Work and Energy | 功和能量考点精讲

📚 Work and Energy | 功和能量

Work and energy form the cornerstone of mechanics, linking forces to motion through scalar quantities that often simplify complex problems. In IB and OCR mathematics, these concepts are applied to constant forces, variable forces requiring integration, and conservation principles. Mastery of work, kinetic energy, potential energy, and power is essential for tackling both straightforward and advanced exam questions. This revision guide breaks down every key idea with clear derivations, practical examples, and targeted exam advice.

功和能量构成了力学的基石，通过标量将力与运动联系起来，常常能简化复杂问题。在 IB 和 OCR 数学中，这些概念被应用于恒力、需要积分处理的变力以及守恒原理。熟练掌握功、动能、势能和功率对于解决基础和进阶的考试题目至关重要。本复习指南通过清晰的推导、实例和有针对性的考试建议，逐一解析每个关键知识点。

1. Work: Definition and Formula | 功：定义与公式

Work is done when a force causes a displacement. For a constant force F acting on an object that moves through a displacement s, the work done W is the product of the force component in the direction of the displacement and the magnitude of the displacement. Mathematically, W = F s cos θ, where θ is the angle between the force and displacement vectors. Work is a scalar quantity and its SI unit is the joule (J). 1 J is the work done by a force of 1 N moving an object 1 m in the direction of the force.

当力引起位移时就做了功。对于作用在物体上的恒力 F，物体移动位移 s，所做的功 W 等于力在位移方向上的分量与位移大小的乘积。数学表达式为 W = F s cos θ，其中 θ 是力与位移矢量之间的夹角。功是标量，其国际单位是焦耳（J）。1 J 等于 1 N 的力在其方向上移动物体 1 m 所做的功。

W = F s cos θ

If the force and displacement are in exactly the same direction (θ = 0°), the formula simplifies to W = F s. When the force is perpendicular to the displacement (θ = 90°), cos 90° = 0, so no work is done. For example, the normal reaction force on an object sliding along a horizontal surface does zero work.

若力与位移方向完全相同（θ = 0°），公式简化为 W = F s。当力垂直于位移时（θ = 90°），cos 90° = 0，因此不做功。例如，物体在水平面上滑动时，法向反作用力做功为零。

2. Work Done by a Constant Force – Examples | 恒力做功示例

Consider a crate pulled along a horizontal floor by a rope exerting a constant force of 50 N at an angle of 30° to the horizontal. If the crate moves 8 m, the work done by the pulling force is W = 50 × 8 × cos 30° = 400 × (√3/2) ≈ 346.4 J. Notice that only the horizontal component of the force contributes to the work.

考虑一个例子：用一根绳索以与水平方向成 30° 角的恒力 50 N 沿着水平地面拉一个板条箱。如果板条箱移动了 8 m，那么拉力所做的功为 W = 50 × 8 × cos 30° = 400 × (√3/2) ≈ 346.4 J。注意，只有力的水平分量对功有贡献。

Work can be positive, negative, or zero. When the angle θ is less than 90°, work is positive, meaning the force is aiding the motion. When θ is greater than 90°, cos θ is negative, so the work done is negative; this occurs when a force opposes the displacement, such as friction or a braking force. Lifting an object at constant speed involves positive work by the lifting force and negative work by gravity.

功可以为正、负或零。当角度 θ 小于 90° 时，功为正，表明力促进运动。当 θ 大于 90° 时，cos θ 为负，所做的功为负；当力阻碍位移时，例如摩擦力或制动力，就会出现负功。以恒定速度提升物体时，提升力做正功，重力做负功。

3. Work Done by a Variable Force | 变力做功

When a force varies with position, you cannot simply multiply force by displacement. Instead, the work done is found by integration. Consider an infinitesimal displacement dx: the tiny amount of work is dW = F dx. The total work done as the object moves from x = a to x = b is the definite integral:

W = ∫ₐᵇ F(x) dx

Geometrically, this is the area under the force–displacement graph between x = a and x = b. This interpretation is especially useful when a force–displacement graph is provided, and you need to estimate the work by counting squares or using the trapezoidal rule.

当力随位置变化时，不能简单地将力与位移相乘。此时，做功可通过积分求得。考虑无穷小位移 dx：微小功为 dW = F dx。物体从 x = a 移动到 x = b 时所做的总功是定积分：W = ∫ₐᵇ F(x) dx。在几何上，这就是力–位移图中 x = a 到 x = b 之间曲线下的面积。当题目提供力–位移图并要求通过数格子或梯形法则估算功时，这种理解尤为有用。

A classic example is the force exerted by an ideal spring, F = kx, where k is the spring constant and x is the extension from the natural length. The work done in stretching the spring from 0 to an extension X is:

W = ∫₀ˣ kx dx = ½ k X²

This result links directly to elastic potential energy, which is stored in the spring. Note that the work done by the spring is equal in magnitude but opposite in sign when the spring is released.

一个典型的例子是理想弹簧的力 F = kx，其中 k 是劲度系数，x 是相对于自然长度的伸长量。将弹簧从 0 拉伸至伸长量 X 所做的功为：W = ∫₀ˣ kx dx = ½ k X²。这一结果直接关联到储存在弹簧中的弹性势能。注意，弹簧释放时所做的功大小相等但符号相反。

4. Kinetic Energy and Derivation | 动能及其推导

Kinetic energy (KE) is the energy possessed by an object due to its motion. It depends on the mass m and the speed v:

KE = ½ m v²

The derivation follows from Newton’s second law and the definition of work. Assume a constant net force F acts on an object of mass m, causing acceleration a. As the object moves through a displacement s, the work done is W = F s = (ma) s. Using the kinematic relation v² = u² + 2as, we get s = (v² − u²)/(2a). Substituting yields W = m a × (v² − u²)/(2a) = ½ m v² − ½ m u². This expression represents the change in kinetic energy.

动能是物体因其运动而具有的能量，取决于质量 m 和速率 v：KE = ½ m v²。推导基于牛顿第二定律和功的定义。假设一个恒定的合外力 F 作用在质量为 m 的物体上，产生加速度 a。物体移动位移 s，所做的功为 W = F s = (ma) s。利用运动学关系式 v² = u² + 2as，得到 s = (v² − u²)/(2a)。代入得 W = m a × (v² − u²)/(2a) = ½ m v² − ½ m u²。该表达式代表了动能的变化量。

Kinetic energy is always non-negative, and like work, it is a scalar with unit joules. The speed v must be in m/s and mass in kg to obtain energy in J. It is often convenient to calculate the kinetic energy before and after a process to apply energy principles.

动能总是非负，和功一样是标量，单位为焦耳。速度 v 必须以 m/s 为单位，质量以 kg 为单位，才能得到以 J 为单位的能量。在运用能量原理时，通常需计算过程始末的动能。

5. Work-Energy Theorem | 动能定理

The work-energy theorem states that the net work done on an object by all forces (external and internal) equals the change in its kinetic energy:

W_net = ΔKE = ½ m v² − ½ m u²

This theorem is extremely powerful because it relates the total work to the speed change without requiring detailed knowledge of the motion’s intermediate stages. It works for constant and variable forces alike, as long as you can compute the net work.

动能定理指出，所有力（外力和内力）对物体所做的净功等于物体动能的变化：W_net = ΔKE = ½ m v² − ½ m u²。这一定理非常强大，因为它将总功与速度变化联系起来，而不需要详细了解运动的中间过程。无论是恒力还是变力，只要能计算出净功，该定理都适用。

When solving problems, identify all forces doing work, compute the work contributed by each (positive or negative), sum them to obtain W_net, and then equate to the change in kinetic energy. For instance, if a car of mass 1000 kg accelerates from rest to 20 m/s under a constant driving force, the net work done equals ½ × 1000 × 20² = 200,000 J, even if the driving force does additional work against friction.

在解题时，先找出所有做功的力，计算每个力所做的功（正或负），求和得到 W_net，然后令其等于动能变化。例如，一辆质量为 1000 kg 的汽车在恒定驱动力下从静止加速到 20 m/s，即使驱动力还需克服摩擦力做功，净功仍等于 ½ × 1000 × 20² = 200,000 J。

6. Gravitational Potential Energy | 重力势能

Gravitational potential energy (GPE) is the energy an object possesses due to its position in a gravitational field. Near the Earth’s surface, the change in GPE when an object of mass m is raised or lowered by a vertical height h is:

ΔGPE = m g h

where g is the acceleration due to gravity (9.8 m/s², often taken as 9.81). The value of GPE at a point depends on the chosen reference level; what matters is the difference in

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Work and Energy | 功和能量考点精讲