功、能与功率

A pole vaulter sprints, plants the pole, and rises over a bar five metres up. No engine lifts her. The speed she brought to the pole becomes bend in the pole, then height. This unit tracks energy as it moves between stores like these, and shows how to account for every joule.

Energy 能量 is the capacity to do work, measured in joules 焦耳 (J). A moving object has kinetic energy 动能:

It depends on the square of the speed, so doubling the speed quadruples the kinetic energy. Kinetic energy is a scalar 标量 and is never negative.

Worked example. A $1500\ \text{kg}$ car travels at $20\ \text{m/s}$. Its kinetic energy is $K=\tfrac12\times1500\times20^2=3.0\times10^{5}\ \text{J}=300\ \text{kJ}$. If it speeds up to $40\ \text{m/s}$ (double), the kinetic energy becomes $4\times$ larger, $1200\ \text{kJ}$ – which is why stopping distance 制动距离 grows so fast with speed.

Work 功 is energy transferred by a force acting over a displacement 位移. For a constant force:

where $\theta$ is the angle between the force and the displacement. Work is positive when the force has a component along the motion (adds energy), negative when it opposes the motion (removes energy), and zero when the force is perpendicular 垂直. On a force–position graph, work is the area under the curve. The work–energy theorem 动能定理 states that the net work equals the change in kinetic energy: $W_{\text{net}}=\Delta K$.

Worked example. A $2.0\ \text{kg}$ block moving at $3.0\ \text{m/s}$ on a frictionless 无摩擦的 floor is pushed by a $5.0\ \text{N}$ force over $4.0\ \text{m}$ in the direction of motion. Find its final speed. The net work is $W=Fd=5.0\times4.0=20\ \text{J}$, and by the work–energy theorem $W=\tfrac12 m(v^2-v_0^2)$:

Potential energy 势能 is stored energy that depends on position or configuration:

• Gravitational potential energy 重力势能 near the surface: $U_g=mgh$ (height $h$ above a reference level 参考面).
• Elastic potential energy 弹性势能 in a spring: $U_s=\tfrac{1}{2}kx^2$.

Potential energy is defined only for conservative forces 保守力 (gravity, springs), for which the stored energy depends on position, not path. Only changes in potential energy matter, so you may put the zero level wherever is convenient.

The $U_g=mgh$ form only holds near a surface where $g$ is roughly constant. The general form, for two spherical masses a distance $r$ apart, is

It is negative and defined to be zero at infinite separation, so gravitational PE rises toward zero as the masses move apart. $mgh$ is simply its near-surface approximation. (This is the form you need for satellites 卫星 and escape speed 逃逸速度.)

The total mechanical energy 机械能 is $E=K+U$. When only conservative forces do work, mechanical energy is conserved 守恒:

Friction 摩擦力 can transfer mechanical energy to thermal energy 热能. Other non-conservative forces 非保守力 can add mechanical energy as well. Total energy is conserved when all stores and transfers across the chosen system boundary are included. Energy bar charts are a good way to track where the energy goes.

Energy bar charts: friction moves one unit into thermal energy, but the total stays four

An energy bar chart 能量柱状图 shows each store at the start and the end. The bars must add to the same total, unless work, heat or another energy transfer crosses the system boundary. Draw the chart before writing an equation: the equation is then just the chart in symbols, $U_{g1}=K_2+\Delta E_{th}$.

Worked example. A ball is released from rest at the top of a frictionless ramp $5.0\ \text{m}$ high. Find its speed at the bottom. All the gravitational potential energy becomes kinetic energy:

The mass cancels, so every object reaches the same speed – exactly the free-fall result, now got from energy. If instead $30\ \text{J}$ were lost to friction, you would subtract it: $mgh-30=\tfrac12 mv^2$.

The most common error in this unit is counting gravity twice. Decide first whether the Earth is inside the system 系统.

• Earth inside the system: gravity is an internal force 内力, so you use gravitational potential energy $U_g=mgh$. Gravity does no "outside" work.
• Earth outside the system: there is no $U_g$; instead gravity does work on the object, $W_g=-mg\,\Delta h$.

Both give the same answer. Using $U_g$ and $W_g$ together counts the same energy twice. For a $10\ \text{kg}$ box lifted $5.0\ \text{m}$ at constant speed, the person does $490\ \text{J}$ of work, not $980\ \text{J}$.

When the force changes, $W=Fd$ no longer works directly. Instead, work is the area under a force–displacement graph. A spring's force grows steadily, $F=kx$, so the graph is a triangle with area $\tfrac12(kx)(x)=\tfrac12kx^2$. That is where the elastic potential energy formula comes from.

Worked example (spring launcher). A spring with $k=200\ \text{N/m}$ is compressed by $0.10\ \text{m}$ and launches a $0.050\ \text{kg}$ ball straight up. Ignore friction and the small height change while the spring extends.

• Energy stored: $U_s=\tfrac12(200)(0.10)^2=1.0\ \text{J}$.
• Launch speed: $\tfrac12(0.050)v^2=1.0$, so $v=\sqrt{40}=6.3\ \text{m/s}$.
• Highest point: $(0.050)(9.8)h=1.0$, so $h=2.0\ \text{m}$.

Worked example (friction on an incline). A $2.0\ \text{kg}$ block starts from rest at the top of a ramp $3.0\ \text{m}$ long and $1.5\ \text{m}$ high. A friction force of $4.0\ \text{N}$ acts along the ramp. Find its speed at the bottom.

Friction turned $12\ \text{J}$ into thermal energy. Without it the block would reach $\sqrt{29.4}=5.4\ \text{m/s}$.

Drawing it as an energy bar chart. Free-response questions often ask for this as bars. Choose the system first: here the block, the Earth and the ramp, so friction's work stays inside as thermal energy $\Delta E_{\text{th}}$. Draw one bar for each energy store at the start and at the end. The totals must match, because no outside force does work on this system.

If the system were the block alone, there would be no $U_g$ bar: gravity would be an outside force doing $+29.4\ \text{J}$ of work, and friction an outside force doing $-12\ \text{J}$. Either choice gives the same speed; what changes is which bars you draw.

Worked example (energy and forces together: a loop). A small cart starts from rest at height $h$ on a frictionless track and goes round a vertical loop of radius $R = 0.40\ \text{m}$. What is the smallest $h$ for which it stays on the track at the top of the loop?

• Forces at the top (from the circular-motion unit): the cart just keeps contact when the normal force 法向力 is zero, so $mg = \dfrac{mv^2}{R}$, giving $v_{\text{top}}^2 = gR$.
• Energy from the start to the top (height $2R$): $mgh = mg(2R) + \tfrac12 m v_{\text{top}}^2 = 2mgR + \tfrac12 mgR$.

Neither idea alone is enough: energy gives the speed at the top, and Newton's second law says how fast it must be. Starting at only $h = 2R$ fails, because the cart would need zero speed at the top, and then it falls off before it gets there.

Power 功率 is the rate of doing work or transferring energy, measured in watts 瓦特 (W):

So the same job done faster requires more power. On an energy–time graph, power is the slope 斜率.

Worked example. A motor lifts a $50\ \text{kg}$ load at a steady $2.0\ \text{m/s}$. Because it moves at constant speed, the lifting force equals the weight, so

Real machines waste some energy, so we quote efficiency 效率 – useful output power divided by total input power. If this motor draws $1400\ \text{W}$ of electrical power to deliver $980\ \text{W}$ of useful lifting, its efficiency is $980/1400=0.70$, or $70\%$; the other $30\%$ becomes heat and sound.

Answer each before reading the answers below. Take $g=9.8\ \text{m/s}^2$.

1. A $0.20\ \text{kg}$ ball is thrown straight up at $12\ \text{m/s}$. Ignoring air resistance 空气阻力, how high does it rise?
2. A spring with $k=400\ \text{N/m}$ is compressed $0.050\ \text{m}$ and launches a $0.10\ \text{kg}$ block along a frictionless floor. Find the block's speed.
3. A $10\ \text{kg}$ box is pushed $5.0\ \text{m}$ across a floor at constant speed against a $30\ \text{N}$ friction force. Find the work done by the person, the net work, and the thermal energy produced.
4. A $60\ \text{kg}$ student climbs stairs $4.0\ \text{m}$ high in $8.0\ \text{s}$. Find the student's average power output for the lifting.
5. A student says: "Lifting the box $5.0\ \text{m}$ needs $980\ \text{J}$: gravity does $-490\ \text{J}$ of work, and the potential energy also rises by $490\ \text{J}$." Correct the student.

1. $\tfrac12mv^2=mgh$, so $h=\dfrac{v^2}{2g}=\dfrac{144}{19.6}=7.3\ \text{m}$. The mass cancels.
2. $\tfrac12kx^2=\tfrac12mv^2$, so $v=x\sqrt{k/m}=0.050\sqrt{4000}=3.2\ \text{m/s}$.
3. Person: $(30)(5.0)=150\ \text{J}$. Net work is zero, because the speed is constant. Thermal energy produced: $150\ \text{J}$.
4. $P=\dfrac{mgh}{t}=\dfrac{(60)(9.8)(4.0)}{8.0}=290\ \text{W}$.
5. The student has counted gravity twice. With the Earth in the system, use only $\Delta U_g=490\ \text{J}$. With the Earth outside, use only gravity's work, $-490\ \text{J}$. Either way the person must supply $490\ \text{J}$.

• Use $W=Fd\cos\theta$: work is zero when the force is perpendicular to the motion, and negative when it opposes it.
• Reach for the work–energy theorem ($W_{\text{net}}=\Delta K$) or energy conservation ($K_1+U_1=K_2+U_2$) instead of forces whenever the path is complicated.
• When friction acts, mechanical energy is not conserved — subtract the energy lost to heat.
• Remember $K\propto v^2$: doubling the speed quadruples the kinetic energy (and, for a constant braking force, the stopping distance).
• Use $P=Fv$ for power at a steady speed; at constant velocity the net force is zero but the power is not.

Common mistakes

• Counting gravity twice: using $U_g$ and the work done by gravity in the same equation. Choose one, depending on whether the Earth is in the system.
• Using $W = Fd$ when the force is at an angle to the motion. Only the component along the displacement does work: $W = Fd\cos\theta$.
• Saying the normal force or a centripetal force 向心力 does work on an object moving in a circle. A force perpendicular to the velocity does no work.
• Writing $\tfrac12 kx$ or $kx^2$ for spring energy. The stored energy is $\tfrac12 kx^2$ (the triangle under $F = kx$).
• Treating energy as a vector. Kinetic and potential energy have no direction, so a ball's speed at a given height is the same whether it moves up or down (no friction).
• Leaving out thermal energy when friction acts. The lost mechanical energy equals $f\,d$ and appears as thermal energy; it does not disappear.