Friction, drag and terminal velocity
Introduced| English | Meaning |
|---|---|
| terminal velocity/ˈtɜːmɪnl vəˈlɒsɪti/ | A constant falling velocity reached when weight is balanced by resistance. |
| air resistance/eə rɪˈzɪstəns/ | A drag force opposing motion through air. |
| friction/ˈfrɪkʃn/ | A force that opposes relative motion between contacting surfaces. |
| drag/dræɡ/ | A resistive force exerted by a fluid on an object moving relative to it. |
| viscous/ˈvɪskəs/ | Describing a fluid that resists relative motion between neighbouring layers. |
| fluid/ˈfluːɪd/ | A substance that can flow, including liquids and gases. |
| upthrust/ˈʌpθrʌst/ | The upward buoyant force exerted by a fluid on an object immersed in it. |
| resultant force/rɪˈzʌltənt fɔːs/ | The single force equivalent to all the forces acting together. |
| thermal energy/ˈθɜːml ˈenədʒi/ | Energy associated with the random motion and interactions of particles in a substance. |
| power/ˈpaʊə/ | The rate of energy transfer or work done. |
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.
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)
Stopping a car
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.
Which best describes a drag (viscous) force?
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Drag is the resistance a fluid (liquid or gas) exerts on something moving through it. Friction is the solid-on-solid case.
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
The drag force on a falling object grows as its speed grows.
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Yes — at rest the drag is zero, and it increases with speed. That is what eventually balances the weight.
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
A steel ball is falling through oil. Which forces act on the ball? Select all that apply.
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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.
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.
Reach terminal velocity
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.
Put the stages of a fall through air in order, from the moment of release.
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Drag starts at zero and builds with speed until it cancels the weight, ending the acceleration.
At terminal velocity, the drag force is equal to the ____.
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Drag = weight, so the resultant force is zero and the speed stays constant.
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".
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.
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?
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At terminal speed $W = U + kv$, so $kv = 0.12 - 0.030 = 0.090\ \text{N}$ and $v = \dfrac{0.090}{0.30} = 0.30\ \dfrac{\text{m}}{\text{s}}$.
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.
As a parachutist falls at terminal velocity, the lost gravitational PE turns mainly into:
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The kinetic energy is constant, so the falling PE cannot become KE — it heats the air (and the parachutist) through drag.
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
A car cruising at a higher steady speed needs more power because the drag is larger.
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Higher speed → larger drag → a larger driving force is needed → more power (power = force × velocity).
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.
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?
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Steady speed, so the driving force is $2000 + 3000 = 5000\ \text{N}$. Power $P = Fv = 5000 \times 20 = 1.0 \times 10^{5}\ \text{W} = 100\ \text{kW}$.
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 $P = Fv$