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Work, Energy and Power · Solutions功、能量与功率 · 详解

Companion to the Practice Set · Mark-by-mark walkthroughs · AP-Feeder / ON / BC / AB styles练习题配套答案 · 逐分讲解 · AP 衔接 / 安 / 卑 / 阿省考风格

EASY易 MEDIUM中 HARD难 🇺🇸 US美 🇨🇦 ON安 🇨🇦 BC卑 🇨🇦 AB阿 AP-style MCQAP 风格选择题 AP-feeder FRQAP 衔接简答题 ON Provincial-style安大略省考风格 BC Provincial-style卑诗省考风格 AB Diploma-style阿尔伯塔毕业考风格 Honors荣誉级
PART I  ·  SHORT RESPONSE · SOLUTIONS第一部分  ·  短答题 · 详解AP MCQ + ON/BC short answer · 18 marksAP 选择题 + 安/卑省考短答 · 共 18 分

Section A · Worked SolutionsA 部分 · 详细解答

Q1EASY易 🇺🇸 US美 AP-style MCQAP 风格选择题 §1 Work done by a force力所做的功 · HS-PS3-1 [3 marks][3 分]

A crate is pushed $4.0$ m by a constant horizontal force of $50$ N. Work done by the force?一个箱子被 $50$ N 的恒定水平力推动 $4.0$ m。该力做了多少功？

Answer:答案：  (B)  $200\ \text{J}$

(a) Apply the work formula with force parallel to displacement套用功的公式（力平行于位移） M1·A1·A1

The force is horizontal and the displacement is horizontal, so $\theta = 0°$ and $\cos 0° = 1$:力水平、位移水平，故 $\theta = 0°$，$\cos 0° = 1$： $$ W \;=\; Fd\cos\theta \;=\; 50 \times 4.0 \times 1 \;=\; 200 \;\text{J}. $$ Option (B).选 (B)。
Why the distractors fail.干扰项分析。
(A) $12.5\ \text{J}$: divides $50/4.0$ instead of multiplying.把 $50$ 除以 $4.0$，而非相乘。
(C) $0\ \text{J}$: would be correct only if the force were perpendicular to the motion ($\theta = 90°$); here it is parallel.只有当力垂直于运动（$\theta = 90°$）时才成立；此处力是平行的。
(D) $54\ \text{J}$: adds force and distance ($50 + 4.0$) instead of multiplying.把力与距离相加（$50 + 4.0$），而非相乘。
Work is the product of force and the displacement along the force, $W = Fd\cos\theta$.功是力与沿力方向位移之积，$W = Fd\cos\theta$。 Only the component of force parallel to the displacement does work. When force and displacement point the same way, $\theta = 0°$ and $W = Fd$ at its maximum. When they are perpendicular ($\theta = 90°$), $\cos 90° = 0$ and no work is done, no matter how large the force; this is why the normal force on a sliding box and the tension on an orbiting satellite both do zero work. The unit is the joule, $1\ \text{J} = 1\ \text{N}\cdot\text{m}$. Always identify $\theta$ as the angle between the force and the displacement, not the angle with any surface.只有力中平行于位移的分量才做功。当力与位移同向时，$\theta = 0°$，$W = Fd$ 取最大值。当两者垂直（$\theta = 90°$）时，$\cos 90° = 0$，无论力多大都不做功；这正是为何滑动箱子受到的法向力、绕轨卫星受到的张力都做零功。单位是焦耳，$1\ \text{J} = 1\ \text{N}\cdot\text{m}$。务必把 $\theta$ 理解为力与位移之间的夹角，而非与任何表面的夹角。
Q2EASY易 🇺🇸 US美 AP-style MCQAP 风格选择题 §2 Kinetic energy动能 · HS-PS3-2 [3 marks][3 分]

A $2.0$ kg ball moves at $6.0$ m/s. Kinetic energy?一个 $2.0$ kg 的球以 $6.0$ m/s 运动。动能为多少？

Answer:答案：  (B)  $36\ \text{J}$

(a) Apply the kinetic-energy formula套用动能公式 M1·A1·A1

$$ KE \;=\; \tfrac{1}{2}mv^2 \;=\; \tfrac{1}{2}(2.0)(6.0)^2 \;=\; \tfrac{1}{2}(2.0)(36) \;=\; 36 \;\text{J.} $$ Option (B).选 (B)。
Why the distractors fail.干扰项分析。
(A) $12\ \text{J}$: forgets to square the speed: $\tfrac{1}{2}(2.0)(6.0) \times 2 = 12$ (uses $v$, not $v^2$).忘记对速率取平方，用 $v$ 代替 $v^2$。
(C) $72\ \text{J}$: drops the factor of $\tfrac{1}{2}$: $mv^2 = 2.0 \times 36 = 72$.漏掉 $\tfrac{1}{2}$ 系数：$mv^2 = 2.0 \times 36 = 72$。
(D) $6.0\ \text{J}$: uses $\tfrac{1}{2}mv$ without squaring and without the mass factor handled correctly.用 $\tfrac{1}{2}mv$ 而未平方，质量因子处理也有误。
Kinetic energy scales with the square of speed: doubling $v$ quadruples $KE$.动能与速率的平方成正比：速率加倍则动能变为四倍。 The single most common $KE$ error is forgetting to square $v$ or dropping the $\tfrac{1}{2}$. The $v^2$ dependence has large real-world consequences: a car at $60$ km/h carries four times the kinetic energy of the same car at $30$ km/h, which is why stopping distance grows so steeply with speed (see Q6). Kinetic energy is a scalar and is always $\ge 0$; direction of motion never enters because $v^2$ erases the sign. Units check: $\mathrm{kg}\cdot(\mathrm{m/s})^2 = \mathrm{kg\cdot m^2/s^2} = \mathrm{J}$.动能最常见的错误就是忘记对 $v$ 取平方或漏掉 $\tfrac{1}{2}$。$v^2$ 的依赖关系有重大现实意义：以 $60$ km/h 行驶的汽车，其动能是同一辆车以 $30$ km/h 行驶时的四倍，这正是刹车距离随速度急剧增长的原因（见 Q6）。动能是标量，恒 $\ge 0$；由于 $v^2$ 抹去符号，运动方向从不进入计算。单位核验：$\mathrm{kg}\cdot(\mathrm{m/s})^2 = \mathrm{kg\cdot m^2/s^2} = \mathrm{J}$。
Q3EASY易 🇨🇦 ON安 ON Provincial-style安大略省考风格 §3 Potential energy势能 · SPH3U D3.2 [4 marks][4 分]

(a) GPE gain lifting a $4.0$ kg book $2.0$ m. (b) EPE of a $k = 300$ N/m spring compressed $0.20$ m. (c) Why a reference level is needed.(a) 抬起 $4.0$ kg 的书升高 $2.0$ m 的重力势能增量。(b) $k = 300$ N/m 弹簧压缩 $0.20$ m 的弹性势能。(c) 为何需要参考面。

Answer:答案：  (a) $\Delta GPE = 78.4\ \text{J}$  ·  (b) $EPE = 6.0\ \text{J}$  ·  (c) only changes in GPE are physical; $h$ is measured from a chosen zero只有 GPE 的变化量有物理意义；$h$ 从选定的零点起算

(a) Gravitational PE gain using $\Delta GPE = mg\Delta h$用 $\Delta GPE = mg\Delta h$ 求重力势能增量 M1·A1

$$ \Delta GPE \;=\; mg\Delta h \;=\; 4.0 \times 9.8 \times 2.0 \;=\; 78.4 \;\text{J.} $$

(b) Elastic PE using $EPE = \tfrac{1}{2}kx^2$用 $EPE = \tfrac{1}{2}kx^2$ 求弹性势能 A1

$$ EPE \;=\; \tfrac{1}{2}kx^2 \;=\; \tfrac{1}{2}(300)(0.20)^2 \;=\; \tfrac{1}{2}(300)(0.040) \;=\; 6.0 \;\text{J.} $$

(c) Why a reference level is required为何需要参考面 A1

Gravitational PE depends on height $h$, but $h$ is measured from a chosen zero level. Only changes in $GPE$ are physically meaningful, so the value of $GPE$ at a point is undefined until a reference level (the floor, a tabletop, etc.) is fixed.重力势能取决于高度 $h$，但 $h$ 是从选定的零点起算的。只有 $GPE$ 的变化量才有物理意义，因此在确定参考面（地面、桌面等）之前，某点的 $GPE$ 数值是未定义的。
$GPE = mgh$ uses the squared-free vertical height; $EPE = \tfrac{1}{2}kx^2$ uses the squared deformation.$GPE = mgh$ 用竖直高度（不平方）；$EPE = \tfrac{1}{2}kx^2$ 用形变量的平方。 Two stored-energy forms appear in this unit. Gravitational PE rises linearly with height, so it has no preferred zero: choose the most convenient reference and stick to it. Elastic PE rises with the square of the compression or extension, so doubling $x$ quadruples the stored energy. A common slip in (b) is to forget to square $x$, or to use $x = 0.20$ m without converting consistently. Note that compression and extension store energy identically because $x^2$ is the same for $\pm x$. The two forms reappear together in Q9, where a spring's $EPE$ converts to $KE$ and then to $GPE$.本单元出现两种储能形式。重力势能随高度线性增长，因此没有偏好的零点：选最方便的参考面并坚持使用。弹性势能随压缩量或拉伸量的平方增长，故 $x$ 加倍则储能变为四倍。(b) 中常见失误是忘记对 $x$ 平方，或未统一单位地使用 $x = 0.20$ m。注意压缩与拉伸储存的能量相同，因为 $\pm x$ 的 $x^2$ 一致。两种形式在 Q9 中再次同时出现：弹簧的 $EPE$ 转化为 $KE$，再转化为 $GPE$。
Q4MEDIUM中 🇺🇸 US美 AP-style MCQAP 风格选择题 §5 Power功率 · HS-PS3-3 [3 marks][3 分]

A motor does $6000$ J of useful work in $30$ s. Average useful power?电机在 $30$ s 内做了 $6000$ J 有用功。平均有用功率为多少？

Answer:答案：  (B)  $200\ \text{W}$

(a) Power is work divided by time, $P = W/t$功率是功除以时间，$P = W/t$ M1·A1·A1

$$ P \;=\; \frac{W}{t} \;=\; \frac{6000}{30} \;=\; 200 \;\text{W.} $$ Option (B).选 (B)。
Why the distractors fail.干扰项分析。
(A) $180{,}000\ \text{W}$: multiplies $W \times t$ instead of dividing.把 $W$ 与 $t$ 相乘，而非相除。
(C) $5.0\ \text{W}$: inverts the ratio, $t / W = 30/6000$ scaled, giving the reciprocal-type error.将比值颠倒，算成 $t / W$ 一类的倒数错误。
(D) $20\ \text{W}$: divides by $300$ instead of $30$ (a decimal-place slip).除以 $300$ 而非 $30$（小数点错位）。
Power is the rate of doing work: the same job done faster needs more power.功率是做功的速率：同样的功做得越快，所需功率越大。 $P = W/t$ measures how quickly energy is transferred, in watts ($1\ \text{W} = 1\ \text{J/s}$). Two motors can do the same $6000$ J of work, but the one that finishes in half the time delivers twice the power. The most frequent exam slip is a time-unit error: if the time had been given as "$0.50$ minutes," it must first be converted to $30$ s before dividing. An equivalent form, $P = Fv$, is handy when a constant force moves at constant speed (used in Q10). Power and energy are different quantities: energy is "how much," power is "how fast."$P = W/t$ 衡量能量转移的快慢，单位为瓦特（$1\ \text{W} = 1\ \text{J/s}$）。两台电机可以做同样的 $6000$ J 功，但用一半时间完成的那台输出两倍的功率。最常见的考试失误是时间单位错误：若时间以"$0.50$ 分钟"给出，必须先换算为 $30$ s 再相除。等价形式 $P = Fv$ 在恒力以恒速运动时很方便（Q10 用到）。功率与能量是不同的量：能量是"多少"，功率是"多快"。
Q5MEDIUM中 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §4 Conservation of energy机械能守恒 · Physics 11 [5 marks][5 分]

A $0.30$ kg ball dropped from rest at $1.8$ m. $g = 9.8\ \text{m/s}^2$. (a) Speed at the ground. (b) KE at impact. (c) Condition for $E_i = E_f$.$0.30$ kg 的球从 $1.8$ m 高处由静止落下，$g = 9.8\ \text{m/s}^2$。(a) 落地速率。(b) 落地动能。(c) $E_i = E_f$ 的条件。

Answer:答案：  (a) $v \approx 5.9\ \text{m/s}$  ·  (b) $KE \approx 5.3\ \text{J}$  ·  (c) only gravity acts (no friction/air resistance)仅有重力作用（无摩擦/空气阻力）

(a) Speed at the ground from $mgh = \tfrac{1}{2}mv^2$由 $mgh = \tfrac{1}{2}mv^2$ 求落地速率 M1·A1

Take the ground as the reference. Initial $KE = 0$ (at rest); at the ground all $GPE$ has become $KE$. The mass cancels:以地面为参考。初动能 $= 0$（静止）；落地时全部 $GPE$ 已转化为 $KE$。质量消去： $$ mgh \;=\; \tfrac{1}{2}mv^2 \;\Longrightarrow\; v \;=\; \sqrt{2gh} \;=\; \sqrt{2 \times 9.8 \times 1.8} \;=\; \sqrt{35.28} \;\approx\; 5.9 \;\text{m/s.} $$

(b) Kinetic energy at impact落地瞬间的动能 M1·A1

By conservation, the impact $KE$ equals the initial $GPE$:由守恒，落地动能等于初始重力势能： $$ KE \;=\; mgh \;=\; 0.30 \times 9.8 \times 1.8 \;=\; 5.29 \;\approx\; 5.3 \;\text{J.} $$ Check: $\tfrac{1}{2}(0.30)(5.94)^2 = 5.29$ J. ✓核验：$\tfrac{1}{2}(0.30)(5.94)^2 = 5.29$ J。✓

(c) State the condition写出条件 A1

Setting initial mechanical energy equal to final requires an isolated system: only conservative forces (here, gravity) do work. With no friction or air resistance, mechanical energy is conserved.将初机械能等于末机械能，要求系统为孤立系统：只有保守力（此处为重力）做功。在无摩擦、无空气阻力时，机械能守恒。
Conservation of energy turns a two-step kinematics problem into one line.能量守恒把两步运动学问题压缩为一行。 A dropped object is the cleanest energy-conservation case: all the gravitational PE at the top converts to kinetic energy at the bottom. Setting $mgh = \tfrac{1}{2}mv^2$ and cancelling $m$ gives $v = \sqrt{2gh}$ directly, with no need for time or the SUVAT chain. Notice the mass cancels, so a heavy ball and a light ball dropped from the same height reach the same speed (ignoring air resistance), a result that surprises many students. The condition in (c) is the whole point: the moment friction or drag enters, mechanical energy is no longer conserved and the extended equation with $W_\text{nc}$ (Q8, Q11) is required.下落物体是最干净的能量守恒情形：顶端的全部重力势能在底端转化为动能。令 $mgh = \tfrac{1}{2}mv^2$ 并消去 $m$，直接得到 $v = \sqrt{2gh}$，无需时间或 SUVAT 链。注意质量消去，因此从同一高度落下的重球与轻球到达相同速率（忽略空气阻力），这一结果常令学生惊讶。(c) 的条件正是关键：一旦摩擦或阻力介入，机械能不再守恒，就需要含 $W_\text{nc}$ 的扩展方程（Q8、Q11）。
PART II  ·  EXTENDED RESPONSE · SOLUTIONS第二部分  ·  简答题 · 详解AP-feeder FRQ + honors · 35 marksAP 衔接简答题 + 荣誉级 · 共 35 分

Section B · Worked SolutionsB 部分 · 详细解答

Q6MEDIUM中 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §2 Work-energy theorem功能定理 · HS-PS3-2 [8 marks][8 分]

A $1000$ kg car at $20$ m/s brakes to a stop in $50$ m on level road. (a) Initial KE. (b) Net work while braking. (c) Braking force. (d) Effect of doubling speed on braking distance.$1000$ kg 的汽车以 $20$ m/s 在水平路面上 $50$ m 内刹停。(a) 初始动能。(b) 刹车合功。(c) 刹车力。(d) 速度加倍对刹车距离的影响。

Answer:答案：  (a) $KE_i = 200{,}000\ \text{J}$  ·  (b) $W_\text{net} = -200{,}000\ \text{J}$  ·  (c) $F = 4000\ \text{N}$  ·  (d) distance becomes $4\times$ ($200$ m)距离变为 $4$ 倍（$200$ m）

(a) Initial kinetic energy初始动能 M1·A1

$$ KE_i \;=\; \tfrac{1}{2}mv^2 \;=\; \tfrac{1}{2}(1000)(20)^2 \;=\; \tfrac{1}{2}(1000)(400) \;=\; 200{,}000 \;\text{J} \;=\; 200 \;\text{kJ.} $$

(b) Net work from the work-energy theorem由功能定理求合功 M1·A1

The car stops, so $KE_f = 0$:车停下，故 $KE_f = 0$： $$ W_\text{net} \;=\; \Delta KE \;=\; KE_f - KE_i \;=\; 0 - 200{,}000 \;=\; -200{,}000 \;\text{J.} $$

(c) Magnitude of the braking force刹车力的大小 M1·A1

The braking force is constant and opposite to the motion, so $|W| = F d$:刹车力恒定且与运动方向相反，故 $|W| = F d$： $$ F \;=\; \frac{|W_\text{net}|}{d} \;=\; \frac{200{,}000}{50} \;=\; 4000 \;\text{N.} $$

(d) Doubling the speed速度加倍 M1·A1

$KE \propto v^2$, so doubling $v$ quadruples the kinetic energy. With the same braking force $F$, the work needed to stop is $F d$, so $d \propto KE \propto v^2$. The braking distance becomes $4 \times 50 = 200$ m.$KE \propto v^2$，故 $v$ 加倍则动能变为四倍。在相同刹车力 $F$ 下，停车所需的功为 $F d$，故 $d \propto KE \propto v^2$。刹车距离变为 $4 \times 50 = 200$ m。
The work-energy theorem replaces a kinematics chain: net work equals the change in kinetic energy.功能定理替代了运动学链：合功等于动能变化。 $W_\text{net} = \Delta KE$ lets you find a force or distance without ever computing acceleration or time. Here, the braking force follows directly from the energy removed divided by the distance. The $v^2$ dependence in part (d) is the safety headline of this whole unit: braking distance scales with the square of speed, so a car going twice as fast needs four times the distance to stop. This is exactly why highway speed limits and following-distance rules grow so steeply with speed, and why the kinetic-energy formula (not momentum, which scales linearly with $v$) is the right tool for stopping-distance reasoning.$W_\text{net} = \Delta KE$ 让你无需计算加速度或时间即可求出力或距离。此处刹车力直接由移除的能量除以距离得到。(d) 中的 $v^2$ 依赖关系是整个单元的安全要点：刹车距离与速度的平方成正比，故速度加倍的汽车需要四倍的距离才能停下。这正是高速公路限速与车距规则随速度急剧增长的原因，也是为何动能公式（而非随 $v$ 线性变化的动量）才是分析刹车距离的正确工具。
Q7MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §4 Conservation of mechanical energy机械能守恒 · SPH3U D3.1 [8 marks][8 分]

$50$ kg child, frictionless slide, top $4.0$ m above bottom, from rest. $g = 9.8\ \text{m/s}^2$, bottom is reference. (a) GPE at top. (b) Speed at bottom. (c) Speed $1.5$ m above bottom. (d) Why mass does not matter.$50$ kg 的孩子，无摩擦滑梯，顶端高于底端 $4.0$ m，从静止出发。$g = 9.8\ \text{m/s}^2$，以底端为参考。(a) 顶端 GPE。(b) 底端速率。(c) 距底端 $1.5$ m 处速率。(d) 为何质量无关。

Answer:答案：  (a) $GPE = 1960\ \text{J}$  ·  (b) $v \approx 8.9\ \text{m/s}$  ·  (c) $v = 7.0\ \text{m/s}$  ·  (d) mass cancels in $mgh = \tfrac{1}{2}mv^2$$mgh = \tfrac{1}{2}mv^2$ 中质量消去

(a) Gravitational PE at the top顶端重力势能 M1·A1

$$ GPE \;=\; mgh \;=\; 50 \times 9.8 \times 4.0 \;=\; 1960 \;\text{J.} $$

(b) Speed at the bottom (conservation)底端速率（守恒） M1·A1·A1

Frictionless slide $\Rightarrow$ all $GPE$ becomes $KE$: $mgh = \tfrac{1}{2}mv^2$. Mass cancels:无摩擦滑梯 $\Rightarrow$ 全部 $GPE$ 转化为 $KE$：$mgh = \tfrac{1}{2}mv^2$。质量消去： $$ v \;=\; \sqrt{2gh} \;=\; \sqrt{2 \times 9.8 \times 4.0} \;=\; \sqrt{78.4} \;\approx\; 8.9 \;\text{m/s.} $$

(c) Speed $1.5$ m above the bottom距底端 $1.5$ m 处的速率 M1·A1

At that point the child has fallen $4.0 - 1.5 = 2.5$ m, so only that drop has converted to $KE$:在该处孩子已下降 $4.0 - 1.5 = 2.5$ m，故仅这段下落转化为 $KE$： $$ v \;=\; \sqrt{2g(\Delta h)} \;=\; \sqrt{2 \times 9.8 \times 2.5} \;=\; \sqrt{49} \;=\; 7.0 \;\text{m/s.} $$

(d) Why the speed is mass-independent为何速率与质量无关 A1

In $mgh = \tfrac{1}{2}mv^2$ the mass $m$ appears on both sides and cancels, leaving $v = \sqrt{2gh}$. The final speed depends only on $g$ and the height dropped.在 $mgh = \tfrac{1}{2}mv^2$ 中，质量 $m$ 在两边都出现并消去，剩下 $v = \sqrt{2gh}$。末速率只取决于 $g$ 与下降高度。
Energy conservation lets you find the speed at any height, using only the vertical drop.能量守恒让你只用竖直下降量就能求出任意高度处的速率。 The power of the conservation law is that the shape of the slide is irrelevant: only the vertical height change matters, because gravity is conservative. Part (c) shows this directly: the speed at any point depends on how far the child has descended, not on the path length along the slide. This is also why mass cancels in part (d), a result Galileo famously demonstrated. A frequent error in (c) is to use the full $4.0$ m instead of the $2.5$ m drop to that point. The same "height drop only" principle drives the roller-coaster analysis in Q12.守恒定律的威力在于滑梯形状无关紧要：因为重力是保守力，只有竖直高度变化才重要。(c) 直接体现了这一点：任意点的速率取决于孩子下降了多少，而非沿滑梯的路径长度。这也是 (d) 中质量消去的原因，正是伽利略著名演示的结果。(c) 中常见错误是用整段 $4.0$ m 而非到该点的 $2.5$ m 下降量。同样的"只看高度下降"原理驱动了 Q12 的过山车分析。
Q8HARD难 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §6 Non-conservative forces非保守力 · Physics 11 [9 marks][9 分]

$2.0$ kg block from rest, ramp top $3.0$ m high, friction $5.0$ N over the $5.0$ m path, $g = 9.8\ \text{m/s}^2$, bottom is reference. (a) GPE at top. (b) Work by friction. (c) KE at bottom. (d) Speed at bottom. (e) Thermal energy.$2.0$ kg 滑块从静止，坡顶高 $3.0$ m，沿 $5.0$ m 路径摩擦力 $5.0$ N，$g = 9.8\ \text{m/s}^2$，以坡底为参考。(a) 顶端 GPE。(b) 摩擦做功。(c) 底端 KE。(d) 底端速率。(e) 热能。

Answer:答案：  (a) $GPE = 58.8\ \text{J}$  ·  (b) $W_\text{fric} = -25\ \text{J}$  ·  (c) $KE_f = 33.8\ \text{J}$  ·  (d) $v \approx 5.8\ \text{m/s}$  ·  (e) $25\ \text{J}$ heat热能

(a) Gravitational PE at the top顶端重力势能 M1·A1

$$ GPE \;=\; mgh \;=\; 2.0 \times 9.8 \times 3.0 \;=\; 58.8 \;\text{J.} $$

(b) Work done by friction摩擦力所做的功 M1·A1

Friction opposes motion ($\theta = 180°$), so its work is negative:摩擦力与运动方向相反（$\theta = 180°$），故其功为负： $$ W_\text{fric} \;=\; -f_k d \;=\; -(5.0)(5.0) \;=\; -25 \;\text{J.} $$

(c) Kinetic energy at the bottom (extended equation)底端动能（扩展方程） M1·A1

With $KE_i = 0$ and $PE_f = 0$:已知 $KE_i = 0$，$PE_f = 0$： $$ KE_f \;=\; KE_i + PE_i + W_\text{nc} - PE_f \;=\; 0 + 58.8 + (-25) - 0 \;=\; 33.8 \;\text{J.} $$

(d) Speed at the bottom底端速率 M1·A1

$$ \tfrac{1}{2}mv^2 \;=\; 33.8 \;\Longrightarrow\; v \;=\; \sqrt{\frac{2 \times 33.8}{2.0}} \;=\; \sqrt{33.8} \;\approx\; 5.8 \;\text{m/s.} $$

(e) Thermal energy generated产生的热能 A1

$25$ J of mechanical energy was removed by friction and converted to thermal energy (heat) in the block and ramp surfaces. It equals $|W_\text{fric}|$.$25$ J 的机械能被摩擦力移除，转化为滑块与斜面表面的热能。它等于 $|W_\text{fric}|$。
When friction acts, mechanical energy is not conserved, but total energy still is: the "missing" energy becomes heat.摩擦作用时机械能不守恒，但总能量仍守恒："消失"的能量变成热。 The extended equation $KE_f + PE_f = KE_i + PE_i + W_\text{nc}$ adds the work of non-conservative forces to the conservation law. Friction always does negative work ($W_\text{nc} < 0$), so it drains mechanical energy. Run the full energy audit: $58.8$ J of PE splits into $33.8$ J of KE plus $25$ J of heat, and $33.8 + 25 = 58.8$ ✓. Note that friction depends on the path length ($5.0$ m), not the vertical drop ($3.0$ m), which is exactly why it cannot be packaged into a potential energy. Compare with the frictionless slide in Q7, where the block would have reached $\sqrt{2(9.8)(3.0)} = 7.7$ m/s instead of $5.8$ m/s.扩展方程 $KE_f + PE_f = KE_i + PE_i + W_\text{nc}$ 在守恒定律上加入非保守力的功。摩擦力恒做负功（$W_\text{nc} < 0$），故它消耗机械能。做完整能量审计：$58.8$ J 势能分为 $33.8$ J 动能加 $25$ J 热能，且 $33.8 + 25 = 58.8$ ✓。注意摩擦力取决于路径长度（$5.0$ m），而非竖直下降（$3.0$ m），这正是它无法被打包成势能的原因。与 Q7 的无摩擦滑梯对比：那里滑块本可达到 $\sqrt{2(9.8)(3.0)} = 7.7$ m/s，而非 $5.8$ m/s。
Q9HARD难Honors荣誉级 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §3 + §4 Spring energy & conservation弹性能与守恒 · HS-PS3-1 [10 marks][10 分]

$k = 800$ N/m spring compressed $0.30$ m launches a $1.5$ kg block on a frictionless flat surface into a frictionless incline. $g = 9.8\ \text{m/s}^2$. (a) EPE stored. (b) Speed leaving spring. (c) Max height on incline. (d) Speed at base if a rough patch does $-9.0$ J. (e) Why it returns at the same speed.$k = 800$ N/m 弹簧压缩 $0.30$ m，在无摩擦水平面上发射 $1.5$ kg 滑块，进入无摩擦斜面。$g = 9.8\ \text{m/s}^2$。(a) 储存的 EPE。(b) 离开弹簧的速率。(c) 斜面最大高度。(d) 若粗糙区做 $-9.0$ J 功的底部速率。(e) 为何以相同速率返回。

Answer:答案：  (a) $EPE = 36\ \text{J}$  ·  (b) $v \approx 6.9\ \text{m/s}$  ·  (c) $h \approx 2.45\ \text{m}$  ·  (d) $v = 6.0\ \text{m/s}$  ·  (e) no friction means no energy is lost无摩擦意味着无能量损失

(a) Elastic PE stored in the spring弹簧储存的弹性势能 M1·A1

$$ EPE \;=\; \tfrac{1}{2}kx^2 \;=\; \tfrac{1}{2}(800)(0.30)^2 \;=\; \tfrac{1}{2}(800)(0.090) \;=\; 36 \;\text{J.} $$

(b) Speed leaving the spring (EPE $\to$ KE)离开弹簧的速率（EPE $\to$ KE） M1·A1

On the flat frictionless surface, all $EPE$ becomes $KE$:在无摩擦水平面上，全部 $EPE$ 转化为 $KE$： $$ 36 \;=\; \tfrac{1}{2}(1.5)v^2 \;\Longrightarrow\; v^2 \;=\; 48 \;\Longrightarrow\; v \;=\; \sqrt{48} \;\approx\; 6.9 \;\text{m/s.} $$

(c) Maximum height on the frictionless incline (KE $\to$ GPE)无摩擦斜面上的最大高度（KE $\to$ GPE） M1·A1

At maximum height the block is momentarily at rest, so all $KE$ has become $GPE$:在最大高度处滑块瞬时静止，故全部 $KE$ 已转化为 $GPE$： $$ 36 \;=\; mgh \;\Longrightarrow\; h \;=\; \frac{36}{1.5 \times 9.8} \;=\; \frac{36}{14.7} \;\approx\; 2.45 \;\text{m.} $$

(d) Speed at the base after a rough patch removes $9.0$ J粗糙区移除 $9.0$ J 后底部的速率 M1·A1

$KE$ at the base $= EPE + W_\text{nc} = 36 + (-9.0) = 27$ J:底部 $KE = EPE + W_\text{nc} = 36 + (-9.0) = 27$ J： $$ 27 \;=\; \tfrac{1}{2}(1.5)v^2 \;\Longrightarrow\; v^2 \;=\; 36 \;\Longrightarrow\; v \;=\; 6.0 \;\text{m/s.} $$

(e) Why the block returns at the same speed (frictionless case)为何滑块以相同速率返回（无摩擦情形） M1·A1

With no non-conservative force anywhere on the path, mechanical energy is conserved at every point. The block rises, converting all $KE$ to $GPE$, then descends, converting all $GPE$ back to $KE$. At the original launch point the $GPE$ is the same as before ($= 0$), so $KE$ and hence speed must also be the same. No energy was lost to heat.由于路径上任何位置都没有非保守力，机械能在每一点都守恒。滑块上升时把全部 $KE$ 转化为 $GPE$，下降时再把全部 $GPE$ 转回 $KE$。在原始发射点 $GPE$ 与之前相同（$= 0$），故 $KE$ 乃至速率也必相同。没有能量转化为热。
A frictionless track is a perfect energy relay: spring PE $\to$ kinetic $\to$ gravitational PE $\to$ kinetic, with the total fixed.无摩擦轨道是完美的能量接力：弹性势能 $\to$ 动能 $\to$ 重力势能 $\to$ 动能，总量不变。 This problem chains all three energy forms of the unit. The $36$ J stored in the spring is the conserved total throughout the frictionless motion; it merely changes wardrobe (elastic, then kinetic, then gravitational, then back). Part (d) shows what a non-conservative force does: it permanently removes $9.0$ J as heat, lowering every subsequent energy total. Part (e) is the conceptual payoff: reversibility is the signature of conservative forces. The moment any friction enters (part d), the motion is no longer reversible and the block returns slower. Always track the running total of mechanical energy; if it changes, a non-conservative force did work.本题串起本单元的三种能量形式。弹簧储存的 $36$ J 是无摩擦运动全程的守恒总量；它只是不断"换装"（弹性、动能、重力势能，再返回）。(d) 展示了非保守力的作用：它把 $9.0$ J 永久地以热的形式移除，降低了此后每一处的能量总量。(e) 是概念上的收获：可逆性是保守力的标志。一旦有任何摩擦介入（(d) 部分），运动就不再可逆，滑块返回时更慢。务必追踪机械能的累计总量；若它改变，说明有非保守力做了功。
PART III  ·  MODELING / APPLIED · SOLUTIONS第三部分  ·  建模与应用 · 详解AB Diploma + Universal · 28 marks阿省毕业考 + 通用题型 · 共 28 分

Section C · Modeling and ApplicationsC 部分 · 建模与应用

Q10MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §5 Power & efficiency (applied)功率与效率（应用） · 20-C2.4k [9 marks][9 分]

Elevator motor lifts a $1200$ kg car at a constant $2.0$ m/s. $g = 9.8\ \text{m/s}^2$. (a) Cable force. (b) Useful power. (c) Efficiency if input is $30$ kW. (d) Rate of energy loss and its form.电梯电机以恒定 $2.0$ m/s 提升 $1200$ kg 轿厢。$g = 9.8\ \text{m/s}^2$。(a) 缆绳力。(b) 有用功率。(c) 输入 $30$ kW 时的效率。(d) 能量损失速率及形式。

Answer:答案：  (a) $F = 11{,}760\ \text{N}$  ·  (b) $P \approx 23{,}500\ \text{W}$  ·  (c) $\eta \approx 78\%$  ·  (d) $6480\ \text{W}$ as heat以热的形式

(a) Upward cable force at constant speed匀速时缆绳的向上力 M1·A1

At constant speed the net force is zero (Newton's first law), so the cable force equals the weight:匀速时合力为零（牛顿第一定律），故缆绳力等于重力： $$ F \;=\; mg \;=\; 1200 \times 9.8 \;=\; 11{,}760 \;\text{N.} $$

(b) Useful power output using $P = Fv$用 $P = Fv$ 求有用功率输出 M1·A1·A1

$$ P \;=\; Fv \;=\; 11{,}760 \times 2.0 \;=\; 23{,}520 \;\text{W} \;\approx\; 23.5 \;\text{kW.} $$

(c) Efficiency效率 M1·A1

$$ \eta \;=\; \frac{P_\text{useful}}{P_\text{input}} \;=\; \frac{23{,}520}{30{,}000} \;=\; 0.784 \;\approx\; 78\%. $$

(d) Rate of energy loss能量损失速率 M1·A1

The "lost" power is the difference between input and useful output:"损失"的功率是输入与有用输出之差： $$ P_\text{lost} \;=\; 30{,}000 - 23{,}520 \;=\; 6480 \;\text{W.} $$ This energy is dissipated as heat (in the motor windings, gears, and friction).这部分能量以热的形式耗散（在电机绕组、齿轮与摩擦中）。
$P = Fv$ shortcut: for a constant force moving at constant speed, power is force times speed.$P = Fv$ 捷径：恒力以恒速运动时，功率等于力乘以速度。 Lifting at constant speed means the upward force exactly balances gravity, so no kinetic energy changes; all the useful power goes into raising gravitational PE at the rate $Fv$. Efficiency compares useful output to total input; the shortfall always appears as heat, never as destroyed energy (first law of thermodynamics). The $6480$ W lost here is why elevator motors need cooling. A frequent slip is to compute power from $mgh/t$ over a full lift; $P = Fv$ gives the instantaneous (and, here, constant) power directly without needing a height or time. Efficiency is the headline design constraint NGSS HS-PS3-3 asks students to optimise.匀速提升意味着向上的力恰好平衡重力，故动能不变；全部有用功率以 $Fv$ 的速率用于增加重力势能。效率比较有用输出与总输入；差额总以热的形式出现，绝不会有能量被消灭（热力学第一定律）。此处损失的 $6480$ W 正是电梯电机需要散热的原因。常见失误是用整段提升的 $mgh/t$ 算功率；$P = Fv$ 无需高度或时间即可直接给出瞬时（此处为恒定）功率。效率是 NGSS HS-PS3-3 要求学生优化的核心设计约束。
Q11MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §6 Energy with friction (applied)含摩擦的能量（应用） · 20-C2.6k [9 marks][9 分]

$70$ kg skier from rest, slope $25$ m high, reaches bottom at $18$ m/s. $g = 9.8\ \text{m/s}^2$, bottom is reference. (a) GPE at top. (b) KE at bottom. (c) Energy lost to friction/drag. (d) Frictionless bottom speed and why it is larger.$70$ kg 滑雪者从静止，斜坡高 $25$ m，底端速率 $18$ m/s。$g = 9.8\ \text{m/s}^2$，以底端为参考。(a) 顶端 GPE。(b) 底端 KE。(c) 摩擦/阻力损失的能量。(d) 无摩擦底端速率及其为何更大。

Answer:答案：  (a) $GPE = 17{,}150\ \text{J}$  ·  (b) $KE = 11{,}340\ \text{J}$  ·  (c) $5810\ \text{J}$  ·  (d) $v \approx 22.1\ \text{m/s}$

(a) Gravitational PE at the top顶端重力势能 M1·A1

$$ GPE \;=\; mgh \;=\; 70 \times 9.8 \times 25 \;=\; 17{,}150 \;\text{J.} $$

(b) Kinetic energy at the bottom底端动能 M1·A1

$$ KE \;=\; \tfrac{1}{2}mv^2 \;=\; \tfrac{1}{2}(70)(18)^2 \;=\; \tfrac{1}{2}(70)(324) \;=\; 11{,}340 \;\text{J.} $$

(c) Energy converted to heat转化为热的能量 M1·A1

The shortfall between the starting $GPE$ and the final $KE$ is the energy dissipated by friction and air resistance:初始 $GPE$ 与末态 $KE$ 之差即为摩擦力与空气阻力耗散的能量： $$ E_\text{lost} \;=\; GPE - KE \;=\; 17{,}150 - 11{,}340 \;=\; 5810 \;\text{J.} $$

(d) Frictionless bottom speed无摩擦底端速率 M1·A1·A1

If frictionless, all $GPE$ becomes $KE$: $mgh = \tfrac{1}{2}mv^2$, mass cancels:若无摩擦，全部 $GPE$ 转化为 $KE$：$mgh = \tfrac{1}{2}mv^2$，质量消去： $$ v \;=\; \sqrt{2gh} \;=\; \sqrt{2 \times 9.8 \times 25} \;=\; \sqrt{490} \;\approx\; 22.1 \;\text{m/s.} $$ It is larger because no mechanical energy is drained to heat, so more of the $GPE$ ends up as $KE$.它更大，因为没有机械能被耗散为热，故更多的 $GPE$ 最终转化为 $KE$。
Friction's energy bill is the gap between the ideal (frictionless) and actual outcome.摩擦的能量账单是理想（无摩擦）与实际结果之间的差额。 This is the canonical AB Physics 20 energy-accounting problem. You never need to know the friction force or the slope length: the heat generated is simply the gravitational PE that did not show up as kinetic energy, $E_\text{lost} = \Delta GPE - \Delta KE = 5810$ J. Part (d) isolates the ideal case as a benchmark, showing how much speed friction cost ($22.1$ vs $18$ m/s). Note again that the frictionless final speed is mass-independent ($\sqrt{2gh}$), but the energy lost to friction does depend on the skier's mass and the surface. Energy audits like this are the surest way to handle systems where you cannot easily resolve all the forces.这是阿省物理 20 经典的能量核算题。你完全不需要知道摩擦力或坡长：产生的热量就是没有转化为动能的那部分重力势能，$E_\text{lost} = \Delta GPE - \Delta KE = 5810$ J。(d) 把理想情形作为基准，显示摩擦让速度损失了多少（$22.1$ 对 $18$ m/s）。再次注意无摩擦末速率与质量无关（$\sqrt{2gh}$），但摩擦损失的能量确实取决于滑雪者的质量与表面。这类能量审计是处理无法轻易分解所有受力的系统的最可靠方法。
Q12HARD难 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §7 Energy modeling (roller-coaster)能量建模（过山车） · HS-PS3-1 [10 marks][10 分]

$500$ kg cart from rest at top of $35$ m hill, frictionless, loop top $20$ m above ground. $g = 9.8\ \text{m/s}^2$, ground is reference. (a) Speed at ground. (b) Speed at top of loop. (c) KE at top of loop. (d) Energy audit at the loop top. (e) Why a real coaster starts higher.$500$ kg 小车从 $35$ m 山顶静止出发，无摩擦，环顶距地面 $20$ m。$g = 9.8\ \text{m/s}^2$，以地面为参考。(a) 地面速率。(b) 环顶速率。(c) 环顶 KE。(d) 环顶能量审计。(e) 为何真实过山车要更高。

Answer:答案：  (a) $v \approx 26.2\ \text{m/s}$  ·  (b) $v \approx 17.1\ \text{m/s}$  ·  (c) $KE = 73{,}500\ \text{J}$  ·  (d) balances to $171{,}500$ J平衡为 $171{,}500$ J  ·  (e) friction drains energy摩擦消耗能量

(a) Speed at ground level地面处速率 M1·A1

All $GPE$ at the top of the hill becomes $KE$ at the ground: $mgh = \tfrac{1}{2}mv^2$, mass cancels:山顶的全部 $GPE$ 在地面处变成 $KE$：$mgh = \tfrac{1}{2}mv^2$，质量消去： $$ v \;=\; \sqrt{2gh} \;=\; \sqrt{2 \times 9.8 \times 35} \;=\; \sqrt{686} \;\approx\; 26.2 \;\text{m/s.} $$

(b) Speed at the top of the loop环顶速率 M1·A1·A1

Only the net drop from the hill top to the loop top ($35 - 20 = 15$ m) converts to $KE$:只有从山顶到环顶的净下降（$35 - 20 = 15$ m）转化为 $KE$： $$ v \;=\; \sqrt{2g(h_i - h_f)} \;=\; \sqrt{2 \times 9.8 \times 15} \;=\; \sqrt{294} \;\approx\; 17.1 \;\text{m/s.} $$

(c) Kinetic energy at the top of the loop环顶动能 M1·A1

$$ KE \;=\; \tfrac{1}{2}mv^2 \;=\; \tfrac{1}{2}(500)(294) \;=\; 73{,}500 \;\text{J.} $$

(d) Energy audit at the loop top环顶能量审计 M1·A1

$GPE$ at the loop top $= mgh_f = 500 \times 9.8 \times 20 = 98{,}000$ J. Total there:环顶 $GPE = mgh_f = 500 \times 9.8 \times 20 = 98{,}000$ J。该处总能量： $$ KE + GPE \;=\; 73{,}500 + 98{,}000 \;=\; 171{,}500 \;\text{J.} $$ Initial $GPE$ at the hill top $= 500 \times 9.8 \times 35 = 171{,}500$ J. They match. ✓山顶初始 $GPE = 500 \times 9.8 \times 35 = 171{,}500$ J。两者相等。✓

(e) Why a real coaster starts higher为何真实过山车从更高处出发 A1

A real track has friction and air resistance, which do negative work and drain mechanical energy as heat. To deliver the same $KE$ at the loop top, the starting hill must store extra $GPE$ to cover those losses, so it must be higher than $35$ m.真实轨道有摩擦力与空气阻力，它们做负功，把机械能以热的形式耗散。为了在环顶提供相同的 $KE$，起始山坡必须储存额外的 $GPE$ 以弥补这些损失，因此必须高于 $35$ m。
An energy audit is a bookkeeping check: at every point, $KE + PE$ (plus heat) must equal the starting total.能量审计是一种记账核验：在每一点，$KE + PE$（加上热）必须等于起始总量。 The roller-coaster is the showcase application of mechanical-energy conservation. On a frictionless track the total $KE + GPE$ is the same constant everywhere, so the speed at any point depends only on the height there: lower is faster, higher is slower. Part (d) makes the bookkeeping explicit and is the single best habit for these problems: if your numbers do not balance, you have a sign error or a missing energy term. Part (e) is the bridge to reality: friction means a real designer must "over-build" the first hill, exactly the kind of constraint NGSS HS-PS3-1 asks students to model computationally. Note the speed at the loop top must also exceed $\sqrt{gr}$ to keep the cart on the track, a circular-motion condition explored in later units.过山车是机械能守恒的典范应用。在无摩擦轨道上，总量 $KE + GPE$ 处处为同一常数，故任意点的速率只取决于该处高度：越低越快，越高越慢。(d) 把记账显式化，是处理这类问题的最佳习惯：若数字不平衡，则存在符号错误或遗漏的能量项。(e) 是通向现实的桥梁：摩擦意味着真实设计者必须"超额建造"第一座山坡，这正是 NGSS HS-PS3-1 要求学生用计算建模的约束类型。注意环顶速率还必须超过 $\sqrt{gr}$ 才能使小车保持在轨道上，这是后续单元探讨的圆周运动条件。