Week 6 — Work, Energy, Conservation of Energy

What this week is about

Week 6 is the energy bookkeeping week. Force and acceleration still exist — but now you stop watching the motion second-by-second and instead draw a boundary around your object, label the energy stored inside, and balance what flows in (work done on the system) against what flows out (work done by the system, e.g. against friction).

Three questions drive the week:

1. How do you put energy into a system? → Work: force times distance, with the right cosine. Sometimes integrate.
2. Where does the energy go? → Into kinetic ($\frac{1}{2}m{v}^2$), gravitational ($mgh$), or spring ($\frac{1}{2}k{x}^2$) energy, or out as friction heat.
3. When is mechanical energy conserved, and when isn’t it? → Only when no non-conservative force acts. Otherwise use the full work–energy balance.

All three reduce to one master equation:

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

See the in-depth note for why this equation is just energy conservation in disguise.

Notes in this week

• Cheatsheet — every formula, table, sign rule, and recipe on one scannable page. Includes the quiz (mixed difficulty, reshuffles every visit).
• In-depth analysis — narrative “why” behind each concept, a worked example per topic, the connection between Newton’s second law and the work–energy theorem, and an exam-style sample.
• Lecture summary — the source reconstruction (formulas, worked examples, tutorial outlines).
• Study guide — what’s directly supported vs inferred, common mistakes, practice questions, confidence report.
• Workshop prep — 5-minute and 20-minute revision plans for Portfolio 5 and Lab Report 1.

Any notes you add to this folder will appear here automatically.

What I need to know before the workshop

• Vector decomposition on an inclined plane ($mg\sin \theta$ down-slope, $mg\cos \theta$ into surface)
• Free-body diagrams with friction
• Kinematics enough to find $v$ from a height drop if needed
• Area under a graph (triangles and rectangles) — this is how variable-force work is computed
• Unit handling: J, N·m, W, and that 1 J = 1 N·m = 1 kg·m²/s²

Assessment relevance

• Laboratory 1 (this week): spring constant from a force–extension graph (slope = $k$), coefficient of kinetic friction from a force balance on a tilted track, and cart displacement down the ramp. Worksheet submission.
• Portfolio 5/6: completed in the workshop class.
• Exam: work–energy balance problems on inclines with friction are a standard exam pattern; expect at least one.

Overview

Week 6 introduces the energy framework: defining a system, accounting for energy flowing in/out as work, and tracking transformations between kinetic, gravitational potential, and spring potential energy. The lecture builds from the basic energy model and the definition of work (area under a force–displacement curve) up to the full work–energy balance for systems containing non-conservative forces such as friction. Week 6 is also a practical week — Laboratory 1 (spring constant, coefficient of kinetic friction, and cart displacement down a track) is run in the lab class, with a worksheet submission, alongside Portfolio 5/6 in the workshop class.

Key concepts

• System: a region defined by a closed boundary that contains the object(s) whose energy we are tracking. Everything outside the boundary is the environment.
• Energy $E$: an abstract scalar quantity representing the capacity to do work. Unit: joule (J). Forms include kinetic, potential (gravitational, elastic), thermal, electrical, chemical, nuclear.
• Basic energy model: energy is transformed without loss within the system and transferred to/from the system by external forces doing work. For an isolated system $\Delta E_{\text{sys}}=0$.
• Work $W$: energy transferred to a system by applying a force over a displacement. Unit: joule (J = N·m). Sign convention: $W>0$ when energy flows into the system, $W<0$ when it flows out.
• Component of force along motion $F_S$: the projection of an applied force $\vec{F}$ onto the direction of displacement. Only this component does work.
• Power $P$: the rate at which energy is transferred (work done per unit time). Unit: watt (W = J/s).
• Kinetic energy $KE$: energy associated with motion. Symbol $KE$, unit J.
• Gravitational potential energy $PE$ (near Earth’s surface): energy stored due to height $h$ above a chosen reference. Unit J.
• Spring/elastic potential energy $P{E}_{\text{spring}}$: energy stored in a spring stretched or compressed by $x$ from its natural length. Spring constant $k$ has units N/m.
• Conservative force: a force that “gives back” the work done against it as kinetic energy (e.g. gravity, ideal spring). Mechanical energy is conserved.
• Non-conservative force: a force that dissipates energy from the mechanical system (e.g. kinetic friction, drag); the work done against it cannot be recovered.
• Mechanical energy: the sum $KE+PE$ (gravitational and elastic).
• Kinetic friction: $F_{\text{fric}}=\mu F_N$ where $\mu$ is the coefficient of kinetic friction (dimensionless) and $F_N$ is the normal force.

Core formulas

System energy balance: $\Delta E_{\text{sys}}=E_{\text{sys},f}-E_{\text{sys},0}=E_{\text{in}}-E_{\text{out}}$

For an isolated system: $E_{\text{sys},f}=E_{\text{sys},0}$.

Work — general (non-constant force, area under force vs. displacement curve): $W={\int }_{s_0}^{s_1}F(s)ds$

Work — constant force parallel to displacement: $W=F_S\cdot (s_1-s_0)=F_S\cdot |s|$

Work — constant force applied at angle $\theta$ to displacement: $W=|F|\cos (\theta )\cdot |s|$

Power: $P=\frac{\Delta W}{\Delta t}$

Kinetic energy: $KE=\frac{1}{2}m{v}^2$

Gravitational potential energy (near Earth’s surface): $PE=mgh$

Elastic (spring) potential energy: $P{E}_{\text{spring}}=\frac{1}{2}k{x}^2$

Kinetic friction force: $F_{\text{fric}}=\mu F_N$

Conservation of mechanical energy (only conservative forces act): $K{E}_{\text{initial}}+P{E}_{\text{initial}}=K{E}_{\text{final}}+P{E}_{\text{final}}$

Work–energy balance (general, including non-conservative forces): $K{E}_{\text{initial}}+P{E}_{\text{initial}}+W_{\text{on-system}}=K{E}_{\text{final}}+P{E}_{\text{final}}+W_{\text{by-system}}$

where $W_{\text{on-system}}$ is energy added by the environment and $W_{\text{by-system}}$ is energy leaving to the environment (e.g. work done against friction).

Worked examples

Example 1 — Work pushing a shopping cart

Problem. What is the work done on a shopping cart if you exert $75\text{N}$ as you push it down a $12\text{m}$ aisle?

Solution. Force is parallel to displacement, so $\theta =0$ and $\cos (0)=1$:

$W=F\cdot s=75\text{N}\times 12\text{m}=900\text{N⋅m}=900\text{J}$

Equivalently with the angle formulation: $W=|F|\cos (\theta )(s_1-s_0)=75\times 1\times 12=900\text{J}$.

Note from the handwritten notes: if instead a test force $F_{\text{test}}$ were applied vertically (perpendicular to the aisle), then $\theta =90°$, $\cos (90°)=0$, so $W=0$. Forces perpendicular to displacement do no work.

Example 2 — Work from a non-constant force graph

Problem. Find the total work done by the force shown (triangular profile rising from $0$ at $0\text{km}$ to $40\text{N}$ at $3\text{km}$ and back to $0$ at $4\text{km}$) as the object moves from $0$ to $4\text{km}$.

Solution. Force is not constant, so work is the area under the force vs. displacement curve. The shape is a triangle with base $b=4\text{km}=4000\text{m}$ and height $h=40\text{N}$:

$W=A_{\text{tri}}=\frac{bh}{2}=\frac{4000\text{m}\times 40\text{N}}{2}=80000\text{J}=80\text{kJ}$

Example 3 — Skateboarder over a hill (work–energy)

Problem. A $70\text{kg}$ skateboarder crosses the top of a hill at $6.0\text{m/s}$ and reaches $12\text{m/s}$ at the bottom. (a) Find the total work done on the skateboarder between top and bottom. (b) Can we determine the height of the hill?

Solution (a). Define the skateboarder as the system. Neglecting friction and drag (lecture notes mark $W_{\text{fric}}$ and $W_{\text{drag}}$ as zero/external), the energy balance becomes:

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

With $P{E}_i$ taken at top and reference $P{E}_f=0$ at the bottom, and no friction/drag, $W_{\text{on-system}}=W_{\text{hill}}=K{E}_f-K{E}_i$:

$W_{\text{hill}}=\frac{1}{2}m{v}_B^2-\frac{1}{2}m{v}_T^2=\frac{1}{2}(70)(12{)}^2-\frac{1}{2}(70)(6{)}^2$ $W_{\text{hill}}=5040-1260=3780\text{J}$

Solution (b). Yes — the work done by the hill is the gravitational potential energy stored at the top:

$P{E}_g=mgh=W_{\text{hill}}$ $3780=70\times 9.8\times h⟹h=5.51\text{m}$

Example 4 — Snowboarder up a ramp with friction

Problem. Alice (with snowboard) has mass $55\text{kg}$ and encounters a $6.0\text{m}$ ramp at $15°$. Coefficient of friction $\mu =0.025$. (a) What initial speed is needed to reach the top at $2.5\text{m/s}$? (b) How much work is done by friction?

Solution. Geometry: $y_1=6\sin (15°)=1.55\text{m}$. Take $y_0=0$.

Free body diagram (rotated to incline axes): $F_N=mg\cos (15°)$.

Friction force and work done against friction:

$F_{\text{fric}}=\mu F_N=\mu mg\cos (15°)$ $W_{\text{fric}}=F_{\text{fric}}\cdot s=55\times 9.8\times \cos (15°)\times 0.025\times 6=78.1\text{J}$

(That is the answer to part b.)

For part (a), apply the work–energy balance:

$\frac{1}{2}m{v}_0^2=\frac{1}{2}m{v}_1^2+mg{y}_1+mg\cos (15°)\mu s$

Solving for $v_0$:

$v_0=\sqrt{v_1^2+2g{y}_1+2g\cos (15°)\mu s}=6.3\text{m/s}$

Things to practise

Tutorial 6 (Wolfson Ch. 6–7) problems with brief solution outlines:

• Exercise 1 — Car/truck same kinetic energy. Subcompact $m_c=950\text{kg}$; truck $m_T=3.2\times 10^4\text{kg}$ at $v_T=20\text{km/h}=5.56\text{m/s}$. Set $\frac{1}{2}{m}_c{v}_c^2=\frac{1}{2}{m}_T{v}_T^2$ and solve $v_c=\sqrt{m_T{v}_T^2/m_c}=32.27\text{m/s}=116.1\text{km/h}$.
• Exercise 2 — Compressing a suspension spring. Uncompressed $50\text{cm}$, compressed to $33\text{cm}$ so $x=0.17\text{m}$; $k=4.1\text{kN/m}=4100\text{N/m}$. Spring force is not constant, so naive $W=Fs$ is wrong. Use the area under the linear $F$-$x$ graph (a triangle) or equivalently elastic PE: $W=\frac{1}{2}k{x}^2=\frac{1}{2}(4100)(0.17{)}^2=59.245\text{J}$.
• Exercise 3 — Sliding a box up a $30°$ ramp. Applied force $F=220\text{N}$ up slope, ${\mu }_k=0.16$, vertical rise $h=1\text{m}$ so distance along slope $s=1/\sin (30°)=2\text{m}$. (a) $W_{\text{APP}}=220\times 2=440\text{J}$. (b) Constant speed so $K{E}_i=K{E}_f$. Work balance: $W_{\text{APP}}=mgh+{\mu }_{k}mg\cos (30°)\cdot s$. Solving: $m=440/[9.8+9.8\cos (30°)(0.16)(2)]=35.16\text{kg}$.
• Exercise 4 — Rocket launch. $m=150000\text{kg}$, thrust $F=4.0\times 10^6\text{N}$, height $h=500\text{m}$. Work by thrust: $W=4\times 10^6\times 500=2\times 10^9\text{J}$. Energy balance $W=\frac{1}{2}m{v}_f^2+mgh$, so $v=\sqrt{(2\times 10^9-150000\cdot 9.8\cdot 500)/(0.5\cdot 150000)}=129.87\text{m/s}$.
• Exercise 5 — Spring-launched package onto a truck. $m=5\text{kg}$, $k=600\text{N/m}$, $x=0.5\text{m}$, ramp height $h=1\text{m}$. (a) Frictionless: $\frac{1}{2}k{x}^2=\frac{1}{2}m{v}^2+mgh$ gives $75=2.5v^2+49$, so $v=\sqrt{26/2.5}=3.2\text{m/s}$. (b) With a $60\text{cm}$ sticky spot, ${\mu }_k=0.3$: package reaches the top if $P{E}_s-P{E}_g>W_{\text{fric}}$, i.e. $26\text{J}>F_N{\mu }_{k}s=5\times 9.8\times 0.3\times 0.6=8.82\text{J}$. Since $26>8.82$, yes, the next package still reaches the truck.

Common pitfalls

• Forgetting the cosine. Only the component of force parallel to motion does work. A vertical lifting force on a horizontally moving cart contributes $W=0$ because $\cos (90°)=0$.
• Applying $W=Fs$ to a non-constant force. For a spring or any variable force, you must integrate (or take the area under the $F$-vs-$s$ graph). Using $F_{\text{max}}\times x$ on a spring overestimates work by exactly a factor of 2.
• Mixing units on a graphed force–displacement problem. Tutorial-style graphs often give displacement in km — convert to metres before reading the area (e.g. $4\text{km}=4000\text{m}$).
• Sign of $W$. Adopt the convention from the energy diagram: $W>0$ when work flows into the system, $W<0$ when it leaves. The balance equation $K{E}_i+P{E}_i+W_{\text{on-system}}=K{E}_f+P{E}_f+W_{\text{by-system}}$ keeps both terms positive on opposite sides — don’t double-count.
• Choosing a reference for $P{E}_g$. Gravitational PE is always relative to a chosen zero height. Pick one (usually the lower of the two states) and stick to it throughout the problem.
• Friction direction. Kinetic friction always opposes motion, so $W_{\text{fric}}$ is energy leaving the system (a $W_{\text{by-system}}$ term), regardless of whether the object is going up or down the slope.
• Normal force on a slope. On an incline at angle $\theta$, $F_N=mg\cos (\theta )$, not $mg$. Forgetting the $\cos \theta$ overestimates friction.
• “Conservation of mechanical energy” misuse. $K{E}_i+P{E}_i=K{E}_f+P{E}_f$ is only valid when no non-conservative forces act. If friction or drag is present, you must use the full work–energy balance.

Source citations

• Lecture 6 slides — Work, Energy, Conservation of Energy (EGD102-Physics/Lecture6_CTP1.pdf): system/energy definitions (slides 9–10), work definitions and graphs (slides 11–14), power (slide 15), Examples 1–4 (slides 16–17, 24–25), types of energy (slide 19), conservative vs non-conservative (slide 21), conservation of mechanical energy (slide 22), work–energy balance (slide 23), Laboratory 1 parts 1–3 (slides 26–28), Week 6 learning activities including Portfolio 5 workshop and the Laboratory worksheet assessment (slide 30).
• Lecture 6 handwritten notes (EGD102-Physics/EGD102 - Lecture6 - Notes.pdf): full worked solutions to Examples 1–4 with FBDs and arithmetic.
• Tutorial 6 (EGD102-Physics/Tutorial 6.pdf): Exercises 1–5 plus repeated reference slides.
• Tutorial 6 Solutions (EGD102-Physics/Tutorial 6_Solutions.pdf): worked solutions including Exercise 2’s three approaches (work formula, graphical, energy) and Exercise 5’s friction check ($26\text{J}>8.82\text{J}$).
• Recommended additional reading: Wolfson, R. (2020) Essential University Physics, Volume 1, Global Edition, 4th Ed. in SI Units, Chapters 6–7.