AP Physics C: Mechanics
Unit 3: Work, Energy, & Power第 3 单元：功、能量与功率
Conservation as a foundational principle of physics, with work as the primary agent of change for energy. Master scalar quantities, integral formulations, and graphical techniques the AP exam demands.
守恒（conservation）是物理学最根本的原则之一，而功（work）是改变能量的主要途径。掌握标量、积分表达式与 AP 考试要求的图像分析技巧。
15–25% Exam Weight考试占分 15–25%
~12–17 Class Periods约 12–17 课时
5 Topics5 个专题
Topic 3.1专题 3.1
Translational Kinetic Energy平动动能
Kinetic energy is the energy an object possesses because it is moving. In AP Physics C: Mechanics, we focus on translational kinetic energy — the energy associated with an object's center-of-mass motion through space (rotational kinetic energy appears in Unit 6).
动能（kinetic energy）是物体由于运动而具有的能量。在 AP Physics C: Mechanics 中，我们重点研究平动动能（translational kinetic energy）——与物体质心（center of mass）在空间中的整体运动相关联的那部分能量（转动动能将在 Unit 6 出现）。
The quantity depends on two things: mass and speed. Because velocity is squared, even modest increases in speed produce large increases in kinetic energy. Doubling the speed quadruples the kinetic energy.
动能只与两个量有关：质量与速率。由于速度是平方出现的，速率的少许增加都会带来动能的大幅上升。速率翻倍，动能变为原来的四倍。
Key Insight
Kinetic energy is a scalar — it has no direction. It is always non-negative because both mass and the square of velocity are non-negative. Two objects moving in opposite directions at the same speed have the same kinetic energy.
关键洞察
动能是标量（scalar）——没有方向。由于质量与速度的平方都非负，动能始终非负。两个以相同速率反向运动的物体动能相同。
Frame Dependence
Different observers may disagree on the kinetic energy of the same object. A passenger sitting still on a train has zero kinetic energy in the train's frame, but may have enormous kinetic energy in the ground frame. Kinetic energy is frame-dependent, even though the laws of physics are the same in all inertial frames.
参考系依赖性
不同观察者对同一物体的动能可能给出不同的数值。坐在火车里不动的乘客在车厢参考系（reference frame）中动能为零，但在地面参考系中可能动能巨大。动能与参考系有关，尽管物理定律在所有惯性系（inertial frame）中都相同。
Worked Example — Comparing Kinetic Energies例题 —— 动能大小比较
A 1200 kg car moves at 15 m/s. A 0.005 kg bullet moves at 900 m/s. Which has greater kinetic energy?
一辆 1200 kg 的汽车以 15 m/s 行驶，一颗 0.005 kg 的子弹以 900 m/s 飞行。哪个动能更大？
Car汽车
$$K_\text{car} = \tfrac{1}{2}(1200)(15)^2 = \tfrac{1}{2}(1200)(225) = 135{,}000\;\text{J} = 135\;\text{kJ}$$
Bullet子弹
$$K_\text{bullet} = \tfrac{1}{2}(0.005)(900)^2 = \tfrac{1}{2}(0.005)(810{,}000) = 2025\;\text{J} \approx 2.0\;\text{kJ}$$
The car has about $67\times$ more KE than the bullet despite being far slower — its much greater mass dominates.尽管汽车慢得多，它的动能仍约为子弹的 $67$ 倍——巨大的质量起了主导作用。
Object A has twice the mass and half the speed of Object B. What is the ratio $K_A / K_B$?物体 A 的质量是物体 B 的两倍，速率是 B 的一半。比值 $K_A / K_B$ 等于多少？
(A) 2
(B) 1
(C) 1/2
(D) 1/4
Correct! Let B have mass $m$ and speed $v$: $K_B = \tfrac{1}{2}mv^2$. Then A has mass $2m$ and speed $v/2$: $K_A = \tfrac{1}{2}(2m)(v/2)^2 = \tfrac{1}{4}mv^2$. So $K_A/K_B = 1/2$. The squared dependence on speed wins over the linear dependence on mass.正确！设 B 的质量为 $m$、速率为 $v$：$K_B = \tfrac{1}{2}mv^2$。则 A 的质量为 $2m$、速率为 $v/2$：$K_A = \tfrac{1}{2}(2m)(v/2)^2 = \tfrac{1}{4}mv^2$。故 $K_A/K_B = 1/2$。速度的平方依赖盖过了质量的线性依赖。
Doubling mass multiplies $K$ by 2, but halving speed multiplies by $(1/2)^2 = 1/4$. Net factor: $2 \times 1/4 = 1/2$. The answer is (C).质量加倍使 $K$ 乘以 2，但速率减半使 $K$ 乘以 $(1/2)^2 = 1/4$。综合因子为 $2 \times 1/4 = 1/2$，故选 (C)。
Topic 3.2专题 3.2
Work功
Work is the mechanism by which energy is transferred into or out of a system by a force acting over a displacement. Forces cause accelerations (Unit 2), but they also cause energy changes — and work is the bridge between force and energy.
功（work）是力沿位移作用时，能量进入或离开系统的机制。力会引起加速度（见 Unit 2），但同时也引起能量变化——而功正是连接力与能量的桥梁。
Work is a scalar quantity. It can be positive (energy flows into the system), negative (energy flows out), or zero (no energy transfer). The sign depends on the angle between the force and displacement.
功是标量。可以为正（能量流入系统）、为负（能量流出），也可以为零（无能量传递）。正负号由力与位移之间的夹角决定。
The Dot Product and Constant-Force Work标量积与恒力做功
When a constant force $\vec{F}$ acts on an object undergoing displacement $\vec{d}$, the work done is the dot product of the two vectors. Only the component of force parallel to the displacement transfers energy. The perpendicular component changes direction without changing speed — think of centripetal force in circular motion.
当恒力 $\vec{F}$ 作用于发生位移 $\vec{d}$ 的物体时，所做的功就是两个矢量的标量积（dot product，又称点积）。只有与位移平行的分量传递能量；垂直分量只改变方向、不改变速率——想想圆周运动中的向心力（centripetal force）。
Sign Convention
When $\theta < 90°$: force has a component along displacement → $W > 0$ (positive work). When $\theta = 90°$: force is perpendicular → $W = 0$. When $\theta > 90°$: force opposes displacement → $W < 0$ (negative work). This is exactly what cosine does.
正负号约定
当 $\theta < 90°$：力沿位移方向有分量 → $W > 0$（正功，positive work）。当 $\theta = 90°$：力垂直于位移 → $W = 0$。当 $\theta > 90°$：力与位移方向相反 → $W < 0$（负功，negative work）。这正是余弦函数的行为。
Work by a Variable Force: Integration变力做功：积分
In AP Physics C, forces are often not constant — springs, gravity at large distances, and many applied forces vary with position. When force depends on position, we must integrate along the path. This is one of the signature calculus applications in the course.
在 AP Physics C 中，力往往不是恒定的——弹簧力、远距离的引力以及许多外加力都会随位置变化。当力依赖于位置时，必须沿路径做积分（integral）。这是本课程最具标志性的微积分应用之一。
Worked Example — Work Done by a Position-Dependent Force例题 —— 位置相关的变力做功
A force $F(x) = 3x^2 + 2$ (N) is applied as an object moves from $x = 1$ m to $x = 4$ m. Find the work done.
物体在 $x = 1$ m 到 $x = 4$ m 之间运动时受到力 $F(x) = 3x^2 + 2$（N）。求所做的功。
$$W = \int_1^4 (3x^2 + 2)\,dx$$
Integrate term by term:逐项积分：
$$W = \bigl[x^3 + 2x\bigr]_1^4$$
At $x = 4$: $64 + 8 = 72$. At $x = 1$: $1 + 2 = 3$.$x = 4$ 时：$64 + 8 = 72$。$x = 1$ 时：$1 + 2 = 3$。
$$W = 72 - 3 = 69\;\text{J}$$
Worked Example — Block on a Rough Incline (Energy Method)例题 —— 粗糙斜面上的滑块（能量方法）
A 3.0 kg block slides 4.0 m down a 30° incline starting from rest. The kinetic friction coefficient is $\mu_k = 0.20$. Find the speed at the bottom using the work–energy theorem.
一个 3.0 kg 的滑块从静止开始沿 30° 斜面下滑 4.0 m，动摩擦系数（coefficient of kinetic friction）$\mu_k = 0.20$。用动能定理（work–energy theorem）求其抵达底端时的速率。
Step 1 — Net force along the slide第 1 步 —— 沿斜面的净力
$$F_\parallel = mg\sin\theta - \mu_k\,mg\cos\theta$$
Step 2 — Net work over $d = 4.0$ m第 2 步 —— $d = 4.0$ m 上所做的净功
$$W_\text{net} = (mg\sin\theta - \mu_k\,mg\cos\theta)\,d$$
$$= (3.0)(9.8)\bigl[\sin 30° - 0.20\cos 30°\bigr](4.0)$$
$$= 29.4\,[0.500 - 0.173](4.0) = 38.5\;\text{J}$$
Step 3 — Apply $\Delta K = W_\text{net}$ with $K_i = 0$第 3 步 —— 代入 $\Delta K = W_\text{net}$（$K_i = 0$）
$$\tfrac{1}{2}mv_f^2 = 38.5 \;\Rightarrow\; v_f = \sqrt{\frac{2(38.5)}{3.0}} \approx 5.07\;\text{m/s}$$
Solving with $\vec{F}=m\vec{a}$ and kinematics gives the same answer, but energy methods skip finding the acceleration explicitly.用 $\vec{F}=m\vec{a}$ 加运动学也能得到同样答案，但能量方法不必显式求出加速度。
The Work–Energy Theorem动能定理
The net work done on an object — by all forces combined — equals the change in that object's kinetic energy. This exact theorem is derivable from Newton's second law via calculus and gives you a powerful alternative to $\vec{F} = m\vec{a}$ for solving problems.
作用在物体上的所有力共同所做的净功等于该物体动能的变化。这条精确定理（work–energy theorem）可由 Newton 第二定律通过微积分推出，是 $\vec{F} = m\vec{a}$ 之外的一种强有力解题工具。
Calculus Derivation — Memorize the Four Lines
This derivation comes up on the FRQ section often enough to be worth memorizing. Start from Newton's second law and integrate along the path.
Step 1 — Newton's second law for the net force along the motion:
$$F_\text{net} = m\,\frac{dv}{dt}$$
Step 2 — Multiply both sides by the displacement and integrate along the path:
$$W_\text{net} = \int_{x_i}^{x_f} F_\text{net}\,dx = \int_{x_i}^{x_f} m\,\frac{dv}{dt}\,dx$$
Step 3 — Swap the order of differentials using the chain rule:
$$\frac{dv}{dt}\,dx = \frac{dx}{dt}\,dv = v\,dv$$
$$W_\text{net} = \int_{v_i}^{v_f} m\,v\,dv = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2$$
Step 4 — Identify the right-hand side as the change in kinetic energy:
$$W_\text{net} = \Delta K$$
The key trick in step 3 is treating $\frac{dv}{dt}\,dx$ as $v\,dv$. If the FRQ asks you to derive the theorem, reproducing these four lines is worth full credit.
微积分推导 —— 把这四步背下来
这条推导在 FRQ 中出现频率足够高，值得背诵。从 Newton 第二定律出发，沿路径积分。
第 1 步 —— 沿运动方向的净力，由 Newton 第二定律：
$$F_\text{net} = m\,\frac{dv}{dt}$$
第 2 步 —— 两边乘以位移，并沿路径积分：
$$W_\text{net} = \int_{x_i}^{x_f} F_\text{net}\,dx = \int_{x_i}^{x_f} m\,\frac{dv}{dt}\,dx$$
第 3 步 —— 利用链式法则交换微分顺序：
$$\frac{dv}{dt}\,dx = \frac{dx}{dt}\,dv = v\,dv$$
$$W_\text{net} = \int_{v_i}^{v_f} m\,v\,dv = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2$$
第 4 步 —— 将右端识别为动能的变化：
$$W_\text{net} = \Delta K$$
第 3 步的关键技巧是把 $\frac{dv}{dt}\,dx$ 改写为 $v\,dv$。如果 FRQ 要求推导这一定理，把这四行写清楚就能拿到满分。
Why It Matters
When you care about speeds rather than accelerations, or when force varies with position, energy methods are often far easier than $\vec{F} = m\vec{a}$. The AP exam tests your ability to choose the best approach.
为什么重要
当题目关心速率而非加速度，或者力随位置变化时，能量方法往往比 $\vec{F} = m\vec{a}$ 简便得多。AP 考试就是要考你能否选对方法。
When friction is present, the energy dissipated equals the friction force times the total path length (not displacement). This energy leaves the mechanical system as thermal energy and sound.
存在摩擦时，耗散的能量等于摩擦力乘以总路程（path length，不是位移）。这部分能量以热能（thermal energy）和声能离开机械系统。
Conservative vs. Nonconservative Forces保守力与非保守力
Conservative forces do work that depends only on starting and ending positions — not on the path taken. A round trip gives zero net work. Gravity, spring forces, and electrostatic forces are conservative. Because their work is path-independent, we can define a potential energy for each.
保守力（conservative force）所做的功只与起点和终点位置有关，与路径无关。沿闭合路径走一圈做功为零。重力、弹簧力、静电力都是保守力。正因其做功与路径无关，我们才能为每种保守力定义一个势能（potential energy）。
Nonconservative forces do path-dependent work. The most common are friction and air resistance. A box sliding back and forth loses energy on every pass — friction's work is always negative and accumulates. You cannot define a potential energy for friction.
非保守力（nonconservative force）所做的功与路径相关，最常见的是摩擦力和空气阻力。来回滑动的木箱每往返一次都损失能量——摩擦力做的功始终为负且累积。摩擦力没有对应的势能。
Exam Insight
The AP exam often asks whether a force is conservative or nonconservative, and what this implies about path independence and potential energy. Remember: potential energies can only be defined for conservative forces.
应试要点
AP 经常考某个力是保守还是非保守，以及由此能否说做功与路径无关、能否定义势能。记住：势能只能对保守力定义。
A block is pulled 5 m across a rough floor by a rope at $30°$ above horizontal with tension 40 N. Kinetic friction is 12 N. What is the net work done on the block?一根绳沿水平方向以上 $30°$ 拉一个木块在粗糙地面上前进 5 m，张力 40 N，动摩擦力 12 N。对木块所做的净功是多少？
(A) 200 J
(B) 113 J
(C) 173 J
(D) 60 J
Correct. $W_T = 40 \times 5 \times \cos 30° \approx 173$ J. $W_f = -12 \times 5 = -60$ J. Normal and gravity are perpendicular → zero work. Net = $173 - 60 = 113$ J.正确。$W_T = 40 \times 5 \times \cos 30° \approx 173$ J；$W_f = -12 \times 5 = -60$ J。法向力与重力均垂直于位移 → 不做功。净功 $= 173 - 60 = 113$ J。
Tension work: $W_T = Fd\cos\theta = 40(5)\cos 30° \approx 173$ J. Friction: $W_f = -60$ J. Net = $113$ J. Answer: (B).张力做功：$W_T = Fd\cos\theta = 40(5)\cos 30° \approx 173$ J；摩擦力做功：$W_f = -60$ J。净功 $= 113$ J，选 (B)。
Topic 3.3专题 3.3
Potential Energy势能
Potential energy is the energy stored in a system by virtue of the positions or configurations of the objects within it. Unlike kinetic energy, potential energy is a property of a system of interacting objects. A ball held above the ground has gravitational PE, but that energy belongs to the ball–Earth system, not the ball alone.
势能（potential energy）是系统因其内部各物体的位置或构型而储存的能量。与动能不同，势能是一个相互作用系统的属性。被高高举起的球具有重力势能，但这部分能量属于"球—地球"系统，而不是球单独所有。
Potential energy is scalar. Its value depends on position, and the observer gets to choose where PE is zero. Only changes in PE matter physically — pick the reference that makes the math simplest.
势能是标量。其数值依赖于位置，且观察者可以自由选定零势能位置。物理上有意义的只是势能的变化——选一个能让计算最简单的参考点即可。
System Thinking
A common misconception: "the ball has potential energy." Strictly, a single isolated object cannot have PE. Potential energy arises from the interaction between objects (ball and Earth, block and spring). The AP exam may test this distinction.
系统观念
一个常见误区："这个球有势能。"严格说来，单独一个孤立物体不会有势能。势能来源于物体之间的相互作用（球与地球、木块与弹簧）。AP 考试有时就会专门考这种区分。
The Force–Potential Energy Relationship力与势能的关系
The deep connection between conservative forces and potential energy runs in both directions. If you know the force, integrate to find PE; if you know PE, differentiate to find force. This reciprocal relationship is one of the most powerful tools in physics.
保守力与势能之间的深层联系是双向的：知道力，积分得到势能；知道势能，求导得到力。这种互逆关系是物理学中最有力的工具之一。
Watch the Negative Sign
Students frequently lose the negative sign in $F = -dU/dx$. This sign is essential: it tells you forces point "downhill" on the PE landscape. If $U$ increases to the right, the force points left.
注意负号
学生经常在 $F = -dU/dx$ 里漏掉负号。这一负号至关重要：它告诉你力总是指向势能"下坡"方向。如果 $U$ 沿 $+x$ 方向增大，那么力指向 $-x$ 方向。
Equilibrium and $U(x)$ Graphs平衡与 $U(x)$ 图像
At any position where $dU/dx = 0$, the force is zero and the system is in equilibrium. Stable equilibrium occurs at a local minimum (valleys) — displaced objects get pushed back. Unstable equilibrium occurs at a local maximum (peaks) — displaced objects get pushed away.
凡是 $dU/dx = 0$ 的位置，力为零，系统处于平衡（equilibrium）。在局部极小值（势能"谷底"）处为稳定平衡（stable equilibrium）——物体偏离后会被推回。在局部极大值（势能"顶峰"）处为不稳定平衡（unstable equilibrium）——物体偏离后会被推得更远。
AP Exam Trap
On a $U(x)$ graph, the force is the negative slope, not the value of $U$. High $U$ with zero slope means zero force. Low $U$ with steep slope means large force. Don't confuse "high potential energy" with "large force."
AP 考试陷阱
在 $U(x)$ 图上，力是斜率的相反数，不是 $U$ 的数值本身。$U$ 很大但斜率为零，力依然为零；$U$ 很小但斜率很陡，力却很大。别把"势能高"误当成"力大"。
Common Potential Energies常见势能形式
Multi-Object Systems
When a system contains more than two objects, the total PE is the sum of every pair's PE. For three masses: $U_{12} + U_{13} + U_{23}$.
多体系统
若系统含两个以上的物体，总势能为所有成对势能之和。对三个质量：$U_{12} + U_{13} + U_{23}$。
Worked Example — Finding Speed from Potential Energy例题 —— 由势能求速率
A spring ($k = 800$ N/m) compressed 0.15 m launches a 0.25 kg ball horizontally. Find the launch speed (ignore friction).
弹簧（$k = 800$ N/m）被压缩 0.15 m，水平弹射一颗 0.25 kg 的球。求出射速率（忽略摩擦）。
// Conservation of energy: spring PE → kinetic energy
// 能量守恒：弹簧势能 → 动能
$$\frac{1}{2}k(\Delta x)^2 + 0 = 0 + \frac{1}{2}mv^2$$
$$v = \sqrt(k(\Delta x)^2 / m)$$
$$v = \sqrt(800 \times (0.15)^2 / 0.25)$$
$$v = \sqrt(800 \times 0.0225 / 0.25)$$
$$v = \sqrt(72) = 8.49\;\text{m/s}$$
A particle moves where $U(x) = 4x^3 - 6x$ (SI units). At which position is it in stable equilibrium?一质点在 $U(x) = 4x^3 - 6x$（SI 单位）势能场中运动。在哪个位置处于稳定平衡？
(A) $x = +\frac{1}{\sqrt{2}}$ m
(B) $x = -\frac{1}{\sqrt{2}}$ m
(C) $x = 0$
(D) No equilibrium exists.不存在平衡位置。
Correct! $dU/dx = 12x^2 - 6 = 0 \Rightarrow x = \pm 1/\sqrt{2}$. Then $d^2U/dx^2 = 24x$. At $x = +1/\sqrt{2}$: positive → local min → stable. At $x = -1/\sqrt{2}$: negative → local max → unstable.正确！$dU/dx = 12x^2 - 6 = 0 \Rightarrow x = \pm 1/\sqrt{2}$。再算 $d^2U/dx^2 = 24x$。在 $x = +1/\sqrt{2}$ 处为正 → 局部极小 → 稳定；在 $x = -1/\sqrt{2}$ 处为负 → 局部极大 → 不稳定。
Set $dU/dx = 12x^2 - 6 = 0$, giving $x = \pm 1/\sqrt{2}$. Check $d^2U/dx^2 = 24x$. Positive at $+1/\sqrt{2}$ → stable. Answer: (A).令 $dU/dx = 12x^2 - 6 = 0$，得 $x = \pm 1/\sqrt{2}$。再看 $d^2U/dx^2 = 24x$：在 $+1/\sqrt{2}$ 处为正 → 稳定。选 (A)。
Topic 3.4专题 3.4
Conservation of Energy能量守恒
Energy conservation is one of the most profound principles in all of science. The total energy of an isolated system never changes — energy can be transformed and transferred, but never created or destroyed.
能量守恒（conservation of energy）是整个科学中最深刻的原理之一：孤立系统的总能量永远不变——能量只能转化或传递，不能被创生或毁灭。
In AP Physics C: Mechanics, we focus on mechanical energy — the sum of kinetic and potential energies. The conservation law takes different forms depending on your system choice:
在 AP Physics C: Mechanics 中，我们重点研究机械能（mechanical energy）——动能与势能之和。守恒律的表达形式取决于系统的选取：
System Choice Matters
A single-object system can have KE but not PE (PE requires interacting objects). By choosing a larger system, you can treat internal forces as conservative and use PE rather than computing work. The AP exam often rewards smart system choices.
系统的选取很关键
单物体系统可以有动能，但不会有势能（势能必须有相互作用的物体）。把系统扩大，就能把原本的内力视为保守力、用势能代替计算功。AP 考试经常奖励会"挑系统"的考生。
When nonconservative forces are present (typically friction), mechanical energy is not conserved — some converts to thermal energy:
当存在非保守力（通常是摩擦力）时，机械能不再守恒——其中一部分会转化为热能：
Worked Example — Roller Coaster with Friction例题 —— 含摩擦的过山车
A 500 kg coaster starts from rest at 40 m high, climbs a 25 m hill. 50,000 J lost to friction. Find speed at the second hill.
一辆 500 kg 的过山车从 40 m 高处静止释放，爬升到 25 m 高的另一座山顶。途中摩擦耗散 50,000 J。求其在第二座山顶处的速率。
Set ground at $h = 0$. System: coaster + Earth. Include friction as work done on the system:取地面为 $h = 0$。系统：过山车 + 地球；把摩擦力做的功并入能量方程：
$$K_i + U_i + W_f = K_f + U_f$$
$$0 + mgh_i + W_f = \tfrac{1}{2}mv_f^2 + mgh_f$$
$$0 + (500)(9.8)(40) + (-50{,}000) = \tfrac{1}{2}(500)\,v_f^2 + (500)(9.8)(25)$$
$$196{,}000 - 50{,}000 = 250\,v_f^2 + 122{,}500$$
$$23{,}500 = 250\,v_f^2 \;\Rightarrow\; v_f = 9.70\;\text{m/s}$$
Worked Example — Loop-the-Loop Minimum Release Height例题 —— 竖直圆环轨道的最小释放高度
A small block is released from rest at height $h$ on a frictionless track that leads into a vertical loop of radius $R$. Find the minimum $h$ (in terms of $R$) for the block to maintain contact with the track at the top of the loop.
在光滑轨道上从高度 $h$ 处静止释放一个小滑块，轨道末端接一个半径为 $R$ 的竖直圆环。求滑块在圆环最高点仍能与轨道保持接触所需的最小 $h$（用 $R$ 表示）。
Step 1 — Centripetal condition at the top of the loop第 1 步 —— 圆环顶端的向心条件
At the threshold of losing contact ($N \to 0$), gravity alone supplies the centripetal force:在恰好脱离接触的临界（法向力 $N \to 0$）时，仅由重力提供向心力（centripetal force）：
$$mg = \frac{m v_\text{top}^2}{R} \;\Rightarrow\; v_\text{top}^2 = gR$$
Step 2 — Energy conservation, release point to top of loop第 2 步 —— 从释放点到圆环顶端的能量守恒
Release from rest at height $h$; top of loop is at height $2R$:从高度 $h$ 处静止释放；圆环顶端高度为 $2R$：
$$mgh = \tfrac{1}{2}m v_\text{top}^2 + mg(2R)$$
$$gh = \tfrac{1}{2}(gR) + 2gR = \tfrac{5}{2}gR$$
$$\boxed{\,h_\text{min} = \tfrac{5}{2}R\,}$$
Common slip: forgetting to add the $2R$ height of the top of the loop to the energy equation.常见错误：忘记把圆环顶端的 $2R$ 高度加进能量方程。
Worked Example — Bungee Jumper Maximum Stretch例题 —— 蹦极的最大伸长
A 70 kg jumper steps off a platform attached to an ideal bungee cord of natural length $L_0 = 15$ m and spring constant $k = 200$ N/m. Use $g = 9.8$ m/s². Find the maximum stretch $\Delta x$ of the cord (the jumper momentarily at rest at the lowest point).
一名 70 kg 的蹦极者从平台跃下，绑着一根理想蹦极绳，自然长度 $L_0 = 15$ m，弹簧常数 $k = 200$ N/m。取 $g = 9.8$ m/s²。求蹦极绳的最大伸长 $\Delta x$（蹦极者在最低点瞬时静止）。
Step 1 — Set up energy states第 1 步 —— 建立能量状态
Choose $U_g = 0$ at the lowest point; initial height = $L_0 + \Delta x$. At the top, $K_i = 0$ and $U_s = 0$ (cord slack). At the bottom, $K_f = 0$ and $U_g = 0$.取最低点处 $U_g = 0$；初始高度为 $L_0 + \Delta x$。在最高点：$K_i = 0$，绳尚松弛 $U_s = 0$。在最低点：$K_f = 0$，$U_g = 0$。
Step 2 — Conservation of energy第 2 步 —— 能量守恒
$$mg(L_0 + \Delta x) = \tfrac{1}{2}k(\Delta x)^2$$
Step 3 — Quadratic in $\Delta x$第 3 步 —— 关于 $\Delta x$ 的二次方程
$$\tfrac{1}{2}k(\Delta x)^2 - mg\,\Delta x - mg L_0 = 0$$
$$100\,(\Delta x)^2 - 686\,\Delta x - 10{,}290 = 0$$
Step 4 — Quadratic formula (positive root)第 4 步 —— 求根公式（取正根）
$$\Delta x = \frac{686 + \sqrt{686^2 + 4{,}116{,}000}}{200} \approx \frac{686 + 2147}{200} \approx 14.2\;\text{m}$$
Sanity check: the jumper falls $L_0 + \Delta x \approx 29$ m total.验算：蹦极者总下落 $L_0 + \Delta x \approx 29$ m，合理。
A pendulum swings from height $h$ above its lowest point. At the lowest point, which statement is correct about the system (bob + Earth)?一个单摆从相对最低点高度为 $h$ 处释放。在最低点处，关于系统（摆球 + 地球）下列哪种说法正确？
(A) KE = $mgh$ and且 PE = $mgh$
(B) KE = 0 and且 PE = $mgh$
(C) KE = $mgh$ and PE = 0 (choosing lowest point as reference)、PE = 0（以最低点为参考）
(D) Total mechanical energy = 0总机械能 = 0
Correct. Setting $U = 0$ at the bottom, all initial gravitational PE ($mgh$) converts to KE there. So $K = mgh$ and $U = 0$.正确。取最低点处 $U = 0$，初始重力势能 $mgh$ 在此处全部转化为动能。故 $K = mgh$，$U = 0$。
At the top: all energy is PE = $mgh$. At the bottom ($U = 0$ there): all PE → KE. So $K = mgh$. Answer: (C).最高点时：全部为势能 $mgh$。最低点（取 $U = 0$）：势能全转为动能，$K = mgh$。选 (C)。
Topic 3.5专题 3.5
Power功率
Power measures how quickly energy is transferred or converted. Two machines might do the same total work, but the faster one delivers more power. The SI unit is the Watt (W = J/s). One horsepower ≈ 746 W.
功率（power）衡量能量被传递或转化的快慢。两台机器即使做的总功相同，做得更快的那台输出更大的功率。SI 单位是瓦特（Watt，W = J/s）。一马力 ≈ 746 W。
A particularly useful derived formula connects power directly to force and velocity:
下面这条派生公式特别有用，它把功率直接与力和速度联系起来：
Terminal Velocity Connection
At terminal velocity, net force is zero and KE doesn't change. All power from the driving force is dissipated by resistance. Common AP setup: $P_\text{engine} = F_\text{drag} \cdot v_\text{terminal}$.
与终端速度的联系
在终端速度（terminal velocity）下，合外力为零，动能不变。驱动力的全部功率都被阻力耗散。AP 常见模型：$P_\text{engine} = F_\text{drag} \cdot v_\text{terminal}$。
Worked Example — Car vs. Quadratic Drag (P = F·v, FRQ-style)例题 —— 汽车与二次阻力（P = F·v，FRQ 风格）
A car of mass $m = 1200$ kg cruises on a level road. Aerodynamic drag is $F_\text{drag} = bv^2$ with $b = 0.80\;\text{N}\cdot\text{s}^2/\text{m}^2$. The engine delivers force $F_\text{eng}$ at the drive wheels. (a) Find the engine force and power needed to cruise at a constant $v = 30$ m/s. (b) If the engine can deliver at most $P_\text{max} = 60$ kW, find the car's top speed. (c) At top speed, what is the instantaneous acceleration if the engine force is suddenly doubled (drag unchanged)?
质量 $m = 1200$ kg 的汽车在平直公路上匀速行驶，空气阻力（aerodynamic drag）$F_\text{drag} = bv^2$，其中 $b = 0.80\;\text{N}\cdot\text{s}^2/\text{m}^2$。发动机在驱动轮上提供力 $F_\text{eng}$。(a) 求保持 $v = 30$ m/s 匀速所需的发动机力与功率；(b) 若发动机最大功率 $P_\text{max} = 60$ kW，求最高时速；(c) 在最高速时，若发动机力突然加倍（阻力不变），瞬时加速度是多少？
(a) Cruising at constant velocity(a) 匀速巡航
At constant speed, net force is zero, so the engine force equals the drag force:匀速时合外力为零，发动机力等于阻力：
$$F_\text{eng} = bv^2 = (0.80)(30)^2 = 720\;\text{N}$$
$$P = F_\text{eng}\,v = (720)(30) = 21{,}600\;\text{W} \approx 21.6\;\text{kW}$$
(b) Top speed at max power(b) 最大功率下的最高速
At top speed the car is still at constant $v$, so drag balances the engine force. The power absorbed by drag at any speed is:达到最高速时车仍以恒定 $v$ 行驶，阻力等于发动机力。任意速度下阻力吸收的功率为：
$$P_\text{drag} = F_\text{drag}\,v = bv^3$$
Setting this equal to the engine's maximum power:令其等于发动机最大功率：
$$P_\text{max} = b\,v_\text{top}^3 \;\Rightarrow\; v_\text{top} = \left(\frac{P_\text{max}}{b}\right)^{1/3}$$
$$v_\text{top} = \left(\frac{60{,}000}{0.80}\right)^{1/3} = (75{,}000)^{1/3} \approx 42.2\;\text{m/s}$$
(c) Instantaneous acceleration after doubling engine force(c) 发动机力加倍后的瞬时加速度
At top speed the drag force equals the original engine force:最高速时阻力等于原发动机力：
$$F_\text{drag} = b\,v_\text{top}^2 = \frac{P_\text{max}}{v_\text{top}} = \frac{60{,}000}{42.2} \approx 1422\;\text{N}$$
Doubling the engine force leaves a net force equal to the drag:发动机力加倍后，合外力就等于阻力本身：
$$F_\text{net} = 2F_\text{drag} - F_\text{drag} = F_\text{drag}$$
$$a = \frac{F_\text{drag}}{m} = \frac{1422}{1200} \approx 1.19\;\text{m/s}^2$$
Takeaway: $P = \vec{F}\cdot\vec{v}$ connects engine output, the drag law, and top speed — each part of this FRQ is one line once you identify which quantity is constant.
要点：$P = \vec{F}\cdot\vec{v}$ 把发动机输出、阻力定律与最高时速串了起来——只要先看清哪个量恒定，FRQ 的每一小题都只是一行计算。
Worked Example — Power Delivered by a Car Engine例题 —— 汽车发动机的瞬时功率
A 1400 kg car accelerates from rest with constant net force 5000 N. What is the instantaneous power at $t = 4$ s?
1400 kg 的汽车以恒定合力 5000 N 从静止加速。$t = 4$ s 时的瞬时功率是多少？
Velocity at $t = 4$ s:$t = 4$ s 时的速度：
$$a = F/m = 5000/1400 = 3.57\;\text{m/s}^2$$
$$v = v_0 + at = 0 + (3.57)(4) = 14.29\;\text{m/s}$$
Power from $P = \vec{F}\cdot\vec{v}$ (force parallel to velocity):由 $P = \vec{F}\cdot\vec{v}$（力与速度方向相同）：
$$P = (5000)(14.29) = 71{,}400\;\text{W} \approx 71.4\;\text{kW}$$
Power increases linearly with time because $v$ increases linearly while $F$ is constant.由于 $F$ 恒定而 $v$ 线性增加，功率随时间线性增长。
Worked Example — Constant-Power Acceleration (Calculus)例题 —— 恒功率加速（微积分）
An engine delivers a constant 60 kW of power to a 1500 kg car starting from rest on a level frictionless road. Find (a) the speed at $t = 8.0$ s and (b) the distance covered in that time.
发动机以恒定 60 kW 功率驱动 1500 kg 的汽车，从静止在平直无摩擦的路面上加速。求 (a) $t = 8.0$ s 时的速率；(b) 这段时间内行驶的距离。
(a) Speed at $t = 8.0$ s(a) $t = 8.0$ s 时的速率
Constant $P$ means $W = Pt$. Apply the work–energy theorem:$P$ 恒定意味着 $W = Pt$。代入动能定理：
$$Pt = \tfrac{1}{2}mv^2 \;\Rightarrow\; v(t) = \sqrt{\tfrac{2Pt}{m}}$$
$$v(8) = \sqrt{\tfrac{2(60{,}000)(8)}{1500}} = \sqrt{640} \approx 25.3\;\text{m/s}$$
(b) Distance covered in 8 s(b) 8 s 内行驶的距离
Integrate $v(t) = \sqrt{2P/m}\;t^{1/2}$ from $0$ to $T$:对 $v(t) = \sqrt{2P/m}\;t^{1/2}$ 从 $0$ 到 $T$ 积分：
$$x(T) = \sqrt{\tfrac{2P}{m}}\int_0^T t^{1/2}\,dt = \sqrt{\tfrac{2P}{m}}\cdot \tfrac{2}{3}\,T^{3/2}$$
$$x(8) = \tfrac{2}{3}\sqrt{\tfrac{2(60{,}000)}{1500}}\,(8)^{3/2} = \tfrac{2}{3}\sqrt{80}\,(22.63) \approx 135\;\text{m}$$
At $t \to 0$ the force $F = P/v$ diverges, which is an idealization. Real engines limit force at low speed via torque-vs-rpm curves.$t \to 0$ 时，力 $F = P/v$ 发散，这是理想化的结果。真实发动机会通过扭矩—转速曲线在低速时限制最大力。
An engine delivers constant power $P$ to a car of mass $m$ starting from rest on a frictionless surface. After time $t$, the speed is proportional to:发动机以恒定功率 $P$ 驱动质量为 $m$ 的汽车，在无摩擦地面上从静止开始运动。经过时间 $t$ 后，速率与下列哪一项成正比？
(A) $t$
(B) $t^2$
(C) $\sqrt[3]{t}$
(D) $\sqrt{t}$
Correct. $P = dW/dt$ so $W = Pt$. Work–energy theorem: $Pt = \frac{1}{2}mv^2$, giving $v = \sqrt{2Pt/m} \propto \sqrt{t}$.正确。$P = dW/dt$，故 $W = Pt$。代入动能定理：$Pt = \frac{1}{2}mv^2$，得 $v = \sqrt{2Pt/m} \propto \sqrt{t}$。
$P$ constant → $W = Pt$. Then $Pt = \frac{1}{2}mv^2 \Rightarrow v \propto \sqrt{t}$. Answer: (D).$P$ 恒定 → $W = Pt$。再代入 $Pt = \frac{1}{2}mv^2 \Rightarrow v \propto \sqrt{t}$。选 (D)。
Exam Preparation考试备考
Exam Strategy考试策略
MCQs test your ability to compare kinetic energies using $v^2$ dependence, determine the sign of work from force–displacement angles, read $U(x)$ graphs to identify equilibrium and turning points, and apply conservation to two-state problems. Many are "conceptual calculus" — asking what happens when you integrate or differentiate. Practice quickly identifying whether a problem is best solved with energy methods or Newton's laws.
Multiple Choice 部分考查：利用 $v^2$ 关系比较动能；由力与位移之间的夹角判断做功正负；从 $U(x)$ 图识别平衡位置与转折点；以及对"两态"问题应用守恒律。许多题目属于"概念性微积分"——问你积分或求导后会得到什么。练就一眼判断"该用能量方法还是 Newton 定律"的本能。
The Mathematical Routines (MR) question often draws from this unit. You'll derive symbolic expressions (e.g., speed at the bottom of a ramp), justify whether a force is conservative, sketch $K$, $U$, $E$ vs. position graphs, and write coherent paragraph-length analyses. Show every step of your calculus — the AP exam awards points for correct integration setup even if algebra has errors.
"数学过程"（Mathematical Routines，MR）题型常出自本单元。你需要推导符号表达式（如滑下斜面后底端的速率）、说明某个力是否保守、画 $K$、$U$、$E$ 关于位置的图像，并写一段连贯的分析。务必把每一步微积分写清楚——即便代数有误，FRQ 也会为正确的积分式给分。
Pitfalls陷阱
Common Mistakes常见错误
Top Point-Losing Errors
1. Confusing force and energy on $U(x)$ graphs. Force is the negative slope of $U(x)$, not $U$ itself.
2. Forgetting the negative sign in $F = -dU/dx$. Without it, forces point uphill instead of downhill.
3. Using displacement instead of path length for friction. Friction dissipates energy proportional to total distance, not net displacement.
4. Applying $\Delta U_g = mg\Delta h$ at planetary scales. This only works near Earth's surface. For orbits, use $U = -Gm_1m_2/r$.
5. Forgetting to square velocity in KE. Writing $K = \frac{1}{2}mv$ instead of $\frac{1}{2}mv^2$ is fatal.
6. Confusing power and work. Work is total energy transferred; power is the rate.
7. Inconsistent $U = 0$ reference. The choice is arbitrary, but you must be consistent throughout the problem.
最容易丢分的错误
1. 在 $U(x)$ 图上把力和能量混淆。力是 $U(x)$ 的负斜率，不是 $U$ 本身。
2. 在 $F = -dU/dx$ 中漏掉负号。漏了负号，力就指向"上坡"而非"下坡"。
3. 摩擦力计算用位移而非路程。摩擦耗散的能量正比于总路程，不是净位移。
4. 在行星尺度上仍用 $\Delta U_g = mg\Delta h$。该式只在近地表成立。轨道问题要用 $U = -Gm_1m_2/r$。
5. 动能漏掉速度的平方。写成 $K = \frac{1}{2}mv$ 而不是 $\frac{1}{2}mv^2$，直接送命。
6. 把功率（power）和功（work）混为一谈。功是总传递能量，功率是变化率。
7. $U = 0$ 参考点不一致。参考点可以任选，但整道题必须前后统一。
Review复习
Flashcards闪卡
Click a card to flip it.
点击卡片可翻面。
Translational Kinetic Energy平动动能
$$K = \tfrac{1}{2}mv^2$$
Scalar, always $\geq 0$, frame-dependent.标量，恒 $\geq 0$，依赖参考系。
Work by a Variable Force变力做功
$$W = \int_a^b \vec{F}\cdot d\vec{r}$$
Area under the $F_\parallel$ vs. $x$ curve.$F_\parallel$ – $x$ 曲线下的面积。
Work–Energy Theorem动能定理
$$\Delta K = W_\text{net}$$
Net work on a particle equals its change in kinetic energy.作用在质点上的净功等于其动能变化。
Force from $U(x)$由 $U(x)$ 求力
$$F_x = -\frac{dU}{dx}$$
Force points toward decreasing potential energy.力指向势能减小的方向。
Elastic Potential Energy弹性势能
$$U_s = \tfrac{1}{2}k(\Delta x)^2$$
Always $\geq 0$; zero at the spring's natural length.恒 $\geq 0$；弹簧自然长度处为零。
Gravitational PE (General)引力势能（一般形式）
$$U_g = -\frac{Gm_1 m_2}{r}$$
Always negative; $U \to 0$ as $r \to \infty$.恒为负；当 $r \to \infty$ 时 $U \to 0$。
Instantaneous Power瞬时功率
$$P = \vec{F}\cdot\vec{v} = Fv\cos\theta$$
Instantaneous rate of energy transfer by a force.力传递能量的瞬时速率。
Stable vs. Unstable Equilibrium稳定与不稳定平衡
$$\frac{dU}{dx} = 0$$
Stable: local minimum of $U(x)$. Unstable: local maximum.稳定：$U(x)$ 的局部极小；不稳定：局部极大。
Assessment测评
Unit Quiz单元测验
Score:得分： 0 / 8
Q1. A spring ($k = 200$ N/m) is compressed 0.3 m and launches a 0.5 kg block on a frictionless surface. What is the block's speed?弹簧（$k = 200$ N/m）被压缩 0.3 m，在无摩擦表面上将 0.5 kg 木块弹出。木块的速率是多少？
(A) 3.0 m/s
(B) 6.0 m/s
(C) 4.2 m/s
(D) 18.0 m/s
$\frac{1}{2}k(\Delta x)^2 = \frac{1}{2}mv^2 \Rightarrow v = \Delta x\sqrt{k/m} = 0.3\sqrt{400} = 6.0$ m/s.由 $\frac{1}{2}k(\Delta x)^2 = \frac{1}{2}mv^2 \Rightarrow v = \Delta x\sqrt{k/m} = 0.3\sqrt{400} = 6.0$ m/s。
$\frac{1}{2}(200)(0.09) = \frac{1}{2}(0.5)v^2 \Rightarrow 9 = 0.25v^2 \Rightarrow v = 6.0$ m/s. Answer: (B).$\frac{1}{2}(200)(0.09) = \frac{1}{2}(0.5)v^2 \Rightarrow 9 = 0.25v^2 \Rightarrow v = 6.0$ m/s。选 (B)。
Q2. A force $F(x) = 5x$ (N) acts from $x = 0$ to $x = 4$ m. The work done is:力 $F(x) = 5x$（N）从 $x = 0$ 作用到 $x = 4$ m，所做的功为：
(A) 20 J
(B) 50 J
(C) 40 J
(D) 80 J
$W = \int_0^4 5x\,dx = \frac{5}{2}x^2\big|_0^4 = \frac{5}{2}(16) = 40$ J.$W = \int_0^4 5x\,dx = \frac{5}{2}x^2\big|_0^4 = \frac{5}{2}(16) = 40$ J。
$W = \frac{5x^2}{2}\big|_0^4 = 40$ J. Answer: (C).$W = \frac{5x^2}{2}\big|_0^4 = 40$ J，选 (C)。
Q3. A 2 kg ball drops from rest at 5 m. Using $g = 10$ m/s², what is its KE just before hitting the ground?2 kg 的小球从 5 m 高处自由下落。取 $g = 10$ m/s²，触地前瞬间的动能是多少？
(A) 100 J
(B) 50 J
(C) 200 J
(D) 10 J
All PE converts to KE: $K = mgh = 2(10)(5) = 100$ J.势能全部转化为动能：$K = mgh = 2(10)(5) = 100$ J。
$K_f = mgh = 100$ J. Answer: (A).$K_f = mgh = 100$ J，选 (A)。
Q4. A motor lifts a 200 kg crate 10 m at constant speed in 8 s. The power output is approximately:电动机以匀速将 200 kg 的集装箱在 8 s 内升高 10 m。输出功率约为：
(A) 1600 W
(B) 2450 W
(C) 19600 W
(D) 250 W
$P = W/\Delta t = mgh/\Delta t = 200(9.8)(10)/8 = 2450$ W.$P = W/\Delta t = mgh/\Delta t = 200(9.8)(10)/8 = 2450$ W。
$P = mgh/t = 19600/8 = 2450$ W. Answer: (B).$P = mgh/t = 19600/8 = 2450$ W，选 (B)。
Q5. The potential energy $U(x) = 8x^2 - 2x^4$. How many equilibrium positions exist?势能 $U(x) = 8x^2 - 2x^4$。有几个平衡位置？
$dU/dx = 16x - 8x^3 = 8x(2 - x^2) = 0$. Solutions: $x = 0, \pm\sqrt{2}$. Three equilibrium positions. At $x = 0$: stable ($d^2U/dx^2 > 0$). At $x = \pm\sqrt{2}$: unstable ($d^2U/dx^2 < 0$).$dU/dx = 16x - 8x^3 = 8x(2 - x^2) = 0$，解为 $x = 0, \pm\sqrt{2}$，共三个平衡位置。$x = 0$ 处 $d^2U/dx^2 > 0$，稳定；$x = \pm\sqrt{2}$ 处 $d^2U/dx^2 < 0$，不稳定。
$8x(2 - x^2) = 0$ gives $x = 0, \pm\sqrt{2}$ — three equilibria. Answer: (D).$8x(2 - x^2) = 0$ 得 $x = 0, \pm\sqrt{2}$——三个平衡位置。选 (D)。
Q6. A $4.0\;\text{kg}$ block slides $5.0\;\text{m}$ across a horizontal surface with $\mu_k = 0.30$. The work done by friction on the block is:$4.0\;\text{kg}$ 的木块在水平面上滑行 $5.0\;\text{m}$，动摩擦系数 $\mu_k = 0.30$。摩擦力对木块所做的功是：
(A) $+58.8\;\text{J}$
(B) $+11.8\;\text{J}$
(C) $-58.8\;\text{J}$
(D) $0\;\text{J}$ (friction is perpendicular to motion)$0\;\text{J}$（摩擦力垂直于运动方向）
$f_k = \mu_k mg = (0.30)(4.0)(9.8) = 11.76\;\text{N}$, opposing the motion. $W_f = -f_k\,d = -(11.76)(5.0) \approx -58.8\;\text{J}$ — friction is non-conservative and removes mechanical energy. Answer (C).$f_k = \mu_k mg = (0.30)(4.0)(9.8) = 11.76\;\text{N}$，方向与运动相反。$W_f = -f_k\,d = -(11.76)(5.0) \approx -58.8\;\text{J}$——摩擦力是非保守力，会消耗机械能。选 (C)。
Kinetic friction always opposes motion ($\theta = 180^{\circ}$), so $W_f = -f_k d < 0$. Magnitude: $\mu_k mg \times d = 0.30 \times 4.0 \times 9.8 \times 5.0 = 58.8\;\text{J}$. Sign is negative.动摩擦力始终与运动反向（$\theta = 180^{\circ}$），故 $W_f = -f_k d < 0$。大小为 $\mu_k mg \times d = 0.30 \times 4.0 \times 9.8 \times 5.0 = 58.8\;\text{J}$，取负号。
Q7. A car of mass $m = 1200\;\text{kg}$ travels at constant velocity $v = 25\;\text{m/s}$ up a $5^{\circ}$ incline. Ignore drag. The instantaneous power delivered by the engine is closest to:质量 $m = 1200\;\text{kg}$ 的汽车以恒定速度 $v = 25\;\text{m/s}$ 沿 $5^{\circ}$ 斜坡上行，忽略空气阻力。发动机输出的瞬时功率最接近：
(A) $5.1\;\text{kW}$
(B) $25.6\;\text{kW}$
(C) $294\;\text{kW}$
(D) $0\;\text{kW}$ (constant velocity)$0\;\text{kW}$（匀速）
At constant velocity the engine's force balances gravity along the slope: $F = mg\sin\theta = (1200)(9.8)\sin 5^{\circ} \approx 1024\;\text{N}$. Power: $P = F v = 1024 \times 25 \approx 25.6\;\text{kW}$. Answer (B). Note that "constant velocity" means $\Delta K = 0$, but the engine still does positive work against gravity.匀速时发动机推力沿斜面与重力分量平衡：$F = mg\sin\theta = (1200)(9.8)\sin 5^{\circ} \approx 1024\;\text{N}$；$P = F v = 1024 \times 25 \approx 25.6\;\text{kW}$。选 (B)。注意"匀速"意味着 $\Delta K = 0$，但发动机仍要对抗重力做正功。
Use $P = \vec F \cdot \vec v$. At constant velocity along the slope, $F = mg\sin\theta$. $P = mg\sin\theta \cdot v = 1200 \times 9.8 \times \sin 5^{\circ} \times 25 \approx 25.6\;\text{kW}$.用 $P = \vec F \cdot \vec v$。沿斜面匀速时 $F = mg\sin\theta$，故 $P = mg\sin\theta \cdot v = 1200 \times 9.8 \times \sin 5^{\circ} \times 25 \approx 25.6\;\text{kW}$。
Q8. A particle moves under a one-dimensional potential $U(x) = \tfrac{1}{2}kx^2$ with $k = 8\;\text{N/m}$. If its total mechanical energy is $E = 4\;\text{J}$, the turning points are at:一质点在一维势能 $U(x) = \tfrac{1}{2}kx^2$（$k = 8\;\text{N/m}$）中运动。若总机械能 $E = 4\;\text{J}$，转折点位置为：
(A) $x = \pm 1.0\;\text{m}$
(B) $x = \pm 0.5\;\text{m}$
(C) $x = \pm 2.0\;\text{m}$
(D) $x = 0$ only仅在 $x = 0$
Turning points are where $K = 0$, i.e. $U(x) = E$. Solve $\tfrac{1}{2}(8)x^2 = 4 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1.0\;\text{m}$. Answer (A).转折点处 $K = 0$，即 $U(x) = E$。解 $\tfrac{1}{2}(8)x^2 = 4 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1.0\;\text{m}$。选 (A)。
Set $U(x) = E$: $\tfrac{1}{2}kx^2 = 4$, so $x^2 = 8/k = 1$, giving $x = \pm 1.0\;\text{m}$.令 $U(x) = E$：$\tfrac{1}{2}kx^2 = 4$，故 $x^2 = 8/k = 1$，得 $x = \pm 1.0\;\text{m}$。
Worked Example — Variable Force, $U(x)$ and Turning Points (FRQ Style)例题 —— 变力、$U(x)$ 与转折点（FRQ 风格）
A particle of mass $m = 0.50\;\text{kg}$ moves along the $x$-axis under a conservative force whose potential energy is $U(x) = x^4 - 8x^2 + 5\;\text{J}$ (with $x$ in metres). The particle is released from rest at $x = 3.0\;\text{m}$. (a) Find the force $F(x)$. (b) Locate every equilibrium of $U(x)$ and classify each as stable or unstable. (c) Find the speed of the particle as it passes through $x = 0$. (d) Determine the turning points of the subsequent motion.
质量 $m = 0.50\;\text{kg}$ 的质点沿 $x$ 轴在保守力作用下运动，势能为 $U(x) = x^4 - 8x^2 + 5\;\text{J}$（$x$ 以米为单位）。该质点从 $x = 3.0\;\text{m}$ 处静止释放。(a) 求力 $F(x)$；(b) 找出 $U(x)$ 的全部平衡位置并判断其稳定性；(c) 求经过 $x = 0$ 时的速率；(d) 找出之后运动的转折点。
Identify明辨
Principle: for a conservative 1-D system, $F(x) = -dU/dx$, equilibria are roots of $dU/dx = 0$ classified by the sign of $d^2U/dx^2$, and energy conservation $E = K + U$ gives speed at any $x$ and turning points at $U(x) = E$.
原理：一维保守系统中，$F(x) = -dU/dx$；平衡位置是 $dU/dx = 0$ 的根，再由 $d^2U/dx^2$ 的符号判定稳定性；能量守恒 $E = K + U$ 给出任意 $x$ 处的速率，转折点满足 $U(x) = E$。
$U(x) = x^4 - 8x^2 + 5$
Mass $m = 0.50\;\text{kg}$质量 $m = 0.50\;\text{kg}$
Released from rest at $x = 3.0\;\text{m}$从 $x = 3.0\;\text{m}$ 静止释放
Set Up建模
Compute the total energy $E$ once at the release point — it is conserved, so the same value gives speed at any $x$ and the locations of the turning points.
在释放点算一次总能量 $E$——它守恒，可直接用于任意 $x$ 处的速率与转折点位置。
$$E = K_0 + U(3.0) = 0 + \bigl[(3)^4 - 8(3)^2 + 5\bigr] = 81 - 72 + 5 = 14\;\text{J}$$
Execute求解
(a) Force(a) 力
$$F(x) = -\frac{dU}{dx} = -(4x^3 - 16x) = -4x(x^2 - 4) = -4x(x-2)(x+2)$$
(b) Equilibria and stability(b) 平衡位置与稳定性
$F = 0 \Rightarrow x = 0,\;\pm 2$. Compute $d^2U/dx^2 = 12x^2 - 16$:$F = 0 \Rightarrow x = 0,\;\pm 2$；再算 $d^2U/dx^2 = 12x^2 - 16$：
$$x = 0: \;d^2U/dx^2 = -16 < 0 \;\Rightarrow\; \text{unstable}$$
$$x = \pm 2: \;d^2U/dx^2 = 48 - 16 = 32 > 0 \;\Rightarrow\; \text{stable}$$
The two stable wells at $x = \pm 2$ flank an unstable peak at $x = 0$ — the classic double-well shape.两侧 $x = \pm 2$ 处的稳定势阱夹着 $x = 0$ 处的不稳定势峰——经典的双势阱（double well）形态。
(c) Speed at $x = 0$(c) $x = 0$ 处的速率
$$\tfrac{1}{2} m v^2 = E - U(0) = 14 - 5 = 9\;\text{J}$$
$$v = \sqrt{\frac{2(9)}{0.50}} = \sqrt{36} = 6.0\;\text{m/s}$$
(d) Turning points(d) 转折点
$$U(x) = E \;\Rightarrow\; x^4 - 8x^2 + 5 = 14 \;\Rightarrow\; x^4 - 8x^2 - 9 = 0$$
Let $u = x^2$. Then $u^2 - 8u - 9 = (u - 9)(u + 1) = 0$, so $u = 9$ (the $u = -1$ root is unphysical). Therefore $x = \pm 3.0\;\text{m}$.令 $u = x^2$，则 $u^2 - 8u - 9 = (u - 9)(u + 1) = 0$，得 $u = 9$（$u = -1$ 不合理）。故 $x = \pm 3.0\;\text{m}$。
$$\boxed{\,x_\text{turn} = \pm 3.0\;\text{m}\,}$$
Evaluate校验
$x = +3.0\;\text{m}$ recovers the release point — the other turning point is at $x = -3.0\;\text{m}$, which is correct because the potential is symmetric about $x = 0$. ✓$x = +3.0\;\text{m}$ 即释放点；另一转折点在 $x = -3.0\;\text{m}$，正确——势能关于 $x = 0$ 对称。✓
The particle has enough energy ($E = 14 > U(0) = 5$) to cross the central barrier at $x = 0$, so it oscillates over the full range $[-3, +3]$ rather than being trapped in a single well. If $E < 5$, it would be confined to one well only. ✓$E = 14 > U(0) = 5$，质点有足够能量翻越中央势垒，故在 $[-3, +3]$ 整个范围内来回振荡，而非被困在单个势阱中。若 $E < 5$，则只能在某一侧势阱内运动。✓
Speed peaks at the bottoms of the wells, $x = \pm 2$, where $U = 16 - 32 + 5 = -11\;\text{J}$ is minimum — speed there is $\sqrt{2(14 + 11)/0.5} = \sqrt{100} = 10\;\text{m/s}$, the global maximum. The $6.0\;\text{m/s}$ at $x = 0$ is correctly less. ✓速率在势阱底部 $x = \pm 2$ 处达到最大，此处 $U = 16 - 32 + 5 = -11\;\text{J}$ 最低——$v = \sqrt{2(14 + 11)/0.5} = \sqrt{100} = 10\;\text{m/s}$。$x = 0$ 处为 $6.0\;\text{m/s}$，确实较小。✓
Worked Example — Spring → Ramp → Friction (Energy Bookkeeping, FRQ Style)例题 —— 弹簧 → 斜面 → 摩擦（能量记账，FRQ 风格）
A spring with $k = 800\;\text{N/m}$ is compressed by $\Delta x = 0.20\;\text{m}$ at the bottom of a frictionless ramp. When released, it launches a $m = 2.0\;\text{kg}$ block up the ramp. Above the spring's release point, the ramp is inclined at $\theta = 30^{\circ}$ and has kinetic-friction coefficient $\mu_k = 0.25$. (a) Find the launch speed. (b) Find the maximum distance $L$ the block slides up the inclined section before momentarily stopping. (c) Determine whether the block slides back down on its own. Use $g = 9.8\;\text{m/s}^2$.
$k = 800\;\text{N/m}$ 的弹簧在一段无摩擦斜面底部被压缩 $\Delta x = 0.20\;\text{m}$。释放后将 $m = 2.0\;\text{kg}$ 的木块沿斜面弹出。在弹簧释放点以上的斜面与水平方向成 $\theta = 30^{\circ}$，动摩擦系数 $\mu_k = 0.25$。(a) 求出射速率；(b) 求木块在斜面上瞬时静止前滑行的最大距离 $L$；(c) 判断木块此后是否会自动滑回。取 $g = 9.8\;\text{m/s}^2$。
Identify明辨
Principle: energy conservation with non-conservative work — spring PE converts entirely to KE on the frictionless launch zone, then KE converts to gravitational PE plus heat as friction acts up the ramp. Resting condition compares the gravity component along the slope to the maximum static friction.
原理：含非保守功的能量守恒——在无摩擦发射段，弹簧势能完全转化为动能；在含摩擦的斜面段，动能转化为重力势能加上摩擦生成的热。是否会滑回则比较沿斜面的重力分量与最大静摩擦。
Spring $k$, compression $\Delta x$弹簧 $k$、压缩 $\Delta x$
Mass $m$质量 $m$
Ramp angle $\theta$, friction $\mu_k$斜面倾角 $\theta$、摩擦 $\mu_k$
Gravity $g$重力 $g$
Set Up建模
Energy conservation across the launch (spring → KE) and across the slide up the ramp (KE → gravitational PE + heat from friction):
分两段写能量方程：发射段（弹簧 → 动能）与上滑段（动能 → 重力势能 + 摩擦热）：
$$\tfrac{1}{2}k(\Delta x)^2 = \tfrac{1}{2}m v_0^2$$
$$\tfrac{1}{2}m v_0^2 = mgL\sin\theta + \mu_k mg\cos\theta\cdot L$$
Execute求解
(a) Launch speed(a) 出射速率
$$v_0 = \Delta x \sqrt{\frac{k}{m}} = 0.20 \sqrt{\frac{800}{2.0}} = 0.20 \times 20 = 4.0\;\text{m/s}$$
(b) Maximum slide distance(b) 最大滑行距离
Solve the second equation for $L$:从第二式解出 $L$：
$$L = \frac{v_0^2}{2g(\sin\theta + \mu_k \cos\theta)} = \frac{(4.0)^2}{2(9.8)(0.500 + 0.25 \times 0.866)}$$
$$L = \frac{16}{2(9.8)(0.7165)} = \frac{16}{14.04} \approx 1.14\;\text{m}$$
(c) Does it slide back?(c) 会不会自动滑回？
At rest, the block slides only if the gravity component along the slope exceeds the maximum static friction. Treating $\mu_s \approx \mu_k = 0.25$ as the bound (typical AP simplification):瞬时静止后，只有当沿斜面的重力分量大于最大静摩擦力时才会下滑。沿用 AP 常用近似 $\mu_s \approx \mu_k = 0.25$：
$$mg\sin\theta \;\;\overset{?}{>}\;\; \mu_s mg\cos\theta$$
$$\tan\theta = \tan 30^{\circ} \approx 0.577 > 0.25 = \mu_s \;\Rightarrow\; \text{yes, it slides back}$$
Evaluate校验
Energy check: $\tfrac{1}{2}k(\Delta x)^2 = \tfrac{1}{2}(800)(0.04) = 16\;\text{J}$; $\tfrac{1}{2}m v_0^2 = \tfrac{1}{2}(2)(16) = 16\;\text{J}$. ✓能量核对：$\tfrac{1}{2}k(\Delta x)^2 = \tfrac{1}{2}(800)(0.04) = 16\;\text{J}$；$\tfrac{1}{2}m v_0^2 = \tfrac{1}{2}(2)(16) = 16\;\text{J}$。✓
Heat dissipated to reach the top: $\mu_k mgL\cos\theta = (0.25)(2)(9.8)(1.14)(0.866) \approx 4.84\;\text{J}$. PE gained: $mgL\sin\theta = (2)(9.8)(1.14)(0.5) \approx 11.16\;\text{J}$. Sum: $11.16 + 4.84 = 16.0\;\text{J}$ — matches the launch KE exactly. ✓上滑过程中耗散的热：$\mu_k mgL\cos\theta = (0.25)(2)(9.8)(1.14)(0.866) \approx 4.84\;\text{J}$。重力势能增量：$mgL\sin\theta = (2)(9.8)(1.14)(0.5) \approx 11.16\;\text{J}$。两者之和 $11.16 + 4.84 = 16.0\;\text{J}$，恰好等于初始动能。✓
Frictionless limit ($\mu_k \to 0$): $L \to v_0^2/(2g\sin\theta) = 16/(9.8) \approx 1.63\;\text{m}$. Friction shortens the slide as expected. ✓无摩擦极限 $\mu_k \to 0$：$L \to v_0^2/(2g\sin\theta) = 16/(9.8) \approx 1.63\;\text{m}$。摩擦确实使滑行距离变短。✓
Unit 3 Formula Quick Reference第 3 单元公式速查
$K = \tfrac{1}{2}mv^2$
$W = Fd\cos\theta$
$W = \int_a^b \vec{F}\cdot d\vec{r}$
$\Delta K = W_\text{net}$
$\Delta U = -\int \vec{F}_{cf}\cdot d\vec{r}$
$F_x = -dU/dx$
$U_s = \tfrac{1}{2}k(\Delta x)^2$
$U_g = -Gm_1 m_2 / r$
$\Delta U_g = mg\Delta y$
$P = \Delta E / \Delta t$
$P = dW/dt$
$P = Fv\cos\theta$