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Solutions详解

Forces and Newton's Laws · Solutions力与牛顿定律 · 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 · 22 marksAP 选择题 + 安/卑省考短答 · 共 22 分

Section A · Worked SolutionsA 部分 · 详细解答

Q1EASY易 🇺🇸 US美 AP-style MCQAP 风格选择题 §4 Third law第三定律 · HS-PS2-1 [3 marks][3 分]

A book rests on a table. Which force is the third-law reaction partner of the book's weight?一本书静放在桌面上。哪个力是书的重力的第三定律反作用配对力？

Answer:答案：  (B)  the book's gravitational pull up on the Earth书向上对地球的引力

(a) Identify the third-law partner by swapping the two objects通过对调两个物体来确定第三定律配对力 M1·A1·A1

A Newton's-third-law pair acts on two different objects and is the same type of force. The given force is "Earth pulls the book down (gravity)." Swap the agent and the object: its partner must be "the book pulls the Earth up (gravity)," equal in magnitude and opposite in direction. That is option (B).牛顿第三定律的一对力作用在两个不同的物体上，且为同一类型的力。所给的力是"地球向下拉书（引力）"。对调施力者与受力者：其配对力必为"书向上拉地球（引力）"，大小相等、方向相反。即选项 (B)。
Why the distractors fail.干扰项分析。
(A): the normal force balances the weight (both act on the book) but is a contact force from the table, not a gravitational pair. It is the classic "equal but not a third-law pair" trap.法向力与重力平衡（二者都作用于书），但它是桌面的接触力，并非引力配对。这是经典的"等大却非第三定律对"陷阱。
(C): the table's weight is an unrelated force on a different object.桌子的重力是作用在另一物体上的无关力。
(D): this is the partner of the table's normal force, not of gravity.这是桌面法向力的配对力，而非重力的配对力。
Third-law pairs act on different objects; balanced forces act on the same object.第三定律的一对力作用于不同物体；平衡力作用于同一物体。 The deepest trap in this unit is calling the normal force the "reaction" to weight. They are equal here only because the book is in equilibrium ($N = mg$ from the first law), but they act on the same object (the book) and are different types of force, so they are not a third-law pair. The true partner of any force is found by reversing the sentence "A pushes/pulls B" into "B pushes/pulls A": the gravitational pull of Earth on the book is partnered by the gravitational pull of the book on Earth. The pair is always equal, opposite, same type, and on two different bodies.本单元最深的陷阱是把法向力当作重力的"反作用"。它们在此相等只因书处于平衡（由第一定律 $N = mg$），但二者作用于同一物体（书）且类型不同，故不是第三定律对。任何力的真正配对力，都是把"A 推/拉 B"这句话反过来变成"B 推/拉 A"：地球对书的引力，配对的是书对地球的引力。这一对力永远等大、反向、同类型，且作用在两个不同物体上。
Q2EASY易 🇺🇸 US美 AP-style MCQAP 风格选择题 §3 Second law ($F=ma$)第二定律（$F=ma$） · HS-PS2-1 [3 marks][3 分]

Net force $36$ N on a $4.0$ kg cart. Magnitude of acceleration?$36$ N 净力作用于 $4.0$ kg 小车。加速度大小为多少？

Answer:答案：  (A)  $9.0\ \text{m/s}^2$

(a) Rearrange Newton's second law to $a = F_{net}/m$把牛顿第二定律变形为 $a = F_{net}/m$ M1·A1·A1

$$ a \;=\; \frac{F_{net}}{m} \;=\; \frac{36}{4.0} \;=\; 9.0 \;\text{m/s}^2. $$ The acceleration points in the direction of the net force. Option (A).加速度沿净力方向。选 (A)。
Why the distractors fail.干扰项分析。
(B) $144\ \text{m/s}^2$: multiplies $F \times m$ instead of dividing.把 $F \times m$ 相乘，而非相除。
(C) $0.11\ \text{m/s}^2$: inverts the ratio to $m/F$.把比值颠倒为 $m/F$。
(D) $40\ \text{m/s}^2$: adds $36 + 4$ instead of applying $F = ma$.把 $36 + 4$ 相加，而非套用 $F = ma$。
Newton's second law links net force, mass and acceleration: $a = F_{net}/m$.牛顿第二定律联系净力、质量与加速度：$a = F_{net}/m$。 This is the literal content of NGSS HS-PS2-1. The word "net" matters: if several forces act, reduce them to a single resultant before dividing by mass. Acceleration is inversely proportional to mass, so for a fixed force a heavier object accelerates less, the quantitative face of inertia. Always check units: $\text{N}/\text{kg} = \text{m/s}^2$, which confirms the formula is the right way up.这正是 NGSS HS-PS2-1 的字面内容。"净"字很关键：若有多个力作用，须先把它们归并为一个合力，再除以质量。加速度与质量成反比，故固定力作用下更重的物体加速更小，这是惯性的定量体现。务必核对单位：$\text{N}/\text{kg} = \text{m/s}^2$，证实公式方向正确。
Q3EASY易 🇨🇦 ON安 ON Provincial-style安大略省考风格 §1 FBD自由体受力图 + §2 First law第一定律 · SPH3U C2 [5 marks][5 分]

$25$ kg crate dragged at constant $1.5$ m/s by horizontal rope. (a) FBD, four forces. (b) Net force + first-law justification. (c) Friction force when rope pulls with $40$ N.$25$ kg 板条箱被水平绳以恒定 $1.5$ m/s 拖动。(a) 自由体受力图，四个力。(b) 净力 + 第一定律论证。(c) 绳以 $40$ N 拉时的摩擦力。

Answer:答案：  (a) weight $mg$ down, normal $N$ up, tension $T$ forward, friction $f$ backward重力 $mg$ 向下、法向力 $N$ 向上、张力 $T$ 向前、摩擦 $f$ 向后  ·  (b) $F_{net} = 0$  ·  (c) $f = 40\ \text{N}$ backward向后

(a) Free-body diagram: the four forces on the crate自由体受力图：箱子受的四个力 A1·A1

Draw the crate as a single dot with four arrows: weight $mg$ pointing straight down, the normal force $N$ from the floor pointing straight up, the rope tension $T$ pointing forward (horizontal), and kinetic friction $f$ pointing backward (opposing the motion).把箱子画成一个点，配四支箭头：重力 $mg$ 竖直向下、地面法向力 $N$ 竖直向上、绳张力 $T$ 水平向前、动摩擦 $f$ 水平向后（与运动方向相反）。

(b) Net force is zero by the first law由第一定律净力为零 M1·A1

The crate moves at constant velocity, so $\vec{a} = 0$. By Newton's first law, constant velocity means the net force is zero:箱子以恒定速度运动，故 $\vec{a} = 0$。由牛顿第一定律，匀速意味着净力为零： $$ \vec{F}_{net} = 0 \quad (\text{vertically } N = mg, \text{ horizontally } T = f). $$

(c) Friction balances the pull摩擦力与拉力平衡 A1

Horizontally $T - f = 0$, so $f = T = 40\ \text{N}$, directed backward (opposite the motion).水平方向 $T - f = 0$，故 $f = T = 40\ \text{N}$，方向向后（与运动方向相反）。
Constant velocity is an equilibrium state: zero net force, not zero force.匀速是一种平衡状态：净力为零，而非没有力。 The crate is moving, yet its acceleration is zero, so the first law applies exactly as it does to an object at rest. Students often assume motion requires a net forward force; it does not. Here the $40\ \text{N}$ pull exists only to cancel the $40\ \text{N}$ of kinetic friction. Note the mass and speed ($25$ kg, $1.5$ m/s) are not needed for the friction force, they are distractors, because at constant velocity friction is set by the horizontal balance, not by $\mu N$ in this part. Drawing the free-body diagram first makes the two independent axis balances ($N = mg$ and $T = f$) obvious.箱子在运动，但加速度为零，故第一定律对它的适用与对静止物体完全相同。学生常以为运动需要净向前力；其实不然。此处 $40\ \text{N}$ 拉力只是为抵消 $40\ \text{N}$ 动摩擦。注意质量与速率（$25$ kg、$1.5$ m/s）对求摩擦力并不需要，是干扰项，因为匀速时摩擦由水平方向的平衡决定，而非本小问中的 $\mu N$。先画自由体受力图，能让两个独立坐标轴的平衡（$N = mg$ 与 $T = f$）一目了然。
Q4MEDIUM中 🇺🇸 US美 AP-style MCQAP 风格选择题 §5 Friction (static vs kinetic)摩擦力（静摩擦与动摩擦） · HS-PS2-1 [4 marks][4 分]

$15$ kg box, $\mu_s = 0.40$, push $50$ N, box does not move. Friction force? ($g = 9.8\ \text{m/s}^2$)$15$ kg 箱子，$\mu_s = 0.40$，推力 $50$ N，箱子不动。摩擦力为多少？（$g = 9.8\ \text{m/s}^2$）

Answer:答案：  (B)  $50\ \text{N}$

(a) Find the static limit, then apply the equilibrium condition先求静摩擦上限，再用平衡条件 M1·A1·A1

First the maximum static friction:先求最大静摩擦： $$ f_{s,\max} \;=\; \mu_s N \;=\; \mu_s mg \;=\; 0.40 \times 15 \times 9.8 \;=\; 58.8 \;\text{N}. $$ The push ($50$ N) is less than $f_{s,\max}$ ($58.8$ N), so the box stays at rest. Static friction adjusts to match the push exactly: with $a = 0$, the horizontal balance gives $f_s = 50\ \text{N}$. Option (B).推力（$50$ N）小于 $f_{s,\max}$（$58.8$ N），故箱子保持静止。静摩擦自动调整以恰好匹配推力：由 $a = 0$，水平平衡给出 $f_s = 50\ \text{N}$。选 (B)。
Why the distractors fail.干扰项分析。
(A) $58.8\ \text{N}$: this is the maximum static friction $f_{s,\max}$, the value friction would reach only at the instant of slipping, not its actual value here.这是最大静摩擦 $f_{s,\max}$，仅在即将滑动的瞬间才达到，并非此处的实际值。
(C) $6.0\ \text{N}$: computes $\mu_s \times$ (push) instead of $\mu_s N$.算成 $\mu_s \times$（推力），而非 $\mu_s N$。
(D) $0\ \text{N}$: wrongly assumes a stationary object has no friction; friction is exactly what keeps it stationary.错误地认为静止物体无摩擦；恰恰是摩擦使它保持静止。
Static friction is an inequality $f_s \le \mu_s N$; it equals the applied push, up to its maximum.静摩擦是不等式 $f_s \le \mu_s N$；它等于外加推力，至多达到其最大值。 The single most common friction error is to plug $\mu_s N$ in whenever a coefficient appears. But $\mu_s N$ is only the ceiling: while the object is stationary, static friction self-adjusts to exactly balance the applied force, so $f_s = F_{push}$ as long as $F_{push} \le f_{s,\max}$. Only at the threshold of motion does $f_s = \mu_s N$. The correct procedure is always: compute $f_{s,\max}$, compare with the push, and only if the push wins does the object slide and kinetic friction $f_k = \mu_k N$ take over.最常见的摩擦错误，就是一看到系数就代入 $\mu_s N$。但 $\mu_s N$ 只是上限：物体静止时，静摩擦自动调整以恰好平衡外加力，故只要 $F_{push} \le f_{s,\max}$，就有 $f_s = F_{push}$。只有在即将运动的临界点才有 $f_s = \mu_s N$。正确步骤永远是：算出 $f_{s,\max}$，与推力比较，只有推力胜出物体才滑动，由动摩擦 $f_k = \mu_k N$ 接管。
Q5MEDIUM中 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §3 Weight vs mass · apparent weight重力与质量 · 视重 · Physics 11 (elevators) [6 marks][6 分]

$60$ kg student on a scale in an elevator. ($g = 9.8\ \text{m/s}^2$) (a) Mass, weight, and the difference. (b) Scale reading when accelerating up at $1.5\ \text{m/s}^2$. (c) Reading at constant velocity.$60$ kg 学生站在电梯内体重秤上。（$g = 9.8\ \text{m/s}^2$）(a) 质量、重力及其区别。(b) 以 $1.5\ \text{m/s}^2$ 向上加速时的读数。(c) 匀速时的读数。

Answer:答案：  (a) $m = 60\ \text{kg}$, $W = 588\ \text{N}$  ·  (b) $N = 678\ \text{N}$  ·  (c) $588\ \text{N}$

(a) Mass, weight, and the distinction质量、重力及其区别 M1·A1

Mass is $m = 60\ \text{kg}$ (a measure of inertia, the same everywhere). Weight is the gravitational force on that mass:质量为 $m = 60\ \text{kg}$（惯性的量度，各处相同）。重力是作用于该质量的引力： $$ W \;=\; mg \;=\; 60 \times 9.8 \;=\; 588 \;\text{N}. $$ Mass (kg) does not change with location; weight (N) does, because it depends on $g$.质量（kg）不随地点改变；重力（N）会，因为它取决于 $g$。

(b) Scale reading $=$ normal force; apply $\vec{F}_{net} = m\vec{a}$ vertically秤读数 $=$ 法向力；竖直应用 $\vec{F}_{net} = m\vec{a}$ M1·A1·A1

Take up as positive. The forces on the student are the scale's normal force $N$ up and weight $mg$ down. With upward acceleration $a$:取向上为正。学生受力为秤的法向力 $N$ 向上、重力 $mg$ 向下。以向上加速度 $a$： $$ N - mg \;=\; ma \;\Longrightarrow\; N \;=\; m(g + a) \;=\; 60(9.8 + 1.5) \;=\; 60 \times 11.3 \;=\; 678 \;\text{N}. $$ The scale reads $678\ \text{N}$, greater than the true weight, the student feels "heavier."秤读数为 $678\ \text{N}$，大于真实重力，学生感觉"更重"。

(c) Constant velocity means equilibrium匀速即平衡 A1

At constant velocity $a = 0$, so $N = mg = 588\ \text{N}$, the scale reads the true weight.匀速时 $a = 0$，故 $N = mg = 588\ \text{N}$，秤显示真实重力。
Apparent weight is the normal force $N = m(g \pm a)$, not the true weight $mg$.视重是法向力 $N = m(g \pm a)$，而非真实重力 $mg$。 A scale never reads weight directly, it reads the force it pushes back with, which is the normal force $N$. In an inertial situation ($a = 0$) that equals $mg$, but under acceleration the first law no longer applies and $N$ shifts: $N = m(g+a)$ accelerating up, $N = m(g-a)$ accelerating down, and $N = 0$ in free fall (weightlessness). The student's true weight $588\ \text{N}$ and mass $60\ \text{kg}$ never change during the ride, only the apparent weight does. Keeping mass and weight separate, and remembering the scale reads $N$, resolves every elevator problem in BC Physics 11.秤从不直接显示重力，它显示自己回推的力，即法向力 $N$。在惯性情形（$a = 0$）下它等于 $mg$，但加速时第一定律不再适用，$N$ 随之改变：向上加速 $N = m(g+a)$，向下加速 $N = m(g-a)$，自由落体时 $N = 0$（失重）。乘梯过程中学生的真实重力 $588\ \text{N}$ 与质量 $60\ \text{kg}$ 始终不变，改变的只有视重。把质量与重力分开，并记住秤读的是 $N$，便能解决 BC Physics 11 中的每道电梯问题。
PART II  ·  EXTENDED RESPONSE · SOLUTIONS第二部分  ·  简答题 · 详解AP-feeder FRQ + honors · 32 marksAP 衔接简答题 + 荣誉级 · 共 32 分

Section B · Worked SolutionsB 部分 · 详细解答

Q6MEDIUM中 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §5 Friction摩擦力 + §3 Second law第二定律 · HS-PS2-1 [8 marks][8 分]

$8.0$ kg box, $\mu_s = 0.50$, $\mu_k = 0.30$, push $50$ N. ($g = 9.8\ \text{m/s}^2$) (a) $N$ and $f_{s,\max}$. (b) Does it slide? (c) Kinetic friction. (d) Acceleration.$8.0$ kg 箱子，$\mu_s = 0.50$，$\mu_k = 0.30$，推力 $50$ N。（$g = 9.8\ \text{m/s}^2$）(a) $N$ 与 $f_{s,\max}$。(b) 是否滑动？(c) 动摩擦。(d) 加速度。

Answer:答案：  (a) $N = 78.4\ \text{N}$, $f_{s,\max} = 39.2\ \text{N}$  ·  (b) yes, it slides是，会滑动  ·  (c) $f_k = 23.52\ \text{N}$  ·  (d) $a \approx 3.31\ \text{m/s}^2$

(a) Normal force and the static limit法向力与静摩擦上限 M1·A1

On level ground the vertical balance gives $N = mg$:平地上竖直平衡给出 $N = mg$： $$ N = mg = 8.0 \times 9.8 = 78.4\ \text{N}, \qquad f_{s,\max} = \mu_s N = 0.50 \times 78.4 = 39.2\ \text{N}. $$

(b) Compare the push with the static limit将推力与静摩擦上限比较 M1·A1

The push $50\ \text{N} > f_{s,\max} = 39.2\ \text{N}$, so static friction cannot hold the box. It breaks free and slides.推力 $50\ \text{N} > f_{s,\max} = 39.2\ \text{N}$，故静摩擦无法保持箱子静止。它挣脱并滑动。

(c) Kinetic friction once sliding滑动后的动摩擦 M1·A1

$$ f_k = \mu_k N = 0.30 \times 78.4 = 23.52\ \text{N (opposing motion).} $$

(d) Acceleration from $\vec{F}_{net} = m\vec{a}$由 $\vec{F}_{net} = m\vec{a}$ 求加速度 M1·A1

$$ a = \frac{F_{push} - f_k}{m} = \frac{50 - 23.52}{8.0} = \frac{26.48}{8.0} \approx 3.31\ \text{m/s}^2. $$
Test against $f_{s,\max}$ first; only after the object slides does $f_k = \mu_k N$ apply.先与 $f_{s,\max}$ 比较；只有物体滑动后才用 $f_k = \mu_k N$。 This problem chains the two faces of friction. The decisive step is (b): you cannot find an acceleration until you know the box is moving. Had the push been below $39.2\ \text{N}$, the answer would be $a = 0$ and friction would equal the push, not $\mu_s N$. Because $\mu_k < \mu_s$, the friction drops the instant motion starts (from a $39.2\ \text{N}$ ceiling to a steady $23.52\ \text{N}$), which is why a pushed object often "jerks" forward once it breaks free. Always reduce to a single net force ($50 - 23.52$) before dividing by mass.本题串联了摩擦的两副面孔。决定性的一步是 (b)：在确认箱子运动之前无法求加速度。若推力低于 $39.2\ \text{N}$，答案将是 $a = 0$，且摩擦等于推力，而非 $\mu_s N$。由于 $\mu_k < \mu_s$，运动一开始摩擦就骤降（从 $39.2\ \text{N}$ 上限降到稳定的 $23.52\ \text{N}$），这正是被推物体挣脱瞬间常"猛地"前冲的原因。务必先归并为单个净力（$50 - 23.52$），再除以质量。
Q7MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §6 Inclined plane (frictionless)斜面（无摩擦） · SPH3U C3 [7 marks][7 分]

$5.0$ kg block released from rest on a frictionless $30^{\circ}$ ramp. ($g = 9.8\ \text{m/s}^2$, $\sin 30^{\circ} = 0.50$, $\cos 30^{\circ} \approx 0.866$) (a) Normal force. (b) Acceleration. (c) Speed after $8.0$ m. (d) Effect of doubling the mass.$5.0$ kg 木块在无摩擦 $30^{\circ}$ 斜面上由静止释放。（$g = 9.8\ \text{m/s}^2$，$\sin 30^{\circ} = 0.50$，$\cos 30^{\circ} \approx 0.866$）(a) 法向力。(b) 加速度。(c) 下滑 $8.0$ m 后的速率。(d) 质量加倍的影响。

Answer:答案：  (a) $N \approx 42.4\ \text{N}$  ·  (b) $a = 4.9\ \text{m/s}^2$  ·  (c) $v \approx 8.85\ \text{m/s}$  ·  (d) no change不变

(a) Normal force perpendicular to the slope垂直坡面的法向力 M1·A1

Tilt the axes ($x$ down the slope, $y$ into the surface). No acceleration into the surface, so $N$ balances the perpendicular weight component:倾斜坐标轴（$x$ 沿坡向下，$y$ 垂直坡面）。垂直坡面无加速度，故 $N$ 平衡重力的垂直分量： $$ N = mg\cos\theta = 5.0 \times 9.8 \times 0.866 \approx 42.4\ \text{N}. $$

(b) Acceleration along the slope沿坡的加速度 M1·A1

Along the slope only gravity's parallel component acts (no friction): $mg\sin\theta = ma$, so沿坡只有重力的平行分量作用（无摩擦）：$mg\sin\theta = ma$，故 $$ a = g\sin\theta = 9.8 \times 0.50 = 4.9\ \text{m/s}^2 \text{ (down the slope).} $$

(c) Speed after $8.0$ m using $v^2 = u^2 + 2as$用 $v^2 = u^2 + 2as$ 求 $8.0$ m 后的速率 M1·A1

$$ v = \sqrt{2as} = \sqrt{2 \times 4.9 \times 8.0} = \sqrt{78.4} \approx 8.85\ \text{m/s}. $$

(d) Effect of doubling the mass质量加倍的影响 A1

The acceleration $a = g\sin\theta$ has no mass in it, so doubling the mass leaves $a$ (and the speed after $8.0$ m) unchanged.加速度 $a = g\sin\theta$ 不含质量，故质量加倍后 $a$（以及 $8.0$ m 后的速率）保持不变。
On a frictionless incline the mass cancels: $a = g\sin\theta$, independent of mass.无摩擦斜面上质量约去：$a = g\sin\theta$，与质量无关。 The key move is tilting the axes so $x$ runs down the slope. Gravity then splits into $mg\sin\theta$ (driving, along $x$) and $mg\cos\theta$ (pressing, along $y$), and the normal force satisfies $N = mg\cos\theta$, which is less than the full weight $mg$. Along the slope, Newton's second law $mg\sin\theta = ma$ has the mass on both sides; it cancels, leaving the incline analogue of free-fall's mass-independence. This is exactly why (d) is "no change," and it generalises to every frictionless ramp regardless of the block.关键一步是倾斜坐标轴使 $x$ 沿坡向下。重力随之分解为 $mg\sin\theta$（驱动，沿 $x$）与 $mg\cos\theta$（压向坡面，沿 $y$），法向力满足 $N = mg\cos\theta$，小于完整重力 $mg$。沿坡方向，牛顿第二定律 $mg\sin\theta = ma$ 两边都有质量；约去后得到斜面版的自由落体质量无关性。这正是 (d) 为"不变"的原因，并对任何无摩擦斜面都成立，与木块无关。
Q8HARD难Honors荣誉级 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §6 Incline + friction斜面 + 摩擦 · HS-PS2-1 (above 1D floor)（超出一维基准） [8 marks][8 分]

$6.0$ kg block slides down a rough $30^{\circ}$ ramp, $\mu_k = 0.20$. ($g = 9.8\ \text{m/s}^2$, $\sin 30^{\circ} = 0.50$, $\cos 30^{\circ} \approx 0.866$) (a) Normal force. (b) Kinetic friction + direction. (c) Acceleration down slope. (d) Compare with frictionless $4.9\ \text{m/s}^2$.$6.0$ kg 木块沿粗糙 $30^{\circ}$ 斜面下滑，$\mu_k = 0.20$。（$g = 9.8\ \text{m/s}^2$，$\sin 30^{\circ} = 0.50$，$\cos 30^{\circ} \approx 0.866$）(a) 法向力。(b) 动摩擦力及方向。(c) 沿坡向下的加速度。(d) 与无摩擦值 $4.9\ \text{m/s}^2$ 比较。

Answer:答案：  (a) $N \approx 50.9\ \text{N}$  ·  (b) $f_k \approx 10.2\ \text{N}$ up the slope沿坡向上  ·  (c) $a \approx 3.2\ \text{m/s}^2$  ·  (d) smaller; friction reduces it较小；摩擦使其减小

(a) Normal force perpendicular to the slope垂直坡面的法向力 M1·A1

Tilt the axes. Perpendicular to the slope there is no acceleration, so倾斜坐标轴。垂直坡面方向无加速度，故 $$ N = mg\cos\theta = 6.0 \times 9.8 \times 0.866 \approx 50.9\ \text{N}. $$

(b) Kinetic friction force动摩擦力 M1·A1

$$ f_k = \mu_k N = 0.20 \times 50.9 \approx 10.2\ \text{N}. $$ It points up the slope, opposing the downhill motion.它沿坡向上，与下滑运动方向相反。

(c) Acceleration along the slope from $\vec{F}_{net} = m\vec{a}$由 $\vec{F}_{net} = m\vec{a}$ 求沿坡加速度 M1·A1·A1

Along the slope, gravity drives ($mg\sin\theta$) and friction opposes ($f_k$):沿坡方向，重力驱动（$mg\sin\theta$），摩擦反向（$f_k$）： $$ ma = mg\sin\theta - f_k = 6.0(9.8)(0.50) - 10.2 = 29.4 - 10.2 = 19.2\ \text{N}, $$ $$ a = \frac{19.2}{6.0} \approx 3.2\ \text{m/s}^2 \text{ (down the slope).} $$

(d) Comparison with the frictionless case与无摩擦情形比较 A1

The rough-ramp value $3.2\ \text{m/s}^2$ is smaller than the frictionless $4.9\ \text{m/s}^2$; friction removes $g\mu_k\cos\theta \approx 1.7\ \text{m/s}^2$ of the acceleration.粗糙斜面值 $3.2\ \text{m/s}^2$ 小于无摩擦值 $4.9\ \text{m/s}^2$；摩擦减去了 $g\mu_k\cos\theta \approx 1.7\ \text{m/s}^2$ 的加速度。
Resolve gravity onto tilted axes; the rough-incline acceleration is $a = g(\sin\theta - \mu_k\cos\theta)$, still mass-independent.把重力分解到倾斜坐标轴；粗糙斜面加速度为 $a = g(\sin\theta - \mu_k\cos\theta)$，仍与质量无关。 Writing the second law symbolically, $ma = mg\sin\theta - \mu_k mg\cos\theta$, the mass cancels on both sides to give $a = g(\sin\theta - \mu_k\cos\theta) = 9.8(0.50 - 0.20 \times 0.866) \approx 3.2\ \text{m/s}^2$, confirming the numerical answer. The friction term carries a $\cos\theta$ because friction scales with $N = mg\cos\theta$, not with the full weight. This is the single most important incline result for AP and IB feeder work: every ramp problem with friction reduces to this one formula once you have tilted the axes and resolved the weight into its parallel and perpendicular components.把第二定律写成符号形式 $ma = mg\sin\theta - \mu_k mg\cos\theta$，两边质量约去，得 $a = g(\sin\theta - \mu_k\cos\theta) = 9.8(0.50 - 0.20 \times 0.866) \approx 3.2\ \text{m/s}^2$，验证了数值答案。摩擦项带有 $\cos\theta$，因为摩擦随 $N = mg\cos\theta$ 缩放，而非随完整重力。这是 AP 与 IB 衔接学习中最重要的斜面结论：一旦倾斜坐标轴并把重力分解为平行与垂直分量，每道含摩擦的斜面题都归结为这一个公式。
Q9HARD难Honors荣誉级 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §7 Connected system连接系统 · HS-PS2-1 (above 1D floor)（超出一维基准） [9 marks][9 分]

$m_A = 3.0$ kg on a frictionless table, rope over a pulley to hanging $m_B = 2.0$ kg, released from rest. ($g = 9.8\ \text{m/s}^2$) (a) FBDs. (b) Acceleration. (c) Tension. (d) Why $T < m_B g$.$m_A = 3.0$ kg 在无摩擦桌面上，绳经滑轮连到悬挂的 $m_B = 2.0$ kg，从静止释放。（$g = 9.8\ \text{m/s}^2$）(a) 受力图。(b) 加速度。(c) 张力。(d) 为何 $T < m_B g$。

Answer:答案：  (b) $a = 3.92\ \text{m/s}^2$  ·  (c) $T = 11.76\ \text{N}$  ·  (d) $B$ accelerates downward, so $T < m_B g$$B$ 向下加速，故 $T < m_B g$

(a) Free-body diagrams for each block每个木块的自由体受力图 A1·A1

Block $A$ (on the table): weight $m_A g$ down, normal $N$ up, tension $T$ horizontal toward the pulley. Block $B$ (hanging): weight $m_B g$ down, tension $T$ up. Both share the same acceleration magnitude $a$ since the rope is inextensible.木块 $A$（桌面上）：重力 $m_A g$ 向下、法向力 $N$ 向上、张力 $T$ 水平指向滑轮。木块 $B$（悬挂）：重力 $m_B g$ 向下、张力 $T$ 向上。因绳不可伸长，二者共享相同的加速度大小 $a$。

(b) Acceleration by treating the two blocks as one system把两个木块视为一个系统求加速度 M1·A1·A1

The only external force driving the system is the weight of $B$ ($m_B g$); tension is internal and cancels. The total inertia is $m_A + m_B$:驱动系统的唯一外力是 $B$ 的重力（$m_B g$）；张力是内力，相互抵消。总惯性为 $m_A + m_B$： $$ a = \frac{m_B g}{m_A + m_B} = \frac{2.0 \times 9.8}{3.0 + 2.0} = \frac{19.6}{5.0} = 3.92\ \text{m/s}^2. $$

(c) Tension from one block's equation由单个木块方程求张力 M1·A1

Apply $\vec{F}_{net} = m\vec{a}$ to block $A$ alone (horizontal): the only horizontal force is $T$, so单独对木块 $A$ 应用 $\vec{F}_{net} = m\vec{a}$（水平方向）：唯一的水平力是 $T$，故 $$ T = m_A a = 3.0 \times 3.92 = 11.76\ \text{N}. $$ Check with $B$: $m_B g - T = 2.0(9.8) - 11.76 = 7.84 = m_B a = 2.0 \times 3.92$. $\checkmark$用 $B$ 验证：$m_B g - T = 2.0(9.8) - 11.76 = 7.84 = m_B a = 2.0 \times 3.92$。$\checkmark$

(d) Why the tension is less than $B$'s weight为何张力小于 $B$ 的重力 M1·A1

Block $B$ accelerates downward, so its net force points down: $m_B g - T = m_B a > 0$, which forces $T < m_B g$. If $T$ equalled the weight, $B$ would be in equilibrium and would not accelerate.木块 $B$ 向下加速，故其净力向下：$m_B g - T = m_B a > 0$，从而迫使 $T < m_B g$。若 $T$ 等于重力，$B$ 将处于平衡而不会加速。
For connected bodies: solve the whole system for $a$, then return to one body for the internal tension.对连接物体：先对整个系统求 $a$，再回到单个物体求内部张力。 The two-step method is the cleanest route through every connected-body problem. Treating $A + B$ as one mass eliminates the unknown tension immediately, because internal forces cancel for the whole system, giving $a$ in one line. Then any single block's equation yields $T$. The result $T < m_B g$ is general: a hanging mass that is accelerating down must have an upward tension smaller than its weight, while the block being pulled along the table feels exactly that tension. Always finish with the cross-check (the other block's equation must give the same $a$), which catches sign and algebra slips.两步法是贯通每道连接物体问题的最简洁路径。把 $A + B$ 当作一个质量可立即消去未知张力，因为内力对整个系统抵消，一行即得 $a$。随后任一木块的方程即给出 $T$。结论 $T < m_B g$ 具有普遍性：向下加速的悬挂质量，其向上张力必小于自身重力，而被拉过桌面的木块恰好感受这一张力。务必以交叉验证收尾（另一木块的方程须给出相同的 $a$），它能捕捉符号与代数失误。
PART III  ·  MODELING / APPLIED · SOLUTIONS第三部分  ·  建模与应用 · 详解AB Diploma + Universal · 25 marks阿省毕业考 + 通用题型 · 共 25 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 Friction (applied)摩擦力（应用） · 20-B1.5k [8 marks][8 分]

$25$ kg sled pulled by constant horizontal $60$ N, $\mu_k = 0.15$, starts from rest. ($g = 9.8\ \text{m/s}^2$) (a) Kinetic friction. (b) Acceleration. (c) Distance in first $4.0$ s. (d) Effect of reducing the pull to $36.75$ N.$25$ kg 雪橇被 $60$ N 的恒定水平力拉动，$\mu_k = 0.15$，从静止出发。（$g = 9.8\ \text{m/s}^2$）(a) 动摩擦力。(b) 加速度。(c) 前 $4.0$ s 内的距离。(d) 拉力减小到 $36.75$ N 的影响。

Answer:答案：  (a) $f_k = 36.75\ \text{N}$  ·  (b) $a = 0.93\ \text{m/s}^2$  ·  (c) $7.44\ \text{m}$  ·  (d) net force becomes zero; constant velocity净力变为零；匀速运动

(a) Kinetic friction from $f_k = \mu_k N = \mu_k mg$由 $f_k = \mu_k N = \mu_k mg$ 求动摩擦力 M1·A1

On level ground the vertical balance gives $N = mg$, so平地上竖直平衡给出 $N = mg$，故 $$ f_k \;=\; \mu_k mg \;=\; 0.15 \times 25 \times 9.8 \;=\; 36.75 \;\text{N (opposing motion).} $$

(b) Acceleration from $\vec{F}_{net} = m\vec{a}$由 $\vec{F}_{net} = m\vec{a}$ 求加速度 M1·A1·A1

Horizontally the pull drives and friction opposes:水平方向拉力驱动、摩擦反向： $$ a \;=\; \frac{F - f_k}{m} \;=\; \frac{60 - 36.75}{25} \;=\; \frac{23.25}{25} \;=\; 0.93 \;\text{m/s}^2. $$

(c) Distance in the first $4.0$ s from rest从静止起前 $4.0$ s 内的距离 M1·A1

$$ s \;=\; \tfrac{1}{2}at^2 \;=\; \tfrac{1}{2}(0.93)(4.0)^2 \;=\; \tfrac{1}{2}(0.93)(16) \;=\; 7.44 \;\text{m.} $$

(d) Reducing the pull to $36.75$ N将拉力减小到 $36.75$ N A1

The pull now equals the kinetic friction, so the net force is zero ($a = 0$): a sled already moving continues at constant velocity instead of speeding up.此时拉力恰好等于动摩擦力，故净力为零（$a = 0$）：已在运动的雪橇将匀速前进，不再加速。
On level ground the driving force fights only kinetic friction; the net of the two, divided by mass, is the acceleration.平地上驱动力只对抗动摩擦力；二者之净力除以质量即加速度。 This is the everyday face of Newton's second law: a steady horizontal pull does not give a steady speed, it gives a steady acceleration, because friction is constant ($f_k = \mu_k mg$) once the sled slides. The decisive step is reducing the two horizontal forces to a single net force ($60 - 36.75 = 23.25\ \text{N}$) before dividing by the mass. Part (d) is the equilibrium boundary: when the pull drops to exactly $f_k$, the net force vanishes and the first law takes over, so a moving sled coasts at constant velocity. Pull any less and friction wins, slowing the sled to a stop.这是牛顿第二定律最日常的一面：稳定的水平拉力不会带来稳定的速度，而是稳定的加速度，因为雪橇一旦滑动，摩擦就保持恒定（$f_k = \mu_k mg$）。决定性的一步是先把两个水平力归并为单个净力（$60 - 36.75 = 23.25\ \text{N}$），再除以质量。(d) 是平衡的临界点：当拉力恰好降到 $f_k$，净力消失，第一定律接管，运动中的雪橇便匀速滑行。拉力再小一点，摩擦取胜，雪橇减速直至停下。
Q11MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §6 Incline + friction (applied)斜面 + 摩擦（应用） · 20-B1.7k [8 marks][8 分]

$60$ kg skier from rest down a $30^{\circ}$ slope, $\mu_k = 0.10$. ($g = 9.8\ \text{m/s}^2$, $\sin 30^{\circ} = 0.50$, $\cos 30^{\circ} \approx 0.866$) (a) Normal force. (b) Acceleration. (c) Speed after $40$ m. (d) One-sentence conclusion.$60$ kg 滑雪者从静止沿 $30^{\circ}$ 坡下滑，$\mu_k = 0.10$。（$g = 9.8\ \text{m/s}^2$，$\sin 30^{\circ} = 0.50$，$\cos 30^{\circ} \approx 0.866$）(a) 法向力。(b) 加速度。(c) 下滑 $40$ m 后的速率。(d) 一句话总结。

Answer:答案：  (a) $N \approx 509\ \text{N}$  ·  (b) $a \approx 4.05\ \text{m/s}^2$  ·  (c) $v = 18\ \text{m/s}$  ·  (d) speeds up steadily down the slope沿坡稳定加速下滑

(a) Normal force perpendicular to the slope垂直坡面的法向力 M1·A1

Tilt the axes ($x$ down the slope). Perpendicular to the slope there is no acceleration, so倾斜坐标轴（$x$ 沿坡向下）。垂直坡面无加速度，故 $$ N \;=\; mg\cos\theta \;=\; 60 \times 9.8 \times 0.866 \;\approx\; 509\ \text{N.} $$

(b) Acceleration from $\vec{F}_{net} = m\vec{a}$ along the slope沿坡由 $\vec{F}_{net} = m\vec{a}$ 求加速度 M1·A1·A1

Gravity drives ($mg\sin\theta$), kinetic friction opposes ($\mu_k mg\cos\theta$). The mass cancels:重力驱动（$mg\sin\theta$），动摩擦反向（$\mu_k mg\cos\theta$）。质量约去： $$ a \;=\; g(\sin\theta - \mu_k\cos\theta) \;=\; 9.8(0.50 - 0.10 \times 0.866) \;=\; 9.8 \times 0.4134 \;\approx\; 4.05\ \text{m/s}^2. $$

(c) Speed after $40$ m using $v^2 = u^2 + 2as$用 $v^2 = u^2 + 2as$ 求 $40$ m 后的速率 M1·A1

$$ v \;=\; \sqrt{2as} \;=\; \sqrt{2 \times 4.05 \times 40} \;=\; \sqrt{324} \;=\; 18 \;\text{m/s.} $$

(d) Conclusion in context结合情境的结论 A1

The skier accelerates uniformly down the slope at about $4.05\ \text{m/s}^2$, reaching $18$ m/s after $40$ m.滑雪者以约 $4.05\ \text{m/s}^2$ 沿坡匀加速下滑，下滑 $40$ m 后达到 $18$ m/s。
A rough incline gives $a = g(\sin\theta - \mu_k\cos\theta)$, still mass-independent; the friction term carries $\cos\theta$ because $f_k = \mu_k N = \mu_k mg\cos\theta$.粗糙斜面给出 $a = g(\sin\theta - \mu_k\cos\theta)$，仍与质量无关；摩擦项带 $\cos\theta$，因为 $f_k = \mu_k N = \mu_k mg\cos\theta$。 Writing the slope equation symbolically, $ma = mg\sin\theta - \mu_k mg\cos\theta$, the mass cancels on both sides, so the $60$ kg figure never enters the acceleration, it only sets the normal force and friction magnitude. The driving term uses $\sin\theta$ (the slope-parallel weight component) and the resisting term uses $\cos\theta$ (because friction scales with the perpendicular normal force). Once Newton's second law has given the acceleration, the speed after a given distance follows in one line from $v^2 = 2as$. This single formula handles every rough-ramp problem in AB Physics 20, AP, and IB feeder work.把沿坡方程写成符号形式 $ma = mg\sin\theta - \mu_k mg\cos\theta$，两边质量约去，故 $60$ kg 从不进入加速度，它只决定法向力与摩擦力的大小。驱动项用 $\sin\theta$（沿坡的重力分量），阻力项用 $\cos\theta$（因为摩擦随垂直方向的法向力缩放）。牛顿第二定律求得加速度后，给定距离后的速率由 $v^2 = 2as$ 一步得出。这一个公式可应对阿省物理 20、AP 与 IB 衔接学习中的每道粗糙斜面题。
Q12HARD难 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §7 Atwood system阿特伍德系统 · HS-PS2-1 [9 marks][9 分]

Atwood machine over a frictionless pulley: $m_1 = 5.0$ kg and $m_2 = 3.0$ kg, released from rest. ($g = 9.8\ \text{m/s}^2$) (a) FBDs + Newton's second law for each. (b) Acceleration. (c) Tension. (d) Why $T$ lies between $m_2 g$ and $m_1 g$.跨过无摩擦滑轮的阿特伍德机：$m_1 = 5.0$ kg，$m_2 = 3.0$ kg，从静止释放。（$g = 9.8\ \text{m/s}^2$）(a) 受力图 + 各自的牛顿第二定律。(b) 加速度。(c) 张力。(d) 为何 $T$ 介于 $m_2 g$ 与 $m_1 g$ 之间。

Answer:答案：  (b) $a = 2.45\ \text{m/s}^2$  ·  (c) $T = 36.75\ \text{N}$  ·  (d) $m_2 g = 29.4\ \text{N} < T < m_1 g = 49\ \text{N}$$m_2 g = 29.4\ \text{N} < T < m_1 g = 49\ \text{N}$

(a) Free-body diagrams and Newton's second law for each mass每个质量块的受力图与牛顿第二定律 A1·A1

Each mass feels its weight down and the rope tension $T$ up. Take the heavier side ($m_1$) as moving down. With a common acceleration magnitude $a$ (inextensible rope):每个质量块都受向下的重力与向上的绳张力 $T$。取较重的一侧（$m_1$）向下运动。因绳不可伸长，二者共享加速度大小 $a$： $$ m_1g - T = m_1 a \qquad (\text{heavier, down}), \qquad T - m_2 g = m_2 a \qquad (\text{lighter, up}). $$

(b) Acceleration by adding the two equations两式相加求加速度 M1·A1·A1

Adding eliminates $T$ (it is internal to the system):相加可消去 $T$（它是系统内力）： $$ a = \frac{(m_1 - m_2)g}{m_1 + m_2} = \frac{(5.0 - 3.0)(9.8)}{5.0 + 3.0} = \frac{19.6}{8.0} = 2.45\ \text{m/s}^2. $$

(c) Tension from one mass's equation由单个质量块方程求张力 M1·A1

Use the lighter mass ($T - m_2 g = m_2 a$):用较轻的质量块（$T - m_2 g = m_2 a$）： $$ T = m_2(g + a) = 3.0(9.8 + 2.45) = 3.0 \times 12.25 = 36.75\ \text{N}. $$ Check with $m_1$: $m_1(g - a) = 5.0(9.8 - 2.45) = 5.0 \times 7.35 = 36.75\ \text{N}$. $\checkmark$用 $m_1$ 验证：$m_1(g - a) = 5.0(9.8 - 2.45) = 5.0 \times 7.35 = 36.75\ \text{N}$。$\checkmark$

(d) Why the tension lies between the two weights为何张力介于两个重力之间 M1·A1

The lighter mass accelerates up, so $T > m_2 g = 29.4\ \text{N}$. The heavier mass accelerates down, so $T < m_1 g = 49\ \text{N}$. Hence $29.4\ \text{N} < T < 49\ \text{N}$, and indeed $T = 36.75\ \text{N}$.较轻的质量块向上加速，故 $T > m_2 g = 29.4\ \text{N}$。较重的质量块向下加速，故 $T < m_1 g = 49\ \text{N}$。因此 $29.4\ \text{N} < T < 49\ \text{N}$，确有 $T = 36.75\ \text{N}$。
For an Atwood machine: the system acceleration is $a = (m_1 - m_2)g/(m_1 + m_2)$, and the single rope tension must lie between the two weights.对阿特伍德机：系统加速度为 $a = (m_1 - m_2)g/(m_1 + m_2)$，唯一的绳张力必介于两个重力之间。 Treating the two masses as one system makes the tension cancel and gives $a$ in one line; only the difference of the weights drives the motion while the sum of the masses sets the inertia. The tension is the same throughout an ideal (massless) rope over a frictionless pulley, so a single $T$ appears in both equations. It cannot equal either weight: if it did, that mass would be in equilibrium and nothing would accelerate. The cross-check (both single-mass equations give the same $T = 36.75\ \text{N}$) is the fastest way to catch a sign slip. As $m_1 \to m_2$, $a \to 0$ and $T \to mg$, recovering static equilibrium.把两个质量块当作一个系统可使张力抵消，一行即得 $a$；驱动运动的只是两重力之差，而决定惯性的是质量之和。理想（无质量）绳跨过无摩擦滑轮时张力处处相等，故两式中只出现一个 $T$。它不可能等于任一重力：若相等，该质量块将处于平衡，便无物体加速。交叉验证（两个单质量方程给出相同的 $T = 36.75\ \text{N}$）是捕捉符号失误最快的办法。当 $m_1 \to m_2$ 时，$a \to 0$、$T \to mg$，回到静态平衡。