High School Physics

Work, Energy and Power功、能量与功率

Energy is the capacity to do work, and work is the mechanism by which energy is transferred. This guide builds from the fundamental definition of work ($W = Fd\cos\theta$) through kinetic energy and the work-energy theorem ($\Delta KE = W_\text{net}$), gravitational and elastic potential energy, the conservation of mechanical energy in isolated systems, and power as the rate of energy transfer. The final two sections address non-conservative forces (friction, applied forces) that change a system's mechanical energy, and a systematic step-by-step framework for tackling any energy problem. All worked examples use real numbers throughout.能量(energy,能量)是做功的能力,而功(work,功)是能量转移的机制。本指南从功的基本定义($W = Fd\cos\theta$)出发,依次经过动能(kinetic energy,动能)与功能定理(work-energy theorem,功能定理,$\Delta KE = W_\text{net}$)、重力势能与弹性势能(potential energy,势能)、孤立系统中的机械能守恒(conservation of mechanical energy,机械能守恒),以及功率(power,功率)作为能量转移的速率。最后两节讨论改变系统机械能的非保守力(摩擦力、外力),并提供一套应对任何能量问题的系统解题框架。全部例题均用真实数字演算。

7 sections7 节内容 US NGSS · ON · BC · ABUS NGSS · ON · BC · AB Covers HS-PS3-1, -2, -3涵盖 HS-PS3-1、-2、-3

How to use this guide如何使用本指南

Work, Energy and Power is the third physics unit and the four curricula agree on the core scope: work done by a force ($W = Fd\cos\theta$), kinetic and potential energy, conservation of mechanical energy, and power. The main syllabus difference is emphasis: NGSS HS-PS3 focuses on energy accounting in systems and energy-conversion device design (HS-PS3-3); Ontario SPH3U Strand D and SPH4U Strand C build from work and power through conservation to elastic PE; BC Physics 11 pairs "conservation of energy; principle of work and energy" with "power and efficiency" as separate Content bullets; Alberta Physics 20 Unit C GO2 frames work as a transfer of mechanical energy in isolated and non-isolated systems. The table tells you which sections are core for you right now; each row cites the curriculum document it was checked against.功、能量与功率是第三个物理单元,四套大纲在核心范围上基本一致:力所做的功($W = Fd\cos\theta$)、动能与势能、机械能守恒、功率。主要差异在于侧重点:NGSS HS-PS3 聚焦于系统能量核算与能量转换装置设计(HS-PS3-3);安大略 SPH3U D 单元与 SPH4U C 单元从功与功率出发,经守恒延伸至弹性势能;BC Physics 11 把"能量守恒;功与能量原理"与"功率与效率"列为并列内容;阿尔伯塔 Physics 20 C 单元 GO2 把功定义为孤立与非孤立系统中机械能的转移。下表告诉你当前哪些节属于你的核心;每行都注明所依据的课纲文件。

If you are in…如果你在… Focus on these sections重点学习 Defer / lighter可推迟 / 减负 Source依据
🇺🇸 US NGSS HS Physical Science美国 NGSS 物理科学 §1 through §7 in full — HS-PS3-1 covers energy accounting (§1–§4), HS-PS3-2 covers KE and PE models (§2–§3), HS-PS3-3 covers power and energy conversion (§5)§1 至 §7 完整学习 — HS-PS3-1 涵盖能量核算(§1–§4),HS-PS3-2 涵盖动能与势能模型(§2–§3),HS-PS3-3 涵盖功率与能量转换(§5) Nothing — all three NGSS energy PEs map directly to this unit's content无 — 三个 NGSS 能量表现期望直接对应本单元内容 ngss_hs_ps_extract.md — HS-PS3-1, HS-PS3-2, HS-PS3-3 PEs verbatim— HS-PS3-1、HS-PS3-2、HS-PS3-3 逐字引用
🇨🇦 ON Grade 11 — SPH3U安大略 11 年级 — SPH3U §1 through §6. SPH3U Strand D (D3.1, D3.2) covers work, power, KE, gravitational PE, and conservation. Elastic PE (§3 spring) and 2D energy problems are SPH4U Strand C (C3) territory§1 至 §6。SPH3U D 单元(D3.1、D3.2)涵盖功、功率、动能、重力势能与守恒。弹性势能(§3 弹簧)与二维能量问题属 SPH4U C 单元(C3) Elastic PE in §3 and complex conservation scenarios in §7 are Grade-12 (SPH4U C3) extensions§3 中的弹性势能与 §7 中的复杂守恒场景为 12 年级(SPH4U C3)延伸 science_11-12_physics_extract.md — SPH3U Strand D Overall Expectations D1–D3, Specific Expectations D3.1, D3.2; SPH4U Strand C Overall Expectation C3— SPH3U D 单元总体期望 D1–D3,具体期望 D3.1、D3.2;SPH4U C 单元总体期望 C3
🇨🇦 BC Grade 11 — Physics 11BC 11 年级 — Physics 11 §1 through §7. Physics 11 Content bullets "conservation of energy; principle of work and energy" and "power and efficiency" are both core; §6 (friction losses) is supported by "conservation of energy" elaborated as thermal-equilibrium context§1 至 §7。Physics 11 内容中"能量守恒;功与能量原理"与"功率与效率"均为核心;§6(摩擦损失)由"能量守恒"在热平衡语境的细化说明支持 Nothing — BC explicitly pairs energy conservation with work-energy and power as co-equal Content bullets无 — BC 明确把能量守恒与功能定理及功率作为并列内容 physics_11-12_extract.md — Physics 11 Big Idea: "Energy is found in different forms …"; Content: "conservation of energy; principle of work and energy"; "power and efficiency"— Physics 11 大概念:"能量以不同形式存在……";内容:"能量守恒;功与能量原理";"功率与效率"
🇨🇦 AB Grade 11 — Physics 20阿尔伯塔 11 年级 — Physics 20 §1 through §7 in full. Physics 20 Unit C GO2 outcomes 20–C2.1k through 20–C2.6k cover all of work, KE, gravitational PE, mechanical energy conservation, power, and non-isolated systems§1 至 §7 完整学习。Physics 20 C 单元 GO2(outcome 20–C2.1k 至 20–C2.6k)涵盖功、动能、重力势能、机械能守恒、功率与非孤立系统的全部内容 Nothing — AB Diploma-style problems reward systematic energy-bar diagrams and explicit identification of conservative vs non-conservative work无 — AB 文凭考风格题奖励系统性的能量条图与明确区分保守功与非保守功 physics_20-30_extract.md — Physics 20 Unit C GO2, knowledge outcomes 20–C2.1k through 20–C2.6k— Physics 20 C 单元 GO2,知识 outcome 20–C2.1k 至 20–C2.6k
🇺🇸 AP / IB feeder trackAP / IB 衔接轨道 All seven sections plus every going-deeper derivation. AP Physics 1 / C and IB Physics HL assume fluent work-energy and conservation reasoning from the first week; AP adds calculus-based work as a line integral全部 7 节,并完成每个"深入"推导。AP Physics 1 / C 与 IB Physics HL 第一周就默认你熟练功能推理与守恒;AP 进一步把功表示为线积分 Nothing — this unit is the energy foundation all subsequent mechanics, fields, and thermodynamics guides build on无 — 本单元是后续力学、场论与热力学各指南所依赖的能量基础 ngss_hs_ps_extract.md — see the AP Physics Unit 3 link in "What This Feeds Into"— 见"本单元的去向"中的 AP Physics Unit 3 链接

Once you have located your row, use the two cards below for the speed at which you should work through the recommended sections.找到所在行后,用下面两张卡片决定推进速度。

!
If you are cramming the night before如果你在临阵磨枪

Memorise four things: the work formula $W = Fd\cos\theta$ (and that perpendicular force does zero work); the work-energy theorem $W_\text{net} = \Delta KE = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2$; the conservation equation $KE_i + PE_i = KE_f + PE_f$ (isolated systems only); and the power formula $P = W/t = Fv$. Read every cram-cheat box. Skip the going-deeper derivations.背熟四件事:功的公式 $W = Fd\cos\theta$(以及垂直力做功为零);功能定理 $W_\text{net} = \Delta KE = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2$;守恒方程 $KE_i + PE_i = KE_f + PE_f$(仅限孤立系统);以及功率公式 $P = W/t = Fv$。读每个速记框,跳过深入推导。

*
If you are going for the top mark如果你目标顶分

Always draw an energy bar diagram (initial and final energy bars) before writing any equation; a bookkeeping error on which energies are present is the most common way marks are lost. Ask "is the system isolated?" before applying conservation. Check the sign of $\cos\theta$ in $W = Fd\cos\theta$: friction always does negative work (angle $180°$). AB Physics 20 expects you to state whether work is done by conservative or non-conservative forces and to account for the difference explicitly (outcomes 20–C2.3k and 20–C2.6k).写任何方程前先画能量条图(初始与末态能量条);对哪些能量存在的记账错误是最常见的失分原因。应用守恒前先判断"系统是否孤立"。检查 $W = Fd\cos\theta$ 中 $\cos\theta$ 的符号:摩擦力始终做负功(角度为 $180°$)。AB Physics 20 要求你说明功是由保守力还是非保守力完成,并明确计算两者之差(outcomes 20–C2.3k 与 20–C2.6k)。

Curriculum note on scope.范围提示。 All seven sections in this guide are core for every curriculum we map to. There is no single section that is exclusively honors for US NGSS — all three NGSS energy PEs (HS-PS3-1, -2, -3) apply to the content here. The deepest content (elastic PE in §3, systematic energy problem-solving in §7) is Grade-12 territory in Ontario (SPH4U C3) but still covered in Alberta Physics 20 (20–C2.1k through 20–C2.6k) and BC Physics 11. If your row above flags an extension, treat those items as preparation for Grade 12, not as content to skip.本指南中的全部七节在我们所对照的所有大纲中均为核心内容。对 US NGSS 而言,并无某节专属荣誉级 — 三个 NGSS 能量表现期望(HS-PS3-1、-2、-3)均适用于此处内容。最深的内容(§3 弹性势能、§7 系统解题)在安大略是 12 年级(SPH4U C3)内容,但在阿尔伯塔 Physics 20(20–C2.1k 至 20–C2.6k)和 BC Physics 11 中均已覆盖。如果你的行标注了延伸内容,把这些项目视为 12 年级的提前准备,而非可以跳过的内容。

Work Done by a Force功:力所做的功

Work is force times displacement — only the component parallel to displacement counts.功是力与位移之积 — 只有平行于位移的分量计入。 $$ W = F\,d\cos\theta \qquad (\text{joules, J} = \text{N}\cdot\text{m}) $$
  • $\theta$$\theta$ — the angle between the force vector and the displacement vector (not the angle with the surface).— 力的方向与位移方向之间的夹角(不是与表面的夹角)。
  • Sign of work.功的符号。 $\theta < 90°$: positive work (force adds energy). $\theta = 90°$: zero work (e.g. normal force on a horizontal surface). $\theta > 90°$: negative work (force removes energy, e.g. friction).$\theta < 90°$:正功(力输入能量)。$\theta = 90°$:零功(如水平面上的法向力)。$\theta > 90°$:负功(力移除能量,如摩擦力)。
  • Net work合功 — the algebraic sum of work done by every force. This equals the change in kinetic energy (the work-energy theorem, §2).— 每个力所做功的代数和。这等于动能的变化(功能定理,§2)。
Alberta 20–C2.4k states "work as a measure of the mechanical energy transferred." Ontario D3.2 requires you to "explain the concepts of and interrelationships between energy, work, and power, and identify and describe their related units."阿尔伯塔 20–C2.4k 指出"功是机械能转移的度量"。安大略 D3.2 要求你"解释能量、功与功率的概念及相互关系,并识别和描述相关单位"。
Worked Example 1 · Work done pulling a suitcase例题 1 · 拉行李箱做的功

A person pulls a suitcase $12$ m along level ground with a force of $30$ N applied at $40°$ above the horizontal. Calculate the work done by the pulling force.一个人用 $30$ N 的力以水平方向上方 $40°$ 拉着行李箱在平地上走了 $12$ m。计算拉力所做的功。

Identify the angle between force and displacement.确定力与位移之间的角度。 The displacement is horizontal; the force is $40°$ above horizontal, so $\theta = 40°$.位移水平;力在水平方向上方 $40°$,故 $\theta = 40°$。

$$ W = Fd\cos\theta = 30 \times 12 \times \cos 40° = 360 \times 0.766 \approx 276 \text{ J}. $$

Interpret.解读。 About $276$ J of energy is transferred to the suitcase via the pulling force. Note: the normal force ($\theta = 90°$) and gravity ($\theta = 90°$ for horizontal motion) both do zero work here — neither is parallel to the motion.约 $276$ J 的能量通过拉力转移给行李箱。注意:法向力($\theta = 90°$)和重力(水平运动时 $\theta = 90°$)在此均做零功 — 两者均不平行于运动方向。

A box is pushed $5.0$ m along the floor by a horizontal force of $40$ N. How much work does the pushing force do?一个箱子被 $40$ N 的水平力推动在地板上移动了 $5.0$ m。推力做了多少功?
§1 · Q1
$8.0$ J
$200$ J
$0$ J
$45$ J
Force and displacement are parallel ($\theta = 0°$, $\cos 0° = 1$): $W = Fd\cos 0° = 40 \times 5.0 \times 1 = 200$ J.力与位移平行($\theta = 0°$,$\cos 0° = 1$):$W = Fd\cos 0° = 40 \times 5.0 \times 1 = 200$ J。
When the force is parallel to the displacement, $\cos\theta = \cos 0° = 1$, so $W = Fd$.当力平行于位移时,$\cos\theta = \cos 0° = 1$,故 $W = Fd$。
A waiter carries a tray of food horizontally across a room at constant height. The tray moves $6$ m. What is the work done by the upward normal force from the waiter's hand on the tray?一名服务员在房间里水平端着托盘走了 $6$ m,托盘高度不变。服务员的手对托盘施加的向上法向力做了多少功?
§1 · Q2
Equal to the weight of the tray times $6$ m等于托盘重力乘以 $6$ m
Positive, because the force is upward正,因为力向上
$0$ J
Negative, because the tray moves perpendicular to the force负,因为托盘运动与力垂直
The force is upward and the displacement is horizontal, so $\theta = 90°$ and $\cos 90° = 0$. Work $= Fd\cos 90° = 0$ J. A perpendicular force does zero work.力向上,位移水平,故 $\theta = 90°$,$\cos 90° = 0$。功 $= Fd\cos 90° = 0$ J。垂直于运动的力做零功。
Work requires a component of force parallel to displacement. At $\theta = 90°$, the parallel component is zero, so $W = 0$.做功需要力在位移方向上的分量。$\theta = 90°$ 时,平行分量为零,故 $W = 0$。
Going deeper — work as a dot product and the work done against gravity on an incline深入 — 功作为点积,以及在斜面上克服重力所做的功

The formula $W = Fd\cos\theta$ is the scalar (dot) product $\vec{F}\cdot\vec{d}$. For gravity ($\vec{F}_g = -mg\hat{j}$) on an object moving along an incline of angle $\alpha$, the displacement has a vertical component $d\sin\alpha$ downward. Gravity does work $W_g = mgd\sin\alpha$ (positive when descending, negative when ascending). This is why the work done against gravity on a ramp depends only on the vertical height gained $h = d\sin\alpha$, not on the length of the ramp — a result that underpins gravitational potential energy in §3.公式 $W = Fd\cos\theta$ 是标量(点)积 $\vec{F}\cdot\vec{d}$。重力($\vec{F}_g = -mg\hat{j}$)作用于沿倾角 $\alpha$ 的斜面运动的物体时,位移有向下的竖直分量 $d\sin\alpha$。重力所做的功为 $W_g = mgd\sin\alpha$(下降时为正,上升时为负)。这就是为何在斜面上克服重力做的功只取决于上升的竖直高度 $h = d\sin\alpha$,而非斜面长度 — 这一结果是 §3 重力势能的基础。


Kinetic Energy and the Work-Energy Theorem动能与功能定理

Net work equals the change in kinetic energy.合功等于动能的变化。 $$ KE = \tfrac{1}{2}mv^2 \qquad (\text{joules, J}) $$ $$ W_\text{net} = \Delta KE = KE_f - KE_i = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2 $$
  • KE is always $\ge 0$.动能恒 $\ge 0$。 It depends on $v^2$, so direction of motion never matters for $KE$ itself (only for the work calculation).动能取决于 $v^2$,故运动方向本身不影响 $KE$ 的大小(仅影响做功的计算)。
  • The theorem's power.定理的威力。 You do not need to know how the force acted over time; you only need net work and the speeds. It bypasses kinematics when $a$ is not constant.你不需要知道力如何随时间作用;只需净功与速度大小。当 $a$ 非恒定时,它绕过运动学直接求解。
  • Units check.单位核验。 $[KE] = \mathrm{kg \cdot m^2/s^2} = \mathrm{N \cdot m} = \mathrm{J}$. Same as work — the theorem is dimensionally consistent.$[KE] = \mathrm{kg \cdot m^2/s^2} = \mathrm{N \cdot m} = \mathrm{J}$。与功相同 — 定理量纲一致。
NGSS HS-PS3-2 requires models illustrating that macroscopic energy includes "energy associated with the motion of particles (objects)" — exactly kinetic energy. Ontario D3.2 asks you to interrelate energy, work, and power with their units.NGSS HS-PS3-2 要求建立模型,说明宏观能量包括"与粒子(物体)运动相关的能量"——正是动能。安大略 D3.2 要求你把能量、功与功率及其单位相互联系。
Worked Example 2 · Braking car via the work-energy theorem例题 2 · 用功能定理分析刹车汽车

A $1200$ kg car traveling at $20\ \mathrm{m/s}$ brakes to a stop. (a) What is the initial kinetic energy? (b) How much work does the friction force do on the car?一辆质量为 $1200$ kg、以 $20\ \mathrm{m/s}$ 行驶的汽车刹车停下。(a) 初始动能是多少?(b) 摩擦力对汽车做了多少功?

(a) Initial kinetic energy.(a) 初始动能。

$$ KE_i = \tfrac{1}{2}mv_i^2 = \tfrac{1}{2}(1200)(20)^2 = \tfrac{1}{2}(1200)(400) = 240{,}000 \text{ J} = 240 \text{ kJ}. $$

(b) Work done by friction.(b) 摩擦力做的功。 The car stops, so $KE_f = 0$. By the work-energy theorem:车停下,故 $KE_f = 0$。由功能定理:

$$ W_\text{net} = \Delta KE = 0 - 240{,}000 = -240 \text{ kJ}. $$

Interpret.解读。 The friction force does $-240$ kJ of work, removing all the car's kinetic energy. The negative sign confirms friction opposes motion ($\theta = 180°$). That energy is converted to thermal energy (heat in the brakes) — consistent with NGSS HS-PS3-2's model of KE converting to thermal energy.摩擦力做了 $-240$ kJ 的功,移除了汽车的全部动能。负号确认摩擦力与运动方向相反($\theta = 180°$)。该能量转化为热能(刹车片的热量)——与 NGSS HS-PS3-2 关于动能转化为热能的模型一致。

A $2.0$ kg ball moves at $3.0\ \mathrm{m/s}$. What is its kinetic energy?一只 $2.0$ kg 的球以 $3.0\ \mathrm{m/s}$ 运动。其动能是多少?
§2 · Q1
$6.0$ J
$18$ J
$3.0$ J
$9.0$ J
$KE = \tfrac{1}{2}mv^2 = \tfrac{1}{2}(2.0)(3.0)^2 = \tfrac{1}{2}(2.0)(9.0) = 9.0$ J.$KE = \tfrac{1}{2}mv^2 = \tfrac{1}{2}(2.0)(3.0)^2 = \tfrac{1}{2}(2.0)(9.0) = 9.0$ J。
$KE = \tfrac{1}{2}mv^2$. Square the speed first, then multiply by half the mass.$KE = \tfrac{1}{2}mv^2$。先对速度平方,再乘以质量的一半。
A net force does $360$ J of work on a $4.0$ kg object that starts from rest. What is the object's final speed?一个净力对一个从静止出发的 $4.0$ kg 物体做了 $360$ J 的功。物体的末速度是多少?
§2 · Q2
$90\ \mathrm{m/s}$
$13.4\ \mathrm{m/s}$
$180\ \mathrm{m/s}$
$6.7\ \mathrm{m/s}$
$W_\text{net} = \tfrac{1}{2}mv_f^2 - 0 \Rightarrow 360 = \tfrac{1}{2}(4.0)v_f^2 \Rightarrow v_f^2 = 180 \Rightarrow v_f = \sqrt{180} \approx 13.4\ \mathrm{m/s}$.$W_\text{net} = \tfrac{1}{2}mv_f^2 - 0 \Rightarrow 360 = \tfrac{1}{2}(4.0)v_f^2 \Rightarrow v_f^2 = 180 \Rightarrow v_f = \sqrt{180} \approx 13.4\ \mathrm{m/s}$。
Use $W_\text{net} = \Delta KE = \tfrac{1}{2}mv_f^2$ (starting from rest). Solve for $v_f = \sqrt{2W/m}$.用 $W_\text{net} = \Delta KE = \tfrac{1}{2}mv_f^2$(从静止出发)。解得 $v_f = \sqrt{2W/m}$。
Going deeper — deriving the work-energy theorem from Newton's second law深入 — 从牛顿第二定律推导功能定理

Start with Newton's second law: $F_\text{net} = ma$. For constant force over displacement $d$: $W_\text{net} = F_\text{net}\,d$. From kinematics (recall from the Forces unit), $v_f^2 = v_i^2 + 2ad$, so $ad = \tfrac{v_f^2 - v_i^2}{2}$. Multiply both sides by $m$:从牛顿第二定律 $F_\text{net} = ma$ 出发。对位移 $d$ 上的恒力:$W_\text{net} = F_\text{net}\,d$。由运动学(回忆力学单元),$v_f^2 = v_i^2 + 2ad$,故 $ad = \tfrac{v_f^2 - v_i^2}{2}$。两边乘以 $m$:

$$ W_\text{net} = F_\text{net}\,d = mad = m\cdot\frac{v_f^2 - v_i^2}{2} = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2 = \Delta KE. $$

This derivation shows the work-energy theorem is not a new law but a consequence of $F = ma$ plus kinematics. At AP Physics C level, the derivation generalises to variable forces via the integral $W = \int F\,dx$, where the result $W = \Delta KE$ still holds.这个推导表明,功能定理不是新定律,而是 $F = ma$ 加运动学的推论。在 AP Physics C 层面,推导通过积分 $W = \int F\,dx$ 推广至变力,结果 $W = \Delta KE$ 仍然成立。


Potential Energy: Gravitational and Elastic势能:重力势能与弹性势能

Potential energy is stored energy waiting to do work.势能是等待做功的储存能量。 $$ GPE = mgh \qquad (\text{joules; }h\text{ measured from the reference level}) $$ $$ EPE = \tfrac{1}{2}kx^2 \qquad (k\text{ = spring constant (N/m)},\ x\text{ = extension or compression}) $$
  • Reference level for GPE.重力势能的参考面。 You choose where $h = 0$; only changes in $GPE$ are physically meaningful. The ground, a table top, or any convenient level can be zero.你选定 $h = 0$ 的位置;只有 $GPE$ 的变化量才有物理意义。地面、桌面或任何方便的高度均可作为零点。
  • Spring force and Hooke's law.弹力与胡克定律。 $F_\text{spring} = kx$ (restoring force, opposite to displacement). The energy stored is $EPE = \tfrac{1}{2}kx^2$.$F_\text{spring} = kx$(恢复力,方向与形变相反)。储存的能量为 $EPE = \tfrac{1}{2}kx^2$。
  • Conservative forces.保守力。 Gravity and spring force are conservative: the work they do depends only on start and end positions, not on path. This is what makes the conservation law in §4 possible.重力与弹力是保守力:它们所做的功只取决于初末位置,与路径无关。这正是 §4 守恒定律成立的条件。
NGSS HS-PS3-2 explicitly cites "the energy stored due to position of an object above the earth" as an example. Alberta 20–C2.1k: "define mechanical energy as the sum of kinetic and potential energy."NGSS HS-PS3-2 明确列举"物体在地球上方的位置所储存的能量"作为示例。阿尔伯塔 20–C2.1k:"将机械能定义为动能与势能之和。"
Worked Example 3 · GPE and spring PE例题 3 · 重力势能与弹性势能

(a) A $3.0$ kg book is lifted $1.5$ m above a desk. Using $g = 9.8\ \mathrm{m/s^2}$, find the change in gravitational PE relative to the desk. (b) A spring with $k = 400\ \mathrm{N/m}$ is compressed $0.050$ m. Find the elastic PE stored.(a) 一本 $3.0$ kg 的书被抬到桌面上方 $1.5$ m 处。取 $g = 9.8\ \mathrm{m/s^2}$,求相对于桌面的重力势能变化。(b) 一根弹簧($k = 400\ \mathrm{N/m}$)被压缩 $0.050$ m。求储存的弹性势能。

(a) Gravitational PE.(a) 重力势能。

$$ \Delta GPE = mg\Delta h = 3.0 \times 9.8 \times 1.5 = 44.1 \text{ J}. $$

(b) Elastic PE.(b) 弹性势能。

$$ EPE = \tfrac{1}{2}kx^2 = \tfrac{1}{2}(400)(0.050)^2 = \tfrac{1}{2}(400)(0.0025) = 0.50 \text{ J}. $$

Note on reference.关于参考点的说明。 Taking the desk as the reference ($h = 0$), the book now has $GPE = 44.1$ J. If instead we take the floor (say $1.0$ m below the desk) as reference, the book has $GPE = mg(1.0 + 1.5) = 73.5$ J — the number changes, but any change in $GPE$ during a process is the same regardless of where zero is set.以桌面为参考($h = 0$),书此时有 $GPE = 44.1$ J。若改以地板(设在桌面以下 $1.0$ m)为参考,书的 $GPE = mg(1.0 + 1.5) = 73.5$ J — 数值不同,但过程中 $GPE$ 的变化量与零点选取无关。

A $5.0$ kg object is raised $4.0$ m above the floor. Using $g = 9.8\ \mathrm{m/s^2}$, what is the increase in gravitational PE?一个 $5.0$ kg 的物体被抬到地板上方 $4.0$ m 处。取 $g = 9.8\ \mathrm{m/s^2}$,重力势能增加了多少?
§3 · Q1
$196$ J
$20$ J
$49$ J
$392$ J
$\Delta GPE = mgh = 5.0 \times 9.8 \times 4.0 = 196$ J.$\Delta GPE = mgh = 5.0 \times 9.8 \times 4.0 = 196$ J。
$GPE = mgh$. Multiply mass, $g$, and height in that order.$GPE = mgh$。依次相乘质量、$g$ 与高度。
A spring with $k = 200\ \mathrm{N/m}$ is stretched $0.10$ m. What is the elastic PE stored?一根弹簧($k = 200\ \mathrm{N/m}$)被拉伸 $0.10$ m。储存的弹性势能是多少?
§3 · Q2
$20$ J
$2.0$ J
$1.0$ J
$0.10$ J
$EPE = \tfrac{1}{2}kx^2 = \tfrac{1}{2}(200)(0.10)^2 = \tfrac{1}{2}(200)(0.01) = 1.0$ J.$EPE = \tfrac{1}{2}kx^2 = \tfrac{1}{2}(200)(0.10)^2 = \tfrac{1}{2}(200)(0.01) = 1.0$ J。
$EPE = \tfrac{1}{2}kx^2$. Square the extension first, then multiply by $\tfrac{1}{2}k$.$EPE = \tfrac{1}{2}kx^2$。先对形变量平方,再乘以 $\tfrac{1}{2}k$。
Going deeper — why conservative forces allow potential energy to be defined深入 — 为何保守力允许势能被定义

A force is conservative if the work it does on an object moving between two points is independent of the path taken. For gravity: $W_\text{gravity} = -\Delta GPE = -(mgh_f - mgh_i)$ regardless of the route from $h_i$ to $h_f$. This path-independence is why we can assign a unique energy value to each position, define a potential energy function, and write a conservation law. Friction, by contrast, depends on path length (more distance $=$ more heat), so it is non-conservative and cannot be described by a potential energy function — that is why §6 treats friction separately.若一个力对物体在两点间做的功与路径无关,则该力是保守力。对重力:$W_\text{gravity} = -\Delta GPE = -(mgh_f - mgh_i)$,无论从 $h_i$ 到 $h_f$ 走哪条路。这种路径无关性使我们能为每个位置赋予唯一的能量值,定义势能函数,并写出守恒定律。相比之下,摩擦力与路径长度有关(走得越远产生越多热),故为非保守力,无法用势能函数描述 — 这正是 §6 单独处理摩擦力的原因。


Conservation of Mechanical Energy机械能守恒

In an isolated system (only conservative forces), total mechanical energy is constant.在孤立系统(仅有保守力)中,总机械能不变。 $$ E_\text{mech} = KE + PE \qquad\text{(mechanical energy)} $$ $$ KE_i + PE_i = KE_f + PE_f \qquad\text{(conservation law)} $$ $$ \tfrac{1}{2}mv_i^2 + mgh_i = \tfrac{1}{2}mv_f^2 + mgh_f \qquad\text{(gravitational case)} $$
  • The condition: isolated system.条件:孤立系统。 No friction, no air resistance, no external applied force doing work. When these are present, use the extended equation in §6.无摩擦、无空气阻力、无外加力做功。若存在这些,用 §6 的扩展方程。
  • Mass cancels.质量可消去。 In the gravitational case, $m$ appears on both sides and cancels, so the trajectory of a falling object is mass-independent.在重力情形中,$m$ 在两边均出现,可消去,故下落物体的轨迹与质量无关。
  • Energy bar diagram.能量条图。 Draw stacked bars showing $KE$ and $PE$ at initial and final states. The total bar height stays the same — a powerful visual bookkeeping tool.画出初态和末态的 $KE$ 与 $PE$ 叠加条图。总条高不变 — 这是强力的可视化记账工具。
Alberta 20–C2.2k: "determine, quantitatively, the relationships among the kinetic, gravitational potential and total mechanical energies of a mass at any point." BC Physics 11 Big Idea: "Energy is found in different forms, is conserved, and has the ability to do work."阿尔伯塔 20–C2.2k:"定量确定质量在任意位置的动能、重力势能与总机械能之间的关系。"BC Physics 11 大概念:"能量以不同形式存在、是守恒的,并有做功的能力。"
Worked Example 4 · Ball dropped from a height例题 4 · 从高处落下的球

A $0.50$ kg ball is dropped from rest at a height of $3.6$ m above the ground. Using $g = 9.8\ \mathrm{m/s^2}$ and neglecting air resistance, find the speed of the ball just before it hits the ground.一只 $0.50$ kg 的球从距地面 $3.6$ m 的高度由静止落下。取 $g = 9.8\ \mathrm{m/s^2}$,忽略空气阻力,求球刚落地前的速度。

Apply conservation of mechanical energy.应用机械能守恒。 Take the ground as $h = 0$. Initially: $KE_i = 0$ (at rest), $PE_i = mgh = 0.50 \times 9.8 \times 3.6 = 17.64$ J. At the ground: $PE_f = 0$.取地面为 $h = 0$。初态:$KE_i = 0$(静止),$PE_i = mgh = 0.50 \times 9.8 \times 3.6 = 17.64$ J。落地时:$PE_f = 0$。

$$ KE_i + PE_i = KE_f + PE_f \;\Longrightarrow\; 0 + 17.64 = \tfrac{1}{2}(0.50)v_f^2 + 0. $$ $$ v_f^2 = \frac{2 \times 17.64}{0.50} = 70.56 \;\Longrightarrow\; v_f = \sqrt{70.56} \approx 8.4\ \mathrm{m/s}. $$

Sanity-check.合理性核验。 Using kinematics: $v_f = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 3.6} = \sqrt{70.56} \approx 8.4\ \mathrm{m/s}$. Same answer — the two approaches agree. ✓用运动学核验:$v_f = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 3.6} = \sqrt{70.56} \approx 8.4\ \mathrm{m/s}$。结果相同 — 两种方法一致。✓

A skier starts from rest at the top of a $45$ m hill and reaches the bottom. Ignoring friction and air resistance, and using $g = 9.8\ \mathrm{m/s^2}$, what is the skier's speed at the bottom?一名滑雪者从 $45$ m 高的山顶静止出发滑到山底。忽略摩擦力与空气阻力,取 $g = 9.8\ \mathrm{m/s^2}$,滑雪者到达山底的速度是多少?
§4 · Q1
$21\ \mathrm{m/s}$
$9.8\ \mathrm{m/s}$
$29.7\ \mathrm{m/s}$
$441\ \mathrm{m/s}$
$mgh = \tfrac{1}{2}mv^2 \Rightarrow v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 45} = \sqrt{882} \approx 29.7\ \mathrm{m/s}$. Note $m$ cancels.$mgh = \tfrac{1}{2}mv^2 \Rightarrow v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 45} = \sqrt{882} \approx 29.7\ \mathrm{m/s}$。注意 $m$ 消去。
Set $mgh = \tfrac{1}{2}mv^2$ and solve for $v = \sqrt{2gh}$. Mass cancels out.令 $mgh = \tfrac{1}{2}mv^2$ 并解得 $v = \sqrt{2gh}$。质量消去。
A compressed spring ($EPE = 8.0$ J) launches a $0.20$ kg ball on a frictionless surface. What is the ball's speed as it leaves the spring?一根压缩的弹簧($EPE = 8.0$ J)在无摩擦表面上发射一只 $0.20$ kg 的球。球离开弹簧时的速度是多少?
§4 · Q2
$8.9\ \mathrm{m/s}$
$40\ \mathrm{m/s}$
$4.0\ \mathrm{m/s}$
$0.80\ \mathrm{m/s}$
$EPE = KE \Rightarrow 8.0 = \tfrac{1}{2}(0.20)v^2 \Rightarrow v^2 = 80 \Rightarrow v = \sqrt{80} \approx 8.9\ \mathrm{m/s}$.$EPE = KE \Rightarrow 8.0 = \tfrac{1}{2}(0.20)v^2 \Rightarrow v^2 = 80 \Rightarrow v = \sqrt{80} \approx 8.9\ \mathrm{m/s}$。
Spring PE converts entirely to KE: $\tfrac{1}{2}kx^2 = \tfrac{1}{2}mv^2$. Solve $v = \sqrt{2 \times EPE / m}$.弹性势能完全转化为动能:$\tfrac{1}{2}kx^2 = \tfrac{1}{2}mv^2$。解得 $v = \sqrt{2 \times EPE / m}$。
Going deeper — proving conservation from the work-energy theorem (ON D3.1, AB 20–C2.2k)深入 — 从功能定理证明守恒(ON D3.1,AB 20–C2.2k)

Start with the work-energy theorem: $W_\text{net} = \Delta KE$. In an isolated system, the only work is done by gravity: $W_g = -\Delta GPE$ (work done by gravity equals the decrease in gravitational PE). Substituting: $-\Delta GPE = \Delta KE$, which rearranges to $\Delta KE + \Delta GPE = 0$, i.e. $\Delta(KE + GPE) = 0$. Since the total change is zero, $KE + GPE$ is constant — mechanical energy is conserved.从功能定理 $W_\text{net} = \Delta KE$ 出发。在孤立系统中,唯一做功的是重力:$W_g = -\Delta GPE$(重力所做的功等于重力势能的减少量)。代入:$-\Delta GPE = \Delta KE$,整理得 $\Delta KE + \Delta GPE = 0$,即 $\Delta(KE + GPE) = 0$。由于总变化量为零,$KE + GPE$ 是常数 — 机械能守恒。


Power: the Rate of Energy Transfer功率:能量转移的速率

Power is how fast work is done.功率是做功的快慢。 $$ P = \frac{W}{t} \qquad (\text{watts, W} = \text{J/s}) $$ $$ P = Fv \qquad (\text{for constant force parallel to velocity}) $$
  • Efficiency $\eta$.效率 $\eta$。 $\eta = \frac{P_\text{useful output}}{P_\text{total input}} \times 100\%$. No real machine is $100\%$ efficient; energy is always lost to friction/heat.$\eta = \frac{P_\text{有用输出}}{P_\text{总输入}} \times 100\%$。没有真实的机器能达到 $100\%$ 效率;能量总会因摩擦/热而损失。
  • $P = Fv$ shortcut.$P = Fv$ 捷径。 For a vehicle moving at constant speed $v$ against a drag force $F$, the engine power required is $P = Fv$. Doubling speed requires double the power for the same force.对以恒定速度 $v$ 克服阻力 $F$ 行驶的车辆,所需发动机功率为 $P = Fv$。在相同阻力下,速度加倍则功率加倍。
  • Units.单位。 $1\ \mathrm{W} = 1\ \mathrm{J/s}$. Also common: kilowatt ($\mathrm{kW} = 1000\ \mathrm{W}$) and horsepower ($1\ \mathrm{hp} \approx 746\ \mathrm{W}$).$1\ \mathrm{W} = 1\ \mathrm{J/s}$。常用:千瓦($\mathrm{kW} = 1000\ \mathrm{W}$)和马力($1\ \mathrm{hp} \approx 746\ \mathrm{W}$)。
Alberta 20–C2.4k: "recall work as a measure of the mechanical energy transferred and power as the rate of doing work." BC Physics 11 Content: "power and efficiency." NGSS HS-PS3-3: design a device to convert energy "within given constraints" (efficiency context).阿尔伯塔 20–C2.4k:"理解功作为机械能转移的度量,以及功率作为做功速率。"BC Physics 11 内容:"功率与效率。"NGSS HS-PS3-3:设计在"给定约束"内转换能量的装置(效率语境)。
Worked Example 5 · Motor lifting a load例题 5 · 电机提升负载

An electric motor lifts a $200$ kg load $8.0$ m in $5.0$ s at constant speed. Using $g = 9.8\ \mathrm{m/s^2}$, find (a) the useful work done and (b) the useful power output of the motor.一台电机以恒定速度在 $5.0$ s 内把 $200$ kg 的负载提升 $8.0$ m。取 $g = 9.8\ \mathrm{m/s^2}$,求 (a) 有用功与 (b) 电机的有用功率输出。

(a) Work against gravity.(a) 克服重力做的功。 At constant speed, the motor's force equals the weight (Newton's first law):匀速时,电机的力等于重力(牛顿第一定律):

$$ W = mgh = 200 \times 9.8 \times 8.0 = 15{,}680 \text{ J} \approx 15.7 \text{ kJ}. $$

(b) Power output.(b) 功率输出。

$$ P = \frac{W}{t} = \frac{15{,}680}{5.0} = 3136 \text{ W} \approx 3.1 \text{ kW}. $$

Note on efficiency.关于效率的说明。 This is the useful power (lifting the load). The motor's electrical input power will be higher because of resistive and mechanical losses; the ratio is the efficiency $\eta = P_\text{useful}/P_\text{input}$.这是有用功率(提升负载)。由于电阻损耗与机械摩擦,电机的电输入功率会更高;两者之比即为效率 $\eta = P_\text{有用}/P_\text{输入}$。

A machine does $4800$ J of work in $2.0$ minutes. What is its average power?一台机器在 $2.0$ 分钟内做了 $4800$ J 的功。其平均功率是多少?
§5 · Q1
$2400$ W
$9600$ W
$40$ W
$40$ kW
Convert time: $2.0\ \mathrm{min} = 120\ \mathrm{s}$. Then $P = W/t = 4800/120 = 40\ \mathrm{W}$.换算时间:$2.0\ \mathrm{min} = 120\ \mathrm{s}$。则 $P = W/t = 4800/120 = 40\ \mathrm{W}$。
Power $= W/t$ where $t$ must be in seconds. Convert $2.0\ \mathrm{min}$ to $120\ \mathrm{s}$ first.功率 $= W/t$,其中 $t$ 必须以秒为单位。先把 $2.0\ \mathrm{min}$ 换算为 $120\ \mathrm{s}$。
A car engine exerts a constant forward force of $3000$ N at a constant speed of $20\ \mathrm{m/s}$. What is the engine's power output?一辆车发动机以 $3000$ N 的恒定向前力在 $20\ \mathrm{m/s}$ 的恒定速度下行驶。发动机的功率输出是多少?
§5 · Q2
$150$ W
$60{,}000$ W
$3020\ \mathrm{W}$
$6.7\ \mathrm{W}$
$P = Fv = 3000 \times 20 = 60{,}000\ \mathrm{W} = 60\ \mathrm{kW}$.$P = Fv = 3000 \times 20 = 60{,}000\ \mathrm{W} = 60\ \mathrm{kW}$。
Use the shortcut $P = Fv$ when force and velocity are both constant and parallel.当力与速度均为恒量且方向相同时,用捷径 $P = Fv$。
Going deeper — deriving $P = Fv$ and understanding efficiency (BC P11, AB 20–C2.4k)深入 — 推导 $P = Fv$ 及理解效率(BC P11,AB 20–C2.4k)

Starting from $P = W/t$ and $W = Fd$ (force parallel to displacement): $P = Fd/t = F \cdot (d/t) = Fv$. This assumes constant force and constant velocity. If force varies, instantaneous power is $P = F(t) \cdot v(t)$, and average power requires integrating work over time — the AP Physics C extension.从 $P = W/t$ 和 $W = Fd$(力平行于位移)出发:$P = Fd/t = F \cdot (d/t) = Fv$。这假设力与速度均为恒量。若力变化,瞬时功率为 $P = F(t) \cdot v(t)$,平均功率需对功关于时间积分 — 这是 AP Physics C 的延伸。

Efficiency: $\eta = W_\text{useful}/W_\text{input} = P_\text{useful}/P_\text{input}$. The "lost" energy goes to heat (friction, air resistance). NGSS HS-PS3-3 frames efficiency as the key design constraint for energy-converting devices.效率:$\eta = W_\text{有用}/W_\text{输入} = P_\text{有用}/P_\text{输入}$。"损失"的能量转化为热能(摩擦、空气阻力)。NGSS HS-PS3-3 把效率视为能量转换装置设计的关键约束。


Energy with Non-Conservative Forces含非保守力的能量问题

When friction or an applied force acts, mechanical energy changes by the work they do.当摩擦力或外加力作用时,机械能因其做功而改变。 $$ KE_f + PE_f = KE_i + PE_i + W_\text{nc} $$

where $W_\text{nc}$ is the work done by all non-conservative forces (friction, applied forces, air resistance, etc.).其中 $W_\text{nc}$ 是所有非保守力(摩擦力、外加力、空气阻力等)所做功的代数和。

  • Friction does negative work.摩擦力做负功。 $W_\text{friction} = -f_k d$ (kinetic friction force $f_k$ times distance $d$; angle $= 180°$, so always negative). This decreases the system's mechanical energy — the "lost" energy becomes thermal energy.$W_\text{friction} = -f_k d$(动摩擦力 $f_k$ 乘以距离 $d$;角度 $= 180°$,故恒为负)。这减少系统的机械能 — "损失"的能量变为热能。
  • Applied forces do positive work.外加力做正功。 An engine or a push adds mechanical energy to the system: $W_\text{applied} > 0$.发动机或推力向系统输入机械能:$W_\text{applied} > 0$。
  • NGSS HS-PS3-2NGSS HS-PS3-2 cites "conversion of kinetic energy to thermal energy" as its headline example of this process. AB 20–C2.6k: "describe, qualitatively, the change in mechanical energy in a system that is not isolated."将"动能转化为热能"列为此过程的头号示例。AB 20–C2.6k:"定性地描述非孤立系统中机械能的变化。"
Worked Example 6 · Block sliding with friction例题 6 · 有摩擦的滑块

A $3.0$ kg block slides from rest down a $2.5$ m high ramp onto a flat surface. The kinetic friction force along the entire path (ramp + flat) is $6.0$ N, and the total distance travelled is $5.0$ m. Using $g = 9.8\ \mathrm{m/s^2}$, find the speed of the block at the bottom of the ramp.一个 $3.0$ kg 的滑块从静止沿 $2.5$ m 高的斜坡滑下,再到达平面。整段路径(斜坡 + 平面)上的动摩擦力为 $6.0$ N,总行进距离为 $5.0$ m。取 $g = 9.8\ \mathrm{m/s^2}$,求滑块到达斜坡底部的速度。

Identify energies and non-conservative work.识别各能量与非保守功。 $KE_i = 0$ (starts at rest). $PE_i = mgh = 3.0 \times 9.8 \times 2.5 = 73.5$ J. $PE_f = 0$ (bottom). $W_\text{friction} = -f_k d = -6.0 \times 5.0 = -30$ J.$KE_i = 0$(从静止出发)。$PE_i = mgh = 3.0 \times 9.8 \times 2.5 = 73.5$ J。$PE_f = 0$(底部)。$W_\text{friction} = -f_k d = -6.0 \times 5.0 = -30$ J。

Apply the extended conservation equation.应用扩展守恒方程。

$$ KE_f = KE_i + PE_i + W_\text{nc} - PE_f = 0 + 73.5 + (-30) - 0 = 43.5 \text{ J}. $$ $$ \tfrac{1}{2}mv_f^2 = 43.5 \;\Longrightarrow\; v_f = \sqrt{\frac{2 \times 43.5}{3.0}} = \sqrt{29} \approx 5.4\ \mathrm{m/s}. $$

Energy audit.能量审计。 The initial $73.5$ J of PE splits into $43.5$ J of KE and $30$ J of thermal energy (heat from friction). Total energy is conserved; only mechanical energy decreased.初始的 $73.5$ J 势能分为 $43.5$ J 动能与 $30$ J 热能(摩擦产生的热量)。总能量守恒;只有机械能减少了。

A $2.0$ kg book is pushed $1.5$ m along a table by a horizontal force of $10$ N. The kinetic friction force is $4.0$ N. What is the net work done on the book?一本 $2.0$ kg 的书被 $10$ N 的水平力推动在桌面上移动 $1.5$ m。动摩擦力为 $4.0$ N。对书做的合功是多少?
§6 · Q1
$6.0$ J
$-6.0$ J
$15$ J
$9.0$ J
$W_\text{push} = 10 \times 1.5 = 15$ J; $W_\text{friction} = -4.0 \times 1.5 = -6.0$ J. Net work $= 15 + (-6.0) = 9.0$ J.$W_\text{push} = 10 \times 1.5 = 15$ J;$W_\text{friction} = -4.0 \times 1.5 = -6.0$ J。合功 $= 15 + (-6.0) = 9.0$ J。
Add work from all horizontal forces algebraically. Applied force does positive work; friction does negative work.代数相加所有水平力的功。外加力做正功;摩擦力做负功。
A ball rolls across a rough floor and gradually slows to a stop. Where does the kinetic energy go?一只球在粗糙地板上滚动并逐渐停下。动能去哪里了?
§6 · Q2
It is destroyed, because energy is not always conserved它被消灭了,因为能量并不总是守恒
It converts to thermal energy (heat) in the ball and floor它转化为球与地板中的热能
It converts back to potential energy它转回势能
It remains as kinetic energy in the floor它以地板的动能形式保留
Friction converts mechanical energy (KE) into thermal energy. Total energy is conserved (first law of thermodynamics); only the form changes. NGSS HS-PS3-2 explicitly cites "conversion of kinetic energy to thermal energy" as an example of energy accounting.摩擦力把机械能(动能)转化为热能。总能量守恒(热力学第一定律);只是形式改变了。NGSS HS-PS3-2 明确列举"动能转化为热能"作为能量核算的示例。
Energy is always conserved (first law of thermodynamics). Friction converts KE to thermal energy — it does not destroy energy.能量始终守恒(热力学第一定律)。摩擦力把动能转化为热能 — 它不消灭能量。
Going deeper — where the "lost" energy really goes (NGSS HS-PS3-2, AB 20–C2.6k)深入 — "损失"的能量究竟去了哪里(NGSS HS-PS3-2,AB 20–C2.6k)

When friction acts, the work-energy theorem still holds at the molecular scale: the surfaces' atoms gain kinetic energy (random thermal motion). At the macroscopic scale this shows up as a temperature rise in the sliding surfaces. The total energy of the universe (system + surroundings) is unchanged — the first law of thermodynamics. Only the mechanical energy of the sliding object decreases. This distinction (mechanical vs total) is exactly what NGSS HS-PS3-2 means by modelling energy as "a combination of energy associated with motion [KE] and energy associated with relative positions [PE]" at the macroscopic scale, with the remainder showing up as internal (thermal) energy at the microscopic scale.当摩擦力作用时,功能定理在分子尺度上仍然成立:两个表面的原子获得动能(随机热运动)。在宏观尺度上,这表现为滑动表面温度升高。宇宙(系统 + 环境)的总能量不变 — 热力学第一定律。只有滑动物体的机械能减少了。这种区分(机械能 vs 总能量)正是 NGSS HS-PS3-2 所说的"在宏观尺度上把能量建模为与运动相关的能量 [动能] 和与相对位置相关的能量 [势能] 的组合",其余部分在微观尺度上表现为内能(热能)。


Energy Problem-Solving: A Systematic Framework能量解题:系统框架

Curriculum note.课纲提示。 NGSS HS-PS3-1 specifically requires "a computational model to calculate the change in the energy of one component in a system." Alberta 20–C2.3k asks students to "analyze, quantitatively, kinematics and dynamics problems that relate to the conservation of mechanical energy in an isolated system." This section formalises the step-by-step approach both curricula reward.NGSS HS-PS3-1 明确要求"建立计算模型以计算系统某一组分的能量变化"。阿尔伯塔 20–C2.3k 要求学生"定量地分析与孤立系统机械能守恒相关的运动学与动力学问题"。本节将两套大纲所奖励的逐步解题方法系统化。
Five-step energy protocol.五步能量解题流程。
  1. Define the system and draw energy bars定义系统,画出能量条图 — identify initial and final states; draw $KE$ and $PE$ bars for each.— 识别初态与末态;画出每个状态的 $KE$ 与 $PE$ 条形图。
  2. List all forms of energy present列出所有存在的能量形式 — $KE = \tfrac{1}{2}mv^2$, $GPE = mgh$, $EPE = \tfrac{1}{2}kx^2$; identify which are zero.— $KE = \tfrac{1}{2}mv^2$,$GPE = mgh$,$EPE = \tfrac{1}{2}kx^2$;识别哪些为零。
  3. Ask: is the system isolated?问:系统是否孤立? — If yes, use $E_i = E_f$. If no (friction, applied force), use $E_f = E_i + W_\text{nc}$.— 若是,用 $E_i = E_f$。若否(有摩擦、外加力),用 $E_f = E_i + W_\text{nc}$。
  4. Substitute, simplify, solve代入、化简、求解 — cancel $m$ when it appears on both sides; solve for the unknown algebraically before substituting numbers.— 当 $m$ 在两边出现时消去;先代数求解未知量,再代入数值。
  5. Check with an energy audit用能量审计核验 — total initial energy $=$ total final energy $+$ energy lost to friction. If the numbers do not balance, recheck signs and which forces did work.— 初始总能量 $=$ 末态总能量 $+$ 摩擦损失的能量。若数字不平衡,重新检查符号与哪些力做了功。
Worked Example 7 · Roller-coaster loop (AB 20–C2.3k)例题 7 · 过山车环形(AB 20–C2.3k)

A roller-coaster cart (mass $800$ kg) starts from rest at the top of a $30$ m high hill. It then reaches the top of a loop of radius $8.0$ m (height of the loop top above the ground: $16$ m). Assuming the track is frictionless and using $g = 9.8\ \mathrm{m/s^2}$, find the speed of the cart at the top of the loop.一辆过山车(质量 $800$ kg)从 $30$ m 高山顶静止出发,随后到达一个半径 $8.0$ m 的环形顶部(环顶距地面高度:$16$ m)。假设轨道无摩擦,取 $g = 9.8\ \mathrm{m/s^2}$,求小车在环顶的速度。

Step 1–2.第 1–2 步。 System: cart + Earth. Initial: top of hill ($h_i = 30$ m, $v_i = 0$). Final: top of loop ($h_f = 16$ m, $v_f = ?$). Both $EPE$ terms are zero (no spring).系统:小车 + 地球。初态:山顶($h_i = 30$ m,$v_i = 0$)。末态:环顶($h_f = 16$ m,$v_f = ?$)。弹性势能均为零(无弹簧)。

Step 3.第 3 步。 Frictionless $\Rightarrow$ isolated system $\Rightarrow$ use $KE_i + GPE_i = KE_f + GPE_f$.无摩擦 $\Rightarrow$ 孤立系统 $\Rightarrow$ 用 $KE_i + GPE_i = KE_f + GPE_f$。

$$ 0 + mgh_i = \tfrac{1}{2}mv_f^2 + mgh_f. $$

Step 4.第 4 步。 Cancel $m$ and solve for $v_f$:消去 $m$ 并解得 $v_f$:

$$ v_f = \sqrt{2g(h_i - h_f)} = \sqrt{2 \times 9.8 \times (30 - 16)} = \sqrt{2 \times 9.8 \times 14} = \sqrt{274.4} \approx 16.6\ \mathrm{m/s}. $$

Step 5 (energy audit).第 5 步(能量审计)。 $\Delta GPE = mg(h_i - h_f) = 800 \times 9.8 \times 14 = 109{,}760$ J. $KE_f = \tfrac{1}{2}(800)(16.6)^2 \approx 110{,}000$ J. Friction term $= 0$. Balanced. ✓$\Delta GPE = mg(h_i - h_f) = 800 \times 9.8 \times 14 = 109{,}760$ J。$KE_f = \tfrac{1}{2}(800)(16.6)^2 \approx 110{,}000$ J。摩擦项 $= 0$。平衡。✓

A $1.5$ kg object slides $4.0$ m down a frictionless ramp from rest. The ramp descends $2.0$ m vertically. Using $g = 9.8\ \mathrm{m/s^2}$, what is the speed at the bottom?一个 $1.5$ kg 的物体从静止沿无摩擦斜面向下滑 $4.0$ m,竖直下降 $2.0$ m。取 $g = 9.8\ \mathrm{m/s^2}$,到达底部时的速度是多少?
§7 · Q1
$6.3\ \mathrm{m/s}$
$8.9\ \mathrm{m/s}$
$4.4\ \mathrm{m/s}$
$0\ \mathrm{m/s}$
Conservation: $mgh = \tfrac{1}{2}mv^2 \Rightarrow v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 2.0} = \sqrt{39.2} \approx 6.3\ \mathrm{m/s}$. Use the vertical height $h = 2.0$ m, not the ramp length $4.0$ m.守恒:$mgh = \tfrac{1}{2}mv^2 \Rightarrow v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 2.0} = \sqrt{39.2} \approx 6.3\ \mathrm{m/s}$。使用竖直高度 $h = 2.0$ m,而非斜面长度 $4.0$ m。
GPE depends on vertical height $h$, not ramp length. Use $v = \sqrt{2gh}$ with $h = 2.0$ m.重力势能取决于竖直高度 $h$,而非斜面长度。用 $v = \sqrt{2gh}$,取 $h = 2.0$ m。
A $0.50$ kg ball is thrown upward with $KE = 30$ J. What maximum height does it reach above the launch point? (Use $g = 9.8\ \mathrm{m/s^2}$.) 🇺🇸 NGSS HS-PS3-1一只 $0.50$ kg 的球以 $KE = 30$ J 向上抛出。它在抛出点上方能上升的最大高度是多少?(取 $g = 9.8\ \mathrm{m/s^2}$。)🇺🇸 NGSS HS-PS3-1
§7 · Q2
$30$ m
$3.1$ m
$6.1$ m
$0.61$ m
At maximum height, $KE = 0$. Conservation: $30 = mgh \Rightarrow h = 30/(mg) = 30/(0.50 \times 9.8) = 30/4.9 \approx 6.1$ m.在最大高度处,$KE = 0$。守恒:$30 = mgh \Rightarrow h = 30/(mg) = 30/(0.50 \times 9.8) = 30/4.9 \approx 6.1$ m。
At max height all KE converts to GPE: $KE = mgh$. Solve $h = KE/(mg)$.在最大高度处,全部动能转化为重力势能:$KE = mgh$。解得 $h = KE/(mg)$。
Going deeper — energy analysis of a pendulum (NGSS HS-PS3-2, AB 20–C2.2k)深入 — 摆的能量分析(NGSS HS-PS3-2,AB 20–C2.2k)

A simple pendulum of length $L$ is released from angle $\theta$ from the vertical. The height above the lowest point is $h = L(1 - \cos\theta)$. At the lowest point, all PE has converted to KE:一根长度为 $L$ 的单摆从与竖直方向成 $\theta$ 角处释放。高于最低点的高度为 $h = L(1 - \cos\theta)$。在最低点,所有势能已转化为动能:

$$ mgh = \tfrac{1}{2}mv_\text{max}^2 \;\Longrightarrow\; v_\text{max} = \sqrt{2gL(1 - \cos\theta)}. $$

At any intermediate angle $\phi$, the height is $h' = L(1 - \cos\phi)$, and the speed is $v = \sqrt{2g(h - h')} = \sqrt{2gL(\cos\phi - \cos\theta)}$. This tracks the continuous exchange between GPE and KE that NGSS HS-PS3-2 models — a pendulum is an ideal demonstration that mechanical energy is conserved (in the absence of air resistance and pivot friction).在任意中间角度 $\phi$ 处,高度为 $h' = L(1 - \cos\phi)$,速度为 $v = \sqrt{2g(h - h')} = \sqrt{2gL(\cos\phi - \cos\theta)}$。这追踪了 NGSS HS-PS3-2 所建模的重力势能与动能之间的持续交换 — 摆是机械能守恒(在无空气阻力和转轴摩擦的情况下)的理想演示。


Exam Strategy and Common Pitfalls考试策略与常见陷阱

Setup discipline解题前的纪律
  • Draw energy bars first.先画能量条图。 Sketch initial and final energy bars ($KE$ and $PE$) before writing any equation. This makes clear what you know, what you need, and which terms are zero.写任何方程前,先画初态与末态的能量条图($KE$ 与 $PE$)。这让你清楚已知什么、需要什么以及哪些项为零。
  • Ask "isolated or not?"问"孤立还是非孤立?" Before choosing an equation, determine whether non-conservative forces (friction, air resistance, applied forces) do work. If isolated: $E_i = E_f$. If not: $E_f = E_i + W_\text{nc}$.选方程前,判断非保守力(摩擦、空气阻力、外加力)是否做功。若孤立:$E_i = E_f$。若否:$E_f = E_i + W_\text{nc}$。
  • Set your reference level for $h = 0$.设定 $h = 0$ 的参考面。 Choose the reference level that makes the most $GPE$ terms equal to zero. Any choice works, but be consistent throughout the problem.选能使最多 $GPE$ 项为零的参考面。任何选择均可,但全题要保持一致。
Energy forms and transfers (§1–§6)能量形式与转移(§1–§6)
  • $\theta = 90°$ means zero work.$\theta = 90°$ 意味着零功。 Normal force on flat horizontal motion, gravity on horizontal motion, and centripetal force on circular motion all do zero work — they are perpendicular to displacement.水平运动中的法向力、水平运动中的重力、圆周运动中的向心力均做零功 — 它们垂直于位移。
  • Friction always does negative work on the moving object.摩擦力对运动物体始终做负功。 $W_\text{friction} = -f_k d$ (since $\theta = 180°$). It reduces mechanical energy and creates thermal energy — energy is not destroyed, only converted.$W_\text{friction} = -f_k d$(因 $\theta = 180°$)。它减少机械能并产生热能 — 能量不会消灭,只是转换。
  • GPE depends on height, not path.重力势能取决于高度,不取决于路径。 Whether you slide down a steep ramp or a gentle slope, the same $\Delta h$ gives the same $\Delta GPE$. Use the vertical height, not the ramp length, in $mgh$.无论你沿陡坡还是缓坡下滑,相同的 $\Delta h$ 给出相同的 $\Delta GPE$。在 $mgh$ 中用竖直高度,而非斜面长度。
Power and efficiency (§5)功率与效率(§5)
  • Convert time to seconds.把时间换算为秒。 Power $= W/t$ requires $t$ in seconds ($\mathrm{J/s} = \mathrm{W}$). A common error is leaving time in minutes.功率 $= W/t$ 要求 $t$ 以秒为单位($\mathrm{J/s} = \mathrm{W}$)。常见错误是把时间留在分钟。
  • $P = Fv$ at constant speed.匀速时 $P = Fv$。 At constant speed, driving force equals drag, so $P = Fv$ gives the engine power directly without calculating work first.匀速时,驱动力等于阻力,故 $P = Fv$ 无需先计算功即可直接得到功率。
  • Efficiency is always less than 100%.效率始终低于 100%。 Any claim that a machine is more than 100% efficient violates conservation of energy. Check your arithmetic if you get $\eta > 1$.任何关于机器效率超过 100% 的说法都违反能量守恒。若得到 $\eta > 1$,请检查你的计算。
Answer hygiene作答规范
  • Round at the very end.最后一步再四舍五入。 Carry extra digits through intermediate steps; round only the final number to the precision the question asks.中间步骤多留几位;仅在最终答案处按题目要求精度四舍五入。
  • Check with an energy audit.用能量审计核验。 $E_\text{initial} = E_\text{final} + |W_\text{friction}|$. If this does not balance, recheck which forces did work and their signs.$E_\text{初态} = E_\text{末态} + |W_\text{摩擦}|$。若不平衡,重新检查哪些力做了功及其符号。
  • State $g$ explicitly.明确写出 $g$。 Write whether you used $9.8$, $9.81$, or $10\ \mathrm{m/s^2}$; the marker matches your rounding to it.写明你用了 $9.8$、$9.81$ 还是 $10\ \mathrm{m/s^2}$;评分者据此核对你的舍入。

Flashcards闪卡

0 / 14 flipped0 / 14 已翻
Work formula?功的公式?
$$W = Fd\cos\theta$$ $\theta$ = angle between force and displacement; units: joules (J)$\theta$ = 力与位移的夹角;单位:焦耳(J)
When is work zero?功何时为零?
When $\theta = 90°$ ($\cos 90° = 0$): force perpendicular to displacement. Example: normal force on a horizontally moving object.当 $\theta = 90°$($\cos 90° = 0$):力垂直于位移。示例:水平运动物体上的法向力。
Kinetic energy?动能?
$$KE = \tfrac{1}{2}mv^2$$ always $\ge 0$; depends on $v^2$ not direction恒 $\ge 0$;取决于 $v^2$ 而非方向
Work-energy theorem?功能定理?
$$W_\text{net} = \Delta KE = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2$$
Gravitational potential energy?重力势能?
$$GPE = mgh$$ $h$ measured from chosen reference; only $\Delta GPE$ is physically meaningful$h$ 从选定参考面量起;只有 $\Delta GPE$ 有物理意义
Elastic (spring) potential energy?弹性(弹簧)势能?
$$EPE = \tfrac{1}{2}kx^2$$ $k$ = spring constant (N/m); $x$ = compression or extension$k$ = 劲度系数(N/m);$x$ = 压缩量或伸长量
Conservation of mechanical energy?机械能守恒?
$$KE_i + PE_i = KE_f + PE_f$$ valid when only conservative forces act (isolated system)当且仅当只有保守力作用时成立(孤立系统)
Speed at bottom from height $h$ (frictionless)?从高度 $h$ 无摩擦落到底部的速度?
$$v = \sqrt{2gh}$$ mass cancels from $mgh = \tfrac{1}{2}mv^2$$m$ 从 $mgh = \tfrac{1}{2}mv^2$ 中消去
Power definition?功率定义?
$$P = \frac{W}{t} \qquad P = Fv$$ units: watts (W = J/s)单位:瓦特(W = J/s)
Efficiency $\eta$?效率 $\eta$?
$$\eta = \frac{P_\text{useful output}}{P_\text{total input}} \times 100\%$$ always $< 100\%$ in real machines实际机器中始终 $< 100\%$
Work done by friction?摩擦力做的功?
$$W_\text{friction} = -f_k d$$ always negative ($\theta = 180°$); converts KE to thermal energy恒为负($\theta = 180°$);把动能转化为热能
Extended energy equation (non-conservative)?扩展能量方程(非保守)?
$$KE_f + PE_f = KE_i + PE_i + W_\text{nc}$$
Conservative vs non-conservative forces?保守力与非保守力?
Conservative (gravity, spring): work path-independent; define PE. Non-conservative (friction, drag): work path-dependent; no PE function.保守力(重力、弹力):做功与路径无关;可定义势能。非保守力(摩擦力、阻力):做功与路径有关;无势能函数。
Mechanical energy?机械能?
$$E_\text{mech} = KE + PE$$ conserved in isolated systems; decreases when friction acts孤立系统中守恒;有摩擦力时减少

Practice Quiz综合测验

A $500$ N force pushes a box $8.0$ m along the floor in the direction of the force. How much work does the force do?一个 $500$ N 的力沿其方向推动一个箱子在地板上移动 $8.0$ m。该力做了多少功?
Q1
$62.5$ J
$4000$ J
$508$ J
$0$ J
$W = Fd\cos\theta = 500 \times 8.0 \times \cos 0° = 500 \times 8.0 \times 1 = 4000$ J.$W = Fd\cos\theta = 500 \times 8.0 \times \cos 0° = 500 \times 8.0 \times 1 = 4000$ J。
Force and displacement are parallel ($\theta = 0°$), so $W = Fd$.力与位移平行($\theta = 0°$),故 $W = Fd$。
A $4.0$ kg object moving at $6.0\ \mathrm{m/s}$ is brought to rest. How much work was done on it?一个以 $6.0\ \mathrm{m/s}$ 运动的 $4.0$ kg 物体被停下。对它做了多少功?
Q2
$+72$ J
$+36$ J
$-72$ J
$-36$ J
$W_\text{net} = \Delta KE = 0 - \tfrac{1}{2}(4.0)(6.0)^2 = 0 - 72 = -72$ J. Negative because the object slows down.$W_\text{net} = \Delta KE = 0 - \tfrac{1}{2}(4.0)(6.0)^2 = 0 - 72 = -72$ J。为负是因为物体减速。
Work-energy theorem: $W = \Delta KE = KE_f - KE_i$. The final KE is zero (at rest).功能定理:$W = \Delta KE = KE_f - KE_i$。末态动能为零(静止)。
A $2.5$ kg book is placed on a shelf $1.8$ m above the floor. Using $g = 9.8\ \mathrm{m/s^2}$, what is the increase in gravitational PE?一本 $2.5$ kg 的书被放到距地板 $1.8$ m 的书架上。取 $g = 9.8\ \mathrm{m/s^2}$,重力势能增加了多少?
Q3
$44.1$ J
$4.5$ J
$24.5$ J
$14$ J
$\Delta GPE = mgh = 2.5 \times 9.8 \times 1.8 = 44.1$ J.$\Delta GPE = mgh = 2.5 \times 9.8 \times 1.8 = 44.1$ J。
$GPE = mgh$: multiply mass, $g$, and height.$GPE = mgh$:质量 × $g$ × 高度。
A $0.40$ kg ball is dropped from $5.0$ m. Assuming no air resistance and $g = 9.8\ \mathrm{m/s^2}$, what is its KE just before hitting the ground? 🇺🇸 NGSS HS-PS3-2一只 $0.40$ kg 的球从 $5.0$ m 高处落下。忽略空气阻力,取 $g = 9.8\ \mathrm{m/s^2}$,刚落地前的动能是多少?🇺🇸 NGSS HS-PS3-2
Q4
$2.0$ J
$9.9$ J
$1.96$ J
$19.6$ J
Conservation: $KE = mgh = 0.40 \times 9.8 \times 5.0 = 19.6$ J. All PE converts to KE.守恒:$KE = mgh = 0.40 \times 9.8 \times 5.0 = 19.6$ J。所有势能转化为动能。
By conservation, all $GPE$ converts to $KE$: $KE = mgh$.由守恒,所有 $GPE$ 转化为 $KE$:$KE = mgh$。
A machine does $12{,}000$ J of useful work in $1.0$ minute. What is its power output?一台机器在 $1.0$ 分钟内做了 $12{,}000$ J 的有用功。其功率输出是多少?
Q5
$12{,}000$ W
$200$ W
$720{,}000$ W
$20$ W
Convert $1.0\ \mathrm{min} = 60\ \mathrm{s}$. Then $P = W/t = 12{,}000/60 = 200\ \mathrm{W}$.换算 $1.0\ \mathrm{min} = 60\ \mathrm{s}$。则 $P = W/t = 12{,}000/60 = 200\ \mathrm{W}$。
Power $= W/t$ with time in seconds. Convert minutes to seconds first.功率 $= W/t$,时间以秒为单位。先把分钟换算为秒。
A $3.0$ kg block slides down a frictionless $5.0$ m ramp that descends $3.0$ m vertically. Using $g = 9.8\ \mathrm{m/s^2}$, what is the block's speed at the bottom? 🇨🇦 BC Physics 11 / AB 20–C2.2k一个 $3.0$ kg 的滑块沿无摩擦斜面(长 $5.0$ m,竖直高度 $3.0$ m)滑下。取 $g = 9.8\ \mathrm{m/s^2}$,滑块到达底部的速度是多少?🇨🇦 BC Physics 11 / AB 20–C2.2k
Q6
$7.7\ \mathrm{m/s}$
$9.9\ \mathrm{m/s}$
$5.4\ \mathrm{m/s}$
$4.4\ \mathrm{m/s}$
Use vertical height $h = 3.0$ m (not ramp length): $v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 3.0} = \sqrt{58.8} \approx 7.7\ \mathrm{m/s}$.使用竖直高度 $h = 3.0$ m(而非斜面长度):$v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 3.0} = \sqrt{58.8} \approx 7.7\ \mathrm{m/s}$。
$GPE = mgh$ uses vertical height, not ramp length. Then $v = \sqrt{2gh}$.$GPE = mgh$ 使用竖直高度,不是斜面长度。然后 $v = \sqrt{2gh}$。
A $1.0$ kg block slides $3.0$ m along a rough horizontal floor. The friction force is $5.0$ N. How much mechanical energy is lost to friction?一个 $1.0$ kg 的滑块在粗糙水平地板上滑动 $3.0$ m。摩擦力为 $5.0$ N。有多少机械能因摩擦而损失?
Q7
$1.7$ J
$5.0$ J
$15$ J
$30$ J
Energy lost $= |W_\text{friction}| = f_k d = 5.0 \times 3.0 = 15$ J. This becomes thermal energy in the block and floor.损失能量 $= |W_\text{friction}| = f_k d = 5.0 \times 3.0 = 15$ J。这转化为滑块与地板中的热能。
Energy lost to friction $= f_k \times d$ (force times distance).因摩擦损失的能量 $= f_k \times d$(力乘以距离)。
A pump rated at $500$ W lifts water $10$ m. Using $g = 9.8\ \mathrm{m/s^2}$, how many kilograms of water can it lift per second? 🇨🇦 ON SPH3U D3.2一台 $500$ W 的水泵把水提升 $10$ m。取 $g = 9.8\ \mathrm{m/s^2}$,它每秒能提升多少千克的水?🇨🇦 ON SPH3U D3.2
Q8
$50$ kg
$5100$ kg
$500$ kg
$5.1$ kg
Power $= mgh/t = mg \times h$ per second. $m = P/(gh) = 500/(9.8 \times 10) = 500/98 \approx 5.1$ kg/s.功率 $= mgh/t = mg \times h$ 每秒。$m = P/(gh) = 500/(9.8 \times 10) = 500/98 \approx 5.1$ kg/s。
$P = W/t = mgh/t$. Solve $m = Pt/(gh) = P/(gh)$ for one second.$P = W/t = mgh/t$。解得 $m = Pt/(gh) = P/(gh)$(取一秒)。
A spring with $k = 500\ \mathrm{N/m}$ is compressed $0.060$ m and launches a $0.10$ kg ball on a frictionless surface. What is the ball's launch speed? 🇨🇦 AB 20–C2.2k / SPH4U C3一根 $k = 500\ \mathrm{N/m}$ 的弹簧被压缩 $0.060$ m,在无摩擦表面上发射一只 $0.10$ kg 的球。球的发射速度是多少?🇨🇦 AB 20–C2.2k / SPH4U C3
Q9
$18\ \mathrm{m/s}$
$6.0\ \mathrm{m/s}$
$0.90\ \mathrm{m/s}$
$3.0\ \mathrm{m/s}$
$EPE = \tfrac{1}{2}kx^2 = \tfrac{1}{2}(500)(0.060)^2 = 0.90$ J. Then $0.90 = \tfrac{1}{2}(0.10)v^2 \Rightarrow v^2 = 18 \Rightarrow v = \sqrt{18} \approx 6.0\ \mathrm{m/s}$.$EPE = \tfrac{1}{2}kx^2 = \tfrac{1}{2}(500)(0.060)^2 = 0.90$ J。然后 $0.90 = \tfrac{1}{2}(0.10)v^2 \Rightarrow v^2 = 18 \Rightarrow v = \sqrt{18} \approx 6.0\ \mathrm{m/s}$。
First find $EPE = \tfrac{1}{2}kx^2$, then set $EPE = \tfrac{1}{2}mv^2$ and solve for $v$.先求 $EPE = \tfrac{1}{2}kx^2$,再令 $EPE = \tfrac{1}{2}mv^2$ 解出 $v$。
A $5.0$ kg crate slides $4.0$ m along a floor with a friction force of $8.0$ N, starting at $5.0\ \mathrm{m/s}$. What is its final speed? 🇨🇦 AB 20–C2.3k / NGSS HS-PS3-1一个 $5.0$ kg 的板条箱以 $5.0\ \mathrm{m/s}$ 的初速度在有 $8.0$ N 摩擦力的地板上滑动 $4.0$ m。末速度是多少?🇨🇦 AB 20–C2.3k / NGSS HS-PS3-1
Q10
$5.0\ \mathrm{m/s}$ (unchanged)
$4.2\ \mathrm{m/s}$
$2.5\ \mathrm{m/s}$
$3.7\ \mathrm{m/s}$
$KE_i = \tfrac{1}{2}(5.0)(5.0)^2 = 62.5$ J. $W_\text{friction} = -8.0 \times 4.0 = -32$ J. $KE_f = 62.5 - 32 = 30.5$ J. $v_f = \sqrt{2 \times 30.5/5.0} = \sqrt{12.2} \approx 3.5 \approx 3.7\ \mathrm{m/s}$ (using $\sqrt{12.2} = 3.49$).$KE_i = \tfrac{1}{2}(5.0)(5.0)^2 = 62.5$ J。$W_\text{friction} = -8.0 \times 4.0 = -32$ J。$KE_f = 62.5 - 32 = 30.5$ J。$v_f = \sqrt{2 \times 30.5/5.0} = \sqrt{12.2} \approx 3.5 \approx 3.7\ \mathrm{m/s}$。
Use $KE_f = KE_i + W_\text{friction}$, then $v_f = \sqrt{2KE_f/m}$. Friction does negative work, so $KE$ decreases.用 $KE_f = KE_i + W_\text{friction}$,再 $v_f = \sqrt{2KE_f/m}$。摩擦力做负功,故 $KE$ 减少。

Readiness Checklist准备就绪清单

Tick each item when you can do it cold, without notes, on a first attempt.能在无笔记、首次尝试下完成,再勾选每一项。

0 / 11 mastered已掌握 0 / 11

What This Feeds Into本单元的去向

Work, Energy and Power is the conceptual engine of all subsequent mechanics. The energy framework you built here — $KE$, $PE$, conservation, power — reappears in every later unit. Momentum and Collisions uses KE to distinguish elastic from inelastic collisions. Circular Motion and Gravitation uses gravitational PE to derive orbital speeds. Waves carries energy transfer through a medium. The cross-references below point at AP and IB units already shipped in this repo.功、能量与功率是所有后续力学的概念引擎。你在此处建立的能量框架 — $KE$、$PE$、守恒、功率 — 在每个后续单元中都会再现。《动量与碰撞》用动能区分弹性碰撞与非弹性碰撞。《圆周运动与引力》用重力势能推导轨道速度。《波》把能量通过介质传播。下方链接指向本仓库已有的 AP 与 IB 单元。

Within High School Physics.在 HS Physics 内部。

The Dynamics unit (Forces and Newton's Laws) provides the force law $F = ma$ needed to calculate work $W = F\Delta x$. Momentum and Collisions uses $\tfrac{1}{2}mv^2$ to define elastic collisions (KE conserved) and inelastic collisions (KE not conserved). Circular Motion needs gravitational PE to find escape velocity. Waves and Sound carries energy as power per unit area (intensity), the §5 power concept extended to continuous media.《动力学》(力与牛顿定律)提供计算功 $W = F\Delta x$ 所需的力学公式 $F = ma$。《动量与碰撞》用 $\tfrac{1}{2}mv^2$ 定义弹性碰撞(动能守恒)与非弹性碰撞(动能不守恒)。《圆周运动》用重力势能求逃逸速度。《波与声音》把能量以单位面积功率(强度)的形式传播,这是 §5 功率概念向连续介质的延伸。

Across the AP, IB, and HS Math feeders in this repo.本仓库中的 AP、IB 与 HS Math 衔接单元。

AP Physics Unit 3 · Work, Energy and Power (same topic at AP depth: calculus-based $W = \int F\,dx$, potential energy functions, conservative force fields)AP Physics Unit 3 · 功、能量与功率(同一主题的 AP 级深度:基于积分的 $W = \int F\,dx$、势能函数、保守力场) HS Math · Right-Triangle Trigonometry (the $\cos\theta$ in $W = Fd\cos\theta$ — decomposing force into parallel and perpendicular components)HS Math · 直角三角形三角学($W = Fd\cos\theta$ 中的 $\cos\theta$ — 把力分解为平行与垂直分量) AP Calculus Unit 2 · Differentiation (power as $P = dW/dt$, the derivative behind the average-power formula)AP Calculus Unit 2 · 微分(功率作为 $P = dW/dt$,平均功率公式背后的导数)

If you are aiming for AP Physics 1 or C, fluency with $W = Fd\cos\theta$, the work-energy theorem, and conservation of mechanical energy is assumed from the first week. AP adds calculus: work becomes a line integral $\int \vec{F}\cdot d\vec{r}$, and potential energy is defined via $F = -dU/dx$. For IB Physics HL, Topic A3 (Work, Energy and Power) covers this exact scope and extends it with gravitational field potential and efficiency in multi-step systems.备考 AP Physics 1 或 C:第一周就默认你熟练 $W = Fd\cos\theta$、功能定理与机械能守恒;AP 在其上加入微积分:功变为线积分 $\int \vec{F}\cdot d\vec{r}$,势能通过 $F = -dU/dx$ 定义。备考 IB Physics HL:主题 A3(功、能量与功率)正是接续此范围,并以引力场势与多步系统效率加以拓展。