gravity is at the middle.

When you draw a free-body diagram 受力图, always put the weight arrow at the centre of gravity.

For an irregular shape, find the centre of gravity by experiment. Hang the object freely from a pin: in equilibrium it settles with its centre of gravity vertically below the pin, because only then does the weight have no moment about the pin. A plumb line 铅垂线 hung from the same pin marks that vertical. Hang the object from a second point and draw the second line; the centre of gravity is where the two lines cross. The same fact answers the exam's "explain why the sheet hangs like this": its centre of gravity is directly below the support, so the weight's line of action passes through the pivot and there is no resultant torque.

For a non-uniform 非均匀 bar, the centre of gravity is not at the middle. Find it by balancing the bar on a pivot, or by taking moments: if the bar's weight $W$ balances a known weight, the distance of its centre of gravity from the pivot is the one moment that makes the two sides equal.

Moment of a force

The moment 力矩 of a force about a point is

where $F$ is the size of the force 力 and $d$ is the perpendicular distance 垂直距离 from the point to the line of action 作用线 of the force. Unit: $\text{N m}$.

If the force acts at angle $\theta$ to a lever arm 力臂 of length $r$ from the pivot 支点, then $d = r\sin\theta$, so $M = F r\sin\theta$. Only the part of the force perpendicular 垂直 to the lever arm makes it turn.

Worked example. A force of $50\ \text{N}$ is applied to the end of a spanner $0.40\ \text{m}$ long, at $30°$ to the spanner. Find the moment of the force about the nut.

A moment is either clockwise 顺时针 or anticlockwise 逆时针 about the chosen point.

Couple and torque

A couple 力偶 is a pair of forces that are:

• equal in size,
• opposite in direction,
• with their lines of action a perpendicular distance apart.

A couple makes the body turn only — its resultant force is zero, so it gives no straight-line acceleration.

The torque 力偶矩 of a couple is the turning effect it makes:

where $F$ is the size of one force and $d$ is the perpendicular distance between the two lines of action. Unit: $\text{N m}$. The torque is the same about any point — a special property of couples.

A common multiple-choice trap: two equal forces in the same direction are not a couple (they have a resultant force and cause translation 平动). A couple needs equal size and opposite direction.

Watch the distance in the formula. For a couple of two forces $F$ each acting a distance $d$ from a central pivot, on opposite sides, the perpendicular distance between the two lines of action is $2d$, so the torque is $F \times 2d = 2Fd$. Taking moments about the pivot gives the same answer, $Fd + Fd$. The two-mark definition wants both parts: the product of one of the forces and the perpendicular distance between the lines of action of the forces.

Conditions for equilibrium

A body is in equilibrium 平衡 when:

1. the resultant force 合力 is zero (no straight-line acceleration), AND
2. the resultant moment about any point is zero (no angular acceleration 角加速度).

Both must hold. A body with no resultant force can still be turning; a body with no resultant moment can still be moving in a straight line.

The classic multiple-choice case: four forces, two equal pairs pointing in opposite directions but not along the same lines. The resultant force is zero, so the centre of mass does not accelerate, but the pairs form couples with a resultant torque, so the object spins faster and faster while its centre of mass moves at constant velocity. Only when both conditions hold is the body in equilibrium.

Principle of moments

For a body that is not turning, the total clockwise moment about any point equals the total anticlockwise moment about the same point. This is the principle of moments 力矩原理.

To solve a balance problem:

1. Choose a pivot — usually where an unknown force acts, so that force drops out (its distance is zero).
2. List every force and its perpendicular distance from the pivot.
3. Set $\sum M_{\text{clockwise}} = \sum M_{\text{anticlockwise}}$.
4. Use $\sum F = 0$ if you need a second equation.

A ruler balanced on a pivot with masses on each side is solved this way. For a heavy uniform rod, remember to include its weight acting at its centre of gravity. Stated for the marks: for a body in equilibrium, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that same point.

Worked example. A uniform beam of length $4.0\ \text{m}$ is pivoted at its centre. A $40\ \text{N}$ weight hangs $1.5\ \text{m}$ from the pivot on one side. How far from the pivot, on the other side, must a $30\ \text{N}$ weight hang to balance the beam?

The beam's own weight acts at the centre (the pivot), so it has no moment. Setting clockwise $=$ anticlockwise moments:

Worked example. A uniform rod of length $L$ and weight $W = 28\ \text{N}$ is hinged to a wall at one end and held horizontal by a wire from its other end. The wire makes $40°$ with the rod. Find the tension $T$ in the wire.

Take moments about the hinge, so the unknown hinge force drops out. The weight acts at the centre, a distance $L/2$ from the hinge: its moment is $W \times L/2$, clockwise. The tension acts at the end, but not at right angles to the rod, so its moment is $T\sin 40° \times L$, anticlockwise (only the component of $T$ perpendicular to the rod turns it). Equate them:

The length $L$ cancels, which is why the question need not give it. If a load hangs from the rod as well, add its moment on the clockwise side. This is the shape of almost every moments question: choose the pivot where the unknown force acts, use the perpendicular component of any angled force, and let unknown lengths cancel.

Worked example. A uniform beam of weight $200\ \text{N}$ and length $9.0\ \text{m}$ rests on two supports, X at $1.0\ \text{m}$ from the left end and Y at $6.5\ \text{m}$. A person of weight $700\ \text{N}$ stands $1.7\ \text{m}$ from Y, beyond it. Find the force from each support.

Two unknown forces, $R_{\text{X}}$ and $R_{\text{Y}}$, so take moments about X to remove $R_{\text{X}}$. The beam's weight acts at its centre, $3.5\ \text{m}$ from X; Y is $5.5\ \text{m}$ from X; the person is $7.2\ \text{m}$ from X:

Then the resultant force is zero: $R_{\text{X}} + R_{\text{Y}} = 200 + 700$, so $R_{\text{X}} = 900 - 1040 = -140\ \text{N}$. A negative answer means the beam would lift off X: with the person that far past Y, X must hold the beam down, not up. Questions often ask where a person can stand before the beam tips; that is the position at which $R_{\text{X}}$ becomes zero.

Vector triangle

Three forces in the same plane that are in equilibrium can be drawn as a closed vector triangle 矢量三角形 — drawn tip-to-tail, the three arrows come back to the start. This is a drawing method instead of splitting into components 分量.

Use the sine rule 正弦定理 or the cosine rule 余弦定理 on the triangle to find unknown sizes or directions, or draw the triangle to scale on graph paper.

You can also split each force into horizontal 水平 and vertical 竖直 components and set $\sum F_{x} = 0$ and $\sum F_{y} = 0$.

To draw the triangle for the marks: draw the weight first (vertical, to scale), then add the other two forces tip to tail in their real directions, so the third arrow closes the triangle; label every side with its force and mark the angles. For a block resting on a slope the three forces are the weight (vertical), the normal contact force (perpendicular to the slope) and friction (along the slope), so the triangle is right-angled with the weight as the hypotenuse. For a picture hanging from a string over a pin, the two equal tensions make an isosceles triangle with the weight. The test of any answer is that the triangle closes; if it does not, the object is not in equilibrium.

Worked example. A pulley of negligible weight is held by a spring. A single cable passes under the pulley, and each side of the cable makes $30°$ with the vertical. The tension in the cable is $T = 60\ \text{N}$. Find the force from the spring and the extension of the spring, given its spring constant is $2000\ \text{N m}^{-1}$.

Because it is one continuous cable, the tension is the same on both sides. The two tensions pull down and outwards; their horizontal components cancel and their vertical components add, so the spring must pull up with $F = 2T\cos 30° = 2 \times 60 \times 0.866 = 104\ \text{N}$. From Hooke's law (topic 6), $x = F / k = 104 / 2000 = 0.052\ \text{m}$. If the angle to the vertical grows, $\cos\theta$ falls, so the spring force and extension fall: the cable is pulling more sideways and less upwards.

Density 密度 is the mass per unit volume:

Unit: $\text{kg m}^{-3}$ (or $\text{g cm}^{-3}$; $1\ \text{g cm}^{-3} = 1000\ \text{kg m}^{-3}$). Density is a scalar 标量.

Some useful densities to know:

• water: $1000\ \text{kg m}^{-3}$
• air at room conditions: $\sim 1.2\ \text{kg m}^{-3}$
• iron / steel: $\sim 7800\ \text{kg m}^{-3}$

Measuring density is a practical question in disguise. For a cuboid, measure the mass on a balance and the three side lengths with a rule or calipers, then $\rho = m / (xyz)$. Because density is a product and quotient, the percentage uncertainty in $\rho$ is the sum of the percentage uncertainties in $m$, $x$, $y$ and $z$ (topic 1). A zero error on the balance, or a rule whose end is worn, is a systematic error; repeating the readings does not remove it, but subtracting the zero reading does.

Worked example. A block has mass $(120 \pm 1)\ \text{g}$ and sides $(4.0 \pm 0.1)\ \text{cm}$, $(3.0 \pm 0.1)\ \text{cm}$ and $(2.0 \pm 0.1)\ \text{cm}$. Find its density and the percentage uncertainty.

$\rho = 120 / (4.0 \times 3.0 \times 2.0) = 5.0\ \text{g cm}^{-3} = 5000\ \text{kg m}^{-3}$. The percentage uncertainties are $0.8\%$, $2.5\%$, $3.3\%$ and $5.0\%$, so the density is uncertain by $12\%$, about $\pm 600\ \text{kg m}^{-3}$. The shortest side contributes most, which is why a question asks which measurement to improve first.

Pressure 压强 is the force per unit area, where the force acts at right angles to the surface:

Unit: $\text{Pa} = \text{N m}^{-2}$. Pressure is a scalar.

The force is the one pressing at right angles to the surface, and the area is the area of contact. A tree trunk of weight $9.0\ \text{kN}$ standing on a post of diameter $0.28\ \text{m}$ presses with $p = F / A = 9000 / (\pi \times 0.14^{2}) = 1.5 \times 10^{5}\ \text{Pa}$; the same weight on a wider post gives a smaller pressure. Convert diameters to radii, and centimetres to metres, before squaring.

Hydrostatic pressure

Take a column of fluid 流体 with density $\rho$, cross-sectional area 横截面积 $A$ and height $\Delta h$. Its weight is

This weight presses down on the area $A$ at the bottom, so the extra pressure at the bottom compared with the top is

This is the hydrostatic pressure 流体静压强 equation. It depends only on the density of the fluid and the depth 深度 — the shape of the container does not matter.

Worked example. Find the extra pressure due to the water at the bottom of a swimming pool $2.5\ \text{m}$ deep. (Water density $1000\ \text{kg m}^{-3}$, $g = 9.81\ \text{m s}^{-2}$.)

For a submarine at depth $h$ below the surface, the pressure from the water is $\rho_{\text{seawater}}\, g\, h$. For the total pressure, add the atmospheric pressure 大气压强 at the surface (about $1.0 \times 10^{5}\ \text{Pa}$).

Upthrust and Archimedes' principle

When an object is submerged 浸没 in a fluid, the pressure at the bottom of the object is greater than the pressure at the top (by $\rho g \Delta h$, where $\Delta h$ is the object's height). This difference gives a net upward force called the upthrust 浮力.

For an object of volume $V$ (the volume of fluid displaced 排开), the upthrust is

This is Archimedes' principle 阿基米德原理: the upthrust on a body in a fluid equals the weight of the fluid it pushes aside.

Worked example. A metal block of volume $2.0 \times 10^{-3}\ \text{m}^{3}$ is fully submerged in water. Find the upthrust on it. (Water density $1000\ \text{kg m}^{-3}$, $g = 9.81\ \text{m s}^{-2}$.)

For a fully submerged object, $V$ is its full volume. For a floating 漂浮 object, $V$ is only the volume below the surface — the object floats when the upthrust on the part below the surface equals its weight.

Whether an object floats at all depends on densities: for a fully submerged object the ratio of upthrust to weight is $\rho_{\text{fluid}} V g / \rho_{\text{object}} V g = \rho_{\text{fluid}} / \rho_{\text{object}}$, so it floats if it is less dense than the fluid and sinks if it is denser. Halving the object's density doubles that ratio; changing its volume or the depth changes nothing.

Worked example. A cylinder of mass $11\ \text{kg}$ and diameter $0.78\ \text{m}$ floats upright in water of density $990\ \text{kg m}^{-3}$. Find the depth $y$ of its base below the surface.

Floating means upthrust $=$ weight. The submerged volume is the base area times the depth, $A y$, with $A = \pi \times 0.39^{2} = 0.478\ \text{m}^{2}$:

Each extra newton of load on the cylinder needs an extra $1 / (\rho g A)$ of depth, so the depth rises in a straight line with the added weight, wherever on the cylinder it is placed.

Force balance with upthrust

A block held under water by a string tied to the bottom of the container is in equilibrium under three vertical forces: weight (down), tension 张力 (down), upthrust (up). Set $F_{\text{upthrust}} = W + T$ to find the tension.

A submerged block hanging from a newton meter 弹簧测力计 reads less than its weight in air, because of the upthrust: reading $= W - F_{\text{upthrust}}$.

Worked example. A cylinder of weight $25.0\ \text{N}$ hangs from a newton meter fully submerged in water, and the meter reads $10.0\ \text{N}$. Find the volume of the cylinder.

The upthrust is the missing $15.0\ \text{N}$, and upthrust $= \rho g V$, so $V = 15.0 / (1000 \times 9.81) = 1.53 \times 10^{-3}\ \text{m}^{3}$. The same method finds the density of the cylinder: $25.0 / (9.81 \times 1.53 \times 10^{-3}) = 1670\ \text{kg m}^{-3}$.

The upthrust depends on the fluid density and the displaced volume, not on the object's material or depth (for an incompressible 不可压缩 fluid). On a planet with smaller $g$, the upthrust is smaller in the same ratio as the weight, so a floating object still floats with the same fraction below the surface.

A definition question is marked against fixed wording. Learn these exactly, and give one answer only.

• Take moments about a chosen pivot; for equilibrium, total clockwise $=$ total anticlockwise moments and the resultant force is zero.
• A body in equilibrium under three forces gives a closed triangle of forces.
• Fluid pressure $= \rho g h$; upthrust $=$ weight of fluid displaced (Archimedes).
• Distinguish mass, weight and density, and give the base unit each time.

Common mistakes

• Resolving with $\cos\theta$ where the geometry needs $\sin\theta$. Draw the triangle and check which side the angle is next to before choosing.
• Substituting the weight of the object where the question is about the upthrust on it. Upthrust is the weight of the fluid displaced, $\rho_{\text{fluid}} V g$.
• Leaving out the weight of a uniform beam. It acts at the centre, and it has a moment about any pivot that is not at the centre.
• Using the distance along the rod for an angled force. The moment needs the perpendicular distance, or the perpendicular component of the force.
• Taking the torque of a couple as $Fd$ with $d$ measured from the pivot. The distance is between the two lines of action, so a symmetric couple gives $2Fd$.

Work done by a force

Work 功 is done when a force moves its point of contact along the line of the force. For the one-mark definition write: work done is the product of the force and the distance moved in the direction of the force. The work done by a constant force $F$ that causes a displacement 位移 $s$ is

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

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).

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 $\cos\alpha$.

Two multiple-choice tests of the definition: no work is done on an object that slides at constant velocity along a frictionless surface (no force along the motion), or on one that is simply held still (no distance moved); work is done when a force lifts it. And when a box is pushed at constant velocity across a rough floor, the work done by the push is $F \times d$ for the whole distance, however the path is split into stages.

Worked example. A block of mass $4.9\ \text{kg}$ is pushed at constant velocity up a rough slope of length $8.0\ \text{m}$ that rises $1.5\ \text{m}$, by a force of $12\ \text{N}$ acting along the slope. Find the work done against friction.

Constant velocity means no change in kinetic energy, so the work done by the push is shared between the gain in gravitational potential energy and the work done against friction. Work by the push: $12 \times 8.0 = 96\ \text{J}$. Gain in GPE, using the vertical rise: $4.9 \times 9.81 \times 1.5 = 72\ \text{J}$. So the work done against friction is $96 - 72 = 24\ \text{J}$, and the friction force is $24 / 8.0 = 3.0\ \text{N}$. Using the slope length in $mgh$ is the commonest error on this question.

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, stated for the marks as: energy cannot be created or destroyed; it can only be transferred from one form to another (so the total energy of a closed system is constant).

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.

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):

The same idea reads a graph. For a ball falling from height $H$ with air resistance negligible, the GPE falls in a straight line with height ($mgh$), the KE rises in a straight line ($mg(H - h)$), and the two add to the constant $mgH$ at every height.

The energy chain can pass through a spring. A block that hits a spring with kinetic energy $110\ \text{J}$ while sliding a little downhill, so that its GPE falls by a further $20\ \text{J}$ before the spring stops it, stores $110 + 20 = 130\ \text{J}$ of elastic potential energy at the greatest compression, if nothing is lost to friction. The spring's force–compression graph then gives the compression: the energy stored is the area under the line, $\tfrac{1}{2} F_{0} x_{0}$ for a spring that obeys Hooke's law (topic 6), so a maximum force of $2600\ \text{N}$ means $x_{0} = 2 \times 130 / 2600 = 0.10\ \text{m}$. At that instant the resultant force on the block is the spring force minus the component of the weight down the slope, and its acceleration follows from $F = ma$ (topic 3).

Efficiency

The efficiency 效率 of a system is

The same idea with power:

Efficiency is always less than 100% in a real system, because some input energy becomes "useless" forms — usually thermal energy. If a total energy $E$ is supplied and an amount $Q$ is wasted, the useful output is $E - Q$ and the efficiency is $(E - Q) / E$; a motor rated at $1.7\ \text{kW}$ input with efficiency $53\%$ delivers $0.53 \times 1.7 = 0.90\ \text{kW}$ of useful output.

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

Worked example. A motor raises a block with a force of $240\ \text{N}$ through a vertical distance of $12\ \text{m}$ in $60\ \text{s}$. The input power to the motor is $900\ \text{W}$. Find the useful output power and the efficiency.

Work done on the block: $W = Fd = 240 \times 12 = 2880\ \text{J}$. Useful output power: $P = W / t = 2880 / 60 = 48\ \text{W}$. Efficiency: $48 / 900 \times 100\% = 5.3\%$. Most of the input goes into thermal energy in the motor and its cable. Turned round, a lift motor of useful output $20\ \text{kW}$ raising $1500\ \text{kg}$ through $20\ \text{m}$ takes $t = mgh / P = 1500 \times 9.81 \times 20 / 20\,000 = 15\ \text{s}$.

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

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 $\Delta t$:

This is one of the most useful results in mechanics. A "derive $P = Fv$" answer needs exactly those three lines: $P = W / t$, $W = Fs$ for a force along the motion, and $s / t = v$; then use it to solve problems in which a vehicle or a load moves at constant speed.

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

• 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.
• A crane raising $600\ \text{kg}$ at a steady $12\ \text{m}$ per minute lifts at $v = 0.20\ \text{m s}^{-1}$, so its useful output power is $mgv = 600 \times 9.81 \times 0.20 = 1.2\ \text{kW}$. Convert the speed to $\text{m s}^{-1}$ first.
• A sailboat pushed by a constant wind force at constant velocity must also feel an equal resistive force from the water: constant velocity means zero resultant force, so the wind's power $Fv$ is all going into work against the water.

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

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. The two-mark derivation wants the same four lines, with the condition stated: the block is raised at constant speed, so the lifting force equals the weight $mg$, and the work done by that force over the vertical distance $\Delta h$ is $mg\Delta h$, which is the gain in gravitational potential energy. A $100\ \text{g}$ object falling $10\ \text{m}$ loses $0.100 \times 9.81 \times 10 = 9.8\ \text{J}$, about $10\ \text{J}$; the mass must be in kilograms.

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

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 $u = 0$,

The work done on the mass is

All this work becomes kinetic energy, so $E_{\text{K}} = \tfrac{1}{2} m v^{2}$. In the exam, state the assumptions as you go: the object starts from rest, the force is constant, so the acceleration is uniform, and $F = ma$ is Newton's second law.

Because $E_{\text{K}} \propto v^{2}$, quadrupling the speed multiplies the kinetic energy by sixteen: an object with $1500\ \text{J}$ at $10\ \text{m s}^{-1}$ has $24\,000\ \text{J}$ at $40\ \text{m s}^{-1}$. The same square is why a stopping force is found from energy: a ball of mass $1.2\ \text{kg}$ moving at $3.0\ \text{m s}^{-1}$ that is stopped by a cushion over $0.020\ \text{m}$ loses $\tfrac{1}{2} \times 1.2 \times 3.0^{2} = 5.4\ \text{J}$, so the average force on it is $5.4 / 0.020 = 270\ \text{N}$: the work done by the cushion equals the kinetic energy lost.

Kinetic energy and momentum

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

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 $(p_{2}^{2} - p_{1}^{2}) / (2m)$.

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 bungee jumper 蹦极者: from the platform to the point where the cord goes taut, GPE becomes KE; from there the cord stretches and takes energy as elastic potential energy, so the KE reaches its maximum where the cord's pull first equals the weight, then falls to zero at the lowest point, where GPE lost $=$ elastic PE stored (plus any thermal energy).

A definition question is marked against fixed wording. Learn these exactly, and give one answer only.

• 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.

Common mistakes

• Giving the change in height as the answer to a change-in-energy question. Finish the calculation: $\Delta E_{\text{P}} = mg\Delta h$.
• Using the distance along a slope as the height in $mg\Delta h$, or using mass where weight is needed. Use the vertical height gained, and check whether the quantity wanted is $m$ in kg or $W$ in N.
• Putting the resistive force into $P = Fv$ for a car that is accelerating. $F$ is the driving force; only at constant speed does it equal the resistive force.
• Efficiency above 100%, or efficiency as output divided by wasted. It is useful output divided by total input, and the total input is always the larger number.
• Reading a force–extension graph's gradient when the question wants the energy. Energy stored is the area under the line, not its slope.

When a force acts on a solid along its length, the object changes shape (deformation 形变). Two cases (treated as one-dimensional here):

• a tensile 拉伸 force stretches the object — it makes an extension 伸长量 $x$,
• a compressive force squeezes the object — it makes a compression 压缩, treated as a negative extension.

The applied force is the load 负载. The change from the natural length is the extension (or compression).

Both are measured from the natural, unstretched length, never from the loaded length: on a diagram of a spring under a box, the distance the spring has shortened is its compression. A spring's own mass is usually said to be negligible 可忽略的, so the only forces on it are the load and the tension it provides.

For many materials at small extensions, the extension is proportional to the load — this is Hooke's law 胡克定律. The constant that links them is the spring constant 劲度系数 $k$:

Unit of $k$: $\text{N m}^{-1}$.

Worked example. A spring stretches by $4.0\ \text{cm}$ when a $2.0\ \text{N}$ load is hung from it. Find its spring constant.

Converting the extension to metres ($4.0\ \text{cm} = 0.040\ \text{m}$):

For the one-mark "state Hooke's law", write it in words: the extension is directly proportional to the applied force (load), provided the limit of proportionality is not exceeded. The spring constant is the force per unit extension; the multiple-choice distractors are "extension per unit force" (that is $1/k$) and $\tfrac{1}{2}Fx$ (the energy stored).

Reading a graph:

• A force–extension ($F$ against $x$) graph has gradient $k$ in the Hooke's-law region.
• An extension–force ($x$ against $F$) graph has gradient $1/k$ in the Hooke's-law region.

A common trap: if a graph plots length $L$ against force, you can still find the spring constant from the gradient (since $L = L_{0} + F/k$, the gradient is $1/k$ — read it off carefully).

Worked example. A spring has length $90\ \text{mm}$ under a force of $0.80\ \text{N}$ and $115\ \text{mm}$ under $1.30\ \text{N}$. Find the spring constant and the unstretched length.

The extra force $0.50\ \text{N}$ produces the extra extension $25\ \text{mm} = 0.025\ \text{m}$, so $k = 0.50 / 0.025 = 20\ \text{N m}^{-1}$. At $0.80\ \text{N}$ the extension is $x = F/k = 0.80 / 20 = 0.040\ \text{m}$, so the unstretched length is $90 - 40 = 50\ \text{mm}$. On a graph of length against force the intercept on the length axis is $L_{0}$, the gradient is $1/k$, and the work done in stretching the spring is the triangle between the line and the level $L = L_{0}$, not the whole area down to the force axis: the rectangle below $L_{0}$ is the unstretched length multiplied by the force, which means nothing.

Limit of proportionality

Hooke's law only holds up to the limit of proportionality 比例极限. Past this point the $F$ against $x$ line curves and is no longer straight. The material may still be elastic 弹性 (it returns to its first length when you remove the load) a little further, then it becomes plastic 塑性.

Springs in series and parallel

You may need to combine spring constants:

• Series 串联 (one spring hangs from another): the same load passes through both, the total extension is the sum, so $\dfrac{1}{k_{\text{total}}} = \dfrac{1}{k_{1}} + \dfrac{1}{k_{2}}$.
• Parallel 并联 (two springs side by side holding the same load): each takes half the load (if they are identical), the extensions are equal, so $k_{\text{total}} = k_{1} + k_{2}$.

Worked example. Springs of spring constant $6.0\ \text{N cm}^{-1}$ and $4.0\ \text{N cm}^{-1}$ are joined end to end and a load of $80\ \text{N}$ hangs from them. Find the total extension.

Each spring carries the full $80\ \text{N}$, so the extensions are $80/6.0 = 13.3\ \text{cm}$ and $80/4.0 = 20\ \text{cm}$: a total of $33\ \text{cm}$ (the same as $k = 2.4\ \text{N cm}^{-1}$ from $1/k = 1/6.0 + 1/4.0$). Three identical springs give the largest combined constant when all three are in parallel ($3k$) and the smallest when all are in series ($k/3$). When a load is shared by two identical springs in parallel, each carries half the force and so stores a quarter of the energy of one spring holding the whole load; the pair together stores half as much as the single spring did. A spring constant in $\text{N cm}^{-1}$ is $100$ times smaller than the same constant in $\text{N m}^{-1}$; give the unit the question asks for.

For a wire of uniform cross-section under a tensile load:

• Stress 应力 $\sigma = \dfrac{F}{A}$, where $F$ is the load and $A$ is the cross-sectional area 横截面积. Unit: $\text{Pa}$.
• Strain 应变 $\varepsilon = \dfrac{x}{L_{0}}$, where $x$ is the extension and $L_{0}$ is the original length. Strain has no unit (it is a ratio of lengths).

For the one-mark definitions: stress is the force per unit cross-sectional area (the force acting normally to the area); strain is the extension per unit original length. The unit of stress, $\text{Pa} = \text{N m}^{-2}$, is $\text{kg m}^{-1}\ \text{s}^{-2}$ in base units. The area of a round wire comes from its diameter $d$: $A = \pi d^{2}/4 = \pi r^{2}$, and using $d$ in place of $r$ makes the area four times too big. Convert $\text{mm}^{2}$ to $\text{m}^{2}$ with $10^{-6}$.

Worked example. A tensile force of $18\ \text{N}$ acts on a wire of cross-sectional area $3.2\ \text{mm}^{2}$. Find the stress.

$\sigma = F/A = 18 / (3.2 \times 10^{-6}) = 5.6 \times 10^{6}\ \text{Pa}$. The same force acts along the whole wire, so a bolt whose diameter is $2d$ at one end and $d$ at the other has four times the stress at the narrow end ($\sigma \propto 1/d^{2}$), and a wire of three times the radius carries one ninth of the stress.

The Young modulus 杨氏模量 is the ratio of stress to strain in the Hooke's-law region:

Unit: $\text{Pa}$ (about $10^{11}$ for metals; e.g. steel $\approx 2.0 \times 10^{11}$ Pa).

Worked example. A steel wire of length $2.0\ \text{m}$ and cross-sectional area $1.5 \times 10^{-7}\ \text{m}^{2}$ stretches by $1.0\ \text{mm}$ under a load of $15\ \text{N}$. Find the Young modulus.

The Young modulus is a property of the material — it does not depend on the wire's shape or size. The spring constant $k$ depends on both the material and the size: $k = EA/L_{0}$.

For the one-mark definition: the Young modulus is the ratio of stress to strain (for a material deformed within its limit of proportionality). Because it belongs to the material, a thicker wire of the same metal has the same Young modulus: it stretches less under the same load because its area is larger, not because the metal is stiffer. A "show that $k = EA/L_{0}$" question wants two lines: $k = F/x$ and $E = FL_{0}/(Ax)$, so $E = kL_{0}/A$. Two things follow: a wire's spring constant stays constant while Hooke's law holds, whatever the force, and a wire of the same metal with twice the diameter has four times the spring constant.

Worked example. A copper wire of length $1.7\ \text{m}$ and diameter $0.64\ \text{mm}$ has Young modulus $1.2 \times 10^{11}\ \text{Pa}$. Find its spring constant.

$A = \pi (0.32 \times 10^{-3})^{2} = 3.2 \times 10^{-7}\ \text{m}^{2}$, so $k = EA/L_{0} = 1.2 \times 10^{11} \times 3.2 \times 10^{-7} / 1.7 = 2.3 \times 10^{4}\ \text{N m}^{-1}$.

On a stress–strain graph the gradient of the straight part is $E$, so of two materials drawn on the same axes the steeper line is the stiffer material, and a wire with double the Young modulus is drawn as a line through the origin with twice the gradient. On a force–extension graph the gradient is $EA/L_{0}$ and the Young modulus is the gradient multiplied by $L_{0}/A$; only a stress–strain graph has a gradient equal to $E$ itself.

Experiment to find the Young modulus of a metal wire

A standard setup:

1. Clamp one end of a long, thin wire to a fixed support. Pass the wire over a pulley 滑轮 at the edge of the bench so it hangs straight down.
2. Measure the original length $L_{0}$ between the clamp and a marker near the pulley, using a metre rule.
3. Measure the diameter 直径 $d$ of the wire at several places with a micrometer 螺旋测微器 and take the average. Work out $A = \pi d^{2}/4$.
4. Hang weights one at a time. Record the load $F$ and the extension $x$ (how far the marker moves against a fixed scale).
5. Plot $F$ against $x$. In the straight region the gradient is $EA/L_{0}$, so $E = \text{gradient} \times L_{0}/A$.

Why a long, thin wire? To make the extension big enough to measure well. Why repeat readings and measure $d$ at several places? To reduce random error 随机误差 and check the wire is uniform.

A "describe an experiment" answer earns its marks for the quantities measured and how ($L_{0}$ with a metre rule; $d$ with a micrometer at several points; the load from the masses or a newton meter; the extension from a marker read against a fixed scale, or a vernier scale), the graph ($F$ against $x$, gradient $EA/L_{0}$, or stress against strain, gradient $E$) and the precautions: safety goggles in case the wire snaps, a small load first to straighten kinks, and readings taken on unloading as well as loading to check the wire stayed elastic.

As the load grows:

1. Elastic and straight (Hooke obeyed) — up to the limit of proportionality. Removing the load returns the object to its first length.
2. Elastic but curved — between the limit of proportionality and the elastic limit 弹性极限. The extension is no longer straight in the load, but on unloading the object still returns to its first length.
3. Plastic — past the elastic limit. On unloading, the object does not return to its first length; a permanent extension stays.

Hooke's law only holds in the straight, elastic region.

In words: elastic deformation means the object returns to its original length (or shape) when the load is removed; plastic deformation means it does not, and a permanent extension remains; the elastic limit is the maximum load (or extension) for which the deformation is still elastic; the limit of proportionality is the point beyond which the extension is no longer proportional to the load. On a graph the limit of proportionality is where the line stops being straight. The elastic limit lies a little beyond it and cannot be read from a loading line alone. So in a "which statement must be correct" item, the end of the straight part is the limit of proportionality, but whether a later point is the elastic limit or the breaking point cannot be told from the shape of the loading line.

On a force–extension graph for a material taken into the plastic region and then unloaded, the loading line and the unloading line are different. The unloading line is parallel to the first Hooke line but shifted to the right (the permanent extension left when the load reaches zero). The area between the loading and unloading lines is the energy turned into thermal energy 热能 in the material.

A rubber band 橡皮筋 also has different loading and unloading curves, but it returns to its original length: its deformation is elastic. The area between the two curves is again energy dissipated as thermal energy (elastic hysteresis 弹性滞后), and the energy recovered on unloading is the area under the lower curve. The two-mark "explain why the work done in stretching the wire is not equal to the energy recovered when the force is removed" answer says that the wire was taken past its elastic limit, so part of the deformation is plastic and a permanent extension remains, and that part of the work done became thermal energy in the wire.

The work done in stretching a material from $0$ to extension $x$, as the load grows from $0$ to $F$, is the area under the force–extension graph.

Why the area: work done is force multiplied by the distance moved, but here the force grows as the material stretches, so the work is the sum of $F\,\Delta x$ over many small extensions, which is the area under the line. Within the limit of proportionality the force rises uniformly from $0$ to $F$, so the average force is $\tfrac{1}{2}F$ and the work is $\tfrac{1}{2}Fx$. On a force–extension graph the gradient is the spring constant.

Hooke's-law material

When Hooke's law holds, the $F$ against $x$ graph is a straight line through the origin. The area under it from $0$ to $x$ is a triangle:

This is the elastic potential energy 弹性势能 stored in a spring or wire stretched within its limit of proportionality. An equal form:

Worked example. A spring of spring constant $50\ \text{N m}^{-1}$ is stretched by $0.20\ \text{m}$, within its limit of proportionality. Find the elastic potential energy stored.

Worked example. A wire of spring constant $2.0 \times 10^{4}\ \text{N m}^{-1}$ is already extended by $2.0\ \text{mm}$. Find the work done to increase its extension to $3.0\ \text{mm}$.

The stored energy rises from $\tfrac{1}{2} k x_{1}^{2}$ to $\tfrac{1}{2} k x_{2}^{2}$: $W = \tfrac{1}{2} \times 2.0 \times 10^{4} \times \left[(3.0 \times 10^{-3})^{2} - (2.0 \times 10^{-3})^{2}\right] = 0.050\ \text{J}$. This is the trapezium under the line between the two extensions, not $\tfrac{1}{2} k (x_{2} - x_{1})^{2}$. Because $E_{\text{P}} \propto x^{2}$ at fixed $k$, doubling the extension of a wire stores four times the energy ($0.65\ \text{J}$ becomes $2.6\ \text{J}$), and a stored energy gives the extension as $x = \sqrt{2E_{\text{P}}/k}$: a spring of $k = 400\ \text{N m}^{-1}$ storing $0.32\ \text{J}$ is compressed by $\sqrt{2 \times 0.32 / 400} = 0.040\ \text{m}$. For two wires joined end to end the tension is the same in both, so the total energy stored is $\tfrac{1}{2}F x_{1} + \tfrac{1}{2}F x_{2}$ with each wire's own extension.

Non-Hooke material

For a graph that is not a straight line (a stretched rubber band, or a spring past its limit of proportionality), find the area by counting grid squares or by using trapezia 梯形. The same idea holds: the area under the force–extension graph is the work done on the material. To estimate the work done up to the breaking point, count the squares under the whole curve (part squares as halves) and multiply by the energy one square represents, the force step multiplied by the extension step; saying that the area was found by counting squares is the "explain your reasoning" mark.

Comparing stored energy

A common multiple-choice case: two materials are stretched by the same force, or by the same extension. Using $E_{\text{P}} = \tfrac{1}{2} F x$:

• same $F$, smaller $k$ (less stiff) → larger $x$ → more energy stored.
• same $x$, larger $k$ (stiffer) → larger $F$ → more energy stored.

A sketch of $E_{\text{P}}$ against extension or compression is a curve through the origin that gets steeper ($E_{\text{P}} \propto x^{2}$), not a straight line. Of two wires of the same length and area under the same load, the one with the smaller Young modulus extends more and so stores more energy.

When a stretched spring is released onto a mass, the elastic potential energy becomes kinetic energy 动能 (and gravitational potential energy if the mass rises). Set $\tfrac{1}{2} k x^{2}$ equal to $\tfrac{1}{2} m v^{2}$ (plus any $mgh$) to find the speed or height.

A definition question is marked against fixed wording. Learn these exactly, and give one answer only.

• Hooke's law ($F = kx$) holds only up to the limit of proportionality.
• Stress $= F/A$, strain $= x/L$, Young modulus $=$ stress$/$strain (gradient of the straight part of the stress-strain graph) — watch the units (Pa).
• Energy stored $=$ area under the force-extension graph $= \frac{1}{2}Fx$ in the elastic region.
• Distinguish elastic (returns to shape) from plastic (permanent) deformation.

Common mistakes

• Using the diameter as the radius in $A = \pi r^{2}$, or leaving an area in $\text{mm}^{2}$. Halve the diameter first; $1\ \text{mm}^{2} = 10^{-6}\ \text{m}^{2}$.
• Reading the work done off a length–force graph as the whole area down to the axis. Only the triangle above $L_{0}$ is work.
• Treating stored energy as proportional to extension. It goes as $x^{2}$: double the extension, four times the energy; and the work done between two extensions is the difference of two $\tfrac{1}{2}kx^{2}$ values.
• Stating Hooke's law without its condition. "Provided the limit of proportionality is not exceeded" is part of the law.
• Swapping the limit of proportionality (the end of the straight line) and the elastic limit (the end of elastic behaviour, a little beyond it).
• Saying a thicker wire has a larger Young modulus. The modulus is the material's; the thicker wire has a larger spring constant.

A wave 波 carries energy 能量 from one place to another without moving matter overall. The particles of the medium 介质 oscillate 振动 about fixed rest positions; only the disturbance (and its energy) propagates 传播. Examples: a transverse wave 横波 on a rope, a longitudinal wave 纵波 on a slinky spring, ripples on water, and sound in air. A wave that travels and carries energy is a progressive wave 行波.

Wave motion 波动 is a series of oscillations of the particles of a medium, each passing the disturbance on to the next: shake one end of a rope, push one end of a spring, or touch the surface of the water in a ripple tank 水波槽, and the oscillation travels away from where it started. For the one-mark definition, a progressive wave transfers energy from one place to another without transferring matter; the particles only oscillate about their rest positions.

Key terms

• displacement 位移 $y$ — how far a particle has moved from its rest position at a moment. A vector 矢量.
• amplitude 振幅 $A$ — the largest displacement from the rest position.
• wavelength 波长 $\lambda$ — the shortest distance along the wave between two points that move in phase 同相 (for example, two next-door crests 波峰).
• period 周期 $T$ — the time for one full oscillation of a particle.
• frequency 频率 $f$ — the number of full oscillations per second; $f = 1/T$. Unit: hertz 赫兹, $\text{Hz}$.
• speed 速率 $v$ — how fast a crest travels along the medium.
• phase difference 相位差 — the fraction of a cycle by which one oscillation leads or lags another. Given in radians 弧度 (a full cycle is $2\pi$) or degrees (a full cycle is $360°$).

Two points one wavelength apart are in phase (phase difference 0 or $2\pi$). Two points half a wavelength apart are exactly out of phase (phase difference $\pi$).

To find the phase difference between two points from a displacement–distance graph, divide their separation by the wavelength and multiply by $360°$ (or $2\pi$): two points $0.50\ \text{m}$ apart on a wave of wavelength $2.0\ \text{m}$ ($v = 600\ \text{m s}^{-1}$, $f = 300\ \text{Hz}$) differ in phase by $0.25 \times 360° = 90°$. Only the fraction of a cycle matters, so $450°$ is the same as $90°$. From two displacement–time graphs, read the time shift between the peaks as a fraction of the period. Which way is a point moving at the instant a graph shows? The whole profile moves along, so each point takes up the displacement that its neighbour on the side the wave comes from has now; a point at a crest or a trough is momentarily at rest.

Worked example. Points R and T on a string are $0.62\ \text{cm}$ apart and a quarter of a cycle out of phase. The wave speed is $0.27\ \text{m s}^{-1}$. Find the frequency.

A quarter of a cycle is a quarter of a wavelength, so $\lambda = 4 \times 0.62 = 2.5\ \text{cm}$ and $f = v / \lambda = 0.27 / 0.025 = 11\ \text{Hz}$.

The wave equation

In one period $T$, the wave moves forward by one wavelength $\lambda$. So speed $= \text{distance} / \text{time} = \lambda / T = \lambda f$:

This comes straight from the definitions of speed, frequency and wavelength, and works for every progressive wave. For the two-mark derivation write both steps: in one period $T$ the wave travels one wavelength $\lambda$, so $v = \lambda / T$; and $f = 1/T$, so $v = f\lambda$. Because $v$ is fixed by the medium, a higher frequency means a shorter wavelength. A period question is the same equation the other way round: light of wavelength $460\ \text{nm}$ has $f = c / \lambda = 6.5 \times 10^{14}\ \text{Hz}$ and $T = 1/f = 1.5 \times 10^{-15}\ \text{s}$.

Reading a CRO trace

A cathode-ray oscilloscope 示波器 (CRO) draws a voltage signal — for sound, the output of a microphone — against time. Two controls matter:

• time-base 时基 (seconds per division across): turns horizontal distance on the screen into time. Read the period $T$ as the distance between two next-door peaks, then $f = 1/T$.
• y-gain 垂直增益 (volts per division up): turns vertical distance into voltage. The amplitude in volts is the peak height from the centre line.

If the time-base is $5\ \text{ms}/\text{div}$ and one full cycle takes $4$ divisions, then $T = 4 \times 5\ \text{ms} = 20\ \text{ms}$ and $f = 50\ \text{Hz}$.

The time-base is the time represented by one division (or one centimetre) across the screen, in $\text{s}\ \text{div}^{-1}$. To set it for a given signal, work backwards: a $2000\ \text{Hz}$ sound has $T = 0.50\ \text{ms}$, so for one cycle to span $2.5\ \text{cm}$ the time-base must be $0.50 / 2.5 = 0.20\ \text{ms cm}^{-1}$. For the amplitude, multiply the peak height in divisions by the y-gain: $2.0$ divisions at $3.5\ \text{mV cm}^{-1}$ is $7.0\ \text{mV}$. When the intensity of the sound is reduced to a quarter at the same frequency, the trace keeps its period and its peaks halve in height, because $I \propto A^{2}$.

Intensity of a wave

A wave carries energy. The intensity 强度 at a point is the power 功率 passing through unit area at right angles to the direction of travel:

Unit: $\text{W m}^{-2}$.

Intensity is proportional to the square of the amplitude:

For a point source 点源 sending out energy equally in all directions, the wavefronts 波前 are spheres; the surface area at distance $r$ is $4\pi r^{2}$, so

Doubling the distance cuts the intensity to a quarter, which means the amplitude is halved (since $I \propto A^{2}$).