AP物理1 功能定理

AP物理1· AP物理1 课程与考试说明（CED）——能量部分· 14 分钟阅读

1. 功能定理的核心定义★★☆☆☆⏱ 3 min

功能定理（也称为功-动能定理，是AP物理1中使用的标准同义表述）是第4单元的核心关系，它将对物体做的功与物体的动能变化关联起来。该定理适用于所有刚体，无论作用力是恒力还是变力，在求解力-位移-速率相关问题时，它通常是比运动学更高效的替代方法。该考点占AP物理1考试总分的20%~25%，在选择题（MCQ）和自由问答题（FRQ）部分均有涉及。

📘 定义

功能定理

W_{\text{net}} = \Delta K = K_f - K_i

作用在刚体上的所有力对物体做的总功，等于该物体的动能变化量，其中 $W_{net}$ 为总功， $K_{f}$ 为末动能， $K_{i}$ 为初动能，所有物理量的单位均为焦耳（J）。

2. 推导过程与核心原理★★★☆☆⏱ 3 min

功能定理并非独立的物理定律，它可以直接由牛顿第二定律和匀加速运动学推导得出，通过对无穷小位移求和，还可以将其适用范围拓展到变力场景。

🔬 推导

目标:

推导恒总力场景下的功能定理

起点:

匀加速运动学与牛顿第二定律

1
从匀加速 $a$ 对应的运动学关系式出发：
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$v_f^2 = v_i^2 + 2a\Delta x$
3
整理公式，解出加速度 $a$ ：
4
$a = \frac{v_f^2 - v_i^2}{2\Delta x}$
5
将其代入牛顿第二定律 $F_{net} = ma$ ：
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$F_{\text{net}} = m\left(\frac{v_f^2 - v_i^2}{2\Delta x}\right)$
7
将等式两边同时乘以位移 $Δ x$ ：
8
$F_{\text{net}}\Delta x = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$
9
等式左侧为总功 $W_{net}$ ，右侧为动能变化量 $Δ K$ ：

结果:

由此得到核心公式 $W_{net} = Δ K$ ，该公式甚至对变力场景也成立。

📐 例题

一名质量为60 kg的滑冰者在无摩擦的水平冰面上滑行，初速率为2.0 m/s。同伴推了滑冰者一下，对滑冰者做的总功为120 J。求滑冰者的末速率是多少？

1
1. 列出所有已知量：
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$m = 60\ \text{kg}, v_i = 2.0\ \text{m/s}, W_{\text{net}} = 120\ \text{J}$
3
1. 写出功能定理表达式：
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$W_{\text{net}} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$
5
1. 整理公式，分离出待求量 $v_{f}$ ：
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$v_f = \sqrt{\frac{2W_{\text{net}}}{m} + v_i^2}$
7
1. 代入数值计算：
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$v_f = \sqrt{\frac{2(120)}{60} + (2.0)^2} = \sqrt{8} \approx 2.8\ \text{m/s}$
9
1. 核验：正的总功会使速率增大，符合常识，因此结果是合理的。

Exam tip:

求解末速率或初速率时，使用功能定理无需先计算加速度，可在选择题中大幅节省解题时间。遇到相关问题时，请优先考虑使用该定理，再尝试使用匀加速运动学公式。

3. 多力场景下的总功计算★★★☆☆⏱ 4 min

应用该定理时最常见的错误是总功计算错误：仅计入了题目提到的某一个力的功，而没有计入所有力的功。总功的计算有两种等价方法，在AP考试中均被认可：

Net force first: Calculate the vector sum of all forces to get $F_{net}$ , then calculate work as $W_{net} = F_{net} Δ x cos θ$ , where $θ$ is the angle between the net force and displacement.
Sum of individual works: Calculate work done by each force separately, then add all work values, keeping track of signs. This is almost always easier for multiple-force problems, since any force perpendicular to displacement automatically contributes zero work and drops out.

功的标准符号约定：如果力存在与位移同向的分量，功为正；如果力的分量与位移反向，功为负；如果力与位移垂直，功为零。摩擦力和空气阻力几乎始终做负功，而水平面上的支持力始终做功为零。

📐 例题

一个质量为5.0 kg的物块从长度为3.0 m、倾角为30°的斜面顶端由静止释放，物块与斜面之间的动摩擦因数为0.20。使用功能定理求解物块滑到斜面底端时的速率。

1
1. 识别所有受力：重力（ $m g$ ）、支持力（ $N$ ）、动摩擦力（ $f_{k}$ ）。支持力与位移方向垂直，因此 $W_{N} = 0$ 。
2
1. 计算重力做的功：竖直下落高度 $Δ y = 3.0 sin 3 0^{\circ} = 1.5 m$ ：
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$W_g = mg\Delta y = (5.0)(9.8)(1.5) = 73.5\ \text{J}$
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1. 计算摩擦力做的功： $N = m g cos 3 0^{\circ} \approx 42.4 N$ ， $f_{k} = μ_{k} N = 8.48 N$ 。摩擦力阻碍运动，因此：
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$W_f = -f_k \Delta x = -(8.48)(3.0) \approx -25.4\ \text{J}$
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1. 计算总功： $W_{net} = 73.5 + 0 - 25.4 = 48.1 J$
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1. 应用动能定理：初始 $K_{i} = 0$ ，因此：
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$v_f = \sqrt{\frac{2W_{\text{net}}}{m}} = \sqrt{\frac{2(48.1)}{5.0}} \approx 4.4\ \text{m/s}$

Exam tip:

如果斜面静止在地面上，无论斜面倾角为多少，支持力做的功始终为零，因为支持力始终与物体沿斜面滑动的位移方向垂直。这一结论可以让你在大多数斜面问题中无需额外计算支持力的功，节省解题时间。

4. 动能定理与机械能守恒定律的对比★★★★☆⏱ 3 min

学生们经常混淆这两个相关概念，但动能定理是更普遍的情况，而机械能守恒定律是由其推导得出的特例。我们可以将所有功分为保守力做的功（ $W_{c}$ ，例如重力、弹簧弹力）和非保守力做的功（ $W_{nc}$ ，例如摩擦力、施加的推力）。对于保守力， $W_{c} = - Δ U$ ，其中 $Δ U$ 为势能的变化量。

W_{\text{net}} = W_c + W_{\text{nc}} = -\Delta U + W_{\text{nc}} = \Delta K

W_{\text{nc}} = \Delta K + \Delta U

当非保守力做功为零（ $W_{nc} = 0$ ）时，该式就简化为我们熟悉的机械能守恒定律表达式 $K_{i} + U_{i} = K_{f} + U_{f}$ 。动能定理的一个核心优势是，如果你可以直接计算所有力做的功，就无需引入势能，从而简化涉及摩擦力等非保守力做功的问题。

📐 例题

将一个质量为0.5 kg的小球从地面以12 m/s的初速度竖直上抛。小球上升过程中空气阻力对其做的负功为2 J，求小球能到达的最大高度。

1
1. 到达最大高度时，末速度为0，因此 $K_{f} = 0$ 。计算初始动能：
2
$K_i = \frac{1}{2}(0.5)(12)^2 = 36\ \text{J}, \Delta K = -36\ \text{J}$
3
1. 总功为重力做功与空气阻力做功之和：
4
$W_{\text{net}} = -mgh - 2 = -36\ \text{J}$
5
1. 整理式子求解 $h$ ：
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$mgh = 34 \implies h = \frac{34}{(0.5)(9.8)} \approx 6.9\ \text{m}$
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该结果与能量守恒框架下的表达式 $W_{nc} = Δ K + Δ U$ 相符，验证了这两种方法之间的联系。

Exam tip:

如果题目要求计算已知位移下的未知平均作用力，请直接使用动能定理。该方法无需整理势能项，可将符号错误的概率降低一半。

5. AP风格概念检测★★★☆☆⏱ 1 min

✓ 快速检测

测试你对核心应用的理解程度：

A 1000 kg car moving at 20 m/s brakes to a stop, leaving a 50 m long skid mark. What is the magnitude of the average friction force that stopped the car?
- 2000 N
- 4000 N
- 8000 N
- 10000 N
显示答案
4000 N —
Using work-energy: $Δ K = - 200000 J = - f d ⟹ f = 200000/50 = 4000 N$ . The other options come from common errors: A is half the correct value, C from doubling kinetic energy, D equals the car's weight.

6. 常见陷阱

错误做法:

Using work done by a single applied force instead of net work, ignoring gravity or friction.

原因:

Students often focus on the force explicitly mentioned in the problem and forget other forces acting on the object.

正确做法:

Before plugging into $W_{net} = Δ K$ , always list all forces acting on the object and confirm their work contributions are included.

错误做法:

Using $W_{net} = K_{f}$ even when the object has non-zero initial kinetic energy.

原因:

Students get used to problems where objects start from rest, memorize the shortcut, and apply it incorrectly to all cases.

正确做法:

Always write out $W_{net} = K_{f} - K_{i}$ explicitly, and plug in $K_{i}$ even if you suspect it is zero, to confirm it is actually zero.

错误做法:

Writing $W_{g} = + m g h$ for an object moving upward to height $h$ .

原因:

Students confuse increasing gravitational potential energy with positive work done by gravity.

正确做法:

Remember work done by gravity is positive when displacement is downward (same direction as gravity) and negative when upward (opposite direction). Draw the force and displacement vectors to check the angle if unsure.

错误做法:

Including work done by internal forces when applying the theorem to a system of multiple objects.

原因:

Students do not distinguish between work done on a single object and work done on a whole system.

正确做法:

For a system of multiple rigid objects, net work only includes work done by external forces; internal forces cancel out for the system's total kinetic energy change.

错误做法:

Using the work-energy theorem to calculate acceleration directly from displacement and initial speed.

原因:

The theorem relates work (a function of displacement) to change in kinetic energy (a function of speed), not the rate of speed change.

正确做法:

Use the theorem to find final speed first, then use kinematics or Newton's second law to find acceleration if needed.

7. 速查表

Category	Formula	Notes
Core Work-Energy Theorem	$W_{net} = Δ K = K_{f} - K_{i}$	Always valid for rigid objects, any force type
Kinetic Energy	$K = \frac{1}{2} m v^{2}$	Always positive, depends on speed not direction
Work done by constant force	$W = F Δ x cos θ$	$θ$ = angle between force and displacement
Net Work Method 1	$W_{net} = F_{net} Δ x cos θ$	Use when net force is known for a single displacement
Net Work Method 2	$W_{net} = W_{1} + W_{2} + ... + W_{n}$	Use for multiple forces; perpendicular forces contribute 0
Work done by a spring	$W_{s} = \frac{1}{2} k (x_{i}^{2} - x_{f}^{2})$	Positive when spring moves to equilibrium from compression/stretch
Work done by gravity	$W_{g} = m g Δ y$	$Δ y$ positive for downward displacement, negative for upward
Relation to energy conservation	$W_{nc} = Δ K + Δ U$	$W_{nc}$ = non-conservative work, $Δ U$ = potential energy change

下一步

The work-energy theorem is the foundational principle for all energy-based problem solving in AP Physics 1, and it is a required prerequisite for all remaining topics in Unit 4: Energy. This theorem simplifies many problems that would require complicated kinematic calculations, and it acts as a clear bridge between force-based and energy-based approaches to motion. Mastery of this theorem will make it much easier to solve complex problems involving springs, rotation, and power that appear later in the AP Physics 1 exam, and it is a core concept regularly tested in both multiple-choice and free-response sections. Next, you will build on this foundation to study conservation of mechanical energy for conservative systems.