Friction, drag and terminal velocity · 摩擦、阻力与终极速度
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| terminal velocity/ˈtɜːmɪnl vəˈlɒsɪti/ | 收尾速度 | shōu wěi sù dù |
| air resistance/eə rɪˈzɪstəns/ | 空气阻力 | kōng qì zǔ lì |
| friction/ˈfrɪkʃn/ | 摩擦力 | mó cā lì |
| drag/dræɡ/ | 阻力 | zǔ lì |
| viscous/ˈvɪskəs/ | 黏性 | nián xìng |
| fluid/ˈfluːɪd/ | 流体 | liú tǐ |
| upthrust/ˈʌpθrʌst/ | 浮力 | fú lì |
| resultant force/rɪˈzʌltənt fɔːs/ | 合力 | hé lì |
| thermal energy/ˈθɜːml ˈenədʒi/ | 热能 | rè néng |
| power/ˈpaʊə/ | 功率 | gōnglǜ |
Why a skydiver stops speeding up
- Jump from a plane and you accelerate — but not forever.
- Soon you fall at a steady speed, the terminal velocity 收尾速度, even though gravity still pulls.
- The reason is air resistance 空气阻力 growing with speed.
为什么跳伞者会停止加速
- 从飞机上跳下,你会加速——但不会永远加速。
- 很快你就以一个稳定的速度下落,即 终极速度(terminal velocity),尽管重力仍在拉你。
- 原因是 空气阻力 随速度增大。
Friction 摩擦力 and drag 阻力
- Friction acts between two solid surfaces, opposing the sliding.
- Drag (a viscous 黏性 force) is the resistance from a fluid 流体 — a liquid or gas.
Newton's third-law pair: R on the book (up) and R-prime on the table (down)
摩擦与阻力
- 摩擦(friction) 作用在两个固体表面之间,阻碍滑动。
- 阻力(drag)(一种黏性力)是来自 流体(液体或气体)的阻碍。
牛顿第三定律的力对:作用在书上的 R(向上)和作用在桌上的 R′(向下)
Stopping a car · 汽车刹车
Friction is what brakes a car. Set a speed and brake — the car keeps moving while the driver reacts, then friction slows it. Double the speed and watch the braking distance quadruple. · 摩擦力使汽车减速。设定速度并刹车——在驾驶员反应期间汽车继续移动，随后摩擦力使其减速。将速度加倍，观察制动距离变为原来的四倍。
Which best describes a drag (viscous) force? · 下面哪个最好地描述了阻力(黏性力)?
Drag is the resistance a fluid (liquid or gas) exerts on something moving through it. Friction is the solid-on-solid case. · 阻力是流体(液体或气体)对穿过它运动的物体施加的阻碍。摩擦则是固体对固体的情形。
Drag grows with speed
- At rest there is no drag.
- The faster you go, the bigger the drag force becomes.
Velocity-time graph for an object falling through air
阻力随速度增大
- 静止时没有阻力。
- 你走得越快,阻力就 越大。
物体在空气中下落的速度–时间图
The drag force on a falling object grows as its speed grows. · 下落物体所受的阻力随它的速度增大而增大。
Yes — at rest the drag is zero, and it increases with speed. That is what eventually balances the weight. · 是的——静止时阻力为零,并随速度增大。这正是最终与重力平衡的原因。
Three forces on a ball falling in a liquid
- Weight $W$ — down, always the same.
- Upthrust 浮力 $U$ — up, the same at every speed (it depends only on the liquid pushed aside).
- Drag $F$ — up, opposite to the motion, and growing as the ball speeds up.
- Exam habit: draw each force as a labelled arrow from the ball, and never add a "motion" arrow.
A free-body diagram shows only the forces on the object — the same idea for a falling ball
液体中下落的小球受三个力
- 重力(weight) $W$——向下,始终不变。
- 浮力(upthrust) $U$——向上,在任何速度下都相同(它只取决于被排开的液体)。
- 阻力(drag) $F$——向上,与运动方向相反,并随小球加速而 增大。
- 考试习惯:每个力都画成 从小球出发 的带标签箭头,绝不要再画一个"运动"箭头。
自由体受力图只画物体所受的力——下落的小球也是同样的思路
A steel ball is falling through oil. Which forces act on the ball? Select all that apply. · 一个钢球正在油中下落。哪些力作用在球上?选出所有正确的。
Three forces: weight down, upthrust up, and drag up (opposite to the motion). There is no separate force of motion — the ball moves because of its velocity, not because a force pushes it along. · 三个力:重力向下,浮力向上,阻力向上(与运动相反)。没有单独的"运动力"——球运动是因为它有速度,不是因为有力推着它走。
Falling to terminal velocity
- Start: only weight (and upthrust) act, so the resultant force is largest and the acceleration is largest.
- Middle: drag grows, the resultant force 合力 shrinks, so the acceleration falls.
- End: upward forces equal the weight → zero resultant force → constant speed.
下落到终极速度
- 开始: 只有重力(和浮力)作用,所以合力最大,加速度也最大。
- 中间: 阻力增大,合力(resultant force) 减小,所以加速度下降。
- 结束: 向上的力等于重力 → 合力为零 → 匀速。
Reach terminal velocity · 达到终端速度
Jump and watch the air-resistance arrow grow until it balances the weight — then the speed is constant. Open the parachute and the much bigger drag drops the diver to a slow, safe terminal velocity. · 跳跃并观察空气阻力箭头增大，直到它与重力平衡——此时速度恒定。打开降落伞后，更大的阻力使跳伞者以缓慢、安全的终端速度下降。
Put the stages of a fall through air in order, from the moment of release. · 把空气中下落的各个阶段按顺序排列,从释放那一刻开始。
Drag starts at zero and builds with speed until it cancels the weight, ending the acceleration. · 阻力从零开始,随速度增大,直到抵消重力,结束加速。
At terminal velocity, the drag force is equal to the ____. · 在终极速度时,阻力等于 ____。
Drag = weight, so the resultant force is zero and the speed stays constant. · 阻力 = 重力,所以合力为零,速度保持不变。
At terminal velocity the acceleration is zero — not the velocity. The object is still moving, and it still loses height. A common exam error is "the forces are balanced so it stops".
在终极速度时,为零的是 加速度——不是速度。物体仍在运动,也仍在下降。一个常见的考试错误是"力平衡了,所以它停下来了"。
Worked example: terminal speed in oil
A small steel ball of weight $0.050\ \text{N}$ falls through oil. The upthrust on it is $0.010\ \text{N}$, and the drag is $F = kv$ with $k = 0.080\ \dfrac{\text{N s}}{\text{m}}$. Find the terminal speed.
- At terminal speed the resultant force is zero: $W = U + F$.
- So $F = 0.050 - 0.010 = 0.040\ \text{N}$.
- Then $kv = 0.040$, giving $v = \dfrac{0.040}{0.080} = 0.50\ \dfrac{\text{m}}{\text{s}}$.
- Check: in a liquid, forgetting the upthrust gives $v = 0.63\ \dfrac{\text{m}}{\text{s}}$ — the wrong answer, and a lost mark.
例题:油中的终极速度
一个重 $0.050\ \text{N}$ 的小钢球在油中下落。它受到的浮力是 $0.010\ \text{N}$,阻力为 $F = kv$,其中 $k = 0.080\ \dfrac{\text{N s}}{\text{m}}$。求终极速度。
- 达到终极速度时 合力为零:$W = U + F$。
- 所以 $F = 0.050 - 0.010 = 0.040\ \text{N}$。
- 于是 $kv = 0.040$,得 $v = \dfrac{0.040}{0.080} = 0.50\ \dfrac{\text{m}}{\text{s}}$。
- 检查: 在液体中,忘记浮力会得到 $v = 0.63\ \dfrac{\text{m}}{\text{s}}$——答案错误,丢一分。
A ball of weight $0.12\ \text{N}$ falls through a liquid. The upthrust is $0.030\ \text{N}$ and the drag is $F = kv$ with $k = 0.30\ \dfrac{\text{N s}}{\text{m}}$. What is its terminal speed, in m/s? · 一个重 $0.12\ \text{N}$ 的球在液体中下落。浮力为 $0.030\ \text{N}$,阻力 $F = kv$,其中 $k = 0.30\ \dfrac{\text{N s}}{\text{m}}$。它的终极速度是多少(单位 m/s)?
At terminal speed $W = U + kv$, so $kv = 0.12 - 0.030 = 0.090\ \text{N}$ and $v = \dfrac{0.090}{0.30} = 0.30\ \dfrac{\text{m}}{\text{s}}$. · 终极速度时 $W = U + kv$,所以 $kv = 0.12 - 0.030 = 0.090\ \text{N}$,$v = \dfrac{0.090}{0.30} = 0.30\ \dfrac{\text{m}}{\text{s}}$。
Energy at terminal velocity
- The kinetic energy is now constant, but the object keeps losing height.
- That lost gravitational PE turns mainly into thermal energy 热能 of the fluid — not into extra speed.
终极速度时的能量
- 动能现在恒定,但物体仍在不断下降。
- 损失的 引力势能 主要转化为流体的 热能——而不是变成更快的速度。
As a parachutist falls at terminal velocity, the lost gravitational PE turns mainly into: · 当跳伞者以终极速度下落时,损失的引力势能主要转化为:
The kinetic energy is constant, so the falling PE cannot become KE — it heats the air (and the parachutist) through drag. · 动能恒定,所以下降损失的势能不能变成动能——它通过阻力把空气(和跳伞者)加热。
Cruising at constant speed
- A car at steady speed has zero resultant force: driving force = total resistance.
- On a slope, the driving force must also balance the component of the weight along the slope, $W\sin\theta$.
- Going faster means more drag, so more driving force — and more power 功率, since $P = Fv$.
On a slope the driving force works against drag and against part of the weight
匀速巡航
- 匀速行驶的汽车合力为零:驱动力 = 总阻力。
- 在 斜坡 上,驱动力还必须平衡重力沿斜坡的分量 $W\sin\theta$。
- 开得更快意味着更大的阻力,所以需要更大的驱动力——以及更大的 功率(power),因为 $P = Fv$。
在斜坡上,驱动力既要对抗阻力,也要对抗一部分重力
A car cruising at a higher steady speed needs more power because the drag is larger. · 以更高的稳定速度巡航的汽车需要更大的功率,因为阻力更大。
Higher speed → larger drag → a larger driving force is needed → more power (power = force × velocity). · 速度更高 → 阻力更大 → 需要更大的驱动力 → 更大的功率(功率 = 力 × 速度)。
Worked example: power up a slope
A car of mass $1200\ \text{kg}$ climbs a slope of $5.0^\circ$ at a steady $25\ \dfrac{\text{m}}{\text{s}}$. The total resistive force is $900\ \text{N}$. Find the useful output power of the engine.
- Steady speed → the driving force $D$ balances everything pulling back.
- Weight component down the slope: $mg\sin\theta = 1200 \times 9.81 \times \sin 5.0^\circ = 1030\ \text{N}$.
- So $D = 900 + 1030 = 1930\ \text{N}$.
- Power: $P = Dv = 1930 \times 25 = 4.8 \times 10^{4}\ \text{W}$ (about $48\ \text{kW}$).
- Check: using $mg$ instead of $mg\sin\theta$ would give a driving force bigger than the car's whole weight — impossible on a gentle slope.
例题:上坡的功率
一辆质量 $1200\ \text{kg}$ 的汽车以 $25\ \dfrac{\text{m}}{\text{s}}$ 的稳定速度爬上 $5.0^\circ$ 的斜坡。总阻力为 $900\ \text{N}$。求发动机的有用输出功率。
- 速度稳定 → 驱动力 $D$ 平衡所有向后拉的力。
- 重力沿斜坡向下的分量:$mg\sin\theta = 1200 \times 9.81 \times \sin 5.0^\circ = 1030\ \text{N}$。
- 所以 $D = 900 + 1030 = 1930\ \text{N}$。
- 功率:$P = Dv = 1930 \times 25 = 4.8 \times 10^{4}\ \text{W}$(约 $48\ \text{kW}$)。
- 检查: 若用 $mg$ 而不是 $mg\sin\theta$,得到的驱动力会比整辆车的重力还大——在缓坡上这不可能。
A lorry climbs a slope at a steady $20\ \dfrac{\text{m}}{\text{s}}$. The total resistive force is $2000\ \text{N}$ and the component of its weight along the slope is $3000\ \text{N}$. What useful output power does the engine deliver, in kW? · 一辆卡车以 $20\ \dfrac{\text{m}}{\text{s}}$ 的稳定速度爬坡。总阻力为 $2000\ \text{N}$,重力沿斜坡的分量为 $3000\ \text{N}$。发动机输出的有用功率是多少(单位 kW)?
Steady speed, so the driving force is $2000 + 3000 = 5000\ \text{N}$. Power $P = Fv = 5000 \times 20 = 1.0 \times 10^{5}\ \text{W} = 100\ \text{kW}$. · 速度稳定,所以驱动力为 $2000 + 3000 = 5000\ \text{N}$。功率 $P = Fv = 5000 \times 20 = 1.0 \times 10^{5}\ \text{W} = 100\ \text{kW}$。
You've got it
- drag grows with speed; friction is between solids, drag is from a fluid
- terminal velocity: upward forces (drag + upthrust) = weight, so zero resultant force and constant speed — zero acceleration, not zero motion
- the lost gravitational PE becomes thermal energy of the fluid, not kinetic energy
- steady speed on a slope: driving force = resistance + $W\sin\theta$, and $P = Fv$
你掌握了
- 阻力随速度增大;摩擦在固体之间,阻力来自流体
- 终极速度:向上的力(阻力 + 浮力)= 重力,所以合力为零、匀速——为零的是 加速度,不是运动
- 损失的引力势能变成 流体的热能,而不是动能
- 斜坡上匀速:驱动力 = 阻力 + $W\sin\theta$,且 $P = Fv$