Work, Energy, and Simple Machines

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Work, Energy, and Simple Machines

TL;DR

You'll learn how "work" has a strict physics meaning: force times distance in the direction of motion. You'll see how energy transfers between forms while the total stays constant, and how simple machines don't create energy — they trade force for distance. By the end, you can calculate work, mechanical advantage, and efficiency for real tools.

1. The Mental Model

Work is just energy changing hands — push something and you hand it energy equal to force times the distance it moves. A machine is a deal-maker: it lets you push with less force if you're willing to push through more distance, but it never gives you extra energy for free. Energy itself just moves between stored (potential) and moving (kinetic) forms, never vanishing. A simple machine doesn't save you effort — it only reshapes how you spend it.

2. The Core Material

What "Work" Actually Means

In everyday speech, "work" means effort. In physics, it means something precise:

$$W = F \times d \times \cos\theta$$

where $F$ is the applied force (newtons), $d$ is the distance the object moves (metres), and $\theta$ is the angle between the force and the direction of motion. Work is measured in joules (J), where 1 J = 1 N·m.

Two things trip people up here. First: if you push on a wall and nothing moves, $d = 0$, so you've done zero work — no matter how tired you feel. Second: if you carry a box at constant height across a room, gravity acts straight down while you move horizontally, so $\theta = 90°$ and $\cos 90° = 0$. Gravity does no work on that box, even though you're sweating.

Example: You push a 20 kg crate 5 m across a floor with a horizontal force of 60 N.
$$W = 60 \text{ N} \times 5 \text{ m} \times \cos 0° = 300 \text{ J}$$

Energy: The Currency Behind the Work

Energy is the capacity to do work, and it comes in forms that convert into each other but never disappear — this is the law of conservation of energy. The two forms you'll use constantly:

Kinetic energy (energy of motion):
$$KE = \frac{1}{2}mv^2$$

Gravitational potential energy (stored energy due to height):
$$PE = mgh$$

where $m$ is mass (kg), $v$ is velocity (m/s), $g \approx 9.8 \text{ m/s}^2$, and $h$ is height (m).

Drop a 2 kg ball from 5 m. At the top, all its energy is potential:
$$PE = 2 \times 9.8 \times 5 = 98 \text{ J}$$

Just before it hits the ground, that 98 J has converted almost entirely into kinetic energy (ignoring air resistance), so you can solve for its speed:
$$98 = \frac{1}{2}(2)v^2 \implies v^2 = 98 \implies v \approx 9.9 \text{ m/s}$$

This swap — potential becoming kinetic — is the same physics behind a rollercoaster, a pendulum, or a car going downhill with the engine off. Nothing is created; it's just changing form.

Simple Machines: Trading Force for Distance

A simple machine (lever, pulley, inclined plane, wheel and axle, wedge, screw) changes the size of the force you need, but because energy is conserved, it must change the distance you push over to compensate. The key measure is mechanical advantage (MA):

$$MA = \frac{F_{output}}{F_{input}} = \frac{\text{load}}{\text{effort}}$$

For a lever, MA depends on where the fulcrum sits:
$$MA = \frac{d_{effort}}{d_{load}}$$

Example — a lever: You place a fulcrum so your effort arm is 2 m long and the load arm is 0.5 m.
$$MA = \frac{2}{0.5} = 4$$
This means a 100 N push on your end can lift a 400 N load. But you'll push your end down 4 times farther than the load rises — the work in roughly equals the work out (ignoring friction), because:
$$W_{in} = F_{effort} \times d_{effort} = 100 \times (4 \times d_{load})$$
$$W_{out} = F_{load} \times d_{load} = 400 \times d_{load}$$
Both equal $400 \times d_{load}$. No free lunch — just a trade.

Real machines lose some energy to friction and heat, so efficiency matters:
$$\text{Efficiency} = \frac{W_{out}}{W_{in}} \times 100\%$$

A pulley system might have 80% efficiency: you put in 500 J but only get 400 J of useful lifting work out — the rest became heat from rope friction.

flowchart TD
    A["You apply input force over input distance"] --> B{"Simple machine<br/>(lever, pulley, ramp)"}
    B --> C["Output force changes<br/>(bigger or smaller than input)"]
    B --> D["Output distance changes<br/>(opposite trend to force)"]
    C --> E["Work out = Work in × efficiency"]
    D --> E
    E --> F["Some energy lost as heat/friction<br/>(efficiency < 100%)"]

Connecting Back: Forces, Acceleration, and Stopping Distance

This ties directly into the rest of your course. Newton's second law, $F = ma$, tells you that a net force causes acceleration. When forces are balanced (net force = 0), an object moves at constant velocity or stays still — no acceleration, as in level flight when lift equals weight and thrust equals drag.

When a car brakes, the work-energy idea explains stopping distance. The brakes do negative work (friction force opposing motion) to remove the car's kinetic energy:
$$F_{friction} \times d_{stop} = \frac{1}{2}mv^2$$

Rearranging:
$$d_{stop} = \frac{mv^2}{2F_{friction}} = \frac{v^2}{2a}$$

Notice $v^2$ — this is why doubling your speed quadruples your stopping distance, not just doubles it. A car going 60 km/h doesn't stop in twice the distance of one going 30 km/h; it needs four times the distance, because kinetic energy depends on velocity squared.

This same $v^2$ relationship explains lift on an aircraft wing: faster airflow over the curved upper surface creates lower pressure (via Bernoulli's principle), and the lift force roughly scales with the square of airspeed. That's why planes need long runways to reach takeoff speed — small increases in speed produce large increases in lift.

3. Worked Example

Problem: A 1200 kg car is traveling at 20 m/s (72 km/h) when the driver slams the brakes. The brakes apply a constant friction force of 6000 N. Find (a) the car's kinetic energy before braking, (b) the stopping distance, and (c) what happens to all that kinetic energy.

Step 1 — Kinetic energy before braking:
$$KE = \frac{1}{2}mv^2 = \frac{1}{2}(1200)(20)^2 = \frac{1}{2}(1200)(400) = 240{,}000 \text{ J}$$

Step 2 — Stopping distance. The brakes must do negative work equal to this KE to bring the car to rest:
$$W_{friction} = F \times d = KE$$
$$6000 \times d = 240{,}000$$
$$d = 40 \text{ m}$$

Step 3 — Where did the energy go? It didn't disappear. Friction between brake pads and rotors converted that 240,000 J of kinetic energy into heat (and a small amount of sound). This is conservation of energy in action — the car didn't lose energy, it transferred it into a form you can't use for motion anymore.

Check with the $v^2$ rule: If this car had been going 40 m/s instead (double the speed), stopping distance would be:
$$d = \frac{(40)^2}{(20)^2} \times 40 = 4 \times 40 = 160 \text{ m}$$
Four times farther — confirming the quadratic relationship between speed and stopping distance.

4. Key Takeaways

4.1 Most Important Concepts

1. Work requires movement in the force's direction — $W = Fd\cos\theta$; no displacement or a perpendicular force means zero work done.
2. Energy is conserved, only converted — potential, kinetic, heat, and sound trade places, but the total stays fixed in a closed system.
3. Machines trade force for distance — mechanical advantage lets you push with less force over a longer distance, but total work in (ideally) equals total work out.
4. Real machines aren't 100% efficient — friction turns some input work into waste heat, so output work is always somewhat less than input work.
5. Stopping distance scales with $v^2$ — doubling speed quadruples the distance needed to stop, because kinetic energy depends on velocity squared.
6. Balanced forces mean zero acceleration — constant velocity (including level flight) happens when all forces cancel out, not when there are no forces at all.
7. Lift depends strongly on speed — because lift relates to airspeed squared, small speed changes near takeoff cause large lift changes.

4.2 Common Misconceptions

• "Doing work means getting tired." Physics work needs displacement in the force's direction; holding a heavy bag still, however exhausting, is zero work in the physics sense.
• "Machines create extra energy." They never do — mechanical advantage reduces the force needed only by increasing the distance over which it's applied; total energy out can't exceed energy in.
• "Balanced forces mean no forces are acting." Balanced just means the forces cancel to a net force of zero; a plane in level flight has four large forces (lift, weight, thrust, drag) all present and balanced in pairs.
• "Stopping distance doubles when speed doubles." Because kinetic energy involves $v^2$, stopping distance actually quadruples — this is a genuinely dangerous misconception for new drivers.

4.3 Compare & Contrast

5. Now Try It

Exam-style question: A cyclist and bike have a combined mass of 80 kg. They coast down a hill, dropping 10 m in height, starting from rest and ignoring air resistance and friction.

(a) Calculate the potential energy lost.
(b) Use conservation of energy to find the cyclist's speed at the bottom of the hill.
(c) The cyclist then brakes on flat ground, and the brakes exert a friction force of 400 N. Calculate the stopping distance.
(d) Explain, using the concept of work, why halving the braking force would more than double the stopping distance's dependence — actually, explain what happens to stopping distance if friction force is halved while speed stays the same.

A top-band answer must:
- Show full working with units at every step (not just final numbers).
- Use $PE = mgh$ and $KE = \frac{1}{2}mv^2$ correctly, setting them equal for part (b).
- Correctly rearrange $W = Fd$ for part (c), getting a numeric distance in metres.
- In part (d), explicitly state that stopping distance is inversely proportional to friction force, so halving the force doubles the distance — and connect this back to the $W = Fd = KE$ relationship.

Success looks like: you get a numeric speed around 14 m/s in part (b), a stopping distance around 20 m in part (c), and a clear written explanation in part (d) linking force, distance, and constant kinetic energy.