IGCSE物理 能量守恒 动能 势能 功率 效率
IGCSE 物理 1.7 把能量、功与功率压在一个板块里考，丢分的地方通常不在计算本身。描述能量转移时的用词、公式代入前的单位处理，才是拉开差距的两处。下面按 Cambridge IGCSE Physics 0625 官方大纲（适用于 2026、2027、2028 年考试，version 2 于 2025 年 12 月发布）1.7.1 到 1.7.4 的顺序展开，每条配一个考场里反复出现的细节。
The 0625 syllabus keeps energy, work and power inside one section, and the marks lost there rarely come from arithmetic. Candidates lose them by naming the wrong store, by using the wrong distance, or by quoting a number with no equation above it. What follows tracks sub-sections 1.7.1 to 1.7.4 of the Cambridge IGCSE Physics syllabus for examination in 2026, 2027 and 2028.
能量板块在 0625 卷面里的位置
大纲 1.7 分成四个小节：1.7.1 能量、1.7.2 功、1.7.3 能源与效率、1.7.4 功率。能量储存与转移属于 Core 内容，Paper 1 与 Paper 3 都会直接问到；动能与重力势能的公式计算属于 Supplement，只出现在 Paper 2 与 Paper 4。按官方大纲的卷面结构，Core 考生考 Paper 1（45 分钟，40 分，占 30%）与 Paper 3（1 小时 15 分钟，80 分，占 50%），Extended 考生考 Paper 2 与 Paper 4，两种路线再加一份 Paper 5 或 Paper 6 实验卷（40 分，占 20%）。理论卷占了半个总分，能量又与力、运动经常串在同一道大题里，把步骤写清楚比抢时间划算。分数比重也提示了复习顺序：结构题的分值远大于选择题，描述类小问却最容易在考场上被跳过。
Cambridge divides this topic into four sub-sections, and only two of them need equations at Extended level. Paper 1 and Paper 3 candidates are asked to name stores and describe transfers in words; Paper 2 and Paper 4 candidates do the same and then substitute into the kinetic and gravitational potential energy equations. One practical paper adds the remaining 20 per cent, and energy questions turn up there too, usually attached to a pendulum or a ramp. Since the theory paper outweighs the multiple choice one, written descriptions deserve as much rehearsal as the equations do.
七个能量储存与四条转移路径
官方 Core 要求能列出七个能量储存：动能（kinetic）、重力势能（gravitational potential）、化学能（chemical）、弹性势能（elastic / strain）、核能（nuclear）、静电势能（electrostatic）与内能（internal / thermal）。能量转移的路径是四条：力做功（mechanical work done）、电流做功（electrical work done）、加热（heating）、波（wave，含电磁波与声波）。答题时用固定句式能稳住分数：energy is transferred from the ... store to the ... store by ...。很多学生以为 heating 是一种能量储存，写成 heat energy is stored in the water，这一句在 Core 卷上就直接丢掉那一分，因为 heating 只是一条路径，储存的名字是 internal store。摆锤从最高点摆到最低点时，重力势能储存减少，动能储存增加，同时有少量能量经由空气阻力与摩擦转到内能储存，这一个例子把储存与路径两套词都用上了。
The syllabus asks for seven named stores, and the list is worth memorising in the syllabus order. Transfers travel along four pathways only, and examiners expect the store and the pathway to be named in the same sentence. Heat is a pathway, not a store, so a sentence about heat being stored earns nothing even when the physics behind it is sound. A pendulum swinging from its highest point gives you the full pattern in one line: potential to kinetic by mechanical work, with a little lost to internal energy through air resistance.
动能公式里的速度平方
Extended 考生要会用 kinetic energy = 1/2 x m x v^2。这条公式只吃质量与速度，质量用 kg，速度用 m/s，答案单位是 J。速度带平方，所以速度翻倍时动能变成四倍，速度变成三倍时动能变成九倍，这类倍数题在 Paper 2 的选择题里几乎每年一套。常见的错误是把 1/2 漏掉，或者把题目给的 cm/s、km/h 直接代进公式。km/h 换成 m/s 要除以 3.6，例如 72 km/h 就是 20 m/s，这一步没做，答案会差出十几倍。拿 1.5 kg 的球以 4.0 m/s 滚动试一次：1/2 x 1.5 x 4.0^2 = 12 J，先算平方再乘，手算顺序对了就不容易出错。
Squaring the speed is what makes this equation worth practising. Double the speed and the kinetic store holds four times as much energy, not twice as much, which is why braking distances grow so quickly with speed. Convert to kilograms and metres per second first; a value left in kilometres per hour gives an answer that is wrong by a factor of about thirteen. A 1.5 kg ball rolling at 4.0 m/s carries 12 J, and the square is best evaluated before any multiplication.
重力势能变化与 g 的取值
重力势能变化写作 change in gravitational potential energy = m x g x change in height。这里的 g 用 9.8 N/kg，如果卷首注明用 10 N/kg 就按 10 算，两套数值的答案都要会。高度差是垂直方向的落差，不是沿斜面走过的路程，这是斜面题最常见的陷阱。举个例子，60 kg 的学生爬上一段高 4.5 m 的楼梯，重力势能增加 60 x 9.8 x 4.5 = 2646 J，若用 g = 10 N/kg 则是 2700 J，答题时先写公式再代入，两个答案都能拿分。水坝蓄水、起重机吊货、电梯上行这三类题共用同一条公式，差别只在高度差的取法。
Height here means vertical height, measured from the lower point to the higher one. A trolley pushed three metres up a slope that rises only one metre gains potential energy for the one metre, not the three. Answers are expected in joules with the equation written out, and both 9.8 N/kg and 10 N/kg are accepted as long as the value matches the one given in the question. Dams, cranes and lifts all reduce to the same substitution once the height difference is identified correctly.
功等于力乘位移
1.7.2 只讲一条关系：work done = force x distance moved in the direction of the force，也等于转移的能量。位移必须沿着力的方向量。提着箱子水平走一段路，手向上的力与水平位移垂直，这份功为零，因为能量没有沿力的方向转移。常见的错误是把走的距离直接乘上重力。摩擦那一类情况更绕：摩擦力与运动方向相反，能量从动能储存转移到物体与地面的内能储存，写答案时说清转移的起点和终点。斜面上拉物块时，拉力沿斜面、重力竖直向下，两个力各算各的功，不要混在一起。
The distance that counts is the distance moved along the line of the force. Carry a suitcase across a level floor and the upward force from your hand does no work on it, since the displacement runs at right angles to the force. Where friction acts, energy leaves the kinetic store and lands in the internal store of the surfaces, and a one-line description of that transfer is usually worth the mark on its own. On a slope the pull acts along the surface while weight acts downwards, so the two forces are treated separately.
功率是能量转移的快慢
功率的定义是单位时间内做的功，也就是单位时间内转移的能量，公式 power = work done / time = energy transferred / time。单位是瓦特，1 W = 1 J/s，1 kW = 1000 W。同一件事，功率大只说明时间短，不说明能量多。两台吊车把同样的钢梁吊到同样高度，做功一样多，功率大的那台先吊完。电压那一章里的 P = I x V 与这里的 P = W / t 说的是同一个量，只是换了已知条件。这类题常问反过来的一步：已知功率与时间求能量，答案就是 W = P x t，把公式倒着念一遍比死记省力。
Power measures how fast energy moves, not how much of it moves. Two cranes lifting identical beams to the same height do the same amount of work, and the more powerful one simply finishes sooner. Watts are joules per second, so a 2000 W kettle transfers about four times as much energy per second as a 500 W blender running on the same supply. Rearranging the same relationship gives energy from power multiplied by time, which covers the electric heater and motor questions.
效率公式与浪费能量的说法
效率有两种写法：efficiency = useful energy output / total energy input x 100%，也可以换成有用功率与输入功率之比。分子只放有用的那一部分，分母放全部输入。算出来大于 100% 就说明分子分母写反了，这在考场上是自查的好信号。所谓浪费掉的能量并没有消失，它转移到了内能储存里，例如灯泡发热、电机摩擦升温。作文式回答里写 energy is dissipated to the internal store of the surroundings，比写 energy is lost 更贴合大纲用词，也更容易拿满分。一台 60 W 的灯如果只有 6 W 变成光，其余 54 W 进了内能储存，效率就是 10%，代入时分子写 6 W 而不是 54 W。
Useful output goes on top and total input underneath; an answer above 100 per cent simply means the two were swapped. Energy that appears wasted has been transferred to the internal store of the surroundings, often by heating or by sound, and it can never be destroyed. Lighting and motors are the usual examples, since both put a noticeable share of their input into warming the room. A lamp that turns 6 W of a 60 W input into light is 10 per cent efficient, and the other 54 W still exists as internal energy.
Sankey 图与多阶段转移题
Supplement 要求会用能量守恒处理多阶段过程，并且能读 Sankey 图。图上的箭头宽度代表能量大小，左右两端的总量相等，从图上读出来的数值直接代入效率公式即可。多阶段题按过程拆开：先是化学能转移到动能，再是动能转移到重力势能，每一步都单独写一次转移，最后用总量守恒对一下账。碰到流程图题，先圈出起点与终点两个储存，再补中间经过的路径。若图上标出的分支有三四条，先找到最宽的那条作为有用输出，剩余分支加总即为损耗。
A Sankey diagram has to balance, because the arrow widths carry the amounts. Trace a multi-stage process one transfer at a time, naming the store that loses energy and the store that gains it, then check the totals add up. Marking tends to reward the named transfers rather than the arithmetic that follows them. Where several branches leave the diagram, the widest one is normally the useful output and the rest are summed as losses.
能源题里的五个比较维度
1.7.3 要求说清从化石燃料、生物燃料、水（含波浪、潮汐与水坝）、地热、核燃料、太阳辐射与风里取得有用能量的方式，并按五项做比较：可再生性、可获得性、可靠性、规模与环境影响。Core 只需要定性判断效率，Supplement 才用得上百分比公式。这五项里最容易漏掉的是可靠性：风的输出时断时续，太阳能只在白天，而化石燃料与核燃料可以按需输出。Supplement 还要求知道太阳辐射是除地热、核能与潮汐之外所有能源的源头，太阳自身的能量来自核聚变，而受控核聚变发电仍处在研究阶段。比较题按维度逐条写，每写一条都点出对应的能源名字，比笼统地说 renewable 更好拿分。
Comparisons here run along five lines: renewability, availability, reliability, scale and environmental impact. Reliability is the one candidates forget, yet it separates wind and solar from fuels that can be burned on demand. Solar radiation, itself released by nuclear fusion in the Sun, is the original source for every resource except geothermal, nuclear and tidal. Answering dimension by dimension, with a named resource attached to each point, is what the mark scheme recognises.
单位、量级与换算清单
能量题有一半的失分来自单位。质量用 kg，速度用 m/s，高度用 m，时间用 s，重力场强度用 N/kg，力用 N，功率用 W。km/h 除以 3.6 得 m/s；g 用 9.8 或题目给的 10 N/kg；1 kW = 1000 W，1 MJ = 1000000 J，1 kW h = 3.6 MJ，最后一条在算电费时会用到。写答案时把单位一起写上，卷面允许的话在公式旁边标出代入值，阅卷老师一眼就能看到思路。若题目给的是厘米、克或分钟，先换成国际单位再代入，这一步写在草稿纸最上面，避免用到一半回头改。遇到 MJ 与 kW h 混用的题目，统一换算成 J 之后再比较，两个量级相差 3.6 倍，凭感觉估会出错。
Most careless errors in this topic are unit errors rather than physics errors. Masses belong in kilograms, speeds in metres per second, heights in metres and power in watts, and the kilowatt hour equals 3.6 million joules. Writing the unit beside every substituted value makes a working mark defensible even when the final figure slips. Convert everything to SI units at the top of your rough work rather than halfway through a substitution.
一套计算流程与两个数字
拿一道两问的题练手：2.0 kg 的球以 6.0 m/s 滚动，动能是 1/2 x 2.0 x 6.0^2 = 36 J；同一只球被举高 3.0 m，重力势能增加 2.0 x 9.8 x 3.0 = 58.8 J，写成 59 J。流程固定为四步：写出公式、代入数值与单位、算出结果、按题目给的数据位数给出答案。题目里给的是 2.0 与 6.0 这样的两位有效数字，答案就保留两位。中间量不要提前四舍五入，58.8 保留两位是 59，若在中间步骤先凑成 60，最后一位就会偏离。若这一问还要求求功率，把算出的能量除以时间再写一次公式，两步之间留出空行，阅卷时更容易找到过程分。
Take a two-part question and work it the same way every time. Write the equation, substitute with units attached, evaluate, then round to match the data given. A ball of 2.0 kg rolling at 6.0 m/s carries 36 J of kinetic energy, and lifting the same ball through 3.0 m adds about 59 J to its gravitational potential store. Reported figures carry no more precision than the numbers they came from, and an extra power part simply repeats the routine with one more line.
实验卷里的能量测量
Paper 5 与 Paper 6 常把能量放进摆或斜面的实验里：量高度用米尺，量时间用秒表，重复几次取平均以减小随机误差。这类题的重点不是算出一个漂亮的数字，而是解释为什么测得的动能小于重力势能的减少量，答案指向摩擦与空气阻力把一部分能量转移到了内能储存。阅卷时看重的是给出理由，只写 human error 那一类空话拿不到分。摆的实验还可以顺带讨论长度对周期的影响，为后面的波与振动内容铺路。数据表里每列都要带单位，有效数字保持一致，最后一列留出计算列并标出所用公式。
Practical papers usually hang energy work on a pendulum or a ramp, and the marks sit in the explanation rather than the readings. Friction and air resistance take a share of the energy into the internal store, which is why the kinetic value measured at the bottom falls short of the potential value at the top. Repeat readings reduce random error, and averaging them is expected in the table rather than added afterwards in prose. Columns carry their units, and the working column shows the equation that produced it.
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