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。
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
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。
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。
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 $m_1\,a$。
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。
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：
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。
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。
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。
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）。
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。
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。
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。
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
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，$\omega \to \infty$。
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 $\cos\theta$。
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。
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。
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。
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。
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 $v_\text{top}^2/R$。