Week 6 Cheatsheet — Work, Energy, Conservation of Energy

How this week breaks down

Energy bookkeeping in four moves: define the system, list its energy stores, account for work crossing the boundary, balance the before vs. after. Skim this once, then revise from the in-depth note.

Concept	What you do
System	Draw a boundary. Anything inside is the system; anything outside is the environment.
Work	Energy crossing the boundary because a force from the environment acts over a displacement.
Energy stores	$KE$, $P{E}_g$, $P{E}_{\text{spring}}$ — all live inside the system.
Balance	$K{E}_i+P{E}_i+W_{\text{on}}=K{E}_f+P{E}_f+W_{\text{by}}$.

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1 — System, energy, and work

Definitions

Term	Definition	Units
System	A region enclosed by a chosen boundary	—
Energy $E$	Scalar; capacity to do work	J
Work $W$	Energy transferred across the boundary by a force	J = N·m
Power $P$	Rate of energy transfer	W = J/s

Sign convention

Situation	Sign
Energy enters the system (e.g. you push the cart)	$W>0$
Energy leaves the system (e.g. friction dissipates)	$W<0$

System energy balance:

$\Delta E_{\text{sys}}=E_{\text{sys},f}-E_{\text{sys},0}=E_{\text{in}}-E_{\text{out}}$

Isolated system: $E_{\text{sys},f}=E_{\text{sys},0}$.

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2 — Work: constant vs. variable force

Case	Formula
Constant force, parallel to motion	$W=F_S\cdot (s_1-s_0)$
Constant force, at angle $\theta$ to motion	$W=\Vert F\Vert \cos \theta \cdot \Vert s\Vert$
Force perpendicular to motion ($\theta =90^{\circ }$)	$W=0$
Variable force	$W={\int }_{s_0}^{s_1}F(s)ds$ = area under $F$-vs-$s$ graph
Spring (linear)	Triangle area $=\frac{1}{2}k{x}^2$

Quick exemplars

Problem	Set-up	Result
Push cart $12\text{m}$ with $75\text{N}$	$W=75\times 12$	$900\text{J}$
Triangular graph, peak $40\text{N}$ at $3\text{km}$, returns to $0$ at $4\text{km}$	$\frac{1}{2}\times 4000\times 40$	$80\text{kJ}$
Spring $k=4100\text{N/m}$, $x=0.17\text{m}$	$\frac{1}{2}(4100)(0.17{)}^2$	$59.25\text{J}$

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3 — Power

$P=\frac{\Delta W}{\Delta t}\text{[W]}$

If a constant force $F$ moves an object at constant speed $v$: $P=Fv$ (useful for engine / drag problems).

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4 — Energy stores

Store	Formula	Notes
Kinetic	$KE=\frac{1}{2}m{v}^2$	Always $\ge 0$; depends on frame
Gravitational PE	$PE=mgh$	$h$ measured from a chosen reference
Elastic (spring) PE	$P{E}_{\text{spring}}=\frac{1}{2}k{x}^2$	$x$ = compression or extension from natural length

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5 — Conservative vs non-conservative forces

Conservative	Non-conservative
Gravity	Kinetic friction
Ideal spring	Drag / air resistance
Work done is recoverable as $KE$	Work done is dissipated (heat)

Conservation of mechanical energy (only conservative forces act):

$K{E}_i+P{E}_i=K{E}_f+P{E}_f$

General work–energy balance (any forces):

$K{E}_i+P{E}_i+W_{\text{on-system}}=K{E}_f+P{E}_f+W_{\text{by-system}}$

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6 — Friction on a slope

Slope at angle $\theta$:

Quantity	Formula
Normal force	$F_N=mg\cos \theta$
Down-slope gravity component	$mg\sin \theta$
Kinetic friction force	$F_{\text{fric}}=\mu F_N=\mu mg\cos \theta$
Friction work over distance $s$	$W_{\text{fric}}=\mu mg\cos \theta \cdot s$ (always energy out)

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Common mistakes

1. Forgetting the cosine. Only the component of $\vec{F}$ along motion does work. Perpendicular forces (normal force, lift) contribute zero.
2. Using $W=Fs$ on a spring. Spring force is linear, not constant. The correct work is $\frac{1}{2}k{x}^2$ — using $F_{\max }x$ doubles it.
3. Forgetting km $\to$ m conversion when reading area off a force–displacement graph.
4. Confusing the sign of $W$. Use the diagram convention: in is positive, out is negative. The balance equation already puts $W_{\text{on}}$ and $W_{\text{by}}$ on opposite sides — don’t double-count.
5. Picking two different $PE$ references within the same problem. Pick zero height once and stick to it.
6. Forgetting $\cos \theta$ in $F_N$ on an incline. $F_N=mg\cos \theta$, not $mg$.
7. Applying conservation of mechanical energy with friction present. $K{E}_i+P{E}_i=K{E}_f+P{E}_f$ is wrong if any non-conservative force acts.

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Key formulas

$W={\int }_{s_0}^{s_1}F(s)dsW=|F|\cos \theta \cdot |s|P=\frac{\Delta W}{\Delta t}$

$KE=\frac{1}{2}m{v}^{2}PE=mghP{E}_{\text{spring}}=\frac{1}{2}k{x}^2$

$F_{\text{fric}}=\mu F_N=\mu mg\cos \theta$

$\boxed{K{E}_i+P{E}_i+W_{\text{on-system}}=K{E}_f+P{E}_f+W_{\text{by-system}}}$

For the why and many more worked examples, see the in-depth note.