Chapter 2: Work and Energy

In everyday English, “work” means effort. You worked hard at the gym. You pulled an all-nighter. You worked all summer at a coffee shop.

Physics is much more particular. To a physicist, you only do work if you push or pull something and it actually moves and the force has at least some component in the direction of motion. Stand still holding a 50 lb weight overhead for an hour and your muscles will be on fire — but in physics terms, you’ve done zero work on that weight, because nothing moved.

This is one of those sections where intuition will betray you. The MCAT writes great trap questions around it. Lock in the formula, learn the three cases (positive, negative, zero), and the rest is plug-and-chug.

The Work Equation

Work is a scalar, not a vector — it can be positive, negative, or zero, but it has no direction.

The whole reason this equation is more nuanced than “force times distance” is the $\cos\theta$ term. That cosine picks out only the part of the force that lines up with the motion. A force perpendicular to motion does zero work; a force directly opposite to motion does negative work.

The Three Cases: Positive, Negative, Zero

The sign of work tells you whether energy is flowing into or out of the object.

Special Cases Worth Memorizing

• $\theta = 0°$ (force in same direction as motion): $\cos 0° = 1$, so $W = Fd$. Maximum work for a given force and distance. Example: pushing a crate in a straight line behind it.
• $\theta = 180°$ (force opposite to motion): $\cos 180° = -1$, so $W = -Fd$. Negative work. Example: kinetic friction always opposes motion, so friction always does negative work on a sliding object.
• $\theta = 90°$ (force perpendicular to motion): $\cos 90° = 0$, so $W = 0$. Examples: the normal force on a box sliding across the floor; the centripetal force on an object in circular motion.

Work Done by Multiple Forces

When several forces act on an object at once, you have two equally valid options:

1. Calculate work for each force separately, then add them: $W_{net} = W_1 + W_2 + W_3 + \ldots$
2. Find the net force first, then compute: $W_{net} = F_{net} \cdot d \cdot \cos\theta$.

Both give the same answer. Pick whichever is faster for the question in front of you.

Gravity’s Work

Gravity deserves special attention because it shows up in nearly every energy problem. The work done by gravity on an object that moves vertically by height $h$ is:

• $W_{gravity} = -mgh$ if the object moves up (gravity opposes the motion).
• $W_{gravity} = +mgh$ if the object moves down (gravity helps the motion).

If the object moves along a diagonal path (like up a ramp), only the vertical component of displacement matters for gravity’s work. Horizontal motion is perpendicular to gravity, so it contributes nothing. This is why a 10 m walk across a flat floor and a 10 m walk across a higher floor both involve zero gravitational work — but climbing stairs to that higher floor does involve work.

Worked Example

You push a 20 kg box across a frictionless floor with a 30 N force at 60° below horizontal (you’re pushing down on a handle), moving it 5 m. How much work do you do?

• Use $W = Fd\cos\theta$ with $\theta = 60°$.
• $W = 30 \times 5 \times \cos 60° = 30 \times 5 \times 0.5 = 75$ J.

Your full 30 N push covered 5 m, but only $30\cos 60° = 15$ N of it actually pointed forward. The other $30\sin 60° \approx 26$ N pressed straight down into the floor and did nothing useful. That’s why the work is 75 J, not 150 J.

The equation $W = Fd\cos\theta$ works perfectly when the force is constant. But what if the force changes as the object moves?

Stretch a rubber band. The first centimeter is easy. The next centimeter is harder. By the time you’ve pulled it 10 cm, you’re really straining. The force isn’t a single number — it’s a different value at every position. You can’t just multiply “the force” by “the distance,” because there’s no single “the force.”

This is where the graphical method comes in. It works for any force — constant, increasing, decreasing, weird-shaped — and the math never goes beyond basic geometry.

The Graphical Method

When force changes with position, work equals the area under the force-vs-displacement (F vs. x) graph.

This is the MCAT-level approach: no calculus required. The MCAT picks problems where the graph is made of straight lines, so the area is always a combination of rectangles, triangles, and trapezoids — shapes you’ve known the area formulas for since middle school.

Common Graph Shapes and Their Areas

Spring Work: The Most Important Triangle

A spring exerts a force proportional to how far it’s stretched or compressed: $F = kx$ (we’ll cover the negative sign and Hooke’s law in §2.5). The force starts at zero when the spring is at its natural length and grows linearly with displacement, so the F vs. x graph is a straight line through the origin.

The area under that line is a triangle:

Notice the $x^2$. Stretching a spring twice as far takes four times as much work, not twice as much. That’s why pulling back a bow string the last few inches is so much harder than the first few.

Reading Complex Graphs

Some MCAT problems give you a piecewise graph — force constant for a while, then changing in a straight line, then maybe constant again. The recipe: break the graph into simple shapes, find each area, and add them up.

For example, if a graph shows a constant 10 N for the first 3 m (rectangle), then a force linearly decreasing from 10 N to 0 N over the next 2 m (triangle):

• Rectangle: $10 \text{ N} \times 3 \text{ m} = 30$ J.
• Triangle: $\tfrac{1}{2} \times 2 \text{ m} \times 10 \text{ N} = 10$ J.
• Total: $30 + 10 = 40$ J.

That’s the entire technique. Identify shapes, sum areas, done.

Positive and Negative Areas

If the force dips below the x-axis (negative force) over some interval, that region contributes negative work — the force was opposing motion in that stretch.

A common scenario: a spring being released from compression. While you compress it, you do positive work on the spring. When you let go and the spring pushes your hand back, the spring does positive work on your hand, and your hand (now resisting) does negative work on the spring (because hand and spring move in the same direction but the spring’s force on the hand is opposite to the hand’s force on the spring — third law again).

When in doubt, always specify which force you’re calculating work for, and on which object.

Worked Example

A bow archer draws back a bowstring that requires linearly increasing force, from 0 N at rest to 200 N at full draw (a draw distance of 0.5 m). How much work does the archer do on the string?

• F vs. x graph: triangle, base = 0.5 m, height = 200 N.
• $W = \tfrac{1}{2}(0.5)(200) = 50$ J.

So 50 J of energy is now stored in the bent bow, ready to fly into the arrow when released. (Roughly the same energy as a small car moving at 1 m/s — modest, but enough to send a light arrow at 60+ m/s.)

A bowling ball rolling at 5 m/s and a tennis ball rolling at 5 m/s have wildly different abilities to knock things over. A car at 60 mph isn’t twice as dangerous as the same car at 30 mph — it’s four times as dangerous. A pitched 90 mph fastball carries enough energy to break a bone; a 30 mph toss doesn’t.

What captures these differences is kinetic energy — the energy something has because it’s moving. The amount depends on mass and speed, but speed matters a lot more than you might guess.

The Kinetic Energy Equation

Two things to internalize:

1. KE depends on $v^2$, not $v$. Doubling the speed quadruples KE. Tripling speed makes KE 9× bigger. This is the single most important fact in this section.
2. KE depends on mass linearly. A 2,000 kg truck at 10 m/s has 2× the KE of a 1,000 kg car at 10 m/s. But the same truck at 20 m/s has 4× the KE of itself at 10 m/s. Speed wins by a wide margin.

This is why highway accidents are catastrophically worse than parking-lot fender-benders. A 40 mph crash isn’t twice as bad as a 20 mph crash — it’s four times as bad. Energy scales with the square of speed.

The Work-Energy Theorem

There’s a direct, beautiful connection between work and kinetic energy: the net work done on an object equals its change in kinetic energy.

Applying the Work-Energy Theorem

The theorem is powerful because it bypasses acceleration and time entirely. If you know the forces and the distance, you can find the final speed without ever computing $a$ or $t$.

Worked example. A 1500 kg car traveling at 20 m/s brakes to a stop over 50 m. What is the average braking force?

• $KE_i = \tfrac{1}{2}(1500)(20)^2 = 300{,}000$ J.
• $KE_f = 0$ (the car stops).
• $W_{net} = KE_f - KE_i = -300{,}000$ J.
• Net work also equals $F \cdot d \cdot \cos 180° = -F \cdot 50$ (force opposes motion).
• So $-F \cdot 50 = -300{,}000$ → $F = 6000$ N.

That’s it — no kinematic equations, no need for time. We jumped straight from energy in to force out.

When Net Work is Zero

If the net work on an object is zero, its speed doesn’t change. This doesn’t mean no forces are acting — it means the positive and negative work cancel exactly.

A car cruising at constant 70 mph on a flat highway is the textbook case. The engine pushes forward (positive work). Air resistance and rolling friction push backward (negative work). They cancel. Net work = 0, KE constant, speed steady. Lots of forces, no change in motion.

Similarly, an object in a circular orbit has gravity acting on it constantly — but gravity does zero work because it’s always perpendicular to motion. KE is constant; speed is constant; orbit radius is constant.

Kinetic Energy is a Scalar

Unlike momentum (which is a vector), kinetic energy has no direction. Two cars moving in opposite directions at the same speed have the same KE — even though their momenta cancel out. This is why head-on collisions can completely stop both cars (momenta cancel) while still releasing the combined KE of both into deformation, heat, and sound.

This scalar nature also makes energy easier to work with than momentum in many problems. You don’t need to break it into x and y components.

Hold a phone above your kitchen counter and let go. It accelerates downward, picks up speed, and lands with a thud. Where did that kinetic energy come from?

It was stored in the phone the moment you lifted it off the counter. Lifting takes effort — your arm muscles do work against gravity. That work doesn’t disappear; it gets stored as gravitational potential energy (gravitational PE). The higher you raise something, the bigger the deposit, and the more kinetic energy it’ll have when it falls.

The same idea explains roller coasters (lift to the top, then convert to speed on the way down), water in a dam (high water = stored energy, released as the water falls through a turbine), and why falling from 30 feet is much more dangerous than falling from 5.

The Equation

This formula applies near Earth’s surface, where $g$ is approximately constant. For objects far from Earth (satellites, spacecraft), gravity weakens with altitude and the MCAT uses a different formula. But for ~95% of MCAT problems, $PE = mgh$ is all you need.

The Reference Point is Arbitrary

This is the part that confuses students most: the value of PE depends on where you set $h = 0$ (your reference point). And that’s fine — because only changes in PE matter, and the changes come out the same no matter where you put zero.

Consider a ball on the edge of a 10 m cliff above a 5 m deep valley:

• Reference = valley floor: PE at cliff edge = $mg(15)$.
• Reference = base of cliff: PE at cliff edge = $mg(10)$.
• Reference = cliff edge: PE at cliff edge = $mg(0) = 0$.

The PE numbers are different. But if the ball falls from cliff edge to cliff base, every reference choice gives the same $\Delta PE = -mg(10)$, and the same final speed at the cliff base.

PE-KE Conversion

When an object falls freely (no friction, no air resistance), gravitational PE converts directly into KE:

• At the top: maximum PE, minimum KE (object is momentarily at rest, KE = 0).
• At the bottom: minimum PE (zero, if you set the reference there), maximum KE.
• In between: PE shrinking, KE growing, total stays constant.

This is the foundation of energy conservation, which we’ll formalize in §2.6.

Negative Heights

If an object is below your reference point, $h$ is negative, and PE is negative. This is mathematically fine. A ball at the bottom of a well, with the reference set at ground level, has negative PE relative to ground level. As it falls deeper, PE becomes more negative (decreases), and KE increases.

The MCAT occasionally tests this. Don’t be thrown by a negative PE value — it just means the object is below the reference you chose.

Quick Calculation Tip

On the MCAT, $g \approx 10$ m/s² makes PE math fast. A 2 kg object at 5 m above the reference: $PE = 2 \times 10 \times 5 = 100$ J. No calculator, no fuss.

To find the speed of a falling object from height $h$ (starting from rest, no friction): set PE = KE, so $mgh = \tfrac{1}{2}mv^2$. Mass cancels (everything falls the same), giving:

$v = \sqrt{2gh}$

This shortcut shows up constantly. A 5 m drop: $v = \sqrt{2(10)(5)} = \sqrt{100} = 10$ m/s. A 20 m drop: $v = \sqrt{2(10)(20)} = \sqrt{400} = 20$ m/s. (Notice that quadrupling the height only doubles the impact speed — because of that $\sqrt{}$.)

Worked Example

A 2 kg ball is dropped from a 5 m balcony onto a soft mat. What’s its speed just before hitting the mat? How much KE does it have at impact?

• $v = \sqrt{2gh} = \sqrt{2(10)(5)} = \sqrt{100} = 10$ m/s.
• $KE = \tfrac{1}{2}mv^2 = \tfrac{1}{2}(2)(100) = 100$ J.

Sanity check: $PE_{top} = mgh = 2(10)(5) = 100$ J. The ball started with 100 J of PE and ended with 100 J of KE — exactly what energy conservation predicts. ✓

Press a Slinky against the floor and let go — it bounces back. Stretch a rubber band, then release — it snaps back hard enough to sting your fingers. Squish the bumper of a car against a wall and it pops back to shape (within reason).

These are all elastic systems. They share a beautifully simple rule, discovered by Robert Hooke in the 1600s: the more you deform an elastic object, the more strongly it pushes back — and that “push-back” force is exactly proportional to how far you’ve deformed it.

That single rule lets you predict spring forces, calculate stored energy, and analyze everything from car suspensions to pole vaulters to the elasticity of human tendons.

Hooke’s Law

The Spring Constant ($k$)

The spring constant tells you how stiff the spring is. Big $k$ = stiff spring (a car suspension); small $k$ = soft spring (a Slinky).

Units of $k$: N/m. If $k = 500$ N/m, the spring exerts 500 N of restoring force for every meter of displacement (or 5 N for every centimeter).

The Negative Sign

The minus sign in $F = -kx$ is the single most important detail of the equation. It’s what makes the spring force a restoring force — always directed back toward the equilibrium position.

• Stretch the spring ($x > 0$) → force is negative (pulls back toward center).
• Compress the spring ($x < 0$) → force is positive (pushes back toward center).

Elastic Potential Energy

A compressed or stretched spring stores energy. That stored energy is elastic potential energy.

This is the same formula we derived graphically in §2.2 (the area of the triangle under $F$ vs. $x$).

Notice the $x^2$. Just like KE depends on $v^2$, elastic PE depends on $x^2$ — doubling the displacement quadruples the stored energy. This is why pulling a bowstring back the last few inches takes so much more effort than the first few.

Symmetry of Elastic PE

Because $x$ is squared, a spring compressed by 3 cm stores exactly the same energy as the same spring stretched by 3 cm. The direction of displacement doesn’t matter — only the magnitude.

This symmetry is also why a spring oscillates back and forth between equal stretch and compression. The energy at the two extremes is identical, so the motion is symmetric.

Springs and Simple Harmonic Motion

When a mass attached to a spring is pulled and released, it oscillates back and forth. Energy continuously trades between KE and elastic PE:

• At maximum stretch (or compression): all PE, zero KE, mass momentarily at rest.
• At equilibrium position (natural length): all KE, zero PE, maximum speed.
• In between: a mix of both, total mechanical energy stays constant.

That oscillation is simple harmonic motion (SHM), covered in detail in the waves and sound chapter. For now: a spring-mass system is the prototype example.

When Hooke’s Law Breaks Down

Hooke’s law only holds within the elastic limit — the range where the material returns to its original shape after you release it. Stretch a spring (or a tendon, or any elastic material) too far and it deforms permanently. Beyond the elastic limit, the F vs. x graph stops being a straight line, and $F = -kx$ no longer applies.

That’s why a Slinky stretched all the way down a flight of stairs never quite returns to its original shape — you exceeded its elastic limit. Same idea for a paper clip you’ve bent too far, or a rubber band that’s been stretched until it goes slack.

Worked Example

A spring with $k = 400$ N/m is compressed 0.1 m and used to launch a 2 kg block on a frictionless surface. What’s the block’s speed when the spring returns to its natural length?

• Stored PE in spring: $PE = \tfrac{1}{2}kx^2 = \tfrac{1}{2}(400)(0.01) = 2$ J.
• All converts to KE: $\tfrac{1}{2}mv^2 = 2$ → $v^2 = 2$ → $v = \sqrt{2} \approx 1.41$ m/s.

The 2 J of stored energy fully transferred into the block’s KE the moment the spring reached its natural length. After that, the spring has no more energy to give and the block coasts away at 1.41 m/s.

A roller coaster grinds slowly up its first hill, pauses for a breathless second at the top, then plunges down — every joule of stored gravitational PE converting into screaming kinetic energy. If you could somehow eliminate friction and air resistance, the coaster could climb back up to exactly the same height on the next hill, without an engine, without any extra push. That’s conservation of energy in action.

This idea — energy doesn’t disappear, it just changes form — is one of the most powerful tools in physics. It lets you skip past acceleration, time, and force-balancing and jump straight from “starting condition” to “ending condition” using a single equation.

Test your prediction below. Set a start height, press Release, and watch the stacked energy bar: blue PE trades for green KE while the total stays pinned. With friction at zero the ball climbs back to its start height on every pass; add friction and the orange heat band grows until the oscillation dies out, or raise the start height above the 5 m hill and watch the ball escape over the top.

Conservative vs. Non-Conservative Forces

The key to applying conservation of energy is knowing which forces play nice with it.

Conservative forces have two defining properties:

1. The work they do is path-independent (only the start and end positions matter, not the route taken).
2. The work done over a closed loop (back to the same starting point) is zero.

| Conservative forces | Non-conservative forces |
|--------------------|------------------------|
| Gravity | Friction |
| Spring (elastic) force | Air resistance |
| Electrostatic force | Tension (in many setups) |
| | Applied push/pull |

Path Independence: What It Really Means

Carry a 1 kg ball from the floor up to a shelf 2 m high. You could:

• Lift it straight up.
• Carry it up a spiral staircase.
• Hike it up a winding mountain trail until you reach 2 m above the floor.

In every case, gravity does the same amount of work: $W_{gravity} = -mgh = -2(10)(2) = -40$ J. The path is irrelevant — only the height difference matters. That’s path independence.

Now slide a box across the floor. A short straight path produces less friction work than a long zigzag path covering more ground. Friction’s work depends entirely on the path length, so friction is non-conservative.

Applying Conservation of Energy

When only conservative forces act, the recipe is short and reliable:

1. Choose the system and identify the initial and final states.
2. Set a reference point for PE (usually the lowest point in the problem = 0).
3. Write $KE_i + PE_i = KE_f + PE_f$.
4. Plug in knowns and solve.

Worked example. A 2 kg ball is dropped from 5 m. What’s its speed just before hitting the ground?

• $KE_i + PE_i = KE_f + PE_f$
• $0 + mgh = \tfrac{1}{2}mv^2 + 0$
• $(2)(10)(5) = \tfrac{1}{2}(2)v^2$
• $100 = v^2$ → $v = 10$ m/s.

Notice mass cancels when the only energies are gravitational PE and KE. That’s exactly why all objects fall at the same rate in a vacuum.

When Non-Conservative Forces Are Present

When friction (or air resistance, or any non-conservative force) does work, mechanical energy is not conserved. Some KE/PE gets siphoned off into heat, sound, or deformation. The equation becomes:

The Round-Trip Test

A quick way to check whether a force is conservative: imagine moving an object in a complete loop back to its starting point. If the force does zero net work over that round trip, it’s conservative.

• Gravity: lift a ball 5 m, then lower it 5 m. Gravity does $-mgh$ on the way up and $+mgh$ on the way down. Net = 0. ✓ Conservative.
• Friction: slide a box 5 m right, then 5 m left back to start. Friction opposes motion both ways, so it does negative work in both directions. Net is not zero — energy was bled into heat in both legs. Non-conservative.

This test instantly classifies any force you encounter.

Two people each carry a 20 kg box up the same flight of stairs. One sprints up in 5 seconds; the other plods up in 30 seconds. They both did the same work — same mass, same height — but most people would agree the sprinter “worked harder” in some sense.

That intuition is what physics calls power: the rate at which work is done. Power isn’t about how much total energy you transfer — it’s about how quickly. The sprinter delivered the same energy in $\frac{1}{6}$ the time, so their power output was 6× higher.

Once you separate “energy” from “energy per second,” a lot of everyday things click. A 100 W bulb and a 60 W bulb both convert electrical energy into light, but the 100 W bulb does it faster — that’s why it’s brighter. A small car engine and a sports car engine can both move the cars forward, but the sports car engine can deliver that energy faster — so it accelerates harder.

The Power Equations

The second formula isn’t a new concept — it’s just an algebraic shortcut. Combine $P = W/t$ with $W = Fd$: $P = Fd/t = F(d/t) = Fv$. Same physics, different inputs. Use whichever the problem makes easy.

Units of Power

For quick estimation: 1 hp ≈ 750 W, so 60 hp ≈ 45 kW. A typical microwave is ~1000 W = 1 kW. A typical car engine peaks at ~150 hp ≈ 110 kW. A fit human can sustain ~200–400 W indefinitely (and briefly hit much higher peaks).

Power and Energy Over Time

Since $P = W/t$, you can rearrange to $W = Pt$ — the energy transferred over a time interval equals power times time. This is exactly how electricity bills are calculated: you pay for energy in kilowatt-hours (kWh), not in watts.

$1 \text{ kWh} = (1000 \text{ W})(3600 \text{ s}) = 3{,}600{,}000 \text{ J} = 3.6 \text{ MJ}$

Run a 100 W bulb for 10 hours = 1 kWh of energy. At ~$0.15 per kWh (US average), that’s about 15 cents.

Applying $P = Fv$

A car engine has to push hard enough to overcome friction and air resistance. At a constant highway speed (no acceleration), the engine’s forward push must exactly cancel the backward resistive forces. The power the engine has to deliver is $P = Fv$, where $F$ is the resistive force and $v$ is the cruising speed.

This means a higher cruising speed needs more power, even if you assume the resistive forces stay the same. And since air resistance actually grows quickly with speed (roughly with $v^2$), the power needed to cruise at 80 mph is several times the power needed at 40 mph. That’s the main reason highway driving uses way more gas than city driving (per mile of distance).

Average vs. Instantaneous Power

$P = W/t$ gives the average power over a time interval. $P = Fv$ (with instantaneous $v$) gives the instantaneous power at one specific moment.

On the MCAT, most problems are about average power, but read the question carefully — “the engine’s peak power” or “the power at the moment v = 30 m/s” both ask for instantaneous values.

Worked Example

A 70 kg person runs up a 5 m staircase in 4 seconds. What’s their average power output?

• Work against gravity: $W = mgh = (70)(10)(5) = 3500$ J.
• Power: $P = W/t = 3500/4 = 875$ W.

That’s about 1.2 hp — a serious sustained burst, but typical for a fit person sprinting upstairs. The body can briefly exceed this, but most adults can only sustain a few hundred watts for any extended period.

You can’t open a door by pushing on the hinge. You push near the handle — as far from the hinge as possible — because that’s where your push has the most rotational effect. Try pushing 1 cm from the hinge and you’ll be amazed at how stiff the door feels. Push at the handle (~80 cm out) and it swings easily. Same door, same hinge, same push — totally different result.

That “rotational effect of a force” is torque. It’s the rotational version of force: where ordinary force makes things accelerate in a straight line, torque makes things spin around a pivot. The MCAT uses torque in seesaws, beams, levers, the human body (every joint!), and any time a problem mentions a hinge, a pivot, or a fulcrum.

The Torque Equation

The product $r\sin\theta$ — or equivalently $F\sin\theta$ times $r$ — is called the lever arm (the perpendicular distance from the pivot to the line of action of the force). You can read the formula two equivalent ways:

1. $\tau = (r\sin\theta) \times F$ = lever arm × full force.
2. $\tau = r \times (F\sin\theta)$ = full distance × perpendicular component of force.

Both give the same answer. Use whichever matches your mental picture of the problem.

Sign Convention for Torque

Torque has a sign that tells you which way it’s spinning:

• Counterclockwise (CCW) = positive (standard convention).
• Clockwise (CW) = negative.

The MCAT may use either convention — what matters is consistency. Pick one for a given problem and stick with it.

Maximizing Torque

Since $\tau = rF\sin\theta$, you maximize torque by making each factor as big as possible:

• $r$ as large as possible (push far from the pivot).
• $F$ as large as possible (push harder).
• $\theta = 90°$ (push perpendicular to the lever arm, since $\sin 90° = 1$).

This is why door handles are placed far from hinges, why long wrenches loosen stubborn bolts, and why a bicycle pedal arm is a foot long, not an inch. All of them are torque-maximizing geometry.

Static Equilibrium

An object is in static equilibrium when it’s not moving and not rotating. That requires two conditions, not one:

The “two conditions, not one” point is what makes equilibrium problems trickier than basic Newton’s-law problems. You can balance forces perfectly and still get rotation; you can balance torques perfectly and still get sliding. You need both.

Interactive Seesaw Simulator

Place two weights on opposite sides of a fulcrum and watch the beam respond. Drag the weights directly on the beam or use the sliders below to adjust masses and distances. Notice how a heavy mass close to the fulcrum can perfectly balance a lighter mass farther away — that’s torque, where the product of mass and distance determines the rotational effect, not mass alone. Hit “Balance It!” to let the math find equilibrium for you.

Solving Equilibrium Problems

Same recipe every time:

1. Draw a free-body diagram showing all forces and where they act on the object.
2. Choose a pivot point. (Any point works — the math is the same. Smart choice = easier math.)
3. Apply $\Sigma F = 0$ (usually as separate x and y component equations).
4. Apply $\Sigma\tau = 0$ about your chosen pivot.
5. Solve the equations simultaneously.

Choosing the Pivot Point Strategically

You can calculate torques about any point — the answer comes out the same. But a smart choice eliminates unknown forces from the torque equation by putting the pivot right where one of those unknowns acts.

If you don’t know the force at a support, make that support your pivot. The torque from that force is automatically zero (because $r = 0$ for any force at the pivot itself), and you never have to solve for it.

Torque in Biological Systems

The human body is full of torque problems. Every joint is a pivot point. Every muscle attaches to a bone at some distance from that joint. Bones act as lever arms.

Take your bicep. The bicep muscle attaches to the forearm only a few cm past the elbow joint, while the weight you’re holding might be 30+ cm out at your hand. That gives the bicep a short lever arm and the weight a long lever arm — so the bicep has to pull with a force several times bigger than the weight, just to hold things steady. (For a 5 kg dumbbell at the hand, the bicep typically pulls with ~15× that — like 750 N or so.)

This is why even moderately heavy gym weights can feel surprisingly hard: the muscle is at a mechanical disadvantage. It’s also why most muscle attachments throughout the body sacrifice mechanical advantage for speed — a small muscle contraction near the joint produces a large, fast motion at the end of the limb.

You need to lift a 200 kg crate onto a 1.5 m high truck bed. You could deadlift it straight up — a couple thousand newtons of force, briefly. Or you could roll it up a 6 m ramp, using only one-quarter the force, but pushing it four times as far.

You do the same total work either way (about 3000 J in both cases). What changes is how that work is split between force and distance. The ramp lets a single human do what would otherwise require a forklift. That trade — give up distance to save force — is the magic of every simple machine, from the inclined plane the Egyptians used to build pyramids to the gear ratios in a modern bicycle.

The Core Principle: Energy Conservation

Every simple machine obeys the same rule: $W_{in} = W_{out}$ (in an ideal, frictionless machine). Since $W = Fd$:

$F_{in} \times d_{in} = F_{out} \times d_{out}$

If the machine reduces the force you need ($F_{in} < F_{out}$), it must increase the distance you move ($d_{in} > d_{out}$). The ratio is fixed by conservation of energy. There’s no free lunch.

Ideal Mechanical Advantage (IMA)

Levers

A lever is a rigid bar that pivots around a fixed point called the fulcrum. There are three classes, distinguished by which of three things — fulcrum, load, effort — sits in the middle.

For levers, $IMA = (\text{distance from effort to fulcrum}) / (\text{distance from load to fulcrum})$. A longer effort arm means greater mechanical advantage. (Why third-class levers always have IMA < 1: the effort sits closer to the fulcrum than the load, so the load arm is longer than the effort arm.)

Pulleys

Pulleys redirect force, and (when combined cleverly) multiply it. The key MCAT shortcut:

Inclined Plane (Ramp)

The ramp is the most intuitive simple machine — you’ve used one every time you’ve walked up a hill instead of a vertical cliff. A gentle slope (long ramp, small rise) takes little force over a long distance. A steep slope (short ramp, big rise) takes more force over a shorter distance.

In the frictionless ideal case, the force needed to push an object up a ramp at constant speed is $F = mg\sin\theta$, where $\theta$ is the ramp angle. (This is the down-slope component of gravity from §1.11.)

Wheel and Axle

A wheel and axle is essentially a rotating lever. The IMA equals the ratio of the wheel radius to the axle radius:

$IMA = \dfrac{R_{wheel}}{R_{axle}}$

A large steering wheel makes it easy to turn a small steering shaft. A doorknob (big radius) makes it easy to turn a small latch mechanism. A screwdriver handle (wide grip) gives big mechanical advantage to the narrow shaft turning the screw.

In §2.9 we assumed every machine was ideal — frictionless, no losses, every joule in equals a joule out. Reality is different. Every real machine wastes some energy.

A typical car engine delivers only about 25% of gasoline’s chemical energy to the crankshaft as work. The rest leaves as waste heat in the exhaust, the radiator, and the engine block. An incandescent light bulb converts about 5% of electrical energy into visible light — the other 95% becomes infrared “heat” radiation. Even your own muscles are only ~20–25% efficient — most of the energy you eat eventually leaves as body heat.

Efficiency tells you what fraction of the energy you put in actually does useful work. Higher efficiency = less waste = lower fuel/electricity costs.

The Efficiency Equation

You can also write this with power: $e = (P_{out}/P_{in}) \times 100\%$, since power is just energy per unit time and the time cancels.

Why Real Machines Are Never 100% Efficient

Every real machine bleeds energy through one (or more) of these channels:

Notice the pattern: almost everything ends up as heat. Sound waves, turbulence, deformation, electrical losses — all of them eventually warm up the surroundings and become “thermal energy” that’s no longer easy to recapture as useful work.

Actual vs. Ideal Mechanical Advantage

Because of friction, the actual force output of a machine is less than the ideal calculation predicts. This gives a second way to express efficiency:

Worked Examples

Example 1. A motor uses 500 J of electrical energy to lift a 30 kg crate 1.5 m. What is the motor’s efficiency? ($g = 10$ m/s².)

• $W_{out} = mgh = (30)(10)(1.5) = 450$ J.
• $e = (450/500) \times 100\% = 90\%$.

The other 50 J became waste heat in the motor windings.

Example 2. A ramp with $IMA = 5$ requires 120 N of force to push a 500 N crate up it. What’s the efficiency?

• $AMA = F_{out}/F_{in} = 500/120 \approx 4.17$.
• $e = (AMA/IMA) \times 100\% = (4.17/5) \times 100\% \approx 83\%$.

The 17% of energy “missing” from the ideal calculation went into friction between the crate and the ramp surface.

Efficiency of Common Systems

The car engine row is brake thermal efficiency: the share of the fuel’s chemical energy that arrives at the crankshaft as useful work. Whole-vehicle figures, sometimes quoted as “tank-to-wheels” efficiency, are lower, because they also subtract drivetrain, idling, and accessory losses. Different denominators, so the two are not interchangeable.

For comparison: a Tesla Model 3’s electric motor + battery system runs around 75–85% efficient, vs. the 20–30% of a gasoline engine. That’s the main reason EVs cost less to fuel per mile.

Cascading Efficiency

When multiple machines are connected in series (the output of one feeds the input of the next), the overall efficiency is the product of the individual efficiencies:

$e_{total} = e_1 \times e_2 \times e_3 \times \ldots$

A 90% efficient motor driving an 80% efficient pump gives an overall system efficiency of $0.9 \times 0.8 = 0.72 = 72\%$. Each stage loses a little, and the losses compound — which is why complex machinery (with many conversion stages) tends to be less efficient than direct ones.

Energy diagrams are one of the most efficient tools in physics — a single PE-vs-position graph can tell you where an object will accelerate, where it will stop and reverse, where the equilibrium points are, and whether each equilibrium is stable or unstable.

The MCAT loves them because they reward big-picture reasoning. Read the graph fluently and you can answer five different questions about the same object’s motion without doing any algebra.

The mental trick that makes the whole thing click: imagine the PE curve as an actual landscape, and visualize a marble rolling on it. Valleys catch the marble (stable equilibrium). Hilltops throw it off in either direction (unstable equilibrium). The marble always rolls downhill on the PE graph — that’s the force.

Reading a PE vs. Position Graph

A potential-energy diagram puts PE on the y-axis and position $x$ on the x-axis. The total mechanical energy ($E_{total}$) is drawn as a horizontal line across the graph, since total energy stays constant when only conservative forces act.

At any position, the kinetic energy is the vertical gap between the total-energy line and the PE curve:

This is the whole secret. Pick a position → look up the height of the PE curve there → subtract from total energy → that’s the KE → so the object is moving fast where PE is low and slow where PE is high.

Turning Points

A turning point is where $KE = 0$ — the object momentarily stops and reverses direction. Graphically, it’s where the PE curve touches the total-energy line.

Between two turning points, the object oscillates back and forth (like a marble rolling in a bowl). The object can’t cross a PE “hill” that rises above its total-energy line — it doesn’t have enough KE to climb over. It’s effectively trapped.

This is exactly why a pendulum swings between two endpoints (its turning points) and not beyond — its mechanical energy isn’t enough to climb any higher.

Three Types of Equilibrium

Equilibrium happens wherever the slope of the PE curve is zero (flat spots). But not all equilibria are equal — the shape of the curve at the flat spot determines whether the equilibrium holds up under a small push.

Force from the PE Curve

There’s a direct relationship between the PE curve and the force on the object:

The negative sign matters. It encodes the “rolls downhill” intuition: force always points toward lower PE — never higher. Steep PE = strong force. Gentle PE = weak force. Flat PE = no force.

You don’t need calculus on the MCAT — just eyeball the slope of the PE curve at the position you’re asked about. Steep down to the right → strong force pointing right. Steep down to the left → strong force pointing left.

Putting It All Together: A Complete Example

Imagine a PE curve with a deep valley at $x = 2$, a small hill at $x = 5$, and a shallow valley at $x = 8$. An object has total energy $E$ shown as a horizontal line cutting through both valleys but just above the hilltop.

What can you say?

1. The object placed in the deep valley near $x = 2$ oscillates between its two turning points (where the PE curve touches the $E$ line).
2. If $E$ is high enough to clear the hill at $x = 5$, the object can travel into the shallow valley at $x = 8$. If not, it’s trapped on one side.
3. At $x = 2$ and $x = 8$ (valley bottoms), the slope is zero → force = 0 → both are equilibrium positions. They’re stable (valleys).
4. At $x = 5$ (hilltop), slope is also zero → force = 0 → equilibrium, but unstable.
5. Maximum KE (and maximum speed) occurs at the deepest point of the deepest valley — biggest gap between $E$ and PE.
6. Speed is zero at every turning point.

Summary: Attacking Energy Diagram Problems

1. Find the total-energy line (horizontal).
2. Find turning points where the PE curve meets the energy line ($KE = 0$).
3. KE at any position = vertical gap between $E_{total}$ and PE.
4. Maximum speed at the deepest valley bottom (largest KE).
5. Equilibrium points where slope = 0 (valleys = stable, hilltops = unstable, flat = neutral).
6. Force always points toward lower PE; steeper slope = bigger force.

The MCAT is an interdisciplinary exam, and energy is one of its favorite crossover topics. A single passage might describe a person climbing stairs and ask you to connect three subjects:

• Physics: $W = mgh$ (the work needed to lift the body’s mass).
• Biology: ATP hydrolysis powers each muscle contraction.
• Biochemistry: glucose oxidation produces the ATP.

This section bridges those worlds. Once you see how the same conservation laws govern both an engine pushing a car and a muscle pulling a tendon, the crossover questions stop feeling like trick questions.

ATP: The Energy Currency

Adenosine triphosphate (ATP) stores energy in its phosphoanhydride bonds. When ATP is hydrolyzed to ADP + $P_i$, it releases roughly 30.5 kJ/mol under standard conditions. Inside actual cells (where concentrations differ from standard state), the real release is closer to 50–54 kJ/mol — significantly more.

Muscle Contraction as Work

When a bicep lifts a weight, the physics work is $W = mgh$. But the biological work involves millions of myosin heads pulling on actin filaments, each driven by one ATP hydrolysis cycle. The bridge between scales:

1. Glucose is oxidized in cellular respiration, producing ATP.
2. ATP binds to myosin heads in the muscle.
3. ATP hydrolysis triggers the myosin “power stroke” — a tiny applied force over a tiny distance (work at the molecular level).
4. Many millions of power strokes happening simultaneously shorten the muscle, producing macroscopic force and shortening.
5. The macroscopic muscle force × the distance the load moves = the physics work done on the load.

So $mgh$ at the gym scale and $\Delta G$ of ATP hydrolysis at the molecular scale are connected by an unbroken chain of energy transfers.

Metabolic Efficiency

The human body converts food energy to mechanical work with an efficiency of roughly 20–25%. The remaining 75–80% becomes body heat. That’s why you get hot during exercise — most of the metabolic energy you’re burning ends up as warmth, not motion.

The two stages multiply (cascading efficiency, §2.10): $0.40 \times 0.50 = 0.20 = 20\%$. Just like any machine, every conversion step shaves off some of the original energy as heat.

Coupled Reactions and Free Energy

In biochemistry, endergonic reactions (positive $\Delta G$, non-spontaneous) are made to proceed by coupling them to ATP hydrolysis (negative $\Delta G$, very spontaneous). The combined $\Delta G$ has to be negative for the coupled reaction to actually go.

Example: phosphorylating glucose costs energy ($\Delta G = +13.8$ kJ/mol — won’t happen by itself). Couple it to ATP hydrolysis ($\Delta G = -30.5$ kJ/mol):

• Combined: $\Delta G = +13.8 + (-30.5) = -16.7$ kJ/mol.
• Now the overall reaction is exergonic and proceeds spontaneously.

This is the biological equivalent of using a heavy falling weight (ATP hydrolysis) to lift a lighter weight (glucose phosphorylation) — the surplus energy from ATP more than pays for the costly reaction. Energy is still conserved; the bookkeeping just spans both reactions.

Energy Storage in the Body

The body stores energy in multiple forms, each with a different energy density and accessibility:

This hierarchy is why athletes train differently for different events. A 100 m sprint is mostly ATP + creatine phosphate (already stored, instantly accessible). A marathon is mostly fat oxidation (vast reserves but slower to mobilize). A 400 m run is the painful in-between zone — too long for ATP/creatine, too fast for full aerobic fat metabolism — so it relies heavily on anaerobic glycolysis, producing lactate.

Biomechanics: The Body as a Machine

The musculoskeletal system is a collection of levers (§§2.8–2.9). Most joints operate as third-class levers: the effort (muscle) is between the fulcrum (joint) and the load (weight in hand). That gives a mechanical advantage less than 1 — the muscle has to pull much harder than the load it’s lifting.

That sounds like bad engineering, but it’s a deliberate trade-off: speed and range of motion. A small contraction of the bicep (a few centimeters) produces a large, fast movement at the hand (tens of centimeters). The body sacrifices brute force for quickness — exactly the opposite of what a crowbar does. Throwing a baseball, swinging a tennis racket, kicking a soccer ball — none of them would be possible with high-MA leverage; you need the speed.