PART I · SHORT RESPONSE第一部分 · 短答题AP-style MCQ + ON/BC short answer · 18 marksAP 风格选择题 + 安/卑省考短答 · 共 18 分
Section A · Short ResponseA 部分 · 短答题
Mix of multiple-choice and short-answer items. For MCQs, circle the letter; show enough work in the margin that a marker could verify. For short-answer items, state units in every answer. Use $g = 9.8\ \text{m/s}^2$ throughout. No calculator on Q1-Q2; calculator permitted on Q3-Q5.本节包含选择题与短答题。选择题请圈出字母答案，并在空白处写下足以让阅卷人核对的过程。短答题每道都要写出单位。全卷取 $g = 9.8\ \text{m/s}^2$。Q1-Q2 不可使用计算器；Q3-Q5 可用计算器。
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 across a level floor by a constant horizontal force of $50$ N. How much work does the pushing force do on the crate?一个箱子被一个 $50$ N 的恒定水平力沿水平地面推动 $4.0$ m。推力对箱子做了多少功？
Use $g = 9.8\ \text{m/s}^2$.取 $g = 9.8\ \text{m/s}^2$。
(a)A $4.0$ kg textbook is lifted $2.0$ m from the floor to a shelf. Compute the gain in gravitational potential energy, with units.一本 $4.0$ kg 的教科书从地面被抬到架子上，升高 $2.0$ m。计算重力势能的增量，并写出单位。[2]
(b)A spring of constant $k = 300$ N/m is compressed by $0.20$ m. Compute the elastic potential energy stored.一根劲度系数 $k = 300$ N/m 的弹簧被压缩 $0.20$ m。计算其储存的弹性势能。[1]
(c)State why a reference level must be chosen before a value of gravitational potential energy is meaningful.说明为何在重力势能的数值有意义之前必须先选定参考面。[1]
A $0.30$ kg ball is dropped from rest at a height of $1.8$ m above the ground. Ignore air resistance and use $g = 9.8\ \text{m/s}^2$.一个 $0.30$ kg 的球从距地面 $1.8$ m 处由静止落下。忽略空气阻力，取 $g = 9.8\ \text{m/s}^2$。
(a)Using energy conservation, find the speed of the ball just before it hits the ground.用能量守恒，求球刚落地前的速率。[2]
(b)State the kinetic energy of the ball at the instant of impact.写出球落地瞬间的动能。[2]
(c)State the condition that allows you to set initial mechanical energy equal to final mechanical energy.写出允许你将初机械能等于末机械能的条件。[1]
PART II · EXTENDED RESPONSE第二部分 · 简答题AP-feeder FRQ + honors · 35 marksAP 衔接简答题 + 荣誉级 · 共 35 分
Section B · Extended ResponseB 部分 · 简答题
Show every step of reasoning. Identify the energy equation used (work-energy theorem, conservation of mechanical energy, or the extended form with $W_\text{nc}$) before substituting values. State units in every final answer. Use $g = 9.8\ \text{m/s}^2$. State your chosen reference level for potential energy at the start of each question. Calculator permitted on Q6-Q9.每一步推理都要写出。在代入数值前先注明所用能量方程（功能定理、机械能守恒，或含 $W_\text{nc}$ 的扩展式）。每个最终答案都要写单位。取 $g = 9.8\ \text{m/s}^2$。每题开头请说明所选的势能参考面。Q6-Q9 可用计算器。
A $1000$ kg car travelling at $20$ m/s brakes and comes to a complete stop in a distance of $50$ m. The road is level.一辆 $1000$ kg 的汽车以 $20$ m/s 行驶，刹车后在 $50$ m 内完全停下。路面水平。
(a)Find the initial kinetic energy of the car.求汽车的初始动能。[2]
(b)Using the work-energy theorem, find the net work done on the car while braking.用功能定理，求刹车过程中对汽车做的合功。[2]
(c)Find the magnitude of the (constant) braking force.求（恒定）刹车力的大小。[2]
(d)If the car had instead been travelling at $40$ m/s with the same braking force, by what factor would the braking distance change? Justify with the work-energy theorem.若汽车改以 $40$ m/s 行驶、刹车力不变，刹车距离会变为原来的几倍？用功能定理论证。[2]
A $50$ kg child starts from rest at the top of a smooth (frictionless) playground slide. The top of the slide is $4.0$ m above the bottom. Take $g = 9.8\ \text{m/s}^2$ and take the bottom as the reference level.一个 $50$ kg 的孩子从一个光滑（无摩擦）滑梯顶端静止出发。滑梯顶端高于底端 $4.0$ m。取 $g = 9.8\ \text{m/s}^2$，以底端为参考面。
(a)Find the gravitational potential energy of the child at the top.求孩子在顶端的重力势能。[2]
(b)Find the speed of the child at the bottom of the slide.求孩子到达滑梯底端的速率。[3]
(c)Find the speed of the child at the point on the slide that is $1.5$ m above the bottom.求孩子在滑梯上距底端 $1.5$ m 处的速率。[2]
(d)Explain why the speed at the bottom does not depend on the mass of the child.解释为何底端速率与孩子的质量无关。[1]
A $2.0$ kg block starts from rest and slides down a ramp whose top is $3.0$ m above the bottom. A constant kinetic friction force of $5.0$ N acts along the entire $5.0$ m path down the ramp. Take $g = 9.8\ \text{m/s}^2$ and the bottom as the reference level.一个 $2.0$ kg 的滑块从静止沿斜坡下滑，坡顶高于坡底 $3.0$ m。沿整段 $5.0$ m 的下坡路径作用着 $5.0$ N 的恒定动摩擦力。取 $g = 9.8\ \text{m/s}^2$，以坡底为参考面。
(a)Find the gravitational potential energy of the block at the top.求滑块在坡顶的重力势能。[2]
(b)Find the work done by friction over the descent.求下坡过程中摩擦力所做的功。[2]
(c)Using the extended energy equation $KE_f + PE_f = KE_i + PE_i + W_\text{nc}$, find the kinetic energy of the block at the bottom.用扩展能量方程 $KE_f + PE_f = KE_i + PE_i + W_\text{nc}$，求滑块到达坡底的动能。[2]
(d)Find the speed of the block at the bottom.求滑块到达坡底的速率。[2]
(e)State how much thermal energy was generated, and where it came from.写出产生了多少热能，以及它来自何处。[1]
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]
A spring of constant $k = 800$ N/m is compressed by $0.30$ m and used to launch a $1.5$ kg block along a frictionless horizontal surface, which then leads smoothly into a frictionless incline. Take $g = 9.8\ \text{m/s}^2$.一根劲度系数 $k = 800$ N/m 的弹簧被压缩 $0.30$ m，用来在无摩擦的水平面上发射一个 $1.5$ kg 的滑块，该水平面平滑地接入一个无摩擦斜面。取 $g = 9.8\ \text{m/s}^2$。
(a)Find the elastic potential energy stored in the compressed spring.求压缩弹簧中储存的弹性势能。[2]
(b)Find the speed of the block as it leaves the spring on the flat surface.求滑块在水平面上离开弹簧时的速率。[2]
(c)Find the maximum vertical height the block rises to on the frictionless incline.求滑块在无摩擦斜面上上升的最大竖直高度。[2]
(d)Suppose instead a rough patch on the flat section does $-9.0$ J of work on the block before it reaches the incline. Find the block's speed at the base of the incline.现假设水平段上一处粗糙区域在滑块到达斜面前对它做了 $-9.0$ J 的功。求滑块在斜面底部的速率。[2]
(e)For the fully frictionless case, use an energy audit to explain why the block returns to the spring with the same speed it left.对完全无摩擦的情形，用能量审计解释为何滑块返回弹簧时的速率与离开时相同。[2]
PART III · MODELING / APPLIED第三部分 · 建模与应用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
Define symbols (with units) at the start of each question. State the energy equation used before substituting. Use $g = 9.8\ \text{m/s}^2$. Conclude each question with a one-sentence answer in context. Calculator permitted throughout Part III.每题开始时定义符号（含单位）。代入数值前先写出所用能量方程。取 $g = 9.8\ \text{m/s}^2$。每题以一句结合情境的完整句子作答。第三部分全程可用计算器。
An elevator motor lifts a fully loaded car of total mass $1200$ kg at a constant speed of $2.0$ m/s. Take $g = 9.8\ \text{m/s}^2$.一台电梯电机以 $2.0$ m/s 的恒定速度提升一节满载、总质量 $1200$ kg 的轿厢。取 $g = 9.8\ \text{m/s}^2$。
(a)Find the upward force the cable must exert on the car while it rises at constant speed.求轿厢匀速上升时缆绳必须施加的向上力。[2]
(b)Find the useful power output of the motor using $P = Fv$.用 $P = Fv$ 求电机的有用功率输出。[3]
(c)The motor draws $30$ kW of electrical power. Find its efficiency.电机消耗 $30$ kW 的电功率。求其效率。[2]
(d)Find the rate at which energy is "lost" by the motor, and state what form it takes.求电机"损失"能量的速率，并说明它以何种形式存在。[2]
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]
A $70$ kg skier starts from rest at the top of a slope $25$ m high and reaches the bottom moving at $18$ m/s. Take $g = 9.8\ \text{m/s}^2$ and the bottom as the reference level.一名 $70$ kg 的滑雪者从一段高 $25$ m 的斜坡顶端静止出发，到达底端时速率为 $18$ m/s。取 $g = 9.8\ \text{m/s}^2$，以底端为参考面。
(a)Find the gravitational potential energy of the skier at the top.求滑雪者在顶端的重力势能。[2]
(b)Find the kinetic energy of the skier at the bottom.求滑雪者在底端的动能。[2]
(c)Find the energy converted to thermal energy by friction and air resistance during the descent.求下坡过程中摩擦力与空气阻力转化为热能的能量。[2]
(d)Find the speed the skier would have reached at the bottom if the slope were frictionless, and comment on why it is larger.求若斜坡无摩擦时滑雪者在底端会达到的速率，并说明它为何更大。[3]
A $500$ kg roller-coaster cart starts from rest at the top of a $35$ m hill on a frictionless track. The track then runs to ground level and up into a vertical loop whose highest point is $20$ m above the ground. Take $g = 9.8\ \text{m/s}^2$ and the ground as the reference level.一辆 $500$ kg 的过山车从无摩擦轨道上一座 $35$ m 高的山顶静止出发。轨道随后下到地面，再进入一个竖直环形，环形最高点距地面 $20$ m。取 $g = 9.8\ \text{m/s}^2$，以地面为参考面。
(a)Find the speed of the cart at ground level (bottom of the hill).求小车在地面（山脚）处的速率。[2]
(b)Find the speed of the cart at the top of the loop.求小车在环顶的速率。[3]
(c)Find the kinetic energy of the cart at the top of the loop.求小车在环顶的动能。[2]
(d)Carry out an energy audit at the top of the loop: show that $KE + GPE$ there equals the initial $GPE$ at the top of the hill.在环顶进行能量审计：证明此处的 $KE + GPE$ 等于山顶的初始 $GPE$。[2]
(e)Explain why a real roller-coaster must start from a hill higher than $35$ m to complete the same loop.解释为何真实的过山车必须从高于 $35$ m 的山坡出发才能完成同样的环形。[1]
🇺🇸 US NGSS美国 NGSSHS-PS3-1 · HS-PS3-2 · HS-PS3-3
🇨🇦 Ontario安大略SPH3U Strand D · D3.1 · D3.2
🇨🇦 British Columbia不列颠哥伦比亚Physics 11: energy, work, power and efficiency物理 11：能量、功、功率与效率
🇨🇦 Alberta阿尔伯塔Physics 20 Unit C · 20-C2.1k · 20-C2.2k · 20-C2.4k · 20-C2.6k
Full Syllabus Map lives in ../Study Guides/Unit_3_Work_Energy_and_Power.html. Note: NGSS frames energy qualitatively and computationally (HS-PS3-1); AB work-energy sits in Physics 20 Unit C, and the Diploma-style framing reflects the standardized-exam structure students meet later in Physics 30.完整大纲对照表见 ../Study Guides/Unit_3_Work_Energy_and_Power.html。注：NGSS 以定性与计算方式构建能量概念（HS-PS3-1）；阿省功能内容位于物理 20 单元 C，毕业考风格反映学生稍后在物理 30 遇到的标准化考试结构。