Each item below restates the prompt and choices, marks the correct letter, and gives a one- to three-sentence justification keyed to the AP topic. Numbers use $g = 9.8~\mathrm{m/s^2}$ unless the problem specifies otherwise.以下每道题均重述题目和选项，标注正确字母，并给出 1 至 3 句对应 AP 主题的简要说明。除题目另有说明外，数值计算取 $g = 9.8~\mathrm{m/s^2}$。
Q1EASY2.1 Newton's First Law2.1 牛顿第一定律No Calculator
A puck slides on a frictionless horizontal surface with constant velocity. Which of the following must be true?一个冰球在无摩擦水平面上以匀速滑行。下列说法中哪一项必然正确？
(A)A net force acts on the puck in the direction of motion.沿运动方向有合力作用于冰球。
(B)The puck experiences no forces at all.冰球不受任何力的作用。
(C)The vector sum of all forces on the puck is zero.冰球所受所有力的矢量和为零。
(D)The puck must be accelerating.冰球一定在加速。
Answer:答案：(C)
Constant velocity means $\vec a = 0$, so by Newton's second law $\sum \vec F = 0$. (A) is the common Aristotelian misconception that "motion requires a force." (B) is wrong, gravity and the normal force still act; they just cancel. (D) contradicts the given constant velocity.匀速意味着 $\vec a = 0$，由牛顿第二定律得 $\sum \vec F = 0$。(A) 是常见的亚里士多德式误解，认为"运动需要力"。(B) 错误，重力与法向力仍存在，只是相互抵消。(D) 与题目所给匀速条件矛盾。
Insight.要点。Constant velocity already implies zero net force; no forward force is needed to keep the puck moving. Separate the kinematic statement ($a=0$) from the force statement ($\sum F=0$).匀速已经意味着合力为零；冰球继续运动不需要向前的力。要把运动学结论（$a=0$）与受力结论（$\sum F=0$）分开。
Q2EASY2.2 Newton's Second Law2.2 牛顿第二定律No Calculator
A net force of $12~\mathrm{N}$ acts on a $3~\mathrm{kg}$ block. The block's acceleration is$12~\mathrm{N}$ 的合力作用于一个 $3~\mathrm{kg}$ 的物块。该物块的加速度为
Insight.要点。Newton's second law is $F_{\text{net}}=ma$, so acceleration is force divided by mass. Always divide by the whole mass, not multiply or subtract, and check that the result has units $m/s^2$.牛顿第二定律是 $F_{\text{net}}=ma$，故加速度等于力除以质量。要除以整个质量，而不是相乘或相减，并检查结果的单位是否为 $m/s^2$。
Q3EASY2.3 Newton's Third Law2.3 牛顿第三定律No Calculator
A car pushes a heavier truck along a road, accelerating both. By Newton's third law, the truck pushes back on the car with a force that is一辆汽车在路上推动一辆更重的卡车，使两者均加速。根据牛顿第三定律，卡车对汽车的反作用力
(A)larger than the car's force on the truck (because the truck is heavier).大于汽车对卡车的力（因为卡车更重）。
(B)smaller than the car's force on the truck (because the car is moving the system).小于汽车对卡车的力（因为汽车在推动整个系统）。
(C)equal in magnitude and opposite in direction to the car's force on the truck.与汽车对卡车的力大小相等、方向相反。
(D)equal only when the system has zero acceleration.仅当系统加速度为零时才与汽车对卡车的力相等。
Answer:答案：(C)
Third-law pairs are always equal in magnitude and opposite in direction, independent of mass, motion, or acceleration. The fact that both vehicles accelerate is consistent: the unequal accelerations come from different forces on different bodies (car-road traction vs. truck-car contact), not from unequal third-law pairs.第三定律作用力对总是大小相等、方向相反，与质量、运动状态或加速度无关。两辆车都在加速是自洽的：加速度不同来源于作用在不同物体上的不同力（汽车与路面的牵引力和卡车与汽车的接触力），而非第三定律力对不等。
Insight.要点。Third-law pairs act on different bodies, so they never cancel in one body's free-body diagram. Equal magnitude survives acceleration; the two vehicles have different accelerations because different net forces act on each.第三定律力对作用在不同物体上，因此不会在单个物体的受力图中相互抵消。两者加速时力的大小仍然相等；两车加速度不同是因为各自所受合力不同。
Q4EASY2.4 Static Friction2.4 静摩擦力No Calculator
A $5.0~\mathrm{kg}$ block sits at rest on a horizontal surface with $\mu_s = 0.40$. A horizontal force of $12~\mathrm{N}$ is applied. Take $g = 10~\mathrm{m/s^2}$. Which statement best describes the situation?一个 $5.0~\mathrm{kg}$ 的物块静止在水平面上，静摩擦系数 $\mu_s = 0.40$。施加 $12~\mathrm{N}$ 的水平力，取 $g = 10~\mathrm{m/s^2}$。下列说法中哪一项最准确描述了该情况？
(A)The block accelerates at $2.4~\mathrm{m/s^2}$.物块以 $2.4~\mathrm{m/s^2}$ 加速。
(B)The block accelerates at $0.4~\mathrm{m/s^2}$.物块以 $0.4~\mathrm{m/s^2}$ 加速。
(C)The block does not move; static friction is $12~\mathrm{N}$ opposing the applied force.物块不动，静摩擦力为 $12~\mathrm{N}$，方向与施力相反。
(D)The block does not move; static friction is exactly $20~\mathrm{N}$.物块不动，静摩擦力恰好为 $20~\mathrm{N}$。
Answer:答案：(C)
Maximum static friction is最大静摩擦力为
$$f_{s,\max} = \mu_s\,m g = 0.40(5.0)(10) = 20~\mathrm{N}$$
Since the applied $12~\mathrm{N}$ is below this threshold, the block stays at rest and static friction adjusts to exactly balance the applied force at $12~\mathrm{N}$. Trap (D) reports the maximum value as if it were always active; (A)/(B) compute kinetic motion that doesn't occur.由于施加的 $12~\mathrm{N}$ 低于该阈值，物块保持静止，静摩擦力自适应调整为恰好平衡施力的 $12~\mathrm{N}$。陷阱选项 (D) 将最大值当作始终有效的静摩擦力；(A)/(B) 计算了实际上不会发生的动态运动。
Insight.要点。Static friction is a constraint force: it equals whatever is needed to prevent slip, up to $\mu_s N$. Below the threshold, use the applied value, not the maximum. Check the threshold before assuming motion.静摩擦力是一种约束力：在防止滑动所需的范围内自适应，最大不超过 $\mu_s N$。低于阈值时取施加力的大小，而不是最大值。应先判断是否越过阈值，再决定是否运动。
Two blocks of masses $m_1 = 2~\mathrm{kg}$ and $m_2 = 3~\mathrm{kg}$ sit in contact on a frictionless horizontal surface. A horizontal force $F = 10~\mathrm{N}$ is applied to $m_1$, pushing both blocks. The contact force between them is质量为 $m_1 = 2~\mathrm{kg}$ 和 $m_2 = 3~\mathrm{kg}$ 的两个物块相互接触，置于无摩擦水平面上。水平力 $F = 10~\mathrm{N}$ 施加于 $m_1$，推动两物块运动。两物块之间的接触力为
(A) $4~\mathrm{N}$
(B) $6~\mathrm{N}$
(C) $10~\mathrm{N}$
(D) $8.3~\mathrm{N}$
Answer:答案：(B)
Treat the pair as one system to get the common acceleration:将两物块视为一个系统求共同加速度：
The contact force is the only horizontal force on $m_2$, so $F_\text{contact} = m_2\,a = 3(2) = 6~\mathrm{N}$. Trap (C) reports the applied force itself; (A) uses $m_1\,a$ on the wrong block.接触力是 $m_2$ 受到的唯一水平力，故 $F_\text{contact} = m_2\,a = 3(2) = 6~\mathrm{N}$。陷阱选项 (C) 将施加力本身作为答案，(A) 对错误物块用了 $m_1\,a$。
Insight.要点。For connected objects, first find the common acceleration by treating all masses as one system, then isolate one object to find an internal force. The contact force is generally smaller than the applied force.处理相连物体时，先把所有质量视为一个系统求共同加速度，再隔离其中一个物体求内力。接触力通常小于外力。
Q6MEDIUM2.6 Inclined Plane2.6 斜面No Calculator
A block slides down a frictionless $30^\circ$ incline. Take $g = 10~\mathrm{m/s^2}$. Its acceleration along the incline is一物块沿无摩擦 $30^\circ$ 斜面下滑，取 $g = 10~\mathrm{m/s^2}$。其沿斜面方向的加速度为
Mass-independent. Trap (B) is $g\cos\theta$ (the normal component, not the slope-parallel one). (C) and (D) ignore the geometry entirely.与质量无关。陷阱选项 (B) 为 $g\cos\theta$（法向分量，而非沿斜面分量）。(C) 和 (D) 完全忽略了几何关系。
Insight.要点。On an incline, decompose weight into $mg\sin\theta$ along the slope and $mg\cos\theta$ into the surface. Acceleration along a frictionless incline is $g\sin\theta$; the cosine belongs to the normal direction.在斜面上，把重力分解为沿斜面的 $mg\sin\theta$ 和垂直斜面的 $mg\cos\theta$。无摩擦斜面加速度为 $g\sin\theta$；余弦分量属于法向。
Q7MEDIUM2.4 Static vs. Kinetic Friction2.4 静摩擦与动摩擦Calculator
A $4.0~\mathrm{kg}$ block sits on a horizontal surface with $\mu_s = 0.30$ and $\mu_k = 0.20$. A horizontal force of $F = 16~\mathrm{N}$ is applied. The block's acceleration is closest to一个 $4.0~\mathrm{kg}$ 的物块置于水平面上，静摩擦系数 $\mu_s = 0.30$，动摩擦系数 $\mu_k = 0.20$。施加水平力 $F = 16~\mathrm{N}$。该物块的加速度最接近
(A)$0$ (the block does not move)$0$（物块不移动）
(B) $2.04~\mathrm{m/s^2}$
(C) $4.0~\mathrm{m/s^2}$
(D) $1.04~\mathrm{m/s^2}$
Answer:答案：(B)
First check whether the block moves:首先判断物块是否运动：
$$f_{s,\max} = \mu_s\,m g = 0.30(4.0)(9.8) = 11.76~\mathrm{N} < F = 16~\mathrm{N}$$
It does. Once moving, friction is kinetic:物块会运动。一旦运动，摩擦力变为动摩擦：
Trap (C) ignores friction altogether; (D) uses $\mu_s$ in the kinetic equation.陷阱选项 (C) 完全忽略摩擦力，(D) 在动摩擦方程中错误地使用了 $\mu_s$。
Insight.要点。Friction has two regimes. Compare the applied force with $\mu_s mg$ first; only if it moves do you switch to $\mu_k mg$. Using $\mu_s$ after sliding starts is the classic error.摩擦分两个阶段：先把外力与 $\mu_s mg$ 比较；只有物体运动后才改用 $\mu_k mg$。开始滑动后仍用 $\mu_s$ 是经典错误。
An object falls through air under gravity and a velocity-dependent drag force opposing motion. At terminal velocity, the object's一物体在重力及与速度有关的阻力（方向与运动相反）共同作用下在空气中下落。在达到终端速度时，该物体的
(A)velocity is zero.速度为零。
(B)acceleration is zero.加速度为零。
(C)acceleration equals $g$.加速度等于 $g$。
(D)acceleration equals the drag force divided by its mass.加速度等于阻力除以质量。
Answer:答案：(B)
"Terminal velocity" means a steady (terminal) speed; drag has grown until it balances gravity, so $\sum F = 0$ and $a = 0$. Trap (A) confuses zero acceleration with zero velocity. Trap (D) drops the gravity contribution."终端速度"是指稳定（终态）速度，此时阻力已增大到与重力平衡，故 $\sum F = 0$，$a = 0$。陷阱选项 (A) 将零加速度与零速度混淆。陷阱选项 (D) 遗漏了重力的贡献。
Insight.要点。Terminal velocity means terminal, or steady, velocity: acceleration is zero because drag has grown to balance gravity. It is not zero velocity and does not mean gravity disappeared.终端速度指稳定不变的终态速度：阻力增大到与重力平衡，加速度为零。它不是零速度，也不表示重力消失。
Q9MEDIUM2.7 Spring Combinations2.7 弹簧组合No Calculator
Two identical springs, each with constant $k$, are connected end-to-end (in series). The equivalent spring constant of the combination is两个相同的弹簧，每个劲度系数均为 $k$，首尾相连（串联）。该组合的等效弹簧系数为
Series-combined springs are softer than either alone. Trap (A) is the parallel result; (D) has the wrong units.串联弹簧比任意单个弹簧更软。陷阱选项 (A) 是并联结果，(D) 的单位不正确。
Insight.要点。Springs in series add reciprocals and are softer than either spring; springs in parallel add directly and are stiffer. Identical series springs therefore have $k_{\text{eq}}=k/2$.弹簧串联时等效劲度系数取倒数相加，比任一弹簧更软；并联时直接相加，更硬。因此相同弹簧串联的 $k_{\text{eq}}=k/2$。
Two point masses separated by a distance $r$ feel a gravitational attraction of magnitude $F$. If the separation is doubled to $2r$ (and the masses are unchanged), the new gravitational force is两个质点相距 $r$，受到大小为 $F$ 的引力。若距离增大为 $2r$（质量不变），新的引力为
(A) $2F$
(B) $F/2$
(C) $F/4$
(D) $F/\sqrt{2}$
Answer:答案：(C)
$F \propto 1/r^2$, so doubling $r$ divides $F$ by $2^2 = 4$:$F \propto 1/r^2$，距离翻倍则 $F$ 除以 $2^2 = 4$：
$$F_\text{new} = \frac{F}{(2)^2} = \frac{F}{4}$$
Trap (B) treats the law as $1/r$; (D) inverts the wrong power.陷阱选项 (B) 将定律误用为 $1/r$，(D) 对错误的幂次取倒数。
Insight.要点。Newton's law of gravitation is inverse-square in distance: changing $r$ to $2r$ changes the force by $1/4$, not $1/2$. Track the exponent before inverting anything.万有引力与距离成平方反比：$r$ 变为 $2r$ 时，力变为 $1/4$，而不是 $1/2$。先看准指数，再取倒数。
A $1200~\mathrm{kg}$ car rounds a flat (unbanked) curve of radius $r = 50~\mathrm{m}$. The static-friction coefficient between tires and road is $\mu_s = 0.50$. The maximum speed without slipping is closest to一辆 $1200~\mathrm{kg}$ 的汽车驶过半径 $r = 50~\mathrm{m}$ 的平坦（无倾斜）弯道，轮胎与路面间的静摩擦系数 $\mu_s = 0.50$。不发生侧滑的最大速度最接近
(A) $9.0~\mathrm{m/s}$
(B) $15.7~\mathrm{m/s}$
(C) $22.0~\mathrm{m/s}$
(D) $49.0~\mathrm{m/s}$
Answer:答案：(B)
At the slip threshold, static friction provides exactly the required centripetal force:在侧滑临界点，静摩擦力恰好提供所需向心力：
$$\mu_s\,m g = \frac{m\,v_{\max}^2}{r} \;\Longrightarrow\; v_{\max} = \sqrt{\mu_s\,g\,r}$$
Mass-independent, trap (C) is $v_{\max}^2$ (correct quantity but missing the square root).结果与质量无关。陷阱选项 (C) 是 $v_{\max}^2$（量纲正确但漏了开方）。
Insight.要点。On a flat curve, static friction is the centripetal force: $\mu_s mg=mv_{\max}^2/r$. Mass cancels, and the answer must be a speed, so remember the square root.在平坦弯道上，静摩擦力提供向心力：$\mu_s mg=mv_{\max}^2/r$。质量消去，答案必须是速度，因此不要忘记开方。
Q12MEDIUM2.6 Slip Threshold on Incline2.6 斜面滑动临界角No Calculator
A block sits at rest on an incline that is slowly tilted upward. The static-friction coefficient between block and incline is $\mu_s$. The block first begins to slide when the incline angle satisfies一物块静止在斜面上，斜面缓慢向上倾斜。物块与斜面之间的静摩擦系数为 $\mu_s$。当斜面角满足下列哪个条件时，物块开始滑动？
(A) $\tan\theta = \mu_s$
(B) $\sin\theta = \mu_s$
(C) $\cos\theta = \mu_s$
(D) $\theta = 45^\circ$ regardless of $\mu_s$
Answer:答案：(A)
At the slip threshold, the slope-parallel gravity component equals the maximum static friction:在滑动临界点，沿斜面方向的重力分量等于最大静摩擦力：
Mass cancels. Trap (D) confuses the universal frictionless answer (always slides) with a specific surface.质量约去。陷阱选项 (D) 将无摩擦情形下的通用答案（始终滑动）与特定表面条件相混淆。
Insight.要点。At the slip threshold, downhill gravity equals maximum static friction: $mg\sin\theta=\mu_s mg\cos\theta$, so $\tan\theta=\mu_s$. This is an angle condition, not a universal motion rule.在滑动临界点，沿坡向下的重力分量等于最大静摩擦力：$mg\sin\theta=\mu_s mg\cos\theta$，故 $\tan\theta=\mu_s$。这是角度条件，不是普遍运动规律。
Q13MEDIUM2.10 Kepler's Third Law2.10 开普勒第三定律No Calculator
A satellite orbits Earth at radius $R$ with period $T$. A second satellite at radius $4R$ has period一颗卫星以半径 $R$ 绕地球运行，周期为 $T$。另一颗卫星运行半径为 $4R$，其周期为
(A) $T$
(B) $2T$
(C) $4T$
(D) $8T$
Answer:答案：(D)
Kepler's third law: $T^2 \propto R^3$, so $T_2/T_1 = (R_2/R_1)^{3/2}$.开普勒第三定律：$T^2 \propto R^3$，故 $T_2/T_1 = (R_2/R_1)^{3/2}$。
$$T_2 = T\,(4)^{3/2} = T\cdot 8 = 8T$$
Trap (C) takes the ratio linearly; (B) takes the square root.陷阱选项 (C) 将比值当成线性关系，(B) 仅取平方根。
Insight.要点。Kepler's third law gives $T\propto r^{3/2}$, so a factor of $4$ in radius gives a factor of $8$ in period. Never treat the ratio as linear or as $r^{1/2}$.开普勒第三定律给出 $T\propto r^{3/2}$，故半径变为 $4$ 倍时，周期变为 $8$ 倍。不要把这个比值当作线性或 $r^{1/2}$ 关系。
Q14HARD2.2 Atwood Machine2.2 阿特伍德机No Calculator
Two masses with $m_1 < m_2$ hang from opposite ends of a light inextensible string passing over a frictionless, massless pulley. The downward acceleration of $m_2$ is$m_1 < m_2$ 的两个质量分别悬挂在一根轻质不可伸长的绳子两端，绳子绕过无摩擦、无质量的滑轮。$m_2$ 向下的加速度为
(A) $\dfrac{(m_2 - m_1)\,g}{m_1 + m_2}$
(B) $\dfrac{(m_2 - m_1)\,g}{m_2}$
(C) $\dfrac{m_2\, g}{m_1 + m_2}$
(D) $g$
Answer:答案：(A)
Newton's second law on each mass (with $T$ the common tension, $a$ the common acceleration magnitude):对每个质量应用牛顿第二定律（$T$ 为共同张力，$a$ 为共同加速度大小）：
Sanity check: $m_1 = m_2$ gives $a = 0$; $m_1 \to 0$ gives $a \to g$ (free fall of $m_2$). Trap (B) divides by only one mass; (D) ignores the pulley constraint.量纲验证：$m_1 = m_2$ 时 $a = 0$；$m_1 \to 0$ 时 $a \to g$（$m_2$ 自由落体）。陷阱选项 (B) 只除以一个质量，(D) 忽略了滑轮约束。
Insight.要点。For an Atwood machine, write Newton's law for each mass and add the equations to eliminate the common tension. The denominator is the total mass, and the limits $m_1=m_2$ and $m_1\to 0$ provide a sanity check.对阿特伍德机，分别对两个质量写牛顿方程，再相加消去共同张力。分母是总质量；$m_1=m_2$ 与 $m_1\to 0$ 两个极限可作为检验。
Q15HARD2.5 Linear Drag, Time Constant2.5 线性阻力，时间常数Calculator
A particle of mass $m$ falls from rest under gravity with linear drag $F_\mathrm{drag} = -bv$. Its velocity satisfies $v(t) = v_t\bigl(1 - e^{-t/\tau}\bigr)$ with terminal velocity $v_t = mg/b$ and time constant $\tau = m/b$. The time at which the speed reaches $0.9\,v_t$ is closest to质量为 $m$ 的质点从静止开始在重力和线性阻力 $F_\mathrm{drag} = -bv$ 作用下下落。其速度满足 $v(t) = v_t\bigl(1 - e^{-t/\tau}\bigr)$，终端速度 $v_t = mg/b$，时间常数 $\tau = m/b$。速度达到 $0.9\,v_t$ 所需时间最接近
Insight.要点。Exponential relaxation is governed by the time constant $\tau=m/b$, not by the time to reach terminal velocity. Useful landmarks: one $\tau$ gives $0.63v_t$, two gives $0.86v_t$, and $\ln 10$ times $\tau$ gives $0.90v_t$.指数趋近由时间常数 $\tau=m/b$ 控制，而不是到达终端速度所需时间。常用参考：一个 $\tau$ 对应 $0.63v_t$，两个对应 $0.86v_t$，$\ln 10$ 个 $\tau$ 对应 $0.90v_t$。
Q16HARD2.9 Conical Pendulum2.9 圆锥摆No Calculator
A bob of mass $m$ swings in a horizontal circle at the end of a string of length $\ell$ that makes a fixed angle $\theta$ with the vertical (a conical pendulum). The angular speed satisfies质量为 $m$ 的摆球悬于长度为 $\ell$ 的绳端，绳与竖直方向成固定角 $\theta$，摆球在水平圆上运动（圆锥摆）。角速度满足
(A) $\omega^2 = g\,\ell\,\sin\theta$
(B) $\omega^2 = g/\ell$
(C) $\omega^2 = \dfrac{g}{\ell\,\cos\theta}$
(D) $\omega^2 = \dfrac{g\,\cos\theta}{\ell}$
Answer:答案：(C)
Let $T$ be the string tension and $r = \ell\sin\theta$ the radius of the horizontal circle. Vertical and radial Newton's laws:设 $T$ 为绳的张力，$r = \ell\sin\theta$ 为水平圆的半径。竖直和径向牛顿方程：
$$T\cos\theta = m g, \qquad T\sin\theta = m\,\omega^2 r = m\,\omega^2 \ell\sin\theta$$
The radial equation gives $T = m\omega^2\ell$. Substitute into the vertical:径向方程得 $T = m\omega^2\ell$。代入竖直方程：
$$m\omega^2\ell\cos\theta = m g \;\Longrightarrow\; \omega^2 = \frac{g}{\ell\cos\theta}$$
Sanity check: $\theta \to 0$ recovers the small-angle pendulum frequency $\sqrt{g/\ell}$; $\theta \to 90^\circ$ pushes $\omega \to \infty$ as the string approaches horizontal.量纲验证：$\theta \to 0$ 时还原小角度摆频率 $\sqrt{g/\ell}$；$\theta \to 90^\circ$ 时绳趋近水平，$\omega \to \infty$。
Insight.要点。In a conical pendulum, vertical balance fixes tension and radial balance fixes $\omega$. The result $\omega^2=g/(\ell\cos\theta)$ grows with $\theta$; a horizontal string is a singular limit.在圆锥摆中，竖直方向平衡确定张力，径向平衡确定 $\omega$。结果 $\omega^2=g/(\ell\cos\theta)$ 随 $\theta$ 增大；绳接近水平是奇异极限。
A $5.0~\mathrm{kg}$ block rests on a frictionless incline of angle $\theta = 25^\circ$. A force $F = 30~\mathrm{N}$ is applied parallel to the slope, directed up the slope. The block's acceleration along the incline is closest to一个 $5.0~\mathrm{kg}$ 的物块静止在 $\theta = 25^\circ$ 的无摩擦斜面上。沿斜面向上施加 $F = 30~\mathrm{N}$ 的力。该物块沿斜面方向的加速度最接近
(A)$1.86~\mathrm{m/s^2}$ up the slope$1.86~\mathrm{m/s^2}$，沿斜面向上
(B)$1.86~\mathrm{m/s^2}$ down the slope$1.86~\mathrm{m/s^2}$，沿斜面向下
(C)$4.14~\mathrm{m/s^2}$ up the slope$4.14~\mathrm{m/s^2}$，沿斜面向上
(D)$0$ (block remains at rest)$0$（物块保持静止）
Answer:答案：(A)
Take up the slope positive. Slope-parallel gravity component is $mg\sin\theta$ down-slope; no friction.取沿斜面向上为正方向。沿斜面方向重力分量 $mg\sin\theta$ 指向下坡，无摩擦。
Positive, so up the slope. Trap (C) accidentally uses $\cos\theta$.结果为正，即沿斜面向上。陷阱选项 (C) 误用了 $\cos\theta$。
Insight.要点。Choose one positive direction along the slope and keep signs consistent. The force law becomes $a=(F-mg\sin\theta)/m$; the slope component uses sine, not cosine.沿斜面选定一个正方向并保持符号一致。受力关系为 $a=(F-mg\sin\theta)/m$；沿斜面分量用正弦，不用余弦。
Q18HARD2.8 Escape vs. Orbital Speed2.8 逃逸速度与轨道速度Calculator
The escape speed from a planet's surface is $v_e = \sqrt{2GM/R}$. Compared with the circular orbital speed $v_o = \sqrt{GM/R}$ at radius $R$, the escape speed satisfies从行星表面的逃逸速度为 $v_e = \sqrt{2GM/R}$。与半径 $R$ 处的圆轨道速度 $v_o = \sqrt{GM/R}$ 相比，逃逸速度满足
So $v_e \approx 1.41\,v_o$, escape is only ~41% faster than orbit, not double. (D) has the wrong units.故 $v_e \approx 1.41\,v_o$，逃逸速度仅比轨道速度快约 41%，并非两倍。(D) 的单位不正确。
Insight.要点。Escape speed is $\sqrt2$ times circular orbital speed at the same radius, about $1.41v_o$, not twice $v_o$. The factor $\sqrt2$ comes from making total energy zero rather than negative.逃逸速度是同一半径圆轨道速度的 $\sqrt2$ 倍，约为 $1.41v_o$，而不是两倍。因子 $\sqrt2$ 来源于使总能量从负值变为零。
Each FRQ walks every part in the canonical AP-style setup → execute → evaluate structure. Numbers use $g = 9.8~\mathrm{m/s^2}$.每道自由解答题均按 AP 标准的建立模型，执行计算，评估结果的结构逐步解析。数值计算取 $g = 9.8~\mathrm{m/s^2}$。
$m_1 = 2.0~\mathrm{kg}$ on a horizontal table with $\mu_k = 0.30$; string over a frictionless pulley to hanging $m_2 = 4.0~\mathrm{kg}$. Released from rest.$m_1 = 2.0~\mathrm{kg}$ 置于水平桌面上，$\mu_k = 0.30$；绳子经无摩擦滑轮连接悬挂的 $m_2 = 4.0~\mathrm{kg}$。从静止开始释放。
(a)Free-body diagrams and Newton's second law.受力图和牛顿第二定律。
$m_1$ (on the table): weight $m_1 g$ down, normal $N$ up, tension $T$ toward the pulley, kinetic friction $f_k = \mu_k N$ opposite the motion.$m_1$（桌面上）：重力 $m_1 g$ 向下，法向力 $N$ 向上，张力 $T$ 指向滑轮，动摩擦力 $f_k = \mu_k N$ 与运动方向相反。
Insight.要点。For a coupled table-and-hanging-mass system, kinetic friction subtracts from the driving weight: $a=(m_2g-\mu_k m_1g)/(m_1+m_2)$. Isolate either mass to cross-check the tension.对于桌面加悬挂质量系统，动摩擦力从驱动重力中扣除：$a=(m_2g-\mu_k m_1g)/(m_1+m_2)$。再隔离任一质量交叉验证张力。
Negative means the assumed direction of motion (up-slope) is inconsistent, the applied force cannot overcome gravity-along-slope plus the friction that would oppose up-slope motion. Conversely, the slope-parallel gravity component ($19.6~\mathrm{N}$) is also less than $F = 25~\mathrm{N}$, so the block doesn't slide down either. With $\mu_s \ge \mu_k$ the block remains static and $a = 0$.结果为负，说明假设的运动方向（向上）不自洽，施加的力不足以克服沿斜面的重力分量及将会阻碍向上运动的摩擦力。反之，沿斜面的重力分量（$19.6~\mathrm{N}$）也小于 $F = 25~\mathrm{N}$，物块也不会向下滑动。由于 $\mu_s \ge \mu_k$，物块保持静止，$a = 0$。
(d)$F$ for constant up-slope velocity. With $a = 0$ and motion up-slope, friction is fully kinetic and opposes motion:使物块以匀速沿斜面向上运动所需的 $F$。当 $a = 0$ 且物块向上运动时，摩擦力为动摩擦力，方向与运动相反：
$$F = m g\sin\theta + \mu_k N = 19.6 + 8.49 \approx 28.1~\mathrm{N}$$
Just $3~\mathrm{N}$ more than the given value, explaining why part (c)'s assumed motion didn't take off.仅比题目给定值多约 $3~\mathrm{N}$，这也解释了为何 (c) 中假设的运动无法实现。
Insight.要点。If an assumed motion direction produces a negative acceleration, that direction is inconsistent. Re-evaluate whether static friction can hold the block; friction direction always opposes the actual or impending motion.若假设的运动方向得到负加速度，说明该方向不自洽。应重新判断静摩擦是否能维持静止；摩擦方向总是与实际或即将发生的运动方向相反。
FRQ 3HARD2.5 Linear Drag, Derivation2.5 线性阻力，推导No Calculator
Particle of mass $m$ falls from rest through a fluid with $\vec F_\mathrm{drag} = -b\vec v$. Downward positive.质量为 $m$ 的质点从静止开始在流体中下落，阻力 $\vec F_\mathrm{drag} = -b\vec v$。取向下为正方向。
(a)Newton's second law with downward as $+$:以向下为正方向应用牛顿第二定律：
$$m\,\frac{dv}{dt} = m g - b v \;\Longleftrightarrow\; \frac{dv}{dt} = g - \frac{b}{m}\,v$$
$v(t)$ starts at $0$ with initial slope $g$, curves over, and approaches $v_t$ asymptotically as $t \to \infty$. The point $(\tau,\,0.63\,v_t)$ sits on the curve; the asymptote at $v = v_t$ is a horizontal dashed line above. By $4\tau$ the curve is within $\approx 2\%$ of $v_t$.$v(t)$ 从 $0$ 出发，初始斜率为 $g$，曲线向上弯曲，当 $t \to \infty$ 时渐近趋向 $v_t$。点 $(\tau,\,0.63\,v_t)$ 在曲线上，渐近线 $v = v_t$ 在上方用水平虚线标注。到 $4\tau$ 时，曲线与 $v_t$ 的偏差约在 $2\%$ 以内。
Insight.要点。The linear-drag equation is separable only after setting the sign convention clearly. Terminal velocity comes from $dv/dt=0$, while the full solution interpolates exponentially from zero to $v_t$.只有先明确正方向，线性阻力方程才能分离变量。终端速度由 $dv/dt=0$ 得到，而完整解是从零按指数规律逼近 $v_t$。
Frictionless track with a vertical loop of radius $R$; block enters the bottom moving horizontally with speed $v_0$.无摩擦轨道包含半径为 $R$ 的竖直圆环，物块以水平速度 $v_0$ 从圆环底部进入。
(a)Minimum entry speed. At the very top of the loop, gravity points toward the center; the normal force vanishes at the threshold $N = 0$, leaving gravity to supply all the centripetal force:最小入口速度。在圆环顶部，重力指向圆心；在临界情况 $N = 0$ 时，重力单独提供所有向心力：
$$m g = \frac{m\,v_\text{top,min}^2}{R} \;\Longrightarrow\; v_\text{top,min}^2 = g R$$
Energy conservation from entry (height $0$) to top (height $2R$) for a frictionless track:无摩擦轨道上从入口（高度 $0$）到顶部（高度 $2R$）的能量守恒：
$$\tfrac{1}{2} m v_{0,\min}^2 = \tfrac{1}{2} m v_\text{top,min}^2 + m g (2R)$$
$$v_{0,\min}^2 = g R + 4 g R = 5 g R \;\Longrightarrow\; v_{0,\min} = \sqrt{5 g R}$$
(b)Speed at the top for $v_0 = 2\,v_{0,\min}$. Then $v_0^2 = 4(5 g R) = 20\,g R$. Energy conservation:当 $v_0 = 2\,v_{0,\min}$ 时顶部速度。此时 $v_0^2 = 4(5 g R) = 20\,g R$。能量守恒：
$$v_\text{top}^2 = 20 g R - 4 g R = 16 g R \;\Longrightarrow\; v_\text{top} = 4\sqrt{g R}$$
(c)Normal force at the top. Both $N$ and gravity point toward the center (downward), so顶部法向力。$N$ 和重力均指向圆心（向下），故
$$N + m g = \frac{m\,v_\text{top}^2}{R} = \frac{m (16 g R)}{R} = 16\,m g$$
$$N = 16 m g - m g = 15\,m g$$
(d)Bottom vs. top.底部与顶部对比。
The normal force at the bottom is larger than at the top. Two effects compound: (i) at the bottom, the normal force must both support gravity and supply the centripetal acceleration, they point opposite ways, so $N_\text{bot} - m g = m v_\text{bot}^2/R$, i.e. $N = m g + m v_\text{bot}^2/R$ (gravity now adds to $N$ rather than helping); and (ii) the block is moving faster at the bottom by energy conservation, so $v_\text{bot}^2/R$ itself is larger than $v_\text{top}^2/R$.底部法向力大于顶部法向力。两个效应叠加：(i) 在底部，法向力既要支撑重力，又要提供向心加速度，二者方向相反，故 $N_\text{bot} - m g = m v_\text{bot}^2/R$，即 $N = m g + m v_\text{bot}^2/R$（重力现在增大了 $N$ 而非减小）；(ii) 根据能量守恒，物块在底部速度更快，因此 $v_\text{bot}^2/R$ 本身也大于 $v_\text{top}^2/R$。
Insight.要点。Vertical-loop problems couple two independent conditions: centripetal force at the top and mechanical energy between two heights. Do not use energy alone; the top still needs $v_{\text{top}}^2=gR$ at the threshold.竖直圆环问题耦合两个独立条件：顶部的向心力条件，以及两高度之间的机械能守恒。不能只用能量；临界时顶部仍需满足 $v_{\text{top}}^2=gR$。