OCR GCSE物理 Forces 受力分析 胡克定律 矢量
OCR 的 GCSE Physics A（J249）把力学内容都放在 Topic P2 Forces 里，拆成 P2.1 运动、P2.2 牛顿定律、P2.3 力的作用三个子单元（OCR 官网 J249 大纲）。这部分在 Paper 1（Foundation）或 Paper 3（Higher）里占的分量不轻，计算题、图像读数和解释题都会考。很多学生以为力学难在公式多，其实 OCR 把大部分公式都印在随卷发的 Equation Sheet 上（OCR 官网 J249 方程表，2026 年 6 月场次版本），真正的失分点是选错公式、单位没统一、把伸长量当成弹簧总长度。下面按考点顺序拆开讲，每个知识点配一个可以直接照着做的算例。
In OCR GCSE Physics A (J249), mechanics sits inside Topic P2 Forces, organised as P2.1 Motion, P2.2 Newton's Laws and P2.3 Forces in action. It carries a substantial share of Paper 1 at Foundation tier and Paper 3 at Higher tier, and it is examined through calculations, graph reading and multi-mark explanation questions. Many students assume the difficulty lies in the number of equations, yet OCR prints most of them on the Equation Sheet issued with the paper, so the marks tend to be lost elsewhere: choosing the wrong relationship, mixing units, or treating the extension of a spring as its total length. The sections below follow the specification order, and each one carries a worked example you can copy.
一、矢量与合力 / Vectors and resultant forces
位移、速度、加速度、力、动量都是矢量，距离、速率、质量、时间、能量是标量。区别不只是有没有方向这么简单，而是矢量相加时方向会改变结果：一辆车绕环岛匀速行驶，速率没变但方向一直在变，所以它一直在加速（OCR P2.1）。合力的求法分两种。同一直线上的力直接带正负号相加，30 N 的驱动力遇上 20 N 的阻力，合力就是向前的 10 N。不共线的力要用平行四边形法则，或者按比例画矢量图，再用尺子量长度、用量角器读角度。比如先向东走 3 m 再向北走 4 m，位移大小是 5 m，方向与正东成 53 度左右，这正好是 3、4、5 直角三角形。Higher tier 会要求画自由体受力图：把作用在同一个物体上的每一条力都画出来，包括重力、法向接触力、摩擦力、驱动力和阻力，箭头的长度表示大小，起点画在物体上。常见的错误是把合力为零当成没有受力：桌上的书受到重力和桌面支持力两条力，大小相等方向相反，合力为零，书才保持静止。
Displacement, velocity, acceleration, force and momentum are vectors, while distance, speed, mass, time and energy are scalars. The distinction is not only about direction: adding vectors changes the answer. A car going round a roundabout at steady speed is still accelerating, because its direction, and therefore its velocity, keeps changing. Resultants are found in two ways. Forces along the same line are added with signs, so a driving force of 30 N against 20 N of drag leaves 10 N forwards. Forces that are not in line need a parallelogram construction or a scaled vector diagram, measured with a ruler and a protractor. Walking 3 m east and then 4 m north gives a displacement of 5 m, about 53 degrees from east, because those numbers form a 3, 4, 5 triangle. At Higher tier a free body diagram must show every force acting on one chosen object: weight, normal contact force, friction, driving force and drag. Arrow length represents magnitude and the arrows are drawn starting from the object. A common error is treating a zero resultant as no forces at all. A book resting on a desk has weight acting downwards and a normal contact force acting upwards; the two are equal and opposite, the resultant is zero, and that is precisely why the book stays still.
二、牛顿三定律与 F = ma / Newton's three laws
牛顿第一定律说的是物体不受合力时保持静止，或保持匀速直线运动。这条定律的考点落在匀速两个字上：高速公路上以 30 m/s 行驶的汽车，驱动力与阻力恰好平衡，合力为零，并不是因为有向前的力它才继续前进。第二定律定量地联系合力、质量和加速度，写成 F = ma。质量用 kg，加速度用 m/s²，算出来的合力单位是 N。一辆 1200 kg 的汽车要获得 2 m/s² 的加速度，需要 2400 N 的合力；同样的车把合力提高到 4800 N，加速度翻倍到 4 m/s²，而如果给这辆车装上 600 kg 的货，同样的 2400 N 只能带来 1.33 m/s²。第三定律说两个物体相互作用时，彼此施加的力大小相等、方向相反，而且作用在两个不同的物体上，所以它们永远不会互相抵消。阅卷时错得最多的是把第三定律的一对力画在同一个物体上，然后写合力为零所以物体不动，这实际上是把相互作用力和第一定律里的平衡力搞混了。动量 p = mv 属于 Higher tier，条件是系统不受外力时，碰撞或爆炸前后的总动量保持不变。
Newton's first law states that an object stays at rest, or keeps moving at constant velocity in a straight line, unless a resultant force acts on it. Exam questions target the word constant: a car cruising at 30 m/s on a motorway has driving force and drag in balance, so the resultant is zero, and it is not the forward force that keeps it moving. The second law links resultant force, mass and acceleration as F = ma. With mass in kilograms and acceleration in metres per second squared, the force comes out in newtons. A car of mass 1200 kg needs a resultant of 2400 N to accelerate at 2 m/s²; raise the resultant to 4800 N and the acceleration doubles, while loading 600 kg of cargo into the same car means the original 2400 N now delivers only 1.33 m/s². The third law says that when two objects interact they push each other with equal and opposite forces, and those forces act on different objects, which is why they never cancel. The most frequent error in scripts is drawing that pair on the same object and then claiming the resultant is zero, which mixes an interaction pair with the balanced forces of the first law. Momentum, written p = mv, is Higher tier only. In a collision or an explosion the total momentum of a closed system is the same before and after, provided no external resultant force acts on it.
三、胡克定律与弹性形变 / Hooke's law and deformation
弹簧的弹力与伸长量成正比，写成 F = kx，k 是弹簧常数，单位 N/m，x 必须是伸长量，也就是拉伸后的长度减去原长，不是弹簧的总长度。这一步最容易出错：原长 12 cm 的弹簧被拉到 15 cm，伸长量是 3 cm，也就是 0.03 m。一条 k = 50 N/m 的弹簧要伸长 0.30 m，需要的力是 50 乘 0.30，等于 15 N。图像上，力对伸长量的图线在弹性范围内是一条过原点的直线，直线的斜率就是 k；超过弹性极限以后图线弯曲，进入非线性区间，胡克定律不再成立，这时再用 F = kx 算出来的值偏大。弹性形变和塑性形变的区别在于卸载后能不能恢复原状：弹簧和橡皮筋在弹性范围内卸载后回到原来的长度，拉过头的弹簧会留下永久变形，不能再按原斜率使用。拉伸过程做的功储存在弹簧里，能量是 0.5kx² 而不是 kx。以 k = 50 N/m、x = 0.30 m 为例，储存的能量是 0.5 乘 50 乘 0.09，等于 2.25 J。原因是力从 0 线性增加到 kx，平均力只有最大值的一半，所以图线下那块三角形的面积才是储存的能量。OCR 的方程表里给了 F = kx 和 0.5kx²，但把伸长量换成总长度、或者忘了平方，仍然是最常见的两个扣分点。
The force needed to stretch a spring is proportional to its extension, written as F = kx, where k is the spring constant in newtons per metre. The symbol x stands for extension, the stretched length minus the original length, not the total length of the spring. A spring with an original length of 12 cm stretched to 15 cm has an extension of 3 cm, which must be written as 0.03 m before substitution. A spring with k = 50 N/m stretched by 0.30 m needs a force of 50 times 0.30, which is 15 N. On a graph of force against extension, the linear region is a straight line through the origin and its gradient gives the spring constant. Beyond the limit of proportionality the line curves, Hooke's law no longer applies, and using F = kx there gives an answer that is too large. Elastic deformation is reversed when the load is removed, while plastic deformation leaves a permanent change of shape, so an overstretched spring no longer follows the original gradient. The energy stored during stretching is 0.5kx², not kx. For k = 50 N/m and x = 0.30 m the stored energy is 0.5 times 50 times 0.09, which is 2.25 J. The force rises steadily from zero to kx, so the average force is half the maximum, and that is why the area of the triangle under the graph gives the stored energy. OCR prints both F = kx and 0.5kx² on the Equation Sheet, yet substituting the total length, or forgetting to square the extension, still accounts for a large share of lost marks.
四、力矩与杠杆 / Moments, levers and gears
力矩描述一个力让物体绕支点转动的效果，M = Fd，其中 d 是支点到力的作用线的垂直距离，不是沿着杆量的距离。力矩的单位是 Nm，不是 N。用一把扳手拧螺母，在离螺母 0.25 m 处垂直施加 60 N 的力，力矩就是 15 Nm。物体平衡的条件是顺时针力矩的总和等于逆时针力矩的总和，这就是力矩原理：跷跷板两端坐着质量不同的两个小孩，靠调整到支点的距离让两边乘积相等，才能保持水平。很多学生以为杠杆省力是凭空少用了力，其实杠杆和齿轮都是力的放大器。在离支点较远的地方施加一个小的力，在靠近支点处得到较大的力，代价是施力的那一点要移动更长的距离，力矩不变，能量也不会凭空增加。齿轮靠齿数比传递转动效果，齿数比等于半径比，小的主动齿轮带动大的从动齿轮时转速变慢、转动力矩变大。题目里出现倾斜的杆时，要先画出力沿垂直方向的分量，再量垂直距离；直接用力乘杆长是很典型的错误，斜杆越长，这样算出来的偏差越大。
A moment measures the turning effect of a force about a pivot, M = Fd, where d is the perpendicular distance from the pivot to the line of action of the force, not the distance measured along a tilted rod. Moments are measured in newton metres. A spanner applying 60 N at 0.25 m from a nut produces a moment of 15 Nm. An object balances when the total clockwise moment equals the total anticlockwise moment, which is the principle of moments. Two children of different mass on a seesaw stay level only by sitting at distances that make the two products equal. Levers and gears act as force multipliers rather than creating energy. A small force applied far from the pivot produces a larger force near it, but the point of application moves further, so the moment is unchanged and no energy is gained. Gears pass on the same idea through the ratio of their teeth, which equals the ratio of their radii; a small driving gear turning a large driven gear slows the rotation and increases the turning effect. When a question shows a rod at an angle, resolve the force into a component perpendicular to the rod first, then measure the perpendicular distance. Multiplying the full force by the length of a slanted rod is a standard scoring error, and the error grows with the angle of the rod.
五、压强与流体中的净力 / Pressure in fluids
压强是作用在单位面积上、垂直于表面的力，p = F/A，单位是帕斯卡，1 Pa 等于 1 N/m²。面积越小压强越大：图钉尖端的面积很小，同样的推力能在木板里扎出孔，而手指那端面积大，压强小，手不会疼。一个 600 N 的人平躺在面积 0.02 m² 的板子上，压强是 600 除以 0.02，等于 30000 Pa，也就是 30 kPa。液压系统利用液体中压强处处相等这一点来放大作用力：小活塞上施加小的力，产生一定压强，这个压强传到面积更大的活塞上，就得到更大的力，汽车刹车和千斤顶都是这个原理。气体压强可以用粒子模型解释。温度升高时粒子平均动能变大，撞击容器壁更频繁也更猛烈，压强随之上升；把一定质量的气体压缩到原来体积的三分之一，分子撞壁的次数增加，压强同样升高（这一条属于 OCR P1 Matter 的内容）。做题时要留意单位换算，cm² 换成 m² 要乘以 10 的负四次方，漏掉这一步答案会相差一万倍。另外，液体中随深度增大的压强 p = hρg 在 OCR 方程表里归在 P1 Matter 部分，而且是 Higher tier 才出现，不要和 P2.3 的 p = F/A 混在一起用错。
Pressure is the force acting normally on unit area, p = F/A, measured in pascals, where 1 Pa equals 1 N/m². A small area gives a large pressure, which is why the sharp point of a drawing pin pierces wood while the flat end is comfortable under your thumb. A person of weight 600 N lying on a board of area 0.02 m² produces a pressure of 600 divided by 0.02, which is 30000 Pa, or 30 kPa. Hydraulic systems exploit the fact that pressure is transmitted equally through a liquid: a small force on a small piston creates a pressure that acts on a larger piston and produces a larger force, as in car brakes and jacks. Gas pressure follows the particle model. Raising the temperature raises the average kinetic energy of the particles, so they strike the container walls more often and harder, and the pressure rises. Compressing a fixed mass of gas to a third of its volume raises the number of collisions per unit area, so the pressure rises as well. Convert areas before substituting, because cm² to m² multiplies by 10^-4 and a missing conversion leaves the answer wrong by a factor of ten thousand. Pressure with depth in a liquid, p = hρg, appears in the OCR Equation Sheet under P1 Matter and is Higher tier only, so it should not be confused with the p = F/A relationship tested in P2.3.
六、计算题与实验题的答题顺序 / Exam technique
力学计算题的步骤是固定的：先写公式，再代入带单位的数值，最后给出答案并保留合理的有效数字。OCR 的 Equation Sheet 把 F = ma、F = kx、W = Fs、M = Fd、p = F/A、W = mg 都给了，所以记不住公式不是借口，难的是从题干里判断该用哪一条。问使弹簧伸长 0.1 m 需要多大的力，用 F = kx；问拉伸过程中储存了多少能量，用 0.5kx²；问重力做了多少功，用 W = Fs，并且 s 要沿力的方向量。统一到 SI 单位是拿分的前提，cm、g、km/h 都要先换算。解释题按分数写点：三分的题至少写三条相互独立的物理信息，例如空气阻力随速度增大、阻力增大到等于重力、合力为零所以速度不再变化，把因果链写完整，比反复说同一句话有效得多。实验题要先写自变量和因变量，比如挂的砝码质量和弹簧的伸长量，再写需要控制不变的量，比如用同一根弹簧、在同一位置读数，最后说明怎样减小误差，比如重复读数取平均值、视线与刻度垂直。很多学生实验题只写步骤不写变量，结果一半的分就这么丢了。
The routine for a mechanics calculation is fixed: write the equation, substitute the values with their units, then state the answer to a sensible number of significant figures. OCR gives F = ma, F = kx, W = Fs, M = Fd, p = F/A and W = mg on the Equation Sheet, so recall is not the barrier; selecting the right relationship is. A question about the force needed to stretch a spring by 0.1 m needs F = kx, one about the energy stored during that stretch needs 0.5kx², and one about the work done against gravity needs W = Fs with the distance measured along the line of the force. Converting everything to SI units first makes the arithmetic safe, because centimetres, grams and kilometres per hour all need changing before substitution. Explanation questions are marked by levels of response, so a three-mark answer should contain three separate physics points arranged as a chain of cause and effect: air resistance increasing with speed, drag growing until it equals weight, and the resultant falling to zero so the velocity stops changing. In practical questions, name the independent variable and the dependent variable, the quantities held constant, and one way of improving reliability such as taking repeat readings and averaging them. Scripts that describe the method but never name the variables typically lose half of the available marks.
备考建议 / Study plan
复习顺序建议跟着大纲走，先过一遍 OCR 官网上 J249 的 P2.1、P2.2、P2.3 条目，把 recall and apply 这类动词和 be able to explain 分开，后者往往就是分层的解答题。做完一套真题后，把错题按公式选错、单位换算错、概念混淆三类归档，两周后再重做一遍，看看哪一类还在丢分。力与伸长量的图线题、力矩图题建议亲手画一次坐标轴，标上单位和刻度，考试时才不会在读数上花时间。如果在同一个点上反复出错，缺的通常不是练习量而是概念，回到课本的受力图和实验过程重读一遍，比再刷十道题更省时间。
Work through the sub-topics in specification order, reading the P2.1, P2.2 and P2.3 statements on the OCR website first, and separating the verbs recall and apply from the verb explain, because the explaining questions carry the level-of-response marks. After each past paper, sort your errors into three piles: the wrong relationship chosen, a unit conversion missed, and a physical idea misunderstood, then redo them two weeks later to see which pile is still growing. Draw the axes, units and scale by hand at least once for the force and extension graph and for a moment diagram, so that reading values does not cost time in the exam. When the same point keeps failing, the missing piece is usually a concept rather than more practice, and rereading the force diagram and the practical write-up is faster than another ten questions.
Cambridge International AS ALevel Physics Coursebook 正版剑桥国际AS ALevel考试物理教材带线上账号 英文原版¥557.0📱 去京东购买京东精选
需要一对一辅导？Need One-on-One Tutoring?
A-Level / GCSE / IB / KS2-KS3 全科辅导
16621398022 同微信
访问 aleveler.com
或截图后用「扫一扫 → 相册」选择本图