Work · 功
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| work/wɜːk/ | 功 | gōng |
| dot product/dɒt ˈprɒdʌkt/ | 点积 | diǎn jī |
The effort that doesn't count
- Lug a heavy suitcase across a flat room and physics says you did no work on it.
- Strange? The force (up) is perpendicular to the motion (sideways).
- In physics, "work" has a precise meaning -- and direction is everything.
- Get it right, and energy problems become easy.
白费力气的努力
- 提着沉重的行李箱横穿平坦的房间,物理学却说你对它没做功。
- 奇怪吗?力(向上)与运动(横向)垂直。
- 在物理学里,"功"有精确的含义——方向就是一切。
- 弄对它,能量问题就变得简单。
Work of a constant force
- Work 功 is energy transferred when a force acts over a displacement:
- The dot product 点积 keeps only the part of the force along the motion.
- Force and motion aligned ($\theta = 0$) gives the most work.
恒力做的功
- 功是力在位移上作用时所转移的能量:
- 点积只保留力沿运动方向的那部分。
- 力与运动同向($\theta = 0$)时做功最多。
A $10\ \text{N}$ force pushes a box $4\ \text{m}$ in the same direction. How much work is done (in J)? · 一个 $10\ \text{N}$ 的力沿同一方向推动箱子移动了 $4\ \text{m}$。做了多少功（单位：J）？
With $\theta = 0$, $W = Fd\cos 0 = 10 \times 4 \times 1 = 40\ \text{J}$. · 利用 $\theta = 0$，可得 $W = Fd\cos 0 = 10 \times 4 \times 1 = 40\ \text{J}$。
A $50\ \text{N}$ force at $60^\circ$ to the motion drags a crate $3\ \text{m}$. Work done (in J)? · 一个大小为 $50\ \text{N}$ 的力以与运动成 $60^\circ$ 的角度拉动箱子移动了 $3\ \text{m}$。做的功（单位：J）？
$W = Fd\cos\theta = 50 \times 3 \times \cos 60^\circ = 50 \times 3 \times 0.5 = 75\ \text{J}$.
The angle sets the sign
- Force along the motion → positive work (energy added).
- Force against the motion → negative work (energy removed, like friction).
- Force perpendicular ($\theta = 90^\circ$) → zero work -- that is the suitcase.
角度决定符号
- 力沿运动方向 → 正功(输入能量)。
- 力逆运动方向 → 负功(移出能量,如摩擦力)。
- 力垂直($\theta = 90^\circ$)→ 零功——那就是行李箱的情形。
You carry a bag horizontally across a room. The work gravity does on the bag is... · 你水平提着一个包穿过房间。重力对包做的功是...
Gravity points down, motion is horizontal -- perpendicular, so $\cos 90^\circ = 0$ and the work is zero. · 重力向下，运动水平——两者垂直，因此 $\cos 90^\circ = 0$，做功为零。
Friction, which opposes motion, does negative work on a sliding object. · 摩擦力阻碍运动，对滑动物体做负功。
The friction force points against the motion ($\theta = 180^\circ$), so its work is negative -- it removes energy. · 摩擦力方向与运动相反（$\theta = 180^\circ$），因此其功为负——它移除了能量。
Select all · 所有 cases where the force does zero · 零 work. · 选择所有力做零功的情况。
Perpendicular force ($\cos 90^\circ = 0$) and no displacement both give zero work. A force along the motion does positive work. · 垂直力（$\cos 90^\circ = 0$）和无位移都会导致零功。沿运动方向的力做正功。
Variable forces: integrate
- When the force changes along the path, sum the tiny bits:
- This is the area under a force-versus-position graph.
- A stretching spring, whose force grows with $x$, needs exactly this.
变力:积分
- 当力沿路径变化时,把微小的部分加起来:
- 这就是力对位置图下的面积。
- 一根被拉伸的弹簧,其力随 $x$ 增大,正需要这个。
Positive, negative or zero work? · 正功、负功还是零功？
Work depends on the angle between the force and the motion. Sort each case. · 功取决于力与运动方向之间的角度。对每种情况进行分类。
For a force that changes with position, work is the ____ under the force-position graph. · 对于随位置变化的力，功是力-位置图线下方的 ____。
$W = \int F\,dx$ is exactly the area under the force-versus-position curve. · $W = \int F\,dx$ 恰好等于力-位置曲线下的面积。
Work moves energy
- Work is how energy is transferred into or out of an object.
- Positive net work speeds it up; negative net work slows it down.
- That is the bridge to the work-energy theorem, $W_{net} = \Delta K$.
功转移能量
- 功是能量被转移进出物体的方式。
- 正的净功使它加速;负的净功使它减速。
- 这就是通向功能定理 $W_{net} = \Delta K$ 的桥梁。
You pull a sled $10\ \text{m}$ with a $20\ \text{N}$ force angled $60^\circ$ above the ground.
- $W = Fd\cos\theta = 20 \times 10 \times \cos 60^\circ = 20 \times 10 \times 0.5 = 100\ \text{J}$.
- Only the along-the-ground part of the pull counts.
你用一个与地面成 $60^\circ$ 的 $20\ \text{N}$ 的力,把雪橇拉了 $10\ \text{m}$。
- $W = Fd\cos\theta = 20 \times 10 \times \cos 60^\circ = 20 \times 10 \times 0.5 = 100\ \text{J}$。
- 只有拉力沿地面的那部分才算数。
Effort is not work. Holding a barbell overhead is exhausting, yet does zero physics-work, because nothing moves. Work needs a force and a displacement along it -- no motion, no work, however tiring.
费力不等于做功。把杠铃举过头顶很累,却做了零物理功,因为没有东西移动。功需要一个力和沿它方向的位移——没有运动,就没有功,无论多累。
Work $W = \vec{F}\cdot\vec{d} = Fd\cos\theta$ counts only the force along the motion (a dot product). Along-motion is positive, against is negative, perpendicular is zero. For a varying force, integrate: $W = \int \vec{F}\cdot d\vec{r}$, the area under the force-position graph.
功 $W = \vec{F}\cdot\vec{d} = Fd\cos\theta$ 只计入沿运动方向的力(一个点积)。同向为正,逆向为负,垂直为零。对于变力,做积分:$W = \int \vec{F}\cdot d\vec{r}$,即力-位置图下的面积。