Work, Energy, and Power

Work, Energy & Power

Part 1 of 4 — Work & the Work-Energy Theorem

What Counts as Work

$W = Fd\cos\theta$

where $\theta$ is the angle between the force and the displacement. Work is a scalar, measured in joules ($1\;\text{J} = 1\;\text{N}\cdot\text{m}$), and it can be positive, negative, or zero:

Three "no work" traps the MCAT reuses constantly:

• A waiter carrying a tray horizontally at constant velocity: his upward support force is perpendicular to the motion — zero work.
• The normal force on a block sliding along a surface: perpendicular — zero work.
• The centripetal force in uniform circular motion: always perpendicular to velocity — zero work, which is why speed stays constant.

Holding a weight stationary does zero work on it ($d = 0$) even though your muscles burn energy — physiological effort is not mechanical work.

Kinetic Energy and the Work-Energy Theorem

$KE = \tfrac{1}{2}mv^2$

$W_{net} = \Delta KE = KE_f - KE_0$

The net work (by all forces together) equals the change in kinetic energy. Because $KE \propto v^2$, doubling speed quadruples kinetic energy — the reason braking distance quadruples too.

This theorem is the fastest route whenever a question links force, distance, and speed without mentioning time.

Worked Example — Pulling a Sled

A $6\;\text{kg}$ sled starts from rest and is pulled $10\;\text{m}$ across snow by a rope exerting $50\;\text{N}$ at $37^\circ$ above the horizontal. Friction opposes with $10\;\text{N}$. ($\cos 37^\circ \approx 0.8$)

Step 1 — Work by the rope. Only the horizontal component moves the sled forward:

$W_{rope} = Fd\cos\theta = (50)(10)(0.8) = 400\;\text{J}$

Step 2 — Work by friction. Friction points opposite the displacement:

$W_{fric} = -(10)(10) = -100\;\text{J}$

Step 3 — Work by gravity and normal force. Both are perpendicular to the horizontal displacement: zero.

Step 4 — Net work and final speed.

$W_{net} = 400 - 100 = 300\;\text{J}$

$W_{net} = \Delta KE \Rightarrow 300 = \tfrac{1}{2}(6)v^2 \Rightarrow v^2 = 100 \Rightarrow v = 10\;\text{m/s}$

No kinematics, no acceleration, no time — the work-energy theorem jumps straight from forces and distance to speed.

Check the sign logic: positive net work speeds the sled up; if friction had done $-400\;\text{J}$ instead, net work would be zero and the sled would end at the same speed it started (here, rest — meaning it never gets going; you would need to check statics first).

Work & the Work-Energy Theorem 🎯

Key Takeaways — Part 1

• $W = Fd\cos\theta$: only the force component along the displacement does work
• Perpendicular forces do zero work: normal force, centripetal force, carrying a load horizontally
• Holding something still is zero mechanical work, whatever your muscles say
• $W_{net} = \Delta KE$ — the shortcut whenever force, distance, and speed appear without time
• $KE \propto v^2$: double the speed, quadruple the energy (and the stopping distance)
• Square speeds before subtracting when computing $\Delta KE$

Work, Energy & Power

Part 2 of 4 — Potential Energy & Conservation of Energy

Stored Energy

Gravitational: $U_g = mgh$ — measured from any reference height you choose; only changes in $U_g$ matter.

Elastic (spring): $U_s = \tfrac{1}{2}kx^2$, where $x$ is the stretch or compression from natural length. Doubling the compression quadruples the stored energy.

Conservative vs. Nonconservative Forces

The work done by gravity depends only on the height change, never on the path — a winding trail and a vertical ladder to the same summit involve identical work against gravity.

Conservation of Mechanical Energy

With only conservative forces doing work:

$KE_0 + U_0 = KE_f + U_f$

When friction or another nonconservative force acts:

$KE_0 + U_0 = KE_f + U_f + E_{lost}$

where $E_{lost}$ appears as heat and sound. Total energy is always conserved; mechanical energy is conserved only in the frictionless idealization.

High-yield consequences:

• Drop from height $h$: landing speed $v = \sqrt{2gh}$ — independent of mass
• Doubling the drop height multiplies speed by $\sqrt{2}$, not 2
• A pendulum or frictionless ramp delivers the same final speed as a straight drop through the same height — the path is irrelevant

Worked Example — Playground Slide with Friction

A $50\;\text{kg}$ child starts from rest atop a $5\;\text{m}$ slide and reaches the bottom moving at $8\;\text{m/s}$. How much energy did friction remove? ($g = 10\;\text{m/s}^2$)

Step 1 — Energy available.

$U_0 = mgh = (50)(10)(5) = 2500\;\text{J}$

Step 2 — Energy that arrived as motion.

$KE_f = \tfrac{1}{2}(50)(8)^2 = (25)(64) = 1600\;\text{J}$

Step 3 — The books must balance.

$E_{lost} = 2500 - 1600 = 900\;\text{J}$

Friction converted $900\;\text{J}$ into heat along the slide. Without friction the child would have arrived at $v = \sqrt{2gh} = \sqrt{100} = 10\;\text{m/s}$.

Worked Example — Spring Launcher

A spring with $k = 200\;\text{N/m}$ is compressed $0.1\;\text{m}$ behind a $0.1\;\text{kg}$ ball on a frictionless horizontal track.

Stored energy:

$U_s = \tfrac{1}{2}(200)(0.1)^2 = \tfrac{1}{2}(200)(0.01) = 1\;\text{J}$

Launch speed: all of it becomes kinetic energy:

$1 = \tfrac{1}{2}(0.1)v^2 \Rightarrow v^2 = 20 \Rightarrow v \approx 4.5\;\text{m/s}$

Estimate the root: $\sqrt{20}$ sits between $\sqrt{16} = 4$ and $\sqrt{25} = 5$, nearer $4.5$ — MCAT answer choices are spaced for exactly this kind of bracketing.

Potential Energy & Conservation 🎯

Key Takeaways — Part 2

• $U_g = mgh$ (only changes matter); $U_s = \tfrac{1}{2}kx^2$ (quadruples when $x$ doubles)
• Conservative forces (gravity, spring, electrostatic) do path-independent work and own a potential energy; friction does not
• Frictionless: $KE_0 + U_0 = KE_f + U_f$; with friction, add an $E_{lost}$ term and balance the books
• $v = \sqrt{2gh}$ for any frictionless descent through height $h$ — mass and path both irrelevant
• Speed scales as $\sqrt{h}$, not $h$
• Energy audits (available minus arrived = lost) are the standard MCAT passage calculation

Work, Energy & Power

Part 3 of 4 — Power, Efficiency & Simple Machines

Power: The Rate of Doing Work

$P = \frac{W}{t} = \frac{\Delta E}{t} \qquad \text{and for a force moving at speed } v: \quad P = Fv$

Units: watts ($1\;\text{W} = 1\;\text{J/s}$). A motor's power rating tells you how FAST it converts energy, not how much work it can ultimately do — a weak motor can lift any weight given a pulley system and enough time.

The $P = Fv$ form explains everyday physics: at highway speed, drag force is large, so maintaining constant $v$ demands high engine power even though $KE$ is not changing.

Efficiency

$e = \frac{\text{useful energy out}}{\text{total energy in}} \times 100\%$

No machine exceeds $100\%$; the shortfall leaves as heat. Human muscle runs near $20$-$25\%$ efficiency — for every $100\;\text{J}$ of metabolic energy spent, only about $20$-$25\;\text{J}$ becomes mechanical work, the rest warming the body (why exercise makes you hot).

Simple Machines: Trading Force for Distance

An ideal machine cannot multiply work or energy — only force (or speed). Mechanical advantage comes from moving the effort through a longer distance:

$F_{in} d_{in} = F_{out} d_{out} \quad \text{(ideal)} \qquad MA = \frac{F_{out}}{F_{in}} = \frac{d_{in}}{d_{out}}$

A ramp $6\;\text{m}$ long rising $1.5\;\text{m}$ has $MA = 4$: one quarter the force, four times the distance, same work. Real machines add friction, so the actual force needed always exceeds the ideal.

Worked Example — A Construction Hoist

A motor lifts a $100\;\text{kg}$ beam $20\;\text{m}$ at constant speed in $40\;\text{s}$. ($g = 10\;\text{m/s}^2$)

Step 1 — Useful work.

$W = mgh = (100)(10)(20) = 20000\;\text{J}$

Step 2 — Useful power output.

$P = \frac{W}{t} = \frac{20000}{40} = 500\;\text{W}$

Step 3 — Electrical input if the motor is 50% efficient.

$P_{in} = \frac{500}{0.5} = 1000\;\text{W}$

Half the electrical energy becomes heat in the motor; only half lifts the beam.

Cross-check with $P = Fv$: the beam rises at $v = 20/40 = 0.5\;\text{m/s}$ and the lifting force at constant speed equals the weight, $1000\;\text{N}$, so $P = (1000)(0.5) = 500\;\text{W}$. Same answer, two routes — a good MCAT self-audit.

Worked Example — Why the Ramp Helps

Loading an $800\;\text{N}$ barrel straight up onto a $1.5\;\text{m}$ platform takes $800\;\text{N}$ of force. Rolling it up a frictionless $6\;\text{m}$ ramp instead:

$MA = \frac{6}{1.5} = 4 \Rightarrow F = \frac{800}{4} = 200\;\text{N}$

Work is identical either way: $(800)(1.5) = 1200\;\text{J}$ vertically, or $(200)(6) = 1200\;\text{J}$ along the ramp. The ramp lowers the force, never the energy bill — with friction, the ramp actually costs MORE total work.

Power, Efficiency & Machines 🎯

Key Takeaways — Part 3

• $P = W/t = \Delta E/t$, and $P = Fv$ for a force pushing at speed $v$; watts are joules per second
• Constant cruising speed still requires power — it feeds drag and friction, not kinetic energy
• Efficiency $=$ useful out / total in; the rest is heat; muscle runs near $20$-$25\%$
• Ideal machines conserve work: less force means proportionally more distance ($MA = d_{in}/d_{out}$)
• Ramp $MA =$ length/height; pulley $MA =$ supporting strands; lever $MA =$ arm ratio
• Cross-check power answers by computing both $W/t$ and $Fv$ when the data allows

Work, Energy & Power

Part 4 of 4 — MCAT Integration: Energy Accounting & Passage Strategy

When to Reach for Energy Instead of Forces

Energy methods shine exactly where kinematics struggles:

• Curved or unknown paths: gravity's work depends only on the height change, so a twisting frictionless waterslide is a one-line problem.
• No time given, none wanted: energy equations never contain $t$; power questions then bring time back in at the end.
• "How much was lost?" questions: compute mechanical energy before and after; the gap is heat.

Force methods win when a question asks about a force, a normal force, or an acceleration at one instant — energy conservation cannot see individual forces.

The Physiology Crossover

MCAT passages love metabolic energy accounting:

• $1$ food Calorie (kcal) $= 4184\;\text{J} \approx 4200\;\text{J}$
• Muscle efficiency $\approx 20\%$: metabolic power $\approx 5\times$ the mechanical power output
• Stair-climb and ergometer experiments measure useful power as $mgh/t$ or from a flywheel's friction; the passage then asks you to scale up to metabolic cost

Scaling Arguments Answer Whole Question Sets

Because $KE = \tfrac{1}{2}mv^2$ and $W = Fd$:

• Same braking force, double the speed: stopping distance quadruples ($d = KE/F$), while stopping time merely doubles ($t = p/F$ — momentum, not energy)
• Same speed, double the mass: both KE and momentum double; distance and time double
• The distance/time contrast above is the classic energy-vs-momentum discrimination — distance pairs with energy, time pairs with momentum (impulse)

Worked Example — A Stair-Climb Metabolic Study (Passage Style)

Exercise physiologists have a $60\;\text{kg}$ subject run up a staircase rising $3\;\text{m}$ in $5\;\text{s}$. Oxygen-uptake measurements independently show the subject's metabolic power during the climb is about $1800\;\text{W}$. ($g = 10\;\text{m/s}^2$)

Step 1 — Useful mechanical work. Only the vertical rise stores energy:

$W = mgh = (60)(10)(3) = 1800\;\text{J}$

Step 2 — Useful power output.

$P_{out} = \frac{1800}{5} = 360\;\text{W}$

Step 3 — Efficiency.

$e = \frac{P_{out}}{P_{met}} = \frac{360}{1800} = 20\%$

Right at the textbook value for muscle — a passage will often make you extract this number from a data table across several subjects.

Step 4 — Fuel cost sanity check. At $1800\;\text{W}$ for $5\;\text{s}$, the climb costs $9000\;\text{J} \approx 2$ food Calories. Climbing stairs is cheap in Calories and expensive in perceived effort — passages exploit that contrast in their discussion paragraphs.

Passage strategy: label every energy number as mechanical out or metabolic in before computing anything. Most wrong answer choices come from dividing the right numbers in the wrong direction (yielding an efficiency of $500\%$) or from using the stair path length instead of the vertical height.

Energy Integration & Passage Strategy 🎯

Key Takeaways — Part 4

• Choose energy methods for curved paths, missing time, and "how much was lost" audits; choose force methods for instantaneous forces and accelerations
• Stair-climb experiments: useful power $= mgh/t$; metabolic power $\approx 5\times$ larger at $20\%$ muscle efficiency
• $1$ food Calorie $\approx 4200\;\text{J}$ — mechanical work is startlingly cheap in Calories
• Same braking force: stopping distance scales with $v^2$ (energy), stopping time with $v$ (momentum)
• Label numbers as mechanical-out vs. metabolic-in before dividing — efficiency above $100\%$ signals an inverted ratio
• Use vertical height, never path length, for gravitational energy