Energy, work and power · 能量、功与功率
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| power/ˈpaʊə/ | 功率 | gōng lǜ |
| potential energy/pəˈtenʃl ˈenədʒi/ | 势能 | shì néng |
| kinetic energy/kɪˈnetɪk ˈenədʒi/ | 动能 | dòng néng |
| conservation of energy/ˌkɒnsəˈveɪʃn ɒv ˈenədʒi/ | 能量守恒 | néng liàng shǒu héng |
| work done/wɜːk dʌn/ | 功 | gōng |
| joule/dʒuːl/ | 焦耳 | jiāo ěr |
| gravitational potential energy/ˌɡrævɪˈteɪʃənl pəˈtenʃl ˈenədʒi/ | 重力势能 | zhòng lì shì néng |
| watt/wɒt/ | 瓦特 | wǎ tè |
The energy that powers 功率 everything
- A roller coaster at the top of its first hill has maximum potential energy 势能. As it plunges down, that energy converts to kinetic energy 动能 — speed.
- Conservation of energy 能量守恒 lets you predict the speed at any point without knowing the path in between.
驱动一切的能量
- 一辆过山车在它第一个坡的顶部有最大的势能。当它向下俯冲时,那能量转换成动能——速度。
- 能量守恒(conservation of energy)让你预测任何点的速度,而不用知道中间的路径。
Work done 功
- Work done by a constant force: $W = Fd\cos\theta$ (measured in joules 焦耳).
- Work is the energy transferred by a force acting through a distance.
Worked example. A 50 N force pulls a box 4 m along the floor at $30^{\circ}$ to the horizontal. $W = 50 \times 4 \times \cos 30^{\circ} = 200 \times 0.866 = 173.2\text{ J}$.
A roller coaster trades potential energy for kinetic energy
做的功
- 一个恒力做的功(work done):$W = Fd\cos\theta$(以焦耳计)。
- 功是一个力作用经过一段距离所转移的能量。
算例。 一个 50 N 的力以与水平 $30^{\circ}$ 的角沿地板拉一个箱子 4 m。$W = 50 \times 4 \times \cos 30^{\circ} = 200 \times 0.866 = 173.2\text{ J}$。
一辆过山车把势能换成动能
A 50 N force pulls a box 4 m along the floor (θ = 0°). Work done = Fd cos θ. Find it (J). · 一个 50 N 的力沿地板拉一个箱子 4 m(θ = 0°)。做的功 = Fd cos θ。求它(J)。
W = 50 × 4 × cos 0° = 50 × 4 × 1 = 200 J. · W = 50 × 4 × cos 0° = 50 × 4 × 1 = 200 J。
Match each quantity to its formula or meaning. · 把每个量与它的公式或含义配对。
Energy is conserved: KE and PE swap, while power is the rate of doing work. · 能量守恒:动能和势能互换,而功率是做功的速率。
Kinetic and potential energy
- Kinetic energy (movement): $\text{KE} = \tfrac12 mv^2$.
- Gravitational potential energy 重力势能 (height): $\text{PE} = mgh$.
KE depends on $v^2$, not $v$. Doubling the speed quadruples the kinetic energy. A car at 60 mph has four times the KE of one at 30 mph — and needs four times the braking distance.
动能与势能
- 动能(kinetic energy,运动):$\text{KE} = \tfrac12 mv^2$。
- 重力势能(gravitational potential energy,高度):$\text{PE} = mgh$。
KE 取决于 $v^2$,不是 $v$。 速度翻倍使动能变成四倍。一辆 60 mph 的车的 KE 是 30 mph 的车的四倍——而且需要四倍的刹车距离。
Conservation of energy · 能量守恒
PE + KE = constant
As the ball falls, potential · 势 energy turns into kinetic energy — but the total · 总 never changes. · 当球下落时,势能变成动能——但总量从不改变。
What is the kinetic energy of a 2 kg object moving at 3 m/s (KE = ½mv²), in J? · 一个以 3 m/s 移动的 2 kg 物体的动能是多少(KE = ½mv²),以 J 计?
KE = ½ × 2 × 3² = ½ × 2 × 9 = 9 J. · KE = ½ × 2 × 3² = ½ × 2 × 9 = 9 J。
Taking g = 10, what is the PE of a 2 kg object raised 5 m (PE = mgh), in J? · 取 g = 10,一个升高 5 m 的 2 kg 物体的 PE 是多少(PE = mgh),以 J 计?
PE = mgh = 2 × 10 × 5 = 100 J. · PE = mgh = 2 × 10 × 5 = 100 J。
Doubling the speed of an object doubles its kinetic energy. · 把一个物体的速度翻倍使它的动能翻倍。
KE = ½mv², so doubling v quadruples KE (2² = 4 times). · KE = ½mv²,所以 v 翻倍使 KE 变成四倍(2² = 4 倍)。
Conservation of energy
- With no friction, total energy is conserved (conservation of energy).
- Loss in PE = gain in KE (for a falling object): $mgh = \tfrac{1}{2}mv^2$.
能量守恒
- 没有摩擦时,总能量守恒(能量守恒)。
- PE 的损失 = KE 的增加(对一个下落的物体):$mgh = \tfrac{1}{2}mv^2$。
A 1 kg ball falls from 5 m (g = 10). Using mgh = ½mv², find the speed at the bottom (m/s). · 一个 1 kg 的球从 5 m 落下(g = 10)。用 mgh = ½mv²,求底部的速度(m/s)。
mgh = ½mv² → v² = 2gh = 2(10)(5) = 100 → v = 10 m/s. · mgh = ½mv² → v² = 2gh = 2(10)(5) = 100 → v = 10 m/s。
Power
- Power is the rate of doing work (in watts 瓦特): $P = \dfrac{W}{t}$.
- For a force pulling along the motion: $P = Fv$.
- Work is the scalar product of force and displacement; differentiating velocity gives the instantaneous acceleration.
功率
- 功率(power)是做功的速率(以瓦特计):$P = \dfrac{W}{t}$。
- 对一个沿运动方向拉的力:$P = Fv$。
- 功是力与位移的数量积(scalar product);对速度求导得瞬时加速度(instantaneous acceleration)。
A car works at 12 kW at 20 m/s. Using P = Fv, what is the driving force, in N? · 一辆车在 20 m/s 时以 12 kW 工作。用 P = Fv,驱动力是多少,以 N 计?
F = P/v = 12000/20 = 600 N. · F = P/v = 12000/20 = 600 N。
You've got it
- work $W = Fd\cos\theta$; KE $= \tfrac12 mv^2$; PE $= mgh$
- with no friction, total energy is conserved
- power = rate of work; $P = Fv$
你掌握了
- 功 $W = Fd\cos\theta$;KE $= \tfrac12 mv^2$;PE $= mgh$
- 没有摩擦时,总能量守恒
- 功率 = 做功的速率;$P = Fv$