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.下方每题均重述题干和选项，标出正确字母，并给出与 AP 主题对应的一到三句解释。
Q1EASY3.1 Kinetic Energy3.1 动能No Calculator
A $2.0~\mathrm{kg}$ block moves at $6.0~\mathrm{m/s}$. Its kinetic energy is一个质量为 $2.0~\mathrm{kg}$ 的滑块以 $6.0~\mathrm{m/s}$ 的速度运动，其动能为
(A) $6~\mathrm{J}$
(B) $12~\mathrm{J}$
(C) $24~\mathrm{J}$
(D) $36~\mathrm{J}$
Answer:答案：(D)
$$K = \tfrac{1}{2} m v^2 = \tfrac{1}{2}(2.0)(6.0)^2 = 36~\mathrm{J}$$
A constant force of magnitude $F = 50~\mathrm{N}$ pulls a block through a displacement of $d = 4.0~\mathrm{m}$. The angle between the force and the displacement is $60^\circ$. The work done by the force is一个大小为 $F = 50~\mathrm{N}$ 的恒力拉动滑块位移 $d = 4.0~\mathrm{m}$，力与位移之间的夹角为 $60^\circ$，该力做的功为
Trap (D) multiplies by $t$; (A) inverts the ratio.干扰项 (D) 乘以 $t$；(A) 颠倒了比值。
Q4EASY3.4 Spring Potential Energy3.4 弹簧弹性势能No Calculator
A spring with $k = 200~\mathrm{N/m}$ is compressed $0.10~\mathrm{m}$ from its natural length. The elastic potential energy stored in the spring is一根弹簧的弹簧常数 $k = 200~\mathrm{N/m}$，从自然长度压缩 $0.10~\mathrm{m}$，弹簧中储存的弹性势能为
(A) $0.5~\mathrm{J}$
(B) $1.0~\mathrm{J}$
(C) $2.0~\mathrm{J}$
(D) $10~\mathrm{J}$
Answer:答案：(B)
$$U_s = \tfrac{1}{2} k (\Delta x)^2 = \tfrac{1}{2}(200)(0.10)^2 = 1.0~\mathrm{J}$$
Trap (C) drops the factor of $\tfrac{1}{2}$; (D) forgets to square $\Delta x$.干扰项 (C) 丢掉了 $\tfrac{1}{2}$ 因子；(D) 忘记对 $\Delta x$ 平方。
A $5.0~\mathrm{kg}$ block moving at $10~\mathrm{m/s}$ along a horizontal surface is brought to rest by friction alone. The total work done on the block by friction is一个质量为 $5.0~\mathrm{kg}$ 的滑块以 $10~\mathrm{m/s}$ 沿水平面运动，仅由摩擦力使其静止，摩擦力对滑块做的总功为
(A) $+250~\mathrm{J}$
(B) $-250~\mathrm{J}$
(C) $+500~\mathrm{J}$
(D) $-500~\mathrm{J}$
Answer:答案：(B)
Work-energy theorem: $W_\text{net} = \Delta K = K_f - K_i$. Friction is the only horizontal force doing work, so动能定理：$W_\text{net} = \Delta K = K_f - K_i$。摩擦力是唯一做功的水平力，故
The minus sign reflects that friction opposes the motion (removes KE). Trap (A) gets the magnitude right but the sign wrong; (D) drops the factor of $\tfrac{1}{2}$.负号反映摩擦力与运动方向相反（减少动能）。干扰项 (A) 数值正确但符号错误；(D) 丢掉了 $\tfrac{1}{2}$ 因子。
Q6MEDIUM3.4 Force from Potential Energy3.4 由势能求力No Calculator
A particle moves along the $x$-axis under a conservative potential $U(x) = 3x^2 - 6x + 2$ (J, with $x$ in m). The force on the particle at $x = 1~\mathrm{m}$ is一个质点在保守势能 $U(x) = 3x^2 - 6x + 2$（J，$x$ 单位为 m）的作用下沿 $x$ 轴运动，质点在 $x = 1~\mathrm{m}$ 处所受的力为
(A) $0$
(B) $-1~\mathrm{N}$
(C) $+1~\mathrm{N}$
(D) $+6~\mathrm{N}$
Answer:答案：(A)
$$F(x) = -\frac{dU}{dx} = -(6x - 6) = 6 - 6x$$
$$F(1) = 6 - 6(1) = 0$$
$x = 1$ is the minimum of $U$ (stable equilibrium), so the force vanishes there. Trap (D) reports $|dU/dx|$ at the wrong point; the sign trap returns $\pm 1$ if $\theta$ is shifted by one unit.$x = 1$ 是 $U$ 的极小值（稳定平衡），故该处力为零。干扰项 (D) 给出错误位置处 $|dU/dx|$ 的值；符号陷阱在 $\theta$ 偏移一个单位时给出 $\pm 1$。
A $1.0~\mathrm{kg}$ block is released from rest on a frictionless ramp at height $h = 5.0~\mathrm{m}$ above the bottom. Take $g = 10~\mathrm{m/s^2}$. Its speed at the bottom of the ramp is一个质量为 $1.0~\mathrm{kg}$ 的滑块从无摩擦斜面上距底部高度 $h = 5.0~\mathrm{m}$ 处由静止释放，取 $g = 10~\mathrm{m/s^2}$，到达斜面底部时的速度为
(A) $5~\mathrm{m/s}$
(B) $10~\mathrm{m/s}$
(C) $25~\mathrm{m/s}$
(D) $50~\mathrm{m/s}$
Answer:答案：(B)
Energy conservation: $m g h = \tfrac{1}{2} m v^2 \Rightarrow v = \sqrt{2 g h}$.能量守恒：$m g h = \tfrac{1}{2} m v^2 \Rightarrow v = \sqrt{2 g h}$。
Q8MEDIUM3.2 Variable Force Integral3.2 变力积分做功Calculator
A force $F(x) = 3x^2 + 2x$ (in N, with $x$ in m) acts on a particle as it moves along the $x$-axis. The work done by this force as the particle moves from $x = 0$ to $x = 2~\mathrm{m}$ is一个力 $F(x) = 3x^2 + 2x$（N，$x$ 单位为 m）作用于质点沿 $x$ 轴运动，质点从 $x = 0$ 运动到 $x = 2~\mathrm{m}$ 过程中该力做的功为
A particle moves under the potential $U(x) = x^3 - 3x$ (J, with $x$ in m). Which statement about its equilibria is correct?一个质点在势能 $U(x) = x^3 - 3x$（J，$x$ 单位为 m）的作用下运动，关于其平衡位置，下列说法正确的是
(A)$x = +1$ is a stable equilibrium; $x = -1$ is unstable.$x = +1$ 是稳定平衡；$x = -1$ 是不稳定平衡。
(B)$x = -1$ is a stable equilibrium; $x = +1$ is unstable.$x = -1$ 是稳定平衡；$x = +1$ 是不稳定平衡。
Stable equilibria sit at local minima of $U$; unstable at local maxima. Trap (B) flips the sign convention.稳定平衡对应 $U$ 的局部极小值，不稳定平衡对应局部极大值。干扰项 (B) 颠倒了符号判别规则。
Q10MEDIUM3.5 Spring Launcher3.5 弹簧发射器No Calculator
A spring with $k = 200~\mathrm{N/m}$ is compressed by $0.30~\mathrm{m}$ and used to launch a $0.50~\mathrm{kg}$ block on a frictionless surface. The block's speed after release is弹簧常数 $k = 200~\mathrm{N/m}$ 的弹簧被压缩 $0.30~\mathrm{m}$，在无摩擦面上弹射一个质量为 $0.50~\mathrm{kg}$ 的滑块，释放后滑块的速度为
(A) $3~\mathrm{m/s}$
(B) $6~\mathrm{m/s}$
(C) $9~\mathrm{m/s}$
(D) $12~\mathrm{m/s}$
Answer:答案：(B)
Energy conservation: $\tfrac{1}{2} k (\Delta x)^2 = \tfrac{1}{2} m v^2$.能量守恒：$\tfrac{1}{2} k (\Delta x)^2 = \tfrac{1}{2} m v^2$。
Trap (D) misses the factor $\sqrt{m}$ in the denominator.干扰项 (D) 分母中漏掉了 $\sqrt{m}$ 因子。
Q11MEDIUM3.3 Instantaneous Power on Incline3.3 斜面上的瞬时功率Calculator
A $1500~\mathrm{kg}$ car climbs a $5^\circ$ incline at constant velocity $v = 20~\mathrm{m/s}$. Drag is negligible. The instantaneous mechanical power delivered by the engine is closest to一辆质量为 $1500~\mathrm{kg}$ 的汽车以匀速 $v = 20~\mathrm{m/s}$ 爬上 $5^\circ$ 的坡道，空气阻力不计，发动机输出的瞬时机械功率最接近
(A) $26~\mathrm{kW}$
(B) $76~\mathrm{kW}$
(C) $147~\mathrm{kW}$
(D) $293~\mathrm{kW}$
Answer:答案：(A)
Constant velocity means engine force balances slope-parallel gravity: $F = m g \sin\theta$. Then匀速运动意味着发动机牵引力与沿坡道方向的重力分量平衡：$F = m g \sin\theta$，因此
$$P = F v = m g \sin\theta\,v = 1500(9.8)\sin 5^\circ\,(20)$$
Trap (C) uses $m g v$ (no $\sin\theta$, i.e. treats road as vertical).干扰项 (C) 使用 $m g v$（无 $\sin\theta$，即将道路视为竖直方向）。
Q12MEDIUM3.5 Conservative vs. Non-Conservative3.5 保守力与非保守力No Calculator
Which of the following forces is not conservative?下列哪个力不是保守力？
(A)Gravity (near Earth's surface)重力（地球表面附近）
(B)Ideal spring force理想弹簧弹力
(C)Universal gravitation万有引力
(D)Kinetic friction滑动摩擦力
Answer:答案：(D)
Conservative forces derive from a potential and have path-independent work. Kinetic friction depends on path length (the longer the path, the more energy dissipated), so it cannot be derived from a $U(x)$, non-conservative. The other three each have a standard potential ($U = mgh$, $U = \tfrac{1}{2}kx^2$, $U = -GMm/r$).保守力来自势能，做功与路径无关。滑动摩擦力取决于路径长度（路径越长，耗散能量越多），因此无法由 $U(x)$ 导出，属于非保守力。其余三项各有标准势能（$U = mgh$，$U = \tfrac{1}{2}kx^2$，$U = -GMm/r$）。
Q13MEDIUM3.1 KE Scaling3.1 动能的比例关系No Calculator
A car's speed is doubled. Its kinetic energy一辆汽车的速度翻倍，其动能
(A)halves减半
(B)stays the same保持不变
(C)doubles翻倍
(D)quadruples变为原来的四倍
Answer:答案：(D)
$K = \tfrac{1}{2} m v^2$, so $K$ scales as $v^2$. Doubling $v$ multiplies $K$ by $4$. Important consequence for safety: braking distance also scales as $v^2$ (constant friction force absorbs $\Delta K$ over the stopping distance).$K = \tfrac{1}{2} m v^2$，故 $K$ 与 $v^2$ 成正比。速度翻倍则动能变为原来的 $4$ 倍。安全驾驶的重要推论：制动距离同样与 $v^2$ 成正比（恒定摩擦力通过制动距离吸收 $\Delta K$）。
Q14HARD3.2 Line Integral of $\vec F \cdot d\vec r$3.2 力的线积分 $\vec F \cdot d\vec r$No Calculator
A particle moves along the straight line from $(0, 0)$ to $(2, 3)$ (in metres) under the force $\vec F = 2y\,\hat\imath + 3x\,\hat\jmath$ (in N). The work done by $\vec F$ along this path is一个质点在力 $\vec F = 2y\,\hat\imath + 3x\,\hat\jmath$（N）的作用下沿直线从 $(0, 0)$ 运动到 $(2, 3)$（单位：m），$\vec F$ 沿该路径所做的功为
A particle moves along the $x$-axis under the potential $U(x) = x^3 - 9x$ (J, with $x$ in m). The locations of the particle's stable equilibrium are一个质点在势能 $U(x) = x^3 - 9x$（J，$x$ 单位为 m）的作用下沿 $x$ 轴运动，质点稳定平衡的位置为
A car of mass $m$ starts from rest on a flat road and is driven by an engine that delivers constant mechanical power $P$. Drag is neglected. The car's speed at time $t$ is一辆质量为 $m$ 的汽车从静止出发在平路上行驶，发动机以恒定机械功率 $P$ 驱动，忽略阻力，汽车在时刻 $t$ 的速度为
(A) $v(t) = \sqrt{Pt/m}$
(B) $v(t) = \sqrt{2Pt/m}$
(C) $v(t) = 2Pt/m$
(D) $v(t) = \bigl(Pt/m\bigr)^{2}$
Answer:答案：(B)
Constant $P$ means $dK/dt = P$, so $K(t) = Pt$ from rest. Then $\tfrac{1}{2} m v^2 = P t$, giving $v = \sqrt{2Pt/m}$. Trap (A) drops the factor of 2; (C) inverts the square root, returning a quantity with units of length per time per time.恒定功率 $P$ 意味着 $dK/dt = P$，从静止开始故 $K(t) = Pt$。则 $\tfrac{1}{2} m v^2 = P t$，得 $v = \sqrt{2Pt/m}$。干扰项 (A) 丢掉了因子 2；(C) 将开方倒置，给出单位为"长度/时间/时间"的量。
Q17HARD3.5 Loop-the-Loop3.5 竖直圆环运动Calculator
A small block is released from rest at height $h$ on a frictionless track that leads into a vertical loop of radius $R$. The minimum value of $h$ such that the block maintains contact with the track at the top of the loop is一个小滑块从无摩擦轨道上高度 $h$ 处由静止释放，进入半径为 $R$ 的竖直圆环，使滑块在圆环顶端保持与轨道接触所需 $h$ 的最小值为
(A) $h = R$
(B) $h = 2R$
(C) $h = \dfrac{5R}{2}$
(D) $h = 3R$
Answer:答案：(C)
Centripetal condition at the top with $N = 0$: $m g = m v_\text{top}^2/R$, so $v_\text{top}^2 = g R$. Energy conservation from height $h$ to the top (height $2R$):顶端 $N = 0$ 时的向心力条件：$m g = m v_\text{top}^2/R$，故 $v_\text{top}^2 = g R$。从高度 $h$ 到顶端（高度 $2R$）的能量守恒：
$$m g h = m g (2R) + \tfrac{1}{2} m v_\text{top}^2 = 2 m g R + \tfrac{1}{2} m g R$$
$$h = 2R + \tfrac{R}{2} = \frac{5R}{2}$$
Trap (B) forgets the kinetic-energy term at the top.干扰项 (B) 忘记了顶端处的动能项。
Q18HARD3.3 Pump Power3.3 水泵功率Calculator
A pump lifts water from a well of depth $h = 30~\mathrm{m}$ and delivers it to ground level at a steady rate of $5.0~\mathrm{kg/s}$. Assuming the water is delivered with negligible exit speed, the minimum mechanical power the pump must supply is closest to一台水泵以 $5.0~\mathrm{kg/s}$ 的稳定速率将水从深度 $h = 30~\mathrm{m}$ 的水井中提升至地面，假设出水口速度可忽略不计，水泵需提供的最小机械功率最接近
(A) $150~\mathrm{W}$
(B) $250~\mathrm{W}$
(C) $1{,}470~\mathrm{W}$
(D) $6{,}000~\mathrm{W}$
Answer:答案：(C)
Energy raised per second equals power. With $dm/dt$ kg lifted height $h$ per second:每秒提升的能量即为功率。每秒提升 $dm/dt$ kg 的水至高度 $h$：
Negligible exit KE means all the input work becomes gravitational PE. Trap (A) drops a factor of $g$; (D) treats $dm/dt$ as mass and uses $mgh$ once.出口动能可忽略，意味着所有输入功均转化为重力势能。干扰项 (A) 丢掉了 $g$ 因子；(D) 将 $dm/dt$ 当作质量，只用了一次 $mgh$。
Each FRQ walks every part in the canonical AP-style setup → execute → evaluate structure. Numbers use $g = 9.8~\mathrm{m/s^2}$.每道自由作答题按照 AP 标准的"建立 → 求解 → 评估"结构逐步演示每个小问，数值计算取 $g = 9.8~\mathrm{m/s^2}$。
FRQ 1MEDIUM3.5 Energy Bookkeeping3.5 能量守恒分析Calculator
$m = 0.50~\mathrm{kg}$ released from rest at $h = 1.2~\mathrm{m}$ on a frictionless curved ramp; floor then frictionless to a spring ($k = 600~\mathrm{N/m}$); on return trip floor has $\mu_k = 0.20$.$m = 0.50~\mathrm{kg}$ 从无摩擦弧形斜面上 $h = 1.2~\mathrm{m}$ 处由静止释放；地面无摩擦直至压缩弹簧（$k = 600~\mathrm{N/m}$）；返回途中地面动摩擦系数 $\mu_k = 0.20$。
(a)Speed at the bottom of the ramp. Energy conservation $m g h = \tfrac{1}{2} m v^2$:斜面底部的速度。能量守恒 $m g h = \tfrac{1}{2} m v^2$：
(b)Maximum spring compression. All KE → spring PE: $\tfrac{1}{2} m v^2 = \tfrac{1}{2} k x_\max^2$, equivalently $m g h = \tfrac{1}{2} k x_\max^2$:弹簧最大压缩量。所有动能转化为弹簧弹性势能：$\tfrac{1}{2} m v^2 = \tfrac{1}{2} k x_\max^2$，等价地 $m g h = \tfrac{1}{2} k x_\max^2$：
$$x_\max = \sqrt{\frac{2 m g h}{k}} = \sqrt{\frac{2(0.50)(9.8)(1.2)}{600}} = \sqrt{0.0196} \approx 0.140~\mathrm{m}$$
(c)Distance on the return trip before friction stops the block. The spring returns all $5.88~\mathrm{J}$ of stored PE to KE, so the block leaves the spring with the same speed $4.85~\mathrm{m/s}$. On the new (frictioned) floor, work-energy theorem with $W_f = -\mu_k m g\,d$:返回途中摩擦力使滑块停止前滑行的距离。弹簧将全部 $5.88~\mathrm{J}$ 的弹性势能还原为动能，故滑块离开弹簧时速度仍为 $4.85~\mathrm{m/s}$。在有摩擦的地面上，由动能定理 $W_f = -\mu_k m g\,d$：
$$\tfrac{1}{2} m v^2 = \mu_k m g\,d \;\Longrightarrow\; d = \frac{v^2}{2\mu_k g} = \frac{23.52}{2(0.20)(9.8)} = \frac{23.52}{3.92} = 6.0~\mathrm{m}$$
(d)Fraction of original mechanical energy dissipated. The block ends at rest on the floor (gravitational PE = 0 relative to its end state), so all of the original $m g h = 5.88~\mathrm{J}$ is gone to friction, fraction $= 1.00$ (100%). The spring/return-trip path simply specifies where the dissipation happens, not how much.原始机械能被耗散的比例。滑块最终静止在地面（重力势能相对末态为 0），故原始 $m g h = 5.88~\mathrm{J}$ 全部转化为摩擦热，比例 $= 1.00$（100%）。弹簧和返程路径只说明耗散发生在何处，而非耗散了多少。
FRQ 2MEDIUM3.4 Potential Energy Diagram3.4 势能曲线分析Calculator
$m = 2.0~\mathrm{kg}$ on the $x$-axis under $U(x) = \tfrac{1}{4} x^4 - 2 x^2 + 3$ (J, $x$ in m).$m = 2.0~\mathrm{kg}$ 的质点在势能 $U(x) = \tfrac{1}{4} x^4 - 2 x^2 + 3$（J，$x$ 单位为 m）的作用下沿 $x$ 轴运动。
(c)Released from rest at $x = 0$. The energy is $E = U(0) = 3~\mathrm{J}$. Solving $U(x) = E$: $\tfrac{1}{4}x^4 - 2x^2 = 0 \Rightarrow x^2(x^2 - 8) = 0$, so the classical turning points are $x = 0$ and $x = \pm 2\sqrt{2}$.从 $x = 0$ 处由静止释放。能量 $E = U(0) = 3~\mathrm{J}$。解 $U(x) = E$：$\tfrac{1}{4}x^4 - 2x^2 = 0 \Rightarrow x^2(x^2 - 8) = 0$，经典转折点为 $x = 0$ 和 $x = \pm 2\sqrt{2}$。
Strictly, the particle sits at an unstable equilibrium and remains at rest there for all time (zero force, zero velocity). With any infinitesimal perturbation, it slides toward one of the wells; from energy conservation the motion is then bounded between $x = 0$ and $x = +2\sqrt{2}$ (or its mirror), with the orbit through $x = 0$ taking infinite time.严格来说，质点位于不稳定平衡处，将永远保持静止（力为零，速度为零）。若有任意无穷小扰动，它将向某个势阱滑动；由能量守恒，运动被限制在 $x = 0$ 与 $x = +2\sqrt{2}$（或对称的另侧）之间，经过 $x = 0$ 的轨道所需时间为无穷大。
(d)Released from rest at $x = 3$. Now $E = U(3) = \tfrac{81}{4} - 18 + 3 = 5.25~\mathrm{J}$. Speed at $x = 0$ from $E = \tfrac{1}{2} m v^2 + U(0)$:从 $x = 3$ 处由静止释放。此时 $E = U(3) = \tfrac{81}{4} - 18 + 3 = 5.25~\mathrm{J}$。由 $E = \tfrac{1}{2} m v^2 + U(0)$ 求 $x = 0$ 处的速度：