AP Physics C: Mechanics

Unit 2: Force & Translational Dynamics第 2 单元：力与平动动力学

From free-body diagrams to orbital mechanics — master every force concept the AP exam demands, with calculus-based rigor throughout.从受力分析图（FBD）到轨道力学——以微积分的严谨贯穿全单元，掌握 AP 考试涉及的每一个力学概念。

20–25% Exam Weight考试占分 20–25% ~15–25 Class Periods约 15–25 课时 10 Topics10 个专题

Topic 2.1专题 2.1

Systems & Center of Mass系统与质心

A system is any collection of objects you choose to analyze together. The power of defining a system is that it lets you decide which forces are "internal" (between objects inside) and which are "external" (from the environment). Internal forces always cancel in pairs by Newton's third law, so only external forces affect the system's overall motion. This seemingly simple idea is one of the most strategically important concepts in mechanics — choosing the right system boundary can turn a five-equation problem into a one-equation problem.

系统（system）是你选定一起分析的一组物体。定义系统的威力在于：它让你决定哪些力是"内力（internal force）"（系统内部物体之间的力）、哪些是"外力（external force）"（来自环境的力）。由于牛顿第三定律，内力总是成对相互抵消，因此真正影响系统整体运动的只有外力。这个看似简单的思想是力学中最具策略价值的概念之一——选对系统边界，能把一道五方程的题变成一方程的题。

When a system's internal structure doesn't matter for the question at hand, you can model the entire system as a single point particle located at the center of mass. This is the position where the mass-weighted average of all positions sits, and it moves as though every external force were applied directly to it.

当题目不关心系统的内部结构时，可以把整个系统当作位于质心（center of mass）处的一个质点。质心就是所有位置的质量加权平均，并且它的运动恰如所有外力都直接作用于它本身。

Key Insight The center of mass of a system moves according to $$\vec{F}_{\text{net,ext}} = M_{\text{total}}\,\vec{a}_{\text{cm}}$$ regardless of what the individual pieces are doing. Even if a firework explodes mid-flight, its center of mass continues along the original parabolic trajectory (ignoring air resistance) because the explosion forces are internal.

关键洞察 系统质心的运动遵循 $$\vec{F}_{\text{net,ext}} = M_{\text{total}}\,\vec{a}_{\text{cm}}$$ 与系统内部各部分如何运动无关。即便烟花（firework）在空中爆炸，只要忽略空气阻力，质心仍然沿原来的抛物线（parabola）继续运动，因为爆炸力都是内力。

Center of Mass for Discrete Particles离散质点系的质心

For a collection of point masses $m_1, m_2, \ldots$ at positions $x_1, x_2, \ldots$, the center of mass along a given axis is

对于位于位置 $x_1, x_2, \ldots$ 的一组质点 $m_1, m_2, \ldots$，某一坐标轴方向上的质心定义为：

Center of Mass — Discrete质心 —— 离散

$$\bar{x}_{\text{cm}} = \frac{\sum m_i \bar{x}_i}{\sum m_i}$$

Generalize to 2D/3D by applying separately to each coordinate axis.对每个坐标轴分别套用即可推广到二维 / 三维。

Center of Mass for Continuous Objects连续体的质心

When you have a continuous mass distribution (like a rod with non-uniform density), you replace the sum with an integral. Split the object into infinitesimal mass elements $dm$, each at position $\vec{r}$:

对连续质量分布（例如非均匀密度的杆），把求和替换为积分（integral）：将物体切分为位置为 $\vec{r}$ 的无穷小质量元 $dm$：

Center of Mass — Continuous质心 —— 连续

$$\vec{r}_{\text{cm}} = \frac{\int \vec{r}\,dm}{\int dm}$$

The trick is expressing $dm$ in terms of the position variable. For a one-dimensional rod with linear mass density $\lambda(x)$, we have $dm = \lambda(x)\,dx$. The linear mass density is defined as:

关键技巧是把 $dm$ 用位置变量表示。对于线密度（linear mass density）为 $\lambda(x)$ 的一维杆，有 $dm = \lambda(x)\,dx$。线密度的定义为：

Linear Mass Density线密度

$$\lambda = \frac{dm}{d\ell}$$

Units: kg/m. For a uniform rod of mass $M$ and length $L$: $\lambda = M/L$.单位：kg/m。对质量为 $M$、长度为 $L$ 的均匀杆，$\lambda = M/L$。

For two-dimensional or three-dimensional objects, you use surface mass density $\sigma$ or volume mass density $\rho$, respectively, and integrate over the appropriate domain. The total mass is always obtained by integrating the density:

对二维或三维物体，分别用面密度（surface mass density）$\sigma$ 或体密度（volume mass density）$\rho$，在相应的区域上积分。总质量始终由密度积分得到：

Total Mass from Density由密度求总质量

$$M_{\text{total}} = \int \rho(\vec{r})\,dV$$

Symmetry Shortcut For any object with a symmetrical mass distribution, the center of mass lies on every line of symmetry. A uniform sphere's center of mass is at its geometric center; a uniform right triangle's center of mass is at $(\bar{x}, \bar{y}) = (b/3, h/3)$ from the right-angle vertex. Use symmetry to eliminate one or two integrals before computing.

对称性捷径 对任何具有对称质量分布的物体，质心都在每一条对称轴（line of symmetry）上。均匀球的质心就是几何中心；均匀直角三角形的质心位于离直角顶点 $(\bar{x}, \bar{y}) = (b/3, h/3)$ 处。在动笔积分之前，先利用对称性消去一两个积分。

Worked Example — Non-Uniform Rod例题 —— 非均匀杆

A rod of length $L = 2$ m has linear mass density $\lambda(x) = 3x$ kg/m, where $x$ is measured from the left end. Find the center of mass.

长 $L = 2$ m 的杆，线密度 $\lambda(x) = 3x$ kg/m（$x$ 从左端量起）。求质心。

Step 1 — Total mass第 1 步 —— 求总质量

$$M = \int_0^L \lambda(x)\,dx = \int_0^2 3x\,dx = 3\left[\tfrac{x^2}{2}\right]_0^2 = 6\;\text{kg}$$

Step 2 — Numerator of $x_\text{cm}$第 2 步 —— 求 $x_\text{cm}$ 的分子

$$\int_0^L x\,\lambda(x)\,dx = \int_0^2 3x^2\,dx = \left[x^3\right]_0^2 = 8\;\text{kg}\cdot\text{m}$$

Step 3 — Divide第 3 步 —— 相除

$$x_\text{cm} = \frac{8}{6} = \tfrac{4}{3}\;\text{m} \approx 1.33\;\text{m}$$

Measured from the left end. Since density increases with $x$, the cm should sit to the right of the geometric center (1.0 m) — and it does. ✓从左端量起。密度随 $x$ 增大，所以质心应当落在几何中心（1.0 m）右侧——结果确实如此。✓

Worked Example — Exploding Firework: CoM Trajectory (FRQ-style)例题 —— 烟花爆炸：质心轨迹（FRQ 风格）

A 3.0 kg firework is launched at 40 m/s at $60°$ above horizontal. At the peak of its trajectory it explodes into two fragments: a 1.0 kg piece and a 2.0 kg piece. Immediately after the explosion, the 1.0 kg fragment is momentarily at rest. (a) Find the velocity of the 2.0 kg fragment just after the explosion. (b) Where does the center of mass land, measured horizontally from the launch point? (Take $g = 10\;\text{m/s}^2$; ignore air resistance.)

一只 3.0 kg 的烟花以 40 m/s、与水平方向成 $60°$ 角发射。轨迹最高点处爆炸为 1.0 kg 与 2.0 kg 两块碎片。爆炸瞬间，1.0 kg 碎片瞬时静止。(a) 求 2.0 kg 碎片爆炸后的速度；(b) 质心在水平方向距发射点多远落地？（取 $g = 10\;\text{m/s}^2$，忽略空气阻力。）

Set up — state at the peak建模 —— 最高点处的状态

At the peak of flight the vertical velocity is zero, so the center-of-mass velocity is horizontal:在最高点，竖直速度为零，故质心速度沿水平方向：

$$v_y = 0, \qquad v_x = v_0\cos 60° = 40(0.5) = 20\;\text{m/s}$$

$$p_x = M\,v_x = (3.0)(20) = 60\;\text{kg}\cdot\text{m/s}$$

(a) Fragment velocity after explosion(a) 爆炸后碎片速度

The explosion forces are internal, so horizontal momentum is conserved:爆炸力是内力，故水平动量（momentum）守恒：

$$p_\text{before} = p_\text{after}: \quad 60 = (1.0)(0) + (2.0)\,v_2$$

$$v_2 = 30\;\text{m/s (horizontal, forward)}$$

(b) Landing point of the center of mass(b) 质心落点

Because the explosion forces are internal, the CoM continues along the original projectile parabola as if nothing had happened:因为爆炸力是内力，质心继续沿原本的抛物线（parabola）运动，仿佛没有发生爆炸：

$$T_\text{flight} = \frac{2v_0\sin\theta}{g} = \frac{2(40)(\sin 60°)}{10} \approx 6.93\;\text{s}$$

$$R_\text{cm} = (v_0\cos\theta)\,T_\text{flight} = (20)(6.93) \approx 138.6\;\text{m}$$

The two fragments hit the ground separately, but the mass-weighted average of their landing positions is still 138.6 m from the launch point.两块碎片各自落地，但它们落点的质量加权平均仍为距发射点 138.6 m。

Why This Is a Classic FRQ "Does the CoM path change?" is a reliable AP question. Answer: NO, so long as the net external force is unchanged. Internal forces — explosions, collisions, muscle forces — cannot move the CoM.

为什么这是经典 FRQ "质心路径会改变吗？"是 AP 反复考查的问题。答案：只要净外力不变就不会。爆炸、碰撞、肌肉力等内力都无法改变质心的运动。

Two particles of mass $m$ and $3m$ are placed at positions $x = 0$ and $x = 4\,\text{m}$, respectively. Where is the center of mass of the system?质量为 $m$ 和 $3m$ 的两个质点分别位于 $x = 0$ 与 $x = 4\,\text{m}$。系统的质心在何处？

$x = 1\,\text{m}$

$x = 2\,\text{m}$

$x = 3\,\text{m}$

$x = 4\,\text{m}$

Correct. $x_{\text{cm}} = \frac{m(0) + 3m(4)}{m + 3m} = \frac{12m}{4m} = 3\,\text{m}$. The center of mass sits three-quarters of the way toward the heavier mass.正确。$x_{\text{cm}} = \frac{m(0) + 3m(4)}{m + 3m} = \frac{12m}{4m} = 3\,\text{m}$。质心位于靠近大质量物体方向的四分之三处。

Not quite. Apply $x_{\text{cm}} = \frac{\sum m_i x_i}{\sum m_i} = \frac{m(0) + 3m(4)}{4m} = 3\,\text{m}$. The cm is always closer to the heavier mass.不对。套用公式 $x_{\text{cm}} = \frac{\sum m_i x_i}{\sum m_i} = \frac{m(0) + 3m(4)}{4m} = 3\,\text{m}$。质心总是更靠近质量大的一方。

Topic 2.2专题 2.2

Forces & Free-Body Diagrams力与受力分析图

A force is a vector quantity describing an interaction between two objects. Every force has an agent (the object exerting the force) and a target (the object experiencing the force). This two-object requirement is absolute: if you cannot name both the agent and the target, you don't have a real force. Contact forces (normal, friction, tension, applied push) arise from interatomic electric interactions when surfaces touch. Non-contact forces (gravity, electromagnetic) act at a distance through fields.

力（force）是描述两个物体相互作用的矢量。每个力都有一个"施力者（agent）"和一个"受力者（target）"。这"两物体"的要求是绝对的：如果叫不出施力者和受力者，那就不是真正的力。接触力（法向力 normal force、摩擦力 friction、张力 tension、外加推力）来自表面接触时的原子级电相互作用；非接触力（引力、电磁力）通过场在远距离作用。

An object or system cannot exert a net force on itself. This is a subtle but crucial point — when you see a car "accelerating itself," the car's engine pushes the tires backward against the road, and the road pushes the tires forward via friction. The external force that actually accelerates the car comes from the ground.

一个物体或系统无法对自己施加净力。这一点微妙但极其关键——汽车看似"自己加速"，其实是发动机驱动轮胎向后蹬地，地面通过摩擦把轮胎向前推，真正让汽车加速的外力来自地面。

Drawing Free-Body Diagrams绘制受力分析图

A free-body diagram (FBD) is your most important problem-solving tool. It isolates a single object, represents it as a dot (its center of mass), and shows every external force acting on it as an arrow originating from that dot. The arrow's direction shows the force direction and relative length indicates relative magnitude.

受力分析图（free-body diagram / FBD）是你最重要的解题工具。它把单个物体单独取出，用一个点表示（即其质心），然后把作用其上的每一个外力画成从该点出发的箭头。箭头方向表示力的方向，相对长度表示相对大小。

Exam Trap On the AP exam, forces in the same direction must be drawn side by side, not overlapping. Components of forces should not appear on the FBD — only the actual forces. Set up your coordinate system after drawing the FBD, choosing one axis parallel to the acceleration direction (e.g., along the incline for inclined plane problems).

考试陷阱 在 AP 考试中，方向相同的力必须并排画，不能互相重叠。FBD 上不画力的分量（components）——只画真实的力。先画完 FBD 再建立坐标系，并尽量让一条坐标轴沿加速度方向（例如斜面问题中沿斜面方向）。

FBD Checklist For every FBD you draw, verify: (1) each force arrow has an identifiable agent, (2) no force appears without a real physical interaction, (3) you haven't included "centripetal force" as a separate force — it's the net inward force, not an additional one, (4) gravity always points straight down, and (5) normal force is perpendicular to the contact surface, not necessarily upward.

FBD 自查清单 每画完一张 FBD，请核对：(1) 每个力箭头都有可指认的施力者；(2) 没有一个力是凭空冒出来的——背后都有真实的物理相互作用；(3) 没有把"向心力（centripetal force）"作为单独的力画上去——它是径向净力，不是新增的一个力；(4) 重力（gravity）总是指向正下方；(5) 法向力（normal force）垂直于接触面，并不一定指向正上方。

Worked Example — Block on an Inclined Plane例题 —— 斜面上的木块

A block of mass $m$ rests on a frictionless incline of angle $\theta$. Find the normal force and the block's acceleration.

质量为 $m$ 的木块静止在角度为 $\theta$ 的无摩擦斜面（inclined plane）上。求法向力和木块的加速度。

FBD forces on the block:
• Weight mg straight down (agent: Earth)
• Normal force N perpendicular to surface (agent: incline)

木块的 FBD 中有两个力：
• 重力 mg 竖直向下（施力者：地球）
• 法向力 N 垂直于斜面（施力者：斜面）

Choose axes: $x$ along the incline (positive down-slope), $y$ perpendicular to the surface.选定坐标轴：$x$ 沿斜面（向下为正），$y$ 垂直于斜面。

$y$-direction (no acceleration perpendicular to the surface):$y$ 方向（垂直方向无加速度）：

$$N - mg\cos\theta = 0 \;\Rightarrow\; N = mg\cos\theta$$

$x$-direction (along the incline):$x$ 方向（沿斜面）：

$$mg\sin\theta = ma \;\Rightarrow\; a = g\sin\theta$$

Limits: $\theta = 0 \Rightarrow a = 0$ (flat) ✓; $\theta = 90° \Rightarrow a = g$ (free fall) ✓.极限检验：$\theta = 0 \Rightarrow a = 0$（水平面） ✓；$\theta = 90° \Rightarrow a = g$（自由落体 free fall） ✓。

A book sits on a table. What is the Newton's third-law pair to the normal force the table exerts on the book?一本书放在桌上。桌面对书施加的法向力的牛顿第三定律对偶力是哪一个？

The gravitational force of Earth on the book地球对书的引力

The force the book exerts on the table书对桌面的力

The gravitational force of the book on Earth书对地球的引力

The normal force of the floor on the table地面对桌子的法向力

Correct. Third-law pairs involve the same two objects and the same type of force. The table pushes the book up; the book pushes the table down. Both are contact (normal) forces between book and table.正确。第三定律对偶力必须涉及同一对物体，且为同一种力。桌子向上推书；书向下压桌子。两者都是书与桌子之间的接触（法向）力。

Not quite. The weight of the book (Earth on book) happens to have the same magnitude, but it involves different objects and a different force type (gravitational). The true pair is the book's force on the table.不对。书所受重力（地球对书）数值上恰好相同，但它涉及不同物体、不同种类的力（引力）。真正的对偶力是书对桌面的力。

Topic 2.3专题 2.3

Newton's Third Law牛顿第三定律

Newton's third law states that when object A exerts a force on object B, then B simultaneously exerts a force on A that is equal in magnitude and opposite in direction:

牛顿第三定律（Newton's third law）指出：当物体 A 对物体 B 施加一个力时，B 同时对 A 施加一个大小相等、方向相反的力：

Newton's Third Law牛顿第三定律

$$\vec{F}_{A \text{ on } B} = -\vec{F}_{B \text{ on } A}$$

Third-law pairs always act on different objects and are always the same type of force.第三定律对偶力始终作用在不同物体上，且始终是同一种力。

A critical consequence: internal forces within a system always cancel pairwise. Only external forces can change the momentum (or accelerate the center of mass) of a system. This is why choosing your system boundary wisely is so powerful.

一个关键推论：系统内部的内力总是成对相互抵消。只有外力能改变系统的动量（或者说让系统质心加速）。这正是聪明地选取系统边界为何如此强大的原因。

Tension in Strings绳中的张力

Tension is the macroscopic result of countless interatomic forces along a string, rope, or cable. For an ideal string (massless and inextensible), the tension is the same at every point. When the string has mass, however, different segments may experience different tensions — a key distinction in more advanced problems.

张力（tension）是绳、索内部无数个原子间作用力的宏观体现。对于理想绳（ideal string，无质量且不可伸长），整根绳上每一点的张力都相同。当绳本身有质量时，不同位置的张力可以不同——这在更高阶题目中是关键的区分。

An ideal pulley has negligible mass and frictionless axle. It simply redirects the tension without changing its magnitude. A non-ideal (massive) pulley will be revisited in Unit 5 with rotational dynamics.

理想滑轮（ideal pulley）的质量可忽略、轴上无摩擦，只改变张力方向、不改变其大小。有质量的非理想滑轮要等到第 5 单元的转动动力学才会再次出现。

Common Confusion In tug-of-war, the rope has the same tension at both ends (for a massless rope). So how does one team win? The winning team pushes harder against the ground. The ground's friction force on the winning team exceeds the ground's friction force on the losing team — the difference in friction forces, not rope tension, determines the outcome.

常见困惑 拔河时，无质量绳两端张力相同。那一方为何能赢？赢的一队对地面蹬得更用力。地面对赢家的摩擦力大于地面对输家的摩擦力——决定胜负的是两队所受地面摩擦力之差，而不是绳子张力。

Worked Example — Atwood Machine例题 —— 阿特伍德机（Atwood machine）

Two masses $m_1$ and $m_2$ ($m_2 > m_1$) hang over an ideal (massless, frictionless) pulley. Find the acceleration and tension.

两个质量 $m_1$ 与 $m_2$（$m_2 > m_1$）挂在理想（无质量、无摩擦）滑轮两侧。求加速度与张力。

Treat the string as ideal: same tension T on both sides.
For m₁ (take up as positive):

把绳视为理想绳：两侧张力 T 相同。
对 m₁（取向上为正）：

$$T − m_1g = m_1a ... (1)$$

For m₂ (take down as positive for m₂ since it descends):对 m₂（因其下降，取向下为正）：

$$m_2g − T = m_2a ... (2)$$

Add (1) and (2):把 (1) 与 (2) 相加：

$$m_2g − m_1g = (m_1 + m_2)a$$

$$a = (m_2 − m_1)g / (m_1 + m_2)$$

Substitute back into (1):代回 (1)：

$$T = m_1(g + a) = m_1g[1 + (m_2 − m_1)/(m_1 + m_2)]$$

$$T = 2m_1m_2g / (m_1 + m_2)$$

Check: if m₁ = m₂, a = 0 and T = m₁g. ✓检验：若 m₁ = m₂，则 a = 0、T = m₁g。✓

Check: if m₂ >> m₁, a → g and T → 2m₁g. ✓检验：若 m₂ ≫ m₁，则 a → g、T → 2m₁g。✓

Topic 2.4专题 2.4

Newton's First Law牛顿第一定律

Newton's first law states: if the net force exerted on a system is zero, the velocity of that system remains constant. An object at rest stays at rest; an object in motion continues in a straight line at constant speed. This is not just a special case of the second law — it also defines what an inertial reference frame is: one in which Newton's first law holds true. Earth's surface is approximately inertial for most problems; a spinning merry-go-round is not.

牛顿第一定律（Newton's first law）：如果系统所受净力为零，则其速度保持不变。静止物体保持静止，运动物体则沿直线匀速继续。它不仅是第二定律的特例——同时也定义了惯性参考系（inertial reference frame）：在其中牛顿第一定律成立。地球表面对大多数题目而言可近似为惯性参考系；旋转的旋转木马则不是。

Translational Equilibrium平动平衡（translational equilibrium）

$$\sum \vec{F}_i = \vec{0}$$

All forces balance → constant velocity (which may be zero).所有力平衡 → 速度恒定（可以是零）。

Forces may be balanced in one direction but unbalanced in another. A ball rolling along a horizontal surface (no friction, no air resistance) has balanced vertical forces ($N = mg$) but if there's a crosswind, there's an unbalanced horizontal force causing the ball to accelerate sideways. Always analyze each dimension independently.

力在某一方向上平衡，并不意味着另一方向也平衡。无摩擦、无空气阻力地面上滚动的球，竖直方向力平衡（$N = mg$）；若有侧风，水平方向就有不平衡的力，使球沿侧向加速。务必对每一坐标方向单独分析。

Equilibrium ≠ Rest Equilibrium means zero net force, which means constant velocity — not necessarily zero velocity. A skydiver falling at terminal velocity is in equilibrium. A hockey puck gliding on frictionless ice at constant velocity is in equilibrium. Don't confuse "not accelerating" with "not moving."

平衡 ≠ 静止 平衡（equilibrium）意味着净力为零，从而速度恒定——但不一定为零。以终极速度（terminal velocity）下落的跳伞员处于平衡。无摩擦冰面上匀速滑行的冰球也处于平衡。不要把"不加速"等同于"不动"。

An elevator moves upward at a constant velocity of $3\,\text{m/s}$. Which statement about the forces on the elevator is correct?电梯以 $3\,\text{m/s}$ 的恒定速度向上运动。下列关于电梯受力的描述哪一项正确？

The tension in the cable is greater than the weight.缆绳张力大于重力。

The weight is greater than the tension in the cable.重力大于缆绳张力。

The tension equals the weight.张力等于重力。

The net force is upward.净力方向向上。

Correct. Constant velocity means zero acceleration, which means zero net force. The upward tension must exactly balance the downward weight.正确。恒速运动意味着加速度为零，即净力为零。向上的张力必须恰好平衡向下的重力。

Not quite. Constant velocity (even upward) means $a = 0$, so by Newton's second law, $F_{\text{net}} = 0$. The tension equals the weight.不对。恒速（哪怕方向向上）意味着 $a = 0$，由牛顿第二定律 $F_{\text{net}} = 0$，所以张力等于重力。

Topic 2.5专题 2.5

Newton's Second Law牛顿第二定律

Newton's second law is the backbone of classical mechanics. It states that the acceleration of a system's center of mass is proportional to the net external force and inversely proportional to the total mass:

牛顿第二定律（Newton's second law）是经典力学的脊梁：系统质心的加速度与净外力（net external force）成正比、与总质量（mass）成反比：

Newton's Second Law牛顿第二定律

$$\vec{a}_{\text{sys}} = \frac{\sum \vec{F}}{m_{\text{sys}}} = \frac{\vec{F}_{\text{net}}}{m_{\text{sys}}}$$

Equivalently: $\vec{F}_{\text{net}} = m\vec{a}$. Direction of $\vec{a}$ is always the direction of $\vec{F}_{\text{net}}$.等价形式：$\vec{F}_{\text{net}} = m\vec{a}$。$\vec{a}$ 的方向始终与 $\vec{F}_{\text{net}}$ 一致。

The problem-solving strategy for any dynamics problem is: (1) identify the system, (2) draw a free-body diagram, (3) choose a coordinate system, (4) write $\sum F_x = ma_x$ and $\sum F_y = ma_y$, (5) solve the resulting system of equations. The calculus connection is that force and acceleration are related to velocity and position through derivatives: $a = dv/dt = d^2x/dt^2$, so Newton's second law becomes a differential equation when forces depend on velocity or position.

所有动力学题目都可用同一套流程：(1) 选定系统；(2) 画 FBD；(3) 建立坐标系；(4) 写出 $\sum F_x = ma_x$ 与 $\sum F_y = ma_y$；(5) 求解联立方程组。微积分上的关键关联：力与加速度通过导数（derivative）与速度、位置相联：$a = dv/dt = d^2x/dt^2$。当力依赖于速度或位置时，牛顿第二定律就变成微分方程。

System vs. Individual Object You can apply $F = ma$ to the entire system (using total mass and only external forces) or to each individual object (using that object's mass and all forces on it, including internal ones). The system approach gives the acceleration of the center of mass quickly; the individual object approach is needed when you want to find internal forces like tension.

系统视角 vs. 单体视角 $F = ma$ 既可以应用于整个系统（用总质量、只考虑外力），也可以分别应用于每个单独的物体（用该物体的质量、考虑作用其上的所有力，包括内力）。系统视角能迅速得到质心加速度；要求张力等内力时则必须用单体视角。

Worked Example — Two Blocks on a Surface例题 —— 平面上的两个木块

An applied force $F$ pushes two blocks ($m_1$ in front, $m_2$ behind) across a frictionless surface. Find the acceleration and the contact force between the blocks.

一个外加力 $F$ 推动两个木块（前为 $m_1$、后为 $m_2$）沿无摩擦面运动。求加速度以及两木块之间的接触力。

System approach (both blocks together):

系统视角（两木块作为一个整体）：

$$F = (m_1 + m_2)a$$

$$a = F / (m_1 + m_2)$$

Individual block (m₂ alone, pushed by contact force N₁₂):单体视角（单独考虑 m₂，受接触力 N₁₂ 推动）：

$$N_1_2 = m_2 \cdot a = m_2F / (m_1 + m_2)$$

Check: N₁₂ < F (the contact force is less than the检验：N₁₂ < F（接触力小于外加力——

applied force — it only needs to accelerate m₂). ✓它只需要让 m₂ 加速）。✓

Check: if m₂ = 0, N₁₂ = 0 (no block to push). ✓检验：若 m₂ = 0，则 N₁₂ = 0（没有后块要推）。✓

F = ma ExplorerF = ma 探索器

See how force, mass, and acceleration are linked. Adjust two to see the third.观察力、质量与加速度的联系。调节其中两个，看第三个如何变化。

Net Force F净力 F 20 N

Mass m质量 m 5 kg

Acceleration加速度

4.0

m/s²

After 1 s1 秒后

4.0

m/s

After 3 s3 秒后

12.0

m/s

Dist in 3 s3 秒内位移

18.0

Topic 2.6专题 2.6

Gravitational Force万有引力

Newton's law of universal gravitation describes the attractive force between any two objects with mass. The force is proportional to each mass and inversely proportional to the square of the distance between their centers:

牛顿万有引力定律（universal gravitation）描述任意两个有质量的物体之间的吸引力。该力与两者质量分别成正比，并与它们质心距离的平方成反比：

Universal Gravitation万有引力

$$|\vec{F}_g| = G\frac{m_1 m_2}{r^2}$$

$G = 6.674 \times 10^{-11}\;\text{N}\cdot\text{m}^2/\text{kg}^2$. Always attractive, always along the line connecting the centers of mass.$G = 6.674 \times 10^{-11}\;\text{N}\cdot\text{m}^2/\text{kg}^2$。始终为吸引力，始终沿两质心连线方向。

Gravitational Field引力场

A field model describes the effect of a non-contact force at various positions in space. The gravitational field strength at distance $r$ from a mass $M$ is:

场模型用于描述非接触力在空间各点上的作用。在距质量 $M$ 为 $r$ 处的引力场强度（gravitational field）为：

Gravitational Field Strength引力场强度

$$|\vec{g}| = \frac{|\vec{F}_g|}{m} = G\frac{M}{r^2}$$

Units: N/kg = m/s². Near Earth's surface, $g \approx 10\;\text{N/kg}$.单位：N/kg = m/s²。地表附近 $g \approx 10\;\text{N/kg}$。

When gravity is the only force on an object, the object's acceleration equals $\vec{g}$ at that location. Weight is the gravitational force on an object near an astronomical body: $W = F_g = mg$.

当物体只受引力时，其加速度就等于该位置的 $\vec{g}$。重力（weight）就是物体在天体附近所受的引力：$W = F_g = mg$。

Apparent Weight & Weightlessness视重与失重

Your apparent weight is the magnitude of the normal force a surface exerts on you. In an accelerating elevator going up, the normal force exceeds $mg$ and you feel heavier; going down, it's less and you feel lighter. In free fall (or orbiting — which is free fall!), the normal force is zero and you feel weightless.

视重（apparent weight）就是表面对你施加的法向力的大小。电梯加速上行时，法向力大于 $mg$，你感到变重；加速下行时则小于 $mg$，你感到变轻。自由下落时（轨道运动本质上也是自由落体！），法向力为零，你感到失重（weightlessness）。

Equivalence Principle An observer in a closed box cannot distinguish between being at rest in a gravitational field and being in a uniformly accelerating reference frame. This principle (which Einstein later made the foundation of general relativity) means that inertial mass and gravitational mass are equivalent — confirmed experimentally to extraordinary precision.

等效原理（equivalence principle） 一位身处封闭盒中的观察者无法区分"自己静止在引力场中"与"自己处于一个匀加速参考系中"。这一原理后来被爱因斯坦发展为广义相对论的基石——它意味着惯性质量与引力质量等价，已被实验以极高精度证实。

Newton's Shell Theorem牛顿壳层定理（`shell theorem`）

A uniform spherical shell of mass: (i) exerts zero net gravitational force on any object inside it, (ii) acts on outside objects as if all its mass were concentrated at its center. For a uniform solid sphere, an object at radius $r$ (inside the sphere) only "feels" the mass enclosed within radius $r$. That enclosed mass is:

均匀的球壳：(i) 对其内部任何物体施加的净引力为零；(ii) 对外部物体的引力等价于"全部质量集中于球心"。对均匀实心球，半径 $r$ 处（在球内）的物体只"感受到"半径 $r$ 之内所包围的质量。这部分质量为：

Partial Mass Inside a Sphere球内半径以内的部分质量

$$m_{\text{partial}} = \rho \frac{4}{3}\pi r_{\text{partial}}^3$$

Substituting into the gravitational force expression shows that inside a uniform sphere, the gravitational force is proportional to distance from the center (a restoring force, like a spring!):

代入引力表达式可发现：在均匀球内部，引力大小与到球心的距离成正比（这是一种回复力 restoring force，如同弹簧！）：

Gravity Inside a Uniform Sphere均匀球内部的引力

$$F_{g,\text{partial}} = -k\,r_{\text{partial}}$$

where $k = \frac{4}{3}\pi G \rho m$. This is linear in $r$, unlike the inverse-square law outside.其中 $k = \frac{4}{3}\pi G \rho m$。这是 $r$ 的线性函数，与球外的平方反比律完全不同。

Worked Example — g on a Mountain例题 —— 山顶上的 g

The gravitational field strength on Earth's surface is $g_0 = 9.8\;\text{m/s}^2$. What is $g$ at altitude $h = R_E$ (one Earth radius above the surface)?

地表引力场强度 $g_0 = 9.8\;\text{m/s}^2$。在高度 $h = R_E$（即地球半径之高）处，$g$ 是多少？

At distance r = R_E + h = 2R_E from Earth's center:

距地心 r = R_E + h = 2R_E：

$$g = GM/(2R_E)^2 = GM/(4R_E^2) = g_0/4$$

$$g = 9.8/4 = 2.45\;\text{m/s}^2$$

Key: distance is measured from the CENTER, not the surface.关键：距离从地心起算，而不是地表。

Doubling the distance → g drops by factor of 4 (inverse-square). ✓距离翻倍 → g 减小为 1/4（平方反比律）。✓

If the distance between two objects is tripled, the gravitational force between them becomes:若两物体之间的距离变为原来的 3 倍，它们之间的引力变为：

$\frac{1}{3}$ of the original原来的 $\frac{1}{3}$

$\frac{1}{9}$ of the original原来的 $\frac{1}{9}$

$3$ times the original原来的 $3$ 倍

$9$ times the original原来的 $9$ 倍

Correct. Gravitational force follows an inverse-square law: $F \propto 1/r^2$. Tripling $r$ means $F \to F/3^2 = F/9$.正确。万有引力满足平方反比律（inverse-square law）：$F \propto 1/r^2$。$r$ 增至 3 倍 → $F \to F/3^2 = F/9$。

Not quite. It's an inverse-square law, not inverse-linear. $F \propto 1/r^2$, so tripling $r$ reduces $F$ by a factor of $9$.不对。这是平方反比律，不是线性反比律。$F \propto 1/r^2$，故 $r$ 变 3 倍，$F$ 减为 1/9。

Topic 2.7专题 2.7

Kinetic & Static Friction动摩擦与静摩擦

Friction is a contact force that opposes relative motion (or the tendency of relative motion) between two surfaces. It comes in two flavors: kinetic friction (surfaces sliding) and static friction (surfaces stationary relative to each other).

摩擦力（friction）是一种接触力，反抗两表面之间的相对运动（或相对运动的趋势）。它分两种：动摩擦（kinetic friction，表面正在滑动）与静摩擦（static friction，表面之间静止）。

Kinetic Friction动摩擦

When two surfaces slide against each other, the kinetic friction force is:

两个表面相互滑动时，动摩擦力为：

Kinetic Friction动摩擦力

$$|\vec{F}_{f,k}| = \mu_k |\vec{F}_N|$$

Direction: opposite to the relative motion of the surfaces. Independent of contact area or speed (for the simple model).方向：与表面相对运动方向相反。在简单模型下，与接触面积或速率无关。

Static Friction静摩擦

Static friction adjusts its magnitude to prevent slipping, up to a maximum value:

静摩擦力的大小会自动调整以防止滑动，但有最大值上限：

Static Friction静摩擦力

$$|\vec{F}_{f,s}| \leq \mu_s |\vec{F}_N|$$

Maximum static friction: $F_{f,s,\max} = \mu_s F_N$. Typically $\mu_s > \mu_k$ for any given surface pair.最大静摩擦：$F_{f,s,\max} = \mu_s F_N$。同一对表面通常满足 $\mu_s > \mu_k$。

The normal force is the perpendicular component of the contact force — it's not always $mg$. On an incline at angle $\theta$, $F_N = mg\cos\theta$. If someone is pressing down on the object, $F_N$ increases. If someone pulls up at an angle, $F_N$ decreases. Always solve for $F_N$ from the perpendicular equilibrium equation.

法向力是接触力的垂直分量——并不总等于 $mg$。在斜角 $\theta$ 的斜面上，$F_N = mg\cos\theta$。若有人向下压物体，$F_N$ 增大；若有人斜向上拉物体，$F_N$ 减小。始终通过"垂直方向力平衡"的方程求解 $F_N$。

Common Mistake Never assume $F_N = mg$. The normal force equals $mg$ only on a horizontal surface with no other vertical force components. On an incline, in an elevator, or with applied forces at an angle, you must derive $F_N$ from the equation of motion perpendicular to the surface.

常见错误 不要默认 $F_N = mg$。法向力等于 $mg$ 只在"水平面 + 没有其他竖直方向力分量"时成立。在斜面、电梯里或者有斜向外加力时，必须从垂直于接触面的运动方程中推出 $F_N$。

Worked Example — Block on Rough Incline例题 —— 粗糙斜面上的木块

A block of mass $m$ slides down a rough incline of angle $\theta$ with coefficient of kinetic friction $\mu_k$. Find the acceleration.

质量为 $m$ 的木块沿角度为 $\theta$、动摩擦系数（coefficient of friction）为 $\mu_k$ 的粗糙斜面下滑。求加速度。

FBD: Weight mg down, Normal N ⊥ surface, Friction f_k up the slope
Axes: x down the incline, y ⊥ to surface

FBD：重力 mg 向下、法向力 N 垂直于斜面、摩擦力 f_k 沿斜面向上
坐标轴：x 沿斜面向下，y 垂直于斜面

Perpendicular direction (no acceleration):垂直方向（无加速度）：

$$N = mg\cos\theta$$

Parallel direction (along the slope):沿斜面方向：

$$mg\sin\theta - f_k = ma$$

$$mg\sin\theta - \mu_k\,mg\cos\theta = ma$$

$$a = g(\sin\theta - \mu_k\cos\theta)$$

If $\mu_k\cos\theta > \sin\theta$ (i.e. $\mu_k > \tan\theta$), the block decelerates — friction exceeds the gravity component along the slope.若 $\mu_k\cos\theta > \sin\theta$（即 $\mu_k > \tan\theta$），木块减速——摩擦力大于重力沿斜面的分量。

Topic 2.8专题 2.8

Spring Forces弹力

An ideal spring has negligible mass and obeys Hooke's law: the force it exerts is proportional to how far it's stretched or compressed from its natural (relaxed) length. The spring force is always directed back toward the equilibrium position — it's a restoring force.

理想弹簧（ideal spring）质量可忽略，并遵守胡克定律（Hooke's law）：弹簧施加的力与其相对自然（松弛）长度的拉伸或压缩量成正比。弹簧力始终指向平衡位置——它是一种回复力（restoring force）。

Hooke's Law胡克定律

$$\vec{F}_s = -k\Delta\vec{x}$$

$k$ = spring constant (N/m). $\Delta x$ = displacement from relaxed length. The negative sign means restoring.$k$ 为劲度系数（spring constant），单位 N/m。$\Delta x$ 是相对松弛长度的位移。负号代表"回复"。

Combinations of Springs弹簧的组合

Arrangement连接方式	Equivalent Spring Constant等效劲度系数	Key Property关键性质
Series (end-to-end)串联（首尾相连）	$$\frac{1}{k_{\text{eq}}} = \frac{1}{k_1} + \frac{1}{k_2} + \cdots$$	Softer than the softest spring比最软的那根还软
Parallel (side-by-side)并联（并排）	$$k_{\text{eq}} = k_1 + k_2 + \cdots$$	Stiffer than the stiffest spring比最硬的那根还硬

Memory Aid Springs combine like resistors, but with the roles swapped: springs in series add like resistors in parallel (reciprocals), and springs in parallel add like resistors in series (direct sum). If you remember circuit rules, you can translate directly.

记忆口诀 弹簧组合与电阻类似，但"角色互换"：串联的弹簧像并联电阻一样取倒数相加；并联的弹簧像串联电阻一样直接相加。把电路里的规则反过来用即可。

Worked Example — Two Springs in Series例题 —— 两根串联弹簧

Two springs with constants $k_1 = 200\;\text{N/m}$ and $k_2 = 300\;\text{N/m}$ are connected in series. A force of $60\;\text{N}$ is applied. Find the total extension.

两根弹簧劲度系数 $k_1 = 200\;\text{N/m}$ 与 $k_2 = 300\;\text{N/m}$ 串联。施加 $60\;\text{N}$ 的力，求总伸长量。

Step 1: Equivalent spring constant

第 1 步：求等效劲度系数

$$1/k_eq = 1/200 + 1/300 = 3/600 + 2/600 = 5/600$$

$$k_eq = 120\;\text{N/m}$$

Step 2: Total extension第 2 步：总伸长量

$$\Delta x = F/k_eq = 60/120 = 0.50\;\text{m}$$

Step 3: Individual extensions第 3 步：各自伸长量

$$\Delta x_1 = F/k_1 = 60/200 = 0.30\;\text{m}$$

$$\Delta x_2 = F/k_2 = 60/300 = 0.20\;\text{m}$$

Total = 0.30 + 0.20 = 0.50 m ✓合计 = 0.30 + 0.20 = 0.50 m ✓

In series, both springs carry the SAME force, but串联时两根弹簧承受同样大的力，但

stretch by different amounts.伸长量各不相同。

Two identical springs of constant $k$ are arranged in parallel. What is the equivalent spring constant?两根劲度系数同为 $k$ 的相同弹簧并联，等效劲度系数是多少？

$2k$

$k/2$

$k$

$4k$

Correct. Parallel springs: $k_{\text{eq}} = k_1 + k_2 = k + k = 2k$.正确。并联弹簧：$k_{\text{eq}} = k_1 + k_2 = k + k = 2k$。

Not quite. Parallel springs add directly: $k_{\text{eq}} = k + k = 2k$. You might be thinking of the series formula (which gives $k/2$).不对。并联弹簧直接相加：$k_{\text{eq}} = k + k = 2k$。你可能把串联公式（结果为 $k/2$）记反了。

Hooke's Law Explorer胡克定律探索器

Stretch or compress a spring and see how force scales linearly with displacement.拉伸或压缩弹簧，观察弹簧力如何随位移线性变化。

Spring constant k劲度系数 k 200 N/m

Displacement Δx位移 Δx 0.30 m

Spring Force弹簧力

−60.0

N (restoring)N（回复力）

PE Stored储存弹性势能

9.0

J (½kΔx²)

State状态

Stretched

Topic 2.9专题 2.9

Resistive Forces阻力

A resistive (drag) force opposes an object's velocity. In the simplest model, it's proportional to speed:

阻力（resistive force / drag）的方向与物体速度方向相反。在最简单的模型中，它与速率成正比：

Linear Drag Force线性阻力

$$\vec{F}_r = -k\vec{v}$$

$k$ is a drag coefficient (units: kg/s). The force always opposes the velocity.$k$ 为阻力系数（单位 kg/s）。该力始终与速度方向相反。

Applying Newton's second law with both gravity and linear drag yields a first-order linear ODE. For an object falling from rest under gravity with drag $F_r = -kv$:

同时考虑重力与线性阻力时，牛顿第二定律给出一个一阶线性常微分方程（ordinary differential equation / ODE）。对一个在 $F_r = -kv$ 阻力下、从静止开始自由下落的物体：

$$m\frac{dv}{dt} = mg - kv$$

Using separation of variables and integrating, the velocity as a function of time is an exponential approach to terminal velocity:

用分离变量法（separation of variables）积分后，速度随时间按指数方式趋向终极速度（terminal velocity）：

$$v(t) = \frac{mg}{k}\left(1 - e^{-kt/m}\right) = v_T\left(1 - e^{-t/\tau}\right)$$

where $v_T = mg/k$ is the terminal velocity and $\tau = m/k$ is the time constant.

其中 $v_T = mg/k$ 是终极速度，$\tau = m/k$ 是时间常数（time constant）。

Terminal Velocity — Linear Drag终极速度 —— 线性阻力

$$v_T = \frac{mg}{k} \qquad (\text{for }F_r = -kv)$$

Reached when $F_{\text{net}} = 0$: drag balances gravity, so $mg = kv_T$. Only valid for the linear drag model.当 $F_{\text{net}} = 0$ 时达到：阻力与重力平衡，故 $mg = kv_T$。仅在线性阻力模型下成立。

Don't Confuse Linear and Quadratic Drag

The terminal velocity formula depends on which drag model applies:

Linear drag (low speed, laminar flow): $F_r = kv \;\Rightarrow\; v_T = \dfrac{mg}{k}$
Quadratic drag (high speed, turbulent flow): $F_r = bv^2 \;\Rightarrow\; v_T = \sqrt{\dfrac{mg}{b}}$

AP Physics C uses whichever model the problem specifies. Read the force law in the prompt before invoking a terminal velocity formula from memory.

别把线性阻力与二次阻力搞混

终极速度公式取决于题目采用哪种阻力模型：

线性阻力（低速、层流 laminar flow）：$F_r = kv \;\Rightarrow\; v_T = \dfrac{mg}{k}$
二次阻力（高速、湍流 turbulent flow）：$F_r = bv^2 \;\Rightarrow\; v_T = \sqrt{\dfrac{mg}{b}}$

AP Physics C 用哪种模型完全看题目说明。在动笔套终极速度公式之前，先看清题干给出的力的形式。

Asymptotic Behavior Position, velocity, and acceleration all exhibit exponential behavior. Velocity rises exponentially toward $v_T$; acceleration starts at $g$ and decays exponentially to zero; position grows linearly at late times (constant $v_T$) but curves at early times. On graphs, recognize the characteristic exponential approach shape — rapid initial change that gradually levels off.

渐近行为 位置、速度、加速度都表现出指数行为：速度按指数趋近 $v_T$；加速度从 $g$ 开始按指数衰减到零；位置在后期随时间线性增长（因为速度恒为 $v_T$），早期则呈现曲线。看图时要认得这种"前期变化迅速、后期逐渐变平"的典型指数趋近形状。

Derivation — Linear Drag ODE to $v(t)$

This is the single most-tested calculus derivation in Unit 2. Object of mass $m$ dropped from rest, drag $\vec{F}_r = -k\vec{v}$ (down-positive).

Step 1 — Newton's second law with down positive:

$$m\,\frac{dv}{dt} = mg - kv$$

Step 2 — Separate variables:

$$\frac{dv}{mg - kv} = \frac{dt}{m}$$

Step 3 — Integrate from rest:

$$\int_0^{v}\!\frac{dv'}{mg - kv'} = \int_0^{t}\!\frac{dt'}{m}$$

$$-\tfrac{1}{k}\bigl[\ln(mg - kv) - \ln(mg)\bigr] = \frac{t}{m}$$

Step 4 — Rearrange and exponentiate:

$$\ln\!\left(\frac{mg - kv}{mg}\right) = -\frac{kt}{m} \;\Longrightarrow\; \frac{mg - kv}{mg} = e^{-kt/m}$$

Step 5 — Solve for $v(t)$:

$$\boxed{\,v(t) = \frac{mg}{k}\bigl(1 - e^{-kt/m}\bigr) = v_T\bigl(1 - e^{-t/\tau}\bigr)\,}$$

Sanity checks: at $t \to 0$, $v \to 0$ ✓; as $t \to \infty$, $v \to mg/k = v_T$ ✓.

Two constants to memorize: terminal velocity $v_T = mg/k$ and time constant $\tau = m/k$. After one $\tau$, the object reaches $(1 - e^{-1}) \approx 63\%$ of $v_T$; after three $\tau$'s, $\approx 95\%$.

推导 —— 从线性阻力 ODE 推出 $v(t)$

这是第 2 单元最常考的微积分推导。质量为 $m$ 的物体从静止下落，阻力 $\vec{F}_r = -k\vec{v}$（取向下为正）。

第 1 步 —— 取向下为正，写牛顿第二定律：

$$m\,\frac{dv}{dt} = mg - kv$$

第 2 步 —— 分离变量（separation of variables）：

$$\frac{dv}{mg - kv} = \frac{dt}{m}$$

第 3 步 —— 从静止开始积分：

$$\int_0^{v}\!\frac{dv'}{mg - kv'} = \int_0^{t}\!\frac{dt'}{m}$$

$$-\tfrac{1}{k}\bigl[\ln(mg - kv) - \ln(mg)\bigr] = \frac{t}{m}$$

第 4 步 —— 整理并取指数：

$$\ln\!\left(\frac{mg - kv}{mg}\right) = -\frac{kt}{m} \;\Longrightarrow\; \frac{mg - kv}{mg} = e^{-kt/m}$$

第 5 步 —— 解出 $v(t)$：

$$\boxed{\,v(t) = \frac{mg}{k}\bigl(1 - e^{-kt/m}\bigr) = v_T\bigl(1 - e^{-t/\tau}\bigr)\,}$$

极限校验：$t \to 0$ 时 $v \to 0$ ✓；$t \to \infty$ 时 $v \to mg/k = v_T$ ✓。

必须记住的两个常数：终极速度 $v_T = mg/k$ 与时间常数 $\tau = m/k$。经过一个 $\tau$，物体达到 $(1 - e^{-1}) \approx 63\%$ 的 $v_T$；三个 $\tau$ 后达到约 95%。

Worked Example — Deriving v(t) with Drag (Full Work)例题 —— 含阻力推导 v(t)（完整过程）

An object of mass $m$ is dropped from rest in a medium where drag force is $\vec{F}_r = -k\vec{v}$. Derive $v(t)$.

质量为 $m$ 的物体在阻力为 $\vec{F}_r = -k\vec{v}$ 的介质中从静止下落。推导 $v(t)$。

Newton's second law with down positive:取向下为正，写牛顿第二定律：

$$m\,\frac{dv}{dt} = mg - kv$$

Separate variables:分离变量：

$$\frac{dv}{mg - kv} = \frac{dt}{m}$$

Integrate from $v = 0$ at $t = 0$:从 $t = 0$ 时 $v = 0$ 开始积分：

$$\int_0^{v}\frac{dv'}{mg - kv'} = \int_0^{t}\frac{dt'}{m}$$

$$-\tfrac{1}{k}\bigl[\ln(mg - kv') \bigr]_0^{v} = \frac{t}{m}$$

$$-\tfrac{1}{k}\bigl[\ln(mg - kv) - \ln(mg)\bigr] = \frac{t}{m}$$

$$\ln\!\left(\frac{mg - kv}{mg}\right) = -\frac{kt}{m}$$

Exponentiate and solve for $v$:取指数，求解 $v$：

$$\frac{mg - kv}{mg} = e^{-kt/m}$$

$$v(t) = \frac{mg}{k}\bigl(1 - e^{-kt/m}\bigr)$$

Limits: at $t \to 0$, $v \to 0$ ✓; as $t \to \infty$, $v \to mg/k = v_T$ ✓.极限校验：$t \to 0$ 时 $v \to 0$ ✓；$t \to \infty$ 时 $v \to mg/k = v_T$ ✓。

Terminal Velocity Explorer终极速度探索器

Watch velocity approach terminal velocity exponentially. Adjust mass and drag coefficient.调节质量与阻力系数，观察速度按指数趋近终极速度的过程。

Mass m质量 m 2.0 kg

Drag coeff k阻力系数 k 0.50 kg/s

Terminal Vel. v_T终极速度 v_T

39.2

m/s

Time Const. τ时间常数 τ

4.0

v at t = τt = τ 时的 v

24.8

m/s (63% of v_T)m/s（v_T 的 63%）

Topic 2.10专题 2.10

Circular Motion圆周运动

An object moving along a curved path must be accelerating, even if its speed is constant, because the direction of velocity is changing. The component of acceleration directed toward the center of the circular path is called centripetal acceleration:

沿曲线运动的物体必定在加速——即便速率不变，因为速度方向在变。指向圆周轨道中心的那一加速度分量称为向心加速度（centripetal acceleration）：

Centripetal Acceleration向心加速度

$$a_c = \frac{v^2}{r}$$

Directed toward the center. $v$ = tangential speed, $r$ = radius of curvature.方向指向圆心。$v$ 为切向速率，$r$ 为曲率半径。

There is no separate "centripetal force." Centripetal acceleration is produced by whatever real forces (or components of forces) happen to point toward the center — tension, gravity, normal force, friction, or some combination. Always write Newton's second law along the radial direction:

不存在单独的"向心力（centripetal force）"。向心加速度是由那些恰好指向圆心的真实力（或其分量）造成的——张力、重力、法向力、摩擦力或它们的组合。沿径向写出牛顿第二定律即可：

$$\sum F_{\text{toward center}} = \frac{mv^2}{r}$$

Key Scenarios几个关键情景

Top of a vertical loop: At the top, both gravity and normal force point toward the center (downward). The minimum speed to maintain contact occurs when $N = 0$, giving:

竖直圆轨道顶端：在顶端，重力与法向力都指向圆心（向下）。保持接触所需的最小速度发生在 $N = 0$ 时：

Minimum Speed at Top of Loop圆轨道顶端最小速度

$$v_{\min} = \sqrt{gr}$$

Banked curves: On a banked surface, components of the normal force and friction can both contribute to the centripetal force. The "ideal" banking angle (no friction needed) satisfies $\tan\theta = v^2/(rg)$.

倾斜弯道（banked curve）：在倾斜路面上，法向力与摩擦力的分量都可以为向心力做贡献。"理想"倾角（不需要摩擦力）满足 $\tan\theta = v^2/(rg)$。

Conical pendulum: The tension's horizontal component provides centripetal acceleration, while its vertical component balances gravity.

圆锥摆（conical pendulum）：张力的水平分量提供向心加速度，竖直分量平衡重力。

Period & Frequency周期与频率

Period of Uniform Circular Motion匀速圆周运动的周期

$$T = \frac{2\pi r}{v} = \frac{1}{f}$$

$T$ = time for one revolution. $f$ = frequency (revolutions per second).$T$ 为转一圈所用时间，$f$ 为频率（每秒圈数）。

Kepler's Third Law (Circular Orbits)开普勒第三定律（圆轨道）

For a satellite in circular orbit, gravitational force provides the centripetal acceleration. Setting $GMm/r^2 = mv^2/r$ and using $T = 2\pi r/v$ gives:

圆轨道上的卫星，引力提供向心加速度。令 $GMm/r^2 = mv^2/r$，再代入 $T = 2\pi r/v$，得开普勒第三定律（Kepler's third law）：

Kepler's Third Law — Circular Orbits开普勒第三定律 —— 圆轨道

$$T^2 = \frac{4\pi^2}{GM}R^3$$

$M$ = mass of central body, $R$ = orbital radius. $T^2 \propto R^3$.$M$ 为中心天体质量，$R$ 为轨道半径。$T^2 \propto R^3$。

Exam Trap — "Centripetal Force" Never write "centripetal force" as a separate force on a free-body diagram. It is not a new force — it is the net radial force produced by real forces. Listing it separately leads to double-counting.

考试陷阱 —— "向心力" 永远不要在 FBD 上把"向心力"列为一项独立的力。它并不是新的力——而是真实力所产生的径向净力。把它当作单独的一项会造成重复计入。

Worked Example — Car on a Banked Curve例题 —— 倾斜弯道上的汽车

A car travels at speed $v$ around a frictionless banked curve of radius $r$ and bank angle $\theta$. Derive the condition for no sliding.

汽车以速度 $v$ 通过半径 $r$、倾角 $\theta$ 的无摩擦倾斜弯道（banked curve）。推导不打滑的条件。

FBD: weight mg downward, normal force N ⊥ to surface
No friction → only N and mg.
Axes: x = horizontal toward center, y = vertical up.

FBD：重力 mg 向下、法向力 N 垂直于路面
无摩擦 → 仅 N 与 mg。
坐标轴：x 取水平指向圆心，y 取竖直向上。

$$y-direction: N cos \theta − mg = 0 \to N = mg/cos \theta$$

$$x-direction (centripetal): N sin \theta = mv^2/r$$

Substitute N:代入 N：

$$(mg/cos \theta) sin \theta = mv^2/r$$

$$mg tan \theta = mv^2/r$$

$$tan \theta = v^2/(rg)$$

This gives the "ideal" banking angle for speed v.这就是速度 v 对应的"理想"倾角。

Faster → need friction inward. Slower → need friction outward.速度更快 → 需要向内的摩擦力；速度更慢 → 需要向外的摩擦力。

A ball on a string moves in a horizontal circle at constant speed. If the string length is doubled while keeping the period the same, the centripetal acceleration:一只系在绳上的小球沿水平圆周匀速运动。若绳长加倍而周期不变，向心加速度：

Halves减为一半

Stays the same不变

Quadruples变 4 倍

Doubles变 2 倍

Correct. With constant $T$, speed $v = 2\pi r/T$ doubles when $r$ doubles. Then $a_c = v^2/r = (2v_0)^2/(2r_0) = 4v_0^2/(2r_0) = 2v_0^2/r_0 = 2a_0$. Centripetal acceleration doubles.正确。$T$ 不变时 $v = 2\pi r/T$，$r$ 加倍意味着 $v$ 也加倍。于是 $a_c = v^2/r = (2v_0)^2/(2r_0) = 2v_0^2/r_0 = 2a_0$，向心加速度变为 2 倍。

Not quite. Write $a_c = 4\pi^2 r / T^2$. With $T$ constant and $r$ doubled, $a_c$ doubles.不对。写成 $a_c = 4\pi^2 r / T^2$。$T$ 不变、$r$ 加倍，$a_c$ 变为 2 倍。

Circular Motion Explorer圆周运动探索器

See how speed and radius set the centripetal acceleration $a_c = v^2/r$.观察速率与半径如何决定向心加速度 $a_c = v^2/r$。

Speed v速率 v 4.0 m/s

Radius r半径 r 2.0 m

a_c = v²/r

8.0

m/s²

Period T周期 T

3.14

Min v (top of loop)圆轨道顶端最小 v

4.47

m/s (√gr)

Exam Prep考试备考

Exam Strategy考试策略

Multiple-Choice Tips

Many MCQs test whether you can identify the correct FBD, apply Newton's second law in component form, or recognize limiting cases (what happens as a variable approaches zero or infinity). Before computing, check limiting cases — they often eliminate 2–3 answers instantly. Pay special attention to questions about apparent weight and the distinction between $F_N$ and $mg$. Questions involving circular motion frequently test whether you incorrectly add "centripetal force" as a separate force.

选择题（Multiple Choice）提示

许多选择题考察你能否：选对 FBD、把牛顿第二定律按分量写出、识别极限情形（当某个量趋于零或无穷大时会发生什么）。在动笔算之前先查极限情形——往往能立刻排除 2–3 个选项。特别留意视重（apparent weight）类题，以及 $F_N$ 与 $mg$ 的区分。圆周运动题经常陷阱式地考你会不会错把"向心力（centripetal force）"作为单独的力加上去。

Free-Response (FRQ) Tips

For the Qualitative/Quantitative Translation (QQT): First make a claim using physical reasoning (no equations), then derive the relevant equation, then connect the two. Show all steps in derivations — partial credit is generous. For experimental design questions, always specify what you measure, what you vary, what you hold constant, and how you extract the desired quantity from a linear graph. When setting up Newton's second law problems, always start with a clearly labeled FBD and state your coordinate system.

自由作答题（FRQ）提示

面对定性/定量转换（Qualitative/Quantitative Translation, QQT）：先用物理推理给出一个结论性论断（不写方程），再推导相关方程，然后把两者衔接起来。推导一定把每一步都写出来——部分分给得很大方。实验设计题永远要写清：测什么、变什么、控制哪些、如何从线性图中读出目标量。建立牛顿第二定律方程之前，永远从一张清楚标注的 FBD 与明确写出的坐标系开始。

Watch Out注意事项

Common Mistakes常见错误

Top Point-Losing Errors

1. Setting $F_N = mg$ automatically. The normal force depends on all perpendicular forces, not just weight. On an incline or with applied forces, you must solve for it.

2. Adding "centripetal force" to the FBD. Centripetal acceleration is caused by existing forces — gravity, tension, normal, friction. It's not a new force.

3. Confusing static and kinetic friction. Static friction is variable ($\leq \mu_s F_N$); kinetic friction is fixed ($= \mu_k F_N$). Use the right one for the situation.

4. Wrong sign in Hooke's law. The spring force is $-k\Delta x$. If $\Delta x$ is positive (stretched), the force is negative (restoring). Keep your signs consistent with your coordinate system.

5. Forgetting that $r$ in universal gravitation is from center to center. For a planet of radius $R$, the surface gravity uses $r = R$, not the altitude above the surface.

6. Using the wrong mass in the Atwood machine. The system mass is $m_1 + m_2$ (always), not $m_2 - m_1$.

7. Not separating variables correctly in resistive-force problems. Practice the ODE $m\,dv/dt = mg - kv$ until the separation-of-variables technique is automatic.

最常见的失分点

1. 不假思索地令 $F_N = mg$。法向力取决于所有垂直方向的力，不仅仅是重力。在斜面或有外加力时，必须解方程求 $F_N$。

2. 把"向心力"画进 FBD。向心加速度是已有的力（重力、张力、法向力、摩擦力）造成的，不是一个新的力。

3. 混淆静摩擦与动摩擦。静摩擦是可变的（$\leq \mu_s F_N$），动摩擦是固定的（$= \mu_k F_N$）。要按情景选对模型。

4. 胡克定律符号写错。弹簧力为 $-k\Delta x$。$\Delta x$ 为正（拉伸）时，力为负（指向回复）。务必让符号与坐标系一致。

5. 忘记万有引力中的 $r$ 是质心到质心。半径为 $R$ 的行星，表面 g 用 $r = R$，而不是相对地表的高度。

6. 阿特伍德机中用错"系统质量"。系统质量始终是 $m_1 + m_2$，不是 $m_2 - m_1$。

7. 阻力问题中分离变量错误。把 ODE $m\,dv/dt = mg - kv$ 的分离变量法练到自动反应。

Review复习

Flashcards闪卡

Click any card to flip it.

点击任意卡片即可翻面。

Newton's Second Law牛顿第二定律

$$\vec{F}_\text{net} = m\vec{a}$$

Net external force = mass × acceleration of the center of mass.

净外力 = 质量 × 质心加速度。

Universal Gravitation万有引力

$$F_g = G\frac{m_1 m_2}{r^2}$$

Attractive, along the line of centers. $r$ is center-to-center distance.

吸引力，沿两质心连线方向。$r$ 是质心到质心的距离。

Centripetal Acceleration向心加速度

$$a_c = \frac{v^2}{r}$$

Always points toward the center of the circular path.

方向始终指向圆周轨道的圆心。

Hooke's Law胡克定律

$$F_s = -k\Delta x$$

Restoring force; proportional to displacement from equilibrium.

回复力；与相对平衡位置的位移成正比。

Terminal Velocity (Linear Drag)终极速度（线性阻力）

$$v_T = \frac{mg}{k}$$

Linear drag $F_r=-kv$. Quadratic drag gives $v_T=\sqrt{mg/b}$.

线性阻力 $F_r=-kv$。若为二次阻力，则 $v_T=\sqrt{mg/b}$。

Kepler's Third Law开普勒第三定律

$$T^2 = \frac{4\pi^2}{GM}R^3$$

$T^2 \propto R^3$ for circular orbits.

圆轨道上满足 $T^2 \propto R^3$。

Kinetic Friction动摩擦

$$f_k = \mu_k F_N$$

Opposes relative sliding; $\mu_k$ depends on the surfaces, not on area.

方向与相对滑动方向相反；$\mu_k$ 取决于接触面材料，与接触面积无关。

Center of Mass (continuous)质心（连续）

$$\vec{r}_\text{cm} = \frac{\int \vec{r}\,dm}{\int dm}$$

Express $dm$ via $\lambda\,dx$, $\sigma\,dA$, or $\rho\,dV$.

把 $dm$ 写成 $\lambda\,dx$、$\sigma\,dA$ 或 $\rho\,dV$。

Assessment测评

Unit Quiz单元测验

Q1. A block of mass $m$ is pulled along a frictionless horizontal surface by a force $F$ applied at angle $\theta$ above horizontal. What is the normal force on the block?第 1 题。质量为 $m$ 的木块沿无摩擦水平面被一个与水平方向成 $\theta$ 角向上的力 $F$ 拉动。木块所受法向力是多少？

$mg$

$mg - F\sin\theta$

$mg + F\sin\theta$

$mg - F\cos\theta$

Correct. Vertical equilibrium: $N + F\sin\theta = mg$, so $N = mg - F\sin\theta$. The upward component of $F$ partially lifts the block, reducing the normal force.正确。竖直方向平衡：$N + F\sin\theta = mg$，故 $N = mg - F\sin\theta$。$F$ 的向上分量部分托起木块，使法向力减小。

Not quite. The applied force has a vertical component $F\sin\theta$ that supports part of the weight. From $N + F\sin\theta = mg$, we get $N = mg - F\sin\theta$.不对。外加力的竖直分量 $F\sin\theta$ 承担了部分重力。由 $N + F\sin\theta = mg$ 得 $N = mg - F\sin\theta$。

Q2. A 2 kg object moves at $4\;\text{m/s}$ in a circle of radius $0.5\;\text{m}$. What is the net centripetal force on the object?第 2 题。2 kg 的物体以 $4\;\text{m/s}$ 沿半径 $0.5\;\text{m}$ 的圆周运动。物体所受净向心力是多少？

$8\;\text{N}$

$16\;\text{N}$

$64\;\text{N}$

$32\;\text{N}$

Correct. $F_c = mv^2/r = 2 \times 16/0.5 = 64\;\text{N}$.正确。$F_c = mv^2/r = 2 \times 16/0.5 = 64\;\text{N}$。

Not quite. $F_c = mv^2/r = 2(4^2)/0.5 = 2 \times 16/0.5 = 64\;\text{N}$.不对。$F_c = mv^2/r = 2(4^2)/0.5 = 2 \times 16/0.5 = 64\;\text{N}$。

Q3. Inside a uniform spherical shell of mass $M$, the gravitational force on a particle of mass $m$ is:第 3 题。在质量为 $M$ 的均匀球壳内部，对质量 $m$ 的质点所受引力是：

Zero everywhere inside壳内处处为零

$GMm/R^2$ at the center球心处为 $GMm/R^2$

Proportional to distance from the center与到球心距离成正比

Inverse-square with distance from the center与到球心距离按平方反比

Correct. Newton's shell theorem: a uniform spherical shell exerts zero net gravitational force on any object inside it. Every pull from one side is exactly cancelled by the pull from the other side.正确。牛顿壳层定理（shell theorem）：均匀球壳对其内部任何物体的净引力为零。一侧的拉力都被对侧的拉力恰好抵消。

Not quite. By Newton's shell theorem, the net gravitational force inside a uniform spherical shell is exactly zero.不对。由牛顿壳层定理，均匀球壳内部净引力恰好为零。

Q4. An object falling through a viscous medium experiences a drag force $\vec{F}_r = -k\vec{v}$. The time constant $\tau$ for the object to approach terminal velocity is:第 4 题。物体下落于粘性介质中，受到阻力 $\vec{F}_r = -k\vec{v}$。物体趋近终极速度的时间常数 $\tau$ 为：

$k/m$

$mg/k$

$k/mg$

$m/k$

Correct. The solution is $v(t) = v_T(1 - e^{-t/\tau})$ where $\tau = m/k$. After one time constant, the object reaches about 63% of terminal velocity.正确。解为 $v(t) = v_T(1 - e^{-t/\tau})$，其中 $\tau = m/k$。经过一个时间常数，物体达到约 63% 的终极速度。

Not quite. From the ODE solution, the exponential decay constant is $k/m$, so the time constant is $\tau = m/k$. Don't confuse this with the terminal velocity $v_T = mg/k$.不对。ODE 解的指数衰减系数为 $k/m$，故时间常数 $\tau = m/k$。注意不要与终极速度 $v_T = mg/k$ 混淆。

Q5. A satellite orbits Earth at radius $R$. If the orbit radius is increased to $4R$, the new orbital period is:第 5 题。卫星绕地球的轨道半径为 $R$。若半径增至 $4R$，新周期是：

$4T$

$8T$

$16T$

$2T$

Correct. By Kepler's third law, $T^2 \propto R^3$. If $R \to 4R$, then $T^2 \to 4^3 T^2 = 64T^2$, so $T \to 8T$.正确。由开普勒第三定律，$T^2 \propto R^3$。$R \to 4R$，故 $T^2 \to 4^3 T^2 = 64T^2$，得 $T \to 8T$。

Not quite. $T^2 \propto R^3$, so $T_{\text{new}}^2/T^2 = (4R)^3/R^3 = 64$, giving $T_{\text{new}} = 8T$.不对。$T^2 \propto R^3$，故 $T_{\text{new}}^2/T^2 = (4R)^3/R^3 = 64$，得 $T_{\text{new}} = 8T$。

Worked Example — Atwood Machine with Friction on a Table (FRQ Style)例题 —— 桌面摩擦的阿特伍德机（FRQ 风格）

Mass $m_1 = 3.0\;\text{kg}$ sits on a horizontal table with kinetic-friction coefficient $\mu_k = 0.20$. A light string runs over a frictionless, massless pulley to a hanging mass $m_2 = 5.0\;\text{kg}$. The system is released from rest. Find (a) the acceleration $a$ of the masses and (b) the tension $T$ in the string. Take $g = 9.8\;\text{m/s}^2$.

质量 $m_1 = 3.0\;\text{kg}$ 置于动摩擦系数 $\mu_k = 0.20$ 的水平桌面上。一根轻绳越过无摩擦、无质量的理想滑轮，接到悬挂物 $m_2 = 5.0\;\text{kg}$。系统从静止释放。求 (a) 两物体的加速度 $a$；(b) 绳中张力 $T$。取 $g = 9.8\;\text{m/s}^2$。

Identify辨识（Identify）

Principle: Newton's second law applied to each mass separately, with the string constraint $|a_1| = |a_2| = a$ (massless inextensible string over a massless pulley). Kinetic friction opposes the motion of $m_1$.

原理：分别对两个质点用牛顿第二定律，并利用绳约束 $|a_1| = |a_2| = a$（无质量不可伸长绳跨过无质量滑轮）。动摩擦力的方向与 $m_1$ 的运动方向相反。

Table mass $m_1$桌上质量 $m_1$ Hanging mass $m_2$悬挂质量 $m_2$ Friction coefficient $\mu_k$摩擦系数 $\mu_k$ Gravity $g$重力加速度 $g$

Set Up建模（Set Up）

Choose "rightward for $m_1$" and "downward for $m_2$" as the positive direction in each FBD — both align with the assumed motion. Friction on $m_1$ is $f_k = \mu_k N = \mu_k m_1 g$, pointing left (opposing the pull).

在各自的 FBD 中选取正方向："$m_1$ 取向右为正"、"$m_2$ 取向下为正"——都与假设运动方向一致。$m_1$ 上的摩擦力 $f_k = \mu_k N = \mu_k m_1 g$，指向左（与拉力相反）。

$$T - \mu_k m_1 g = m_1 a \qquad m_2 g - T = m_2 a$$

Execute执行（Execute）

(a) Add the two equations to eliminate $T$(a) 两式相加消去 $T$

$$m_2 g - \mu_k m_1 g = (m_1 + m_2)\,a$$

$$a = \frac{(m_2 - \mu_k m_1)\,g}{m_1 + m_2} = \frac{(5.0 - 0.20 \times 3.0)(9.8)}{8.0} = \frac{4.4 \times 9.8}{8.0} \approx 5.39\;\text{m/s}^2$$

(b) Back-substitute for $T$(b) 代回求 $T$

$$T = m_1 a + \mu_k m_1 g = m_1 (a + \mu_k g) = 3.0\,(5.39 + 0.20 \times 9.8)$$

$$T = 3.0 \times 7.35 \approx 22.1\;\text{N}$$

Evaluate评估（Evaluate）

Bound: $T < m_2 g = 49\;\text{N}$, otherwise $m_2$ would not accelerate downward. ✓上界：$T < m_2 g = 49\;\text{N}$，否则 $m_2$ 不会向下加速。✓

Bound: $T > \mu_k m_1 g = 5.88\;\text{N}$, otherwise $m_1$ would not slide. ✓下界：$T > \mu_k m_1 g = 5.88\;\text{N}$，否则 $m_1$ 不会滑动。✓

Limit $\mu_k \to 0$: $a \to m_2 g/(m_1 + m_2) = 6.13\;\text{m/s}^2$ — the standard frictionless Atwood result. Friction reduces $a$ as expected. ✓极限 $\mu_k \to 0$：$a \to m_2 g/(m_1 + m_2) = 6.13\;\text{m/s}^2$——即经典无摩擦阿特伍德机的结果。摩擦力按预期使 $a$ 减小。✓

Sanity check the system condition: motion requires $m_2 > \mu_k m_1$, i.e. $5.0 > 0.6$. Comfortably satisfied. ✓系统启动条件校验：需要 $m_2 > \mu_k m_1$，即 $5.0 > 0.6$，明显满足。✓

Worked Example — Block on Incline with Applied Force at an Angle (FRQ Style)例题 —— 斜面上加力的木块（FRQ 风格）

A block of mass $m = 4.0\;\text{kg}$ rests on an incline of angle $\theta = 30^{\circ}$ with kinetic-friction coefficient $\mu_k = 0.25$. An applied force $F = 30\;\text{N}$ is directed parallel to the incline, pulling the block up the slope. Find the block's acceleration along the incline. Take $g = 9.8\;\text{m/s}^2$.

质量 $m = 4.0\;\text{kg}$ 的木块置于倾角 $\theta = 30^{\circ}$、动摩擦系数 $\mu_k = 0.25$ 的斜面上。外加力 $F = 30\;\text{N}$ 沿斜面向上拉动木块。求木块沿斜面方向的加速度。取 $g = 9.8\;\text{m/s}^2$。

Identify辨识（Identify）

Principle: Newton's second law in incline coordinates ($x$ along the slope, $y$ perpendicular). Friction opposes motion, so it points down-slope when the block moves up. Normal force comes from the perpendicular force balance.

原理：在斜面坐标系（$x$ 沿斜面，$y$ 垂直于斜面）中应用牛顿第二定律。摩擦力方向与运动方向相反——木块沿斜面向上运动时，摩擦力沿斜面向下。法向力由垂直方向力平衡求得。

Mass $m$质量 $m$ Incline angle $\theta$斜面倾角 $\theta$ Friction coefficient $\mu_k$摩擦系数 $\mu_k$ Applied force $F$ (up-slope)外加力 $F$（沿斜面向上）

Set Up建模（Set Up）

In incline coordinates the gravity vector splits into $-mg\sin\theta$ (down-slope) and $-mg\cos\theta$ (into the surface). Perpendicular balance gives $N = mg\cos\theta$, so $f_k = \mu_k mg\cos\theta$ (down-slope, opposing the assumed up-slope motion).

在斜面坐标系中，重力矢量分解为 $-mg\sin\theta$（沿斜面向下）与 $-mg\cos\theta$（垂直压向斜面）。垂直方向力平衡得 $N = mg\cos\theta$，于是 $f_k = \mu_k mg\cos\theta$（沿斜面向下，与假设的上行运动方向相反）。

$$F - mg\sin\theta - \mu_k mg\cos\theta = m a$$

Execute执行（Execute）

Solve symbolically, then plug in先写出符号解，再代入数值

$$a = \frac{F - mg(\sin\theta + \mu_k\cos\theta)}{m}$$

$$mg\sin\theta = (4.0)(9.8)(0.500) = 19.6\;\text{N}$$

$$\mu_k mg\cos\theta = (0.25)(4.0)(9.8)(0.866) = 8.49\;\text{N}$$

$$a = \frac{30 - 19.6 - 8.49}{4.0} = \frac{1.91}{4.0} \approx 0.48\;\text{m/s}^2$$

Evaluate评估（Evaluate）

$a > 0$ confirms our assumption that the block accelerates up the slope; the applied force exceeds the combined opposition of gravity and friction. ✓$a > 0$ 证实了"木块沿斜面向上加速"的假设——外加力大于重力沿斜面分量加摩擦力之和。✓

Marginal case: if $F$ had been just $mg\sin\theta + \mu_k mg\cos\theta = 28.1\;\text{N}$, the block would slide at constant velocity ($a = 0$). The chosen $F = 30\;\text{N}$ is barely above that threshold. ✓临界情形：若 $F$ 恰为 $mg\sin\theta + \mu_k mg\cos\theta = 28.1\;\text{N}$，木块将匀速滑动（$a = 0$）。本题取 $F = 30\;\text{N}$ 刚好略高于该阈值。✓

Common slip: signing friction the wrong way. Friction always opposes the actual (or assumed) motion, not the applied force — so when the block is being pulled up, $f_k$ adds to gravity in the down-slope direction. ✓常见错误：把摩擦力的方向画反。摩擦力始终与实际（或假设的）运动方向相反，而不是与外加力相反——故木块被向上拉动时，$f_k$ 与重力分量同向，都指向斜面下方。✓

Unit 2: Force & Translational Dynamics第 2 单元：力与平动动力学

Systems & Center of Mass系统与质心

Center of Mass for Discrete Particles离散质点系的质心

Center of Mass for Continuous Objects连续体的质心

Forces & Free-Body Diagrams力与受力分析图

Drawing Free-Body Diagrams绘制受力分析图

Newton's Third Law牛顿第三定律

Tension in Strings绳中的张力

Newton's First Law牛顿第一定律

Newton's Second Law牛顿第二定律

Gravitational Force万有引力

Gravitational Field引力场

Apparent Weight & Weightlessness视重与失重

Newton's Shell Theorem牛顿壳层定理（shell theorem）

Kinetic & Static Friction动摩擦与静摩擦

Kinetic Friction动摩擦

Static Friction静摩擦

Spring Forces弹力

Combinations of Springs弹簧的组合

Resistive Forces阻力

Circular Motion圆周运动

Key Scenarios几个关键情景

Period & Frequency周期与频率

Kepler's Third Law (Circular Orbits)开普勒第三定律（圆轨道）

Exam Strategy考试策略

Common Mistakes常见错误

Flashcards闪卡

Unit Quiz单元测验

Newton's Shell Theorem牛顿壳层定理（`shell theorem`）