Unit A.3: Work, Energy and Power单元 A.3：功、能量与功率

When the force changes with position, sum work over infinitesimal displacements:

当力随位置变化时，对无穷小位移求和：

Graphically this is exactly the area under the $F$–$s$ curve. For a spring obeying $F = k x$, the area under the straight line from $0$ to $x$ is a triangle, $W = \tfrac12 (k x)(x) = \tfrac12 k x^2$ — which is where elastic PE comes from (A3.3). The constant-force result $W = F s$ is the special case where $F$ is flat and the area is a rectangle.

几何上这正是 $F$–$s$ 曲线下的面积。对满足 $F = k x$ 的弹簧，从 $0$ 到 $x$ 直线下的面积是三角形，$W = \tfrac12 (k x)(x) = \tfrac12 k x^2$——这正是弹性势能的来源（A3.3）。恒力结果 $W = F s$ 是 $F$ 为常数、面积为矩形的特例。

For a constant net force $F$ over displacement $s$, Newton's second law gives $F = m a$. The suvat relation $v^2 = u^2 + 2 a s$ rearranges to $a s = \tfrac12 (v^2 - u^2)$. Multiply by $m$:

对恒定合力 $F$ 作用于位移 $s$，牛顿第二定律给出 $F = m a$。suvat 关系 $v^2 = u^2 + 2 a s$ 整理为 $a s = \tfrac12 (v^2 - u^2)$。两边乘 $m$：

The HL integral form $W = \int F\,ds$ recovers the same result for a variable force, because $F\,ds = m\,\dfrac{dv}{dt}\,ds = m v\,dv$, which integrates to $\tfrac12 m v^2$.

HL 的积分形式 $W = \int F\,ds$ 对变力给出同样结果，因为 $F\,ds = m\,\dfrac{dv}{dt}\,ds = m v\,dv$，积分得 $\tfrac12 m v^2$。

The spring force grows from $0$ to $k x$ as it is stretched, so the average force is $\tfrac12 k x$. Work done against it is average force times distance: $\bar{F} x = (\tfrac12 k x)(x) = \tfrac12 k x^2$. This is exactly the triangular area under the $F = k x$ line from A3.1.

弹簧力在拉伸过程中从 $0$ 增长到 $k x$，故平均力为 $\tfrac12 k x$。克服它所做的功为平均力乘距离：$\bar{F} x = (\tfrac12 k x)(x) = \tfrac12 k x^2$。这正是 A3.1 中 $F = k x$ 直线下的三角形面积。

A common slip is to write $W = F x = k x \cdot x = k x^2$, double the true value — that would only hold if the force were constant at $k x$ the whole way.

常见错误是写成 $W = F x = k x \cdot x = k x^2$，是真值的两倍——只有当力全程恒为 $k x$ 时才成立。

An energy bar chart draws one column per store (kinetic, gravitational, elastic, thermal) at each instant. The conservation rule is visual: the total height of all bars is identical at every instant. In the slide example, the "before" chart is one tall PE bar ($736\ \mathrm{J}$); the "after" chart is a shorter $E_k$ bar ($375\ \mathrm{J}$) plus a thermal bar ($361\ \mathrm{J}$) that sum to the same total.

能量条形图在每一时刻为每种形式（动能、重力势能、弹性势能、热能）画一根柱。守恒规则是可视的：所有柱的总高度在每个时刻都相同。在滑道例子中，"之前"图是一根高 PE 柱（$736\ \mathrm{J}$）；"之后"图是一根较矮的 $E_k$ 柱（$375\ \mathrm{J}$）加上一根热能柱（$361\ \mathrm{J}$），两者之和等于同一总高。

Bar charts are markschemed in Paper 2: an examiner can credit a correct chart even if the arithmetic slips, because it shows you tracked every store.

条形图在 Paper 2 中可计分：即便算术出错，考官也能给出正确条形图的分，因为它显示你追踪了每一种能量形式。

Over a small time $\Delta t$ a constant force $F$ moves its point of application a distance $\Delta s = v\,\Delta t$ (with $F$ along the motion). The work is $\Delta W = F\,\Delta s = F v\,\Delta t$, so

在小时间 $\Delta t$ 内，恒力 $F$ 使作用点移动 $\Delta s = v\,\Delta t$（$F$ 沿运动方向）。功为 $\Delta W = F\,\Delta s = F v\,\Delta t$，故

This is why a car needs much more power at high speed even to hold the same speed: drag rises with $v$, and power is $F v$, so power demand climbs faster than linearly with speed.

这就是为何汽车在高速下即便维持同一速度也需大得多的功率：阻力随 $v$ 增大，而功率为 $F v$，故功率需求随速度的增长快于线性。

In two dimensions momentum is conserved componentwise: $\sum m u_x = \sum m v_x$ and $\sum m u_y = \sum m v_y$. Kinetic energy, a scalar, uses the speed-squared $v^2 = v_x^2 + v_y^2$ for each body. A classic HL result: when an object strikes an identical stationary object elastically and they separate, the two move off at $90^\circ$ to each other (verifiable by combining the momentum and energy conditions).

在二维中动量分量守恒：$\sum m u_x = \sum m v_x$ 与 $\sum m u_y = \sum m v_y$。动能作为标量，对每个物体用速率平方 $v^2 = v_x^2 + v_y^2$。一个经典 HL 结论：当物体弹性撞上一个相同的静止物体并分开后，两者以相互 $90^\circ$ 飞出（联立动量与能量条件可验证）。

In every case the work–energy bookkeeping must close: any kinetic energy not present after an inelastic collision must reappear as an equal amount of thermal/internal energy.

任何情形下功能账目都须闭合：非弹性碰撞后缺失的动能必以等量的热能/内能重新出现。