Week 6 In-Depth — Why Energy Bookkeeping Works

Read the cheatsheet first. This note explains why the energy balance works, walks through a substantial worked example per topic, and ties Newton’s second law to the work–energy theorem.

1 — Why the system boundary matters

Energy is a bookkeeping quantity. It is not a substance — you cannot see joules sloshing around. The whole framework only works because we draw a boundary and agree on:

• What lives inside the boundary (the system).
• What lives outside (the environment).
• How energy crosses the boundary (only via work, in our purely mechanical setting; later weeks will add heat).

This sounds pedantic. It is not. Without a boundary you cannot say whether $W$ is positive or negative — because “in” and “out” only make sense once you’ve picked a side. The single most common student mistake is failing to define the system on paper before writing the energy equation.

The basic energy model in one sentence

The energy inside the system can be transformed between forms (kinetic to gravitational, gravitational to kinetic, kinetic to thermal via friction). It is transferred to or from the environment only through work crossing the boundary. Inside the system, nothing is lost — that is the conservation principle. The only “losses” are deliberate exports of energy to the environment.

If you take the system to include everything friction touches, friction’s work becomes an internal energy transformation (mechanical $\to$ thermal) rather than a loss. The numbers don’t change, but your mental picture does: nothing is ever destroyed, just moved.

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2 — Why work is the area under $F$ vs $s$

Start from Newton’s second law projected onto the direction of motion:

$F_S=m\frac{dv}{dt}.$

Multiply both sides by velocity $v=ds/dt$:

$F_S\cdot v=mv\frac{dv}{dt}=\frac{d}{dt}\left(\frac{1}{2}m{v}^2\right)=\frac{d(KE)}{dt}.$

Integrate over time from $t_0$ to $t_1$:

${\int }_{t_0}^{t_1}{F}_S\cdot vdt=\Delta KE.$

But $vdt=ds$, so the left-hand side is exactly

${\int }_{s_0}^{s_1}{F}_S(s)ds=\Delta KE.$

This is the work–energy theorem. The integral on the left is “work done by the net force in the direction of motion.” It equals the change in kinetic energy. That’s why work is the area under $F$ versus $s$: it falls straight out of Newton’s second law combined with the definition of velocity.

A constant force is just a special case — the area becomes a rectangle, $W=F_S\cdot \Delta s$.

Worked example with all the steps — pushing a cart

Problem. A horizontal $75\text{N}$ force is applied to a cart over $12\text{m}$ of aisle. The applied force is parallel to motion.

Setup. Define the system as just the cart. The applied force is from the environment (you) crossing the boundary, so it does work on the system.

$W_{\text{applied}}=|F|\cos \theta \cdot |s|=75\times \cos (0^{\circ })\times 12=900\text{J}.$

Twist. Suppose a $30\text{N}$ downward “test force” were applied while the cart moves horizontally. Then $\theta =90^{\circ }$ and $\cos \theta =0$:

$W_{\text{test}}=30\times \cos (90^{\circ })\times 12=0\text{J}.$

The test force changes the normal force (and therefore friction, if any) but contributes zero work directly because its direction is perpendicular to the displacement. Only the component of force along motion does work.

Worked example — variable force from a graph

A force versus displacement graph rises linearly from $0$ at $s=0$ to a peak of $40\text{N}$ at $s=3\text{km}$, then drops back to $0$ at $s=4\text{km}$.

Solution. Since $F$ varies, $W$ is the area under the curve. The shape is a triangle:

$W=\frac{1}{2}\cdot b\cdot h.$

The trap is the unit. Read the base in metres:

$b=4\text{km}=4000\text{m},h=40\text{N},$ $W=\frac{1}{2}\times 4000\times 40=80000\text{J}=80\text{kJ}.$

If you’d used $b=4$, you’d get $80\text{J}$ — off by a factor of 1000. Always convert before computing area.

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3 — Why $\frac{1}{2}k{x}^2$ instead of $kx\cdot x$

Spring force obeys Hooke’s law: $F(x)=kx$ where $x$ is displacement from natural length. The force is not constant — it grows linearly with $x$.

Naive (wrong) calculation: at maximum compression $x$, the force is $F_{\max }=kx$, so $W=F_{\max }x=k{x}^2$. This overcounts by exactly a factor of 2 because the force was smaller for most of the compression.

Correct: $W$ is the area under the $F$ vs $x$ graph, which is a triangle:

$W=\frac{1}{2}\cdot x\cdot kx=\frac{1}{2}k{x}^2.$

Equivalently, integrate: ${\int }_0^{x}ksds=\frac{1}{2}k{x}^2$. Both give the elastic potential energy stored in the spring.

Worked example — compressing a suspension spring

A car suspension spring is $50\text{cm}$ natural length and compresses to $33\text{cm}$ under load; $k=4.1\text{kN/m}=4100\text{N/m}$.

Compression $x=0.50-0.33=0.17\text{m}$.

Three equivalent approaches (all in Tutorial 6 solutions):

1. Force-area (triangle). Plot $F(s)=ks$ for $0\le s\le 0.17$. Area $=\frac{1}{2}(0.17)(4100\times 0.17)=59.245\text{J}$.
2. Integration. ${\int }_0^{0.17}(4100s)ds=4100\cdot \frac{1}{2}(0.17{)}^2=59.245\text{J}$.
3. Energy formula. $P{E}_{\text{spring}}=\frac{1}{2}k{x}^2=0.5\times 4100\times 0.0289=59.245\text{J}$.

They are the same calculation written three ways. The energy formula is just the integral pre-evaluated.

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4 — Why mechanical energy is conserved (when it is)

Take a system with only conservative forces acting. Gravity and ideal springs are conservative because the work they do on a moving object can be fully recovered if the object returns to its starting position. That’s actually the definition of “conservative” — path-independent work, recoverable as kinetic energy.

For each conservative force, we define a potential energy function so that the work done by the force equals $-\Delta PE$. Sum over all conservative forces:

$W_{\text{cons}}=-(\Delta P{E}_g+\Delta P{E}_{\text{spring}})=-\Delta PE.$

The work–energy theorem says $W_{\text{net}}=\Delta KE$. If only conservative forces act, $W_{\text{net}}=W_{\text{cons}}=-\Delta PE$, so

$\Delta KE=-\Delta PE⟹\Delta (KE+PE)=0⟹K{E}_i+P{E}_i=K{E}_f+P{E}_f.$

That is conservation of mechanical energy — a direct consequence of (a) Newton II, (b) the definitions of potential energy, and (c) the absence of non-conservative forces.

What breaks when friction shows up

Kinetic friction does negative work on the moving object regardless of direction (it always opposes motion). There is no potential-energy function for friction because the work done is path-dependent: a long meandering path dissipates more than a straight one. Friction’s “lost” energy becomes heat — it leaves the mechanical system.

To handle friction we go back to the full balance:

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

Friction is a $W_{\text{by-system}}$ term — energy leaving as heat. Applied forces from outside are $W_{\text{on-system}}$ terms.

Worked example — skateboarder over a hill

A $70\text{kg}$ skateboarder is at the top of a hill moving at $6.0\text{m/s}$ and reaches $12\text{m/s}$ at the bottom. Find the height of the hill, neglecting friction and drag.

Setup. System = skateboarder. With no friction or drag, $W_{\text{on}}=W_{\text{by}}=0$ (the only force from outside is gravity, which we account for via $P{E}_g$, not as a $W$ term). Take $P{E}_g=0$ at the bottom of the hill.

$K{E}_i+P{E}_i=K{E}_f+P{E}_f$ $\frac{1}{2}m{v}_T^2+mgh=\frac{1}{2}m{v}_B^2+0$ $h=\frac{v_B^2-v_T^2}{2g}=\frac{144-36}{19.6}=5.51\text{m}.$

Equivalently: the kinetic-energy gain $\Delta KE=\frac{1}{2}(70)(144-36)=3780\text{J}$ matches the gravitational PE released $mgh=70\times 9.8\times 5.51\approx 3780\text{J}$. Energy is just rearranged within the system.

Worked example — snowboarder up a ramp with friction

Alice (with snowboard) has mass $m=55\text{kg}$. She slides up a $s=6.0\text{m}$ ramp inclined at $\theta =15^{\circ }$, with $\mu =0.025$. She reaches the top at $v_1=2.5\text{m/s}$. Find the initial speed $v_0$ and the work done by friction.

Geometry. Vertical rise: $y_1=s\sin \theta =6\sin (15^{\circ })=1.55\text{m}$.

Normal force. $F_N=mg\cos \theta$ (resolving gravity perpendicular to slope).

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

Friction work (out of the system):

$W_{\text{fric}}=F_{\text{fric}}\cdot s=0.025\times 55\times 9.8\times \cos (15^{\circ })\times 6=78.1\text{J}.$

That answers part (b): $W_{\text{fric}}\approx 78\text{J}$, leaving the system as heat.

Energy balance. Taking $P{E}_g=0$ at the bottom of the ramp and recognising $W_{\text{fric}}$ as a $W_{\text{by-system}}$ term:

$\frac{1}{2}m{v}_0^2+0=\frac{1}{2}m{v}_1^2+mg{y}_1+W_{\text{fric}}.$

Cancel $m$ (which appears in every term):

$\frac{1}{2}{v}_0^2=\frac{1}{2}{v}_1^2+g{y}_1+g\cos \theta \cdot \mu s.$

Plug in:

$\frac{1}{2}{v}_0^2=\frac{1}{2}(2.5{)}^2+9.8\times 1.55+9.8\times \cos (15^{\circ })\times 0.025\times 6$ $\frac{1}{2}{v}_0^2=3.125+15.19+1.420=19.74$ $v_0=\sqrt{2\times 19.74}=\sqrt{39.47}\approx 6.28\text{m/s}.$

So Alice needs $v_0\approx 6.3\text{m/s}$ at the bottom of the ramp to reach the top at $2.5\text{m/s}$, having dumped $\approx 78\text{J}$ to heat along the way.

Three takeaways from this problem:

1. The mass $m$ cancelled. On an inclined plane with friction proportional to weight, the answer depends only on geometry and $\mu$ — not on how heavy you are.
2. Friction work is always a $W_{\text{by-system}}$ term, regardless of direction. Even sliding down the ramp later, friction would still drain energy from the mechanical system.
3. The same $W_{\text{fric}}$ would appear on a flat surface or any slope — what changes is $F_N$, not the structure of the equation.

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5 — Power and the rate question

Power answers a different question from work: not “how much energy?” but “how quickly?”.

$P=\frac{\Delta W}{\Delta t}.$

For a constant force pushing an object at constant speed, $\Delta W=F\Delta s$ and $\Delta s/\Delta t=v$, so $P=Fv$. This is the form used for engines and motors — torque × angular velocity in rotational settings.

A practical consequence: doubling the speed at which you push a cart with the same force doubles the power you need, but the total work done in covering the same distance is unchanged. Power tells you how strenuous a task is; work tells you the total energy required.

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6 — How these all connect

Concept	Role
System & boundary	Defines what counts as “in” and “out”.
Work	The only way energy crosses the boundary (in mechanics).
Energy stores ($KE$, $P{E}_g$, $P{E}_{\text{spring}}$)	Forms energy takes inside the system.
Conservative forces	Their work becomes potential energy; no net loss.
Non-conservative forces (friction)	Their work leaves the mechanical system as heat.
Work–energy balance	The accountant’s ledger: in = out + $\Delta$(stored).

If you remember nothing else, remember the master equation:

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

Every other equation in this week is a special case (conservation when $W$‘s are zero; spring PE when $PE=\frac{1}{2}k{x}^2$; etc.).

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7 — Exam-style sample, end-to-end

A $5\text{kg}$ package is launched horizontally by a spring of constant $k=600\text{N/m}$, compressed by $0.5\text{m}$. After leaving the spring it climbs a ramp of vertical rise $h=1.0\text{m}$ to land on a truck. (a) Find the launch speed and the speed at the top of the ramp, assuming no friction. (b) A patch on the ramp has ${\mu }_k=0.3$ over $0.6\text{m}$ of slope. Will the package still make it to the top?

Setup.

Item	Value
Mass	$m=5\text{kg}$
Spring constant	$k=600\text{N/m}$
Compression	$x=0.5\text{m}$
Ramp rise	$h=1.0\text{m}$
Friction patch	$s=0.6\text{m}$, ${\mu }_k=0.3$

System = package. $P{E}_g=0$ at floor level.

Energy at launch. Initially all energy is in the spring:

$P{E}_{\text{spring}}=\frac{1}{2}k{x}^2=\frac{1}{2}(600)(0.5{)}^2=75\text{J}.$

Part (a) — no friction. Conservation of mechanical energy from compressed-spring state to top-of-ramp:

$\frac{1}{2}k{x}^2=\frac{1}{2}m{v}_{\text{top}}^2+mgh$ $75=2.5v_{\text{top}}^2+5(9.8)(1)$ $75-49=2.5v_{\text{top}}^2⟹v_{\text{top}}=\sqrt{26/2.5}\approx 3.2\text{m/s}.$

Part (b) — with the friction patch. Friction does work by the system over $0.6\text{m}$ of slope. On a slope the normal force is $F_N=mg\cos \theta$, but the question models the friction patch by giving us the path length directly and treating the ramp incline as small enough that $F_N\approx mg$ on the patch (a standard simplification — check the question wording):

$W_{\text{fric}}={\mu }_{k}mg\cdot s=0.3\times 5\times 9.8\times 0.6=8.82\text{J}.$

Energy needed to reach the top with $v_{\text{top}}\ge 0$:

$\text{required}=mgh+W_{\text{fric}}=49+8.82=57.82\text{J}.$

The spring supplied $75\text{J}$. Since $75\text{J}>57.82\text{J}$, yes, the package still reaches the top.

The excess energy ($75-57.82=17.18\text{J}$) becomes kinetic energy at the top: $v=\sqrt{2\times 17.18/5}\approx 2.62\text{m/s}$.

Answer. (a) $v_{\text{top}}\approx 3.2\text{m/s}$ frictionless. (b) Yes — spring energy $75\text{J}$ exceeds the $57.8\text{J}$ needed, so the package reaches the truck.

That layout — define system, account for stores, identify $W$ terms, balance — is the template for every work–energy question on the paper.

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Where to look next

• The cheatsheet for fast revision and the quiz.
• The lab brief (Laboratory 1) in this week’s PDF set for the spring-constant + kinetic-friction experiments.
• Whatever you add to this folder — Lecture Atlas picks up any new Markdown note automatically.