Non-conservative forces — friction, air drag: Worked examples

Before anything, one promise about symbols. We reuse these throughout, so pin them down now.

Definition The cast of symbols (all earned before use)

$m$ — mass, in kilograms (kg). "How much stuff." When two masses appear at once we write $m_{b}$ (bullet) and $m_{B}$ (block).
$g = 9.8 m/s^{2}$ — strength of gravity's pull per kg.
$μ_{k}$ — coefficient of kinetic friction, a pure number (no units). Bigger = grippier. Governs sliding friction. See Friction — static & kinetic.
$μ_{s}$ — coefficient of static friction, the "won't-budge" grip before sliding starts. Always $μ_{s} \geq μ_{k}$ . It decides whether motion begins at all.
$N$ — normal force: the push a surface gives perpendicular (at $9 0^{\circ}$ ) to itself, in newtons (N).
$L$ — path length: the total distance a wheel-counter would tick off along the actual route, NOT the straight-line displacement. Metres (m).
$Δ$ (delta) — "change in": $Δ X = X_{final} - X_{start}$ .
$K E = \frac{1}{2} m v^{2}$ — kinetic energy, energy of motion. Joules (J).
$U$ — potential energy, stored energy of position ( $U = m g h$ for height $h$ ). See Conservative forces & potential energy.
$E_{m ec h} = K E + U$ — total mechanical energy.
$W_{n c}$ — work done by all non-conservative forces (friction, drag, collisions) added together.
$W_{f r i c}$ — the part of $W_{n c}$ done specifically by kinetic friction: $W_{f r i c} = - μ_{k} N L$ . It is one contribution to $W_{n c}$ ; on a flat friction-only floor the two are equal.
$Q$ — heat generated, $Q = - W_{n c}$ (the stolen energy reappears as warmth). Joules (J).
$b$ — linear drag coefficient, units N per (m/s), so that $F_{d} = b v$ at low speed.
$C_{d}$ — drag coefficient, a pure shape number (how "streamlined" — a sphere is about $0.5$ ).
$ρ$ (rho) — air density, mass of air per cubic metre, $kg/m^{3}$ (about $1.2$ at sea level).
$A$ — frontal area, the cross-section the body pushes through air, in $m^{2}$ .

The one master equation, derived in the parent note:

The scenario matrix

Every non-conservative problem in this chapter lands in one of these cells. The rightmost column names the example that nails it.

Cell	Situation class	What's special / degenerate	Example
A	Flat floor, friction stops motion	pure $Δ K E$ , $Δ U = 0$	Ex 1
B	Closed loop, round trip	displacement $= 0$ but $W \neq = 0$ (the loop test)	Ex 2
C	Rough incline, block slides	$tan θ > μ_{k}$ : sign of bracket positive	Ex 3
D	Rough incline, degenerate	$tan θ \leq μ_{s}$ : block won't move — what the formula says	Ex 4
E	Collision then friction	momentum first, energy second (two-tool problem)	Ex 5
F	Quadratic air drag, terminal velocity	limiting value as $v \to v_{t}$ , $a \to 0$	Ex 6
G	Linear (Stokes) drag, low speed	$F_{d} = b v$ regime, different terminal velocity	Ex 7
H	Real-world word problem	skid marks → speed (forensics)	Ex 8
I	Exam twist	block slides up, both gravity & friction oppose	Ex 9
J	Both at once — friction AND air drag	two thieves in one problem	Ex 10

We'll do them in order. Every number is machine-checked at the bottom.

Worked example Ex 1 — Cell A: flat floor, friction alone stops a puck

A hockey puck of mass $m = 0.2 kg$ slides at $v_{0} = 6 m/s$ on rough ice, $μ_{k} = 0.10$ . How far before it stops?

Forecast: Guess the stopping distance before reading on — is it a few centimetres, or many metres? (Ice is slippery, so I bet it's large.)

Height never changes, so $Δ U = 0$ . Why this step? The floor is flat; no rise or fall, so gravity's potential energy is untouched. This collapses the master equation to $W_{n c} = Δ K E$ .
Write the friction work. On flat ground $N = m g$ , so $W_{f r i c} = - μ_{k} m g L$ . Why? Normal force just balances weight here; nothing else pushes on the surface.
Set stolen energy = lost kinetic energy. Final speed is $0$ : $- μ_{k} m g L = \frac{1}{2} m v_{0}^{2} - \frac{1}{2} m v_{0}^{2}_{start} = 0 - \frac{1}{2} m v_{0}^{2}$ $\Rightarrow μ_{k} m g L = \frac{1}{2} m v_{0}^{2}$ Why? $Δ K E = K E_{f} - K E_{i}$ , and the puck ends at rest.
Mass cancels! Divide both sides by $m$ , then divide by $μ_{k} g$ : $L = \frac{v _{0}^{2}}{2 μ _{k} g} = \frac{6 ^{2}}{2 ( 0.10 ) ( 9.8 )} = 18.4 m$ Why this step? $m$ appears on both sides (in $K E$ and in $W_{f r i c}$ ), so it algebraically cancels — that's why stopping distance doesn't care about mass. We isolate $L$ to answer the question.

Verify: Units: $\frac{( m/s ) ^{2}}{( m/s ^{2} )} = m$ ✔. Mass dropped out — heavier puck stores more KE but also feels more friction, exactly cancelling. Large distance, as forecast (slippery ice, small $μ_{k}$ ).

Worked example Ex 2 — Cell B: the closed-loop test (displacement zero, work not)

Push the same puck ( $m = 0.2$ , $μ_{k} = 0.10$ ) $3 m$ forward, then $3 m$ back to the exact start. How much mechanical energy is lost over the loop? Contrast with what gravity would do.

Forecast: Net displacement is zero. A "fair" force gives zero work over a round trip. Does friction?

Total path length, not displacement. $L = 3 + 3 = 6 m$ . Why this step? This is the whole point of non-conservative forces. Friction flips to oppose you on the way back, so it steals on both legs — you sum the distances.
Compute the loop work. $W_{f r i c} = - μ_{k} m g L = - (0.10) (0.2) (9.8) (6) = - 1.176 J$ Why this step? We use the same $W_{f r i c} = - μ_{k} m g L$ as Ex 1, but now $L$ is the whole loop's length — this single number is what we compare against gravity's zero.
Compare to gravity around the same loop. Gravity would give $- m g Δ h$ going up and exactly $+ m g Δ h$ coming down: net $0$ . Friction gives $- 1.176 J \neq = 0$ . Why? Gravity's sign depends on displacement direction (up vs down); friction's sign depends on velocity direction — and velocity opposes friction on every leg.

Verify: $\oint F \cdot d r = - 1.176 J \neq = 0$ confirms friction is non-conservative. This is the defining loop-integral criterion from Conservative forces & potential energy.

Figure — Non-conservative forces — friction, air drag

Reading the figure: the white block sits on the slope. Its weight $m g$ (the straight-down white arrow) is split into two chalk arrows: a blue one pointing down the slope ( $m g sin θ$ ) and a pink one pressing into the slope ( $m g cos θ$ ). The yellow arrow up the slope is friction $f_{k}$ , opposing the dashed "slides down" motion arrow. The yellow arc at the base marks $θ = 3 0^{\circ}$ . (Each arrow is printed with its own label, so direction + text identify it without relying on colour.)

Forecast: Frictionless would give $v = 2 g L sin θ$ . With friction, will it be a little slower or does it not move at all? (Check: $tan 3 0^{\circ} = 0.577 > 0.2$ , so it moves.)

Resolve gravity on the tilted plane. As the figure shows, weight $m g$ splits into a piece along the slope, $m g sin θ$ (blue arrow, pulls it down-slope), and a piece into the slope, $m g cos θ$ (pink arrow). Why this step? Only the along-slope piece does the sliding; the into-slope piece sets the normal force and hence friction. Splitting into these two perpendicular directions lets us treat each independently.

Normal force from the perpendicular balance. The block doesn't sink into the slope, so the surface's normal push $N$ must exactly cancel the into-slope weight: $N = m g cos θ$ . Hence friction magnitude $f_{k} = μ_{k} m g cos θ$ (the yellow up-slope arrow). Why this step? Friction is $μ_{k} N$ , so we need $N$ first; the perpendicular force balance (no motion into the slope) is what pins $N$ down.

Height dropped. Sliding $L$ along the slope lowers height by $L sin θ$ , so $Δ U = - m g L sin θ$ . Why this step? The master equation needs $Δ U = m g Δ h$ ; the geometry of the ramp converts "distance along slope" into "vertical drop" via $sin θ$ . It's negative because the block falls.

Feed the master equation $W_{n c} = Δ K E + Δ U$ : $- μ_{k} m g cos θ L = \frac{1}{2} m v^{2} - 0 + (- m g L sin θ)$ Why this step? Now every piece is known: $W_{n c}$ is the friction work (step 2), $Δ U$ is step 3, and $Δ K E = \frac{1}{2} m v^{2}$ since it starts from rest. Substituting turns the physics into one solvable equation. Cancel $m$ and rearrange: $v = 2 g L (sin θ - μ_{k} cos θ)$

Plug in the numbers: $sin 3 0^{\circ} = 0.5$ , $cos 3 0^{\circ} = 0.866$ : $v = 2 (9.8) (4) (0.5 - 0.2 \cdot 0.866) = 78.4 \times 0.3268 = 5.06 m/s$ Why this step? The symbolic formula is general; substituting the specific given values ( $g, L, θ, μ_{k}$ ) is what turns it into the one concrete number the question asks for, and lets us sanity-check it against the frictionless case below.

Verify: Frictionless would give $2 (9.8) (4) (0.5) = 6.26 m/s$ ; friction correctly makes it slower, $5.06 < 6.26$ ✔. Units of $m/s^{2} \cdot m = m/s$ ✔.

Worked example Ex 4 — Cell D: the degenerate incline (won't move at all)

Same setup but $θ = 1 0^{\circ}$ , $μ_{k} = 0.4$ , and (this matters) $μ_{s} = 0.5$ . What does the sliding-speed formula give, and what does it mean?

Forecast: $tan 1 0^{\circ} = 0.176 < 0.4$ . My prediction: the kinetic formula breaks — a negative inside the square root — and static friction never even lets it start.

First ask the RIGHT question: does it start moving? Motion begins only if the along-slope gravity pull beats the maximum static friction: $m g sin θ > μ_{s} m g cos θ$ , i.e. $tan θ > μ_{s}$ . Why this step? Whether a block starts sliding is governed by the static coefficient $μ_{s}$ , not the kinetic $μ_{k}$ . Using $μ_{k}$ here would be the classic slip-up; $μ_{s} \geq μ_{k}$ is what actually holds a resting block. Here $tan 1 0^{\circ} = 0.176 < 0.5 = μ_{s}$ , so it never budges.
Now see what the kinetic formula says if we (wrongly) tried to apply it. Test the bracket $sin θ - μ_{k} cos θ$ : $sin 1 0^{\circ} - 0.4 cos 1 0^{\circ} = 0.1736 - 0.4 (0.9848) = 0.1736 - 0.3939 = - 0.220$ Why this step? This bracket is the net along-slope energy budget per metre once sliding. A negative value is the formula's own way of warning "there is no forward energy to give," consistent with step 1's static verdict.
Interpret the negative root. $v^{2} = 2 g L (- 0.220) < 0$ has no real solution. Why? A real speed can't square to a negative number. Physically the block is held by static friction and never accelerates; the kinetic-energy formula simply doesn't apply, and it flags that by going imaginary.
State both thresholds cleanly. Dividing $sin θ > μ cos θ$ by the positive $cos θ$ (which keeps the inequality's direction) collapses two trig terms into one clean test: $starts sliding if tan θ > μ_{s}; once sliding, speeds up if tan θ > μ_{k}$ Why this step? Dividing by $cos θ$ turns two terms into a single glance-able $tan θ$ , and separating $μ_{s}$ (start) from $μ_{k}$ (keep going) keeps the two physical questions distinct.

Verify: $tan 1 0^{\circ} = 0.176 < 0.4 = μ_{k} < 0.5 = μ_{s}$ , so the block neither starts nor could sustain sliding — perfectly consistent with statics. No numeric speed to report; the "answer" is no motion.

Worked example Ex 5 — Cell E: bullet embeds, then block skids (two tools)

A bullet of mass $m_{b} = 10 g = 0.01 kg$ moving at $v_{b} = 400 m/s$ embeds in a block of mass $m_{B} = 2 kg$ . The combined lump then slides on a floor with $μ_{k} = 0.4$ . Find (a) speed just after impact, (b) how far it skids, (c) heat from friction.

Forecast: Which conserves in the collision — energy or momentum? (Trap: it's momentum, not energy — the embed is violently inelastic.)

Collision: use momentum, NOT energy. With $m_{b}$ the bullet mass and $m_{B}$ the block mass, see Inelastic collisions. $m_{b} v_{b} = (m_{b} + m_{B}) v \Rightarrow 0.01 (400) = (2.01) v \Rightarrow v = 1.990 m/s$ Why this step? In an embed, huge KE becomes heat/deformation, so mechanical energy is not conserved. But no external horizontal impulse acts during the brief impact, so momentum is conserved.
Sliding phase: now use energy. Flat floor, so $Δ U = 0$ , $W_{n c} = Δ K E$ : $- μ_{k} (m_{b} + m_{B}) g d = 0 - \frac{1}{2} (m_{b} + m_{B}) v^{2}$ Mass cancels: $d = \frac{v ^{2}}{2 μ _{k} g} = \frac{1.99 0 ^{2}}{2 ( 0.4 ) ( 9.8 )} = 0.505 m$ . Why? Once moving as one lump, only friction acts along the flat floor — same machinery as Ex 1.
Heat from the skid $Q = - W_{n c} = \frac{1}{2} (m_{b} + m_{B}) v^{2} = \frac{1}{2} (2.01) (1.990)^{2} = 3.98 J$ . Why? All the post-collision KE is what friction converts to heat over the skid.

Verify: Skid distance $0.505 m$ and heat $3.98 J$ ; the embedding collision also dumped huge heat (see Heat & first law of thermodynamics). Units: $Q$ in J ✔, $d$ in m ✔.

Reading the figure: the blue horizontal line is the constant weight $m g$ ; the pink upward-curving parabola is the drag force $\frac{1}{2} C_{d} ρ A v^{2}$ , which grows with speed. They cross at the yellow dot — the yellow dashed line drops from it to the speed axis marking $v_{t}$ . Left of the cross, drag < weight (still accelerating); right of it, drag > weight (would decelerate). The crossing is where acceleration is exactly zero. (Both curves are also labelled in words on the plot, so the "flat weight line vs rising drag curve" story reads without colour.)

Forecast: As speed rises, drag rises like $v^{2}$ . Eventually it must catch gravity. Roughly $50 m/s$ ? Let's see.

Why "terminal" exists. As the figure shows, at low $v$ the flat blue weight line sits above the pink drag curve, so the diver accelerates. As $v$ grows, the $v^{2}$ drag curve climbs steeply and crosses the weight line at the yellow dot. Why this step? The crossing is where net force $= 0$ ; past it, drag would exceed weight and slow the diver, so speed settles exactly at that crossing.

Set forces equal at terminal speed $v_{t}$ (where $a = 0$ ): $m g = \frac{1}{2} C_{d} ρ A v_{t}^{2} \Rightarrow v_{t} = \frac{2 m g}{C _{d} ρ A}$ Why this step? "Terminal" means the speed stops changing, i.e. acceleration is zero, i.e. the two forces balance. Setting weight $=$ drag is the algebraic statement of that balance, which we then solve for $v_{t}$ by dividing and square-rooting.

Plug in the numbers: $v_{t} = \frac{2 ( 70 ) ( 9.8 )}{0.7} = 1960 = 44.3 m/s$ Why this step? We substitute the given $m$ , $g$ , and the lumped constant $C_{d} ρ A = 0.7$ into the derived formula precisely so we can compare the concrete result against real skydiver speeds — a sanity anchor the symbolic form can't give.

Limiting behaviour. Acceleration $a = g - \frac{C _{d} ρ A}{2 m} v^{2}$ . At $v = v_{t}$ : $a = 9.8 - \frac{0.7}{140} (1960) = 9.8 - 9.8 = 0$ ✔. Why? Confirms our balance: the diver stops speeding up. After this, all released gravitational PE goes straight to heating the air — no more KE gain. See Terminal velocity & projectile with drag.

Verify: $v_{t} = 44.3 m/s$ ( $\approx 160$ km/h, a realistic belly-to-earth value). At $v_{t}$ , $a = 0$ exactly. Units check: inside the root we have $\frac{kg \cdot m/s ^{2}}{kg/m} = \frac{m ^{2}}{s ^{2}}$ , whose square root is $m/s$ ✔ — the reason we do this check is to catch any slip in the lumped-constant units before trusting the number.

Worked example Ex 7 — Cell G: linear (Stokes) drag at low speed

A tiny $2 \times 1 0^{- 8} kg$ mist droplet falls through still air with linear drag $F_{d} = b v$ , $b = 3.0 \times 1 0^{- 8} N per (m/s)$ . Find its terminal velocity, and contrast the regime with Ex 6.

Forecast: Same balance idea as Ex 6, but the drag law is now $b v$ (first power of $v$ ), not $v^{2}$ . Will $v_{t}$ be tiny (droplets drift) or fast?

Pick the right drag law. For a slow, small object viscous shear dominates, so $F_{d} = b v$ (linear), not the quadratic law. Why this step? The two regimes from the parent note aren't interchangeable: a fog droplet moves at millimetres-per-second where air behaves smoothly (laminar), so the linear model is the correct tool here.
Balance weight against linear drag at terminal speed ( $a = 0$ ): $m g = b v_{t} \Rightarrow v_{t} = \frac{m g}{b}$ Why this step? "Terminal" again means acceleration zero, so downward weight equals upward drag. Because drag is now linear in $v$ , solving is a plain division — no square root, unlike Ex 6.
Plug in the numbers: $v_{t} = \frac{( 2 \times 1 0 ^{- 8} ) ( 9.8 )}{3.0 \times 1 0 ^{- 8}} = \frac{1.96 \times 1 0 ^{- 7}}{3.0 \times 1 0 ^{- 8}} = 6.53 m/s$ Why this step? Substituting the droplet's mass and the given $b$ turns the general balance into a concrete drift speed, which we then judge for physical plausibility.

Verify: $v_{t} = 6.53 m/s$ . Units check: $\frac{kg \cdot m/s ^{2}}{N/(m/s)} = \frac{N}{N/(m/s)} = m/s$ ✔ — we run it because the linear-drag coefficient's odd units (N per m/s) are exactly where mistakes hide. Note the structural contrast: linear drag gives $v_{t} \propto m$ , whereas quadratic drag (Ex 6) gave $v_{t} \propto m$ — the tool you pick changes the physics. See Terminal velocity & projectile with drag.

Worked example Ex 8 — Cell H: forensic word problem (skid marks → speed)

A car leaves $25 m$ of straight skid marks then stops. Road $μ_{k} = 0.8$ . How fast was it going when it locked its brakes?

Forecast: Skid marks are a friction-only, flat-floor problem — same shape as Ex 1, run backward. Faster than a residential $50$ km/h limit?

Flat road, brakes locked → kinetic friction stops it. With height unchanged $Δ U = 0$ , so the master equation collapses to $W_{n c} = Δ K E$ : $- μ_{k} m g L = 0 - \frac{1}{2} m v_{0}^{2}$ Why this step? Locked wheels slide, so it's kinetic friction (not rolling), and a level road means gravity's PE doesn't change — exactly the Ex 1 setup, letting us equate friction work to lost KE.
Mass cancels, solve for the initial speed: $v_{0} = 2 μ_{k} g L = 2 (0.8) (9.8) (25) = 392 = 19.8 m/s$ Why this step? Skid length $L$ is exactly the friction path length; we invert Ex 1's logic (solve for $v_{0}$ instead of $L$ ) to recover the speed that produced the marks. $m$ cancels, so the answer needs no knowledge of the car's mass.
Convert for a human: $19.8 m/s \times 3.6 = 71.3 km/h$ . Why this step? Skid physics gives SI m/s, but speed limits are in km/h; multiplying by $3.6$ (since $1 m/s = 3.6 km/h$ ) makes the result directly comparable to the posted limit, which is the point of the forensic question.

Verify: $19.8 m/s = 71.3 km/h$ — over a $50$ km/h limit, so the driver was speeding. Independent of car mass (why forensics loves this trick). Units ✔.

Reading the figure: same ramp as Ex 3, but now the dashed motion arrow points up the slope. Crucially the yellow friction arrow has flipped to point down-slope (it opposes the upward velocity), so it now sits alongside the blue gravity component $m g sin θ$ — both drain energy together. The pink $m g cos θ$ still presses into the slope. (As before, every arrow is word-labelled and direction-tagged so the picture reads in greyscale.)

Forecast: Going up, BOTH gravity and friction oppose motion — so it should stop sooner than a frictionless launch. Trap: friction now points down-slope, not up.

Direction check (the twist). Moving up-slope, velocity points up, so friction points down-slope — same direction as gravity's along-slope component (see the flipped yellow arrow in the figure). Why this step? Friction opposes velocity, not gravity. On the way up, both drain the block's KE together, which will show up as a plus sign inside the bracket.

Set up the energy budget. Rising a distance $L$ raises height by $L sin θ$ , so $Δ U = + m g L sin θ$ (positive — it climbs). Friction work is $W_{n c} = - μ_{k} m g cos θ L$ , and it ends at rest so $Δ K E = 0 - \frac{1}{2} m v_{0}^{2}$ . Why this step? Every term of the master equation must be pinned to this specific geometry before we can solve: $Δ U$ is positive here (unlike the sliding-down Ex 3), and $Δ K E$ is negative because the block loses all its launch speed.

Master equation, substitute the three terms: $W_{n c} - μ_{k} m g cos θ L = Δ K E - \frac{1}{2} m v_{0}^{2} + Δ U m g L sin θ$ Cancel $m$ , gather $L$ on one side: $\frac{1}{2} v_{0}^{2} = g L (sin θ + μ_{k} cos θ) \Rightarrow L = \frac{v _{0}^{2}}{2 g ( s i n θ + μ _{k} c o s θ )}$ Why the plus sign? Both energy drains — the gravity term $sin θ$ and the friction term $μ_{k} cos θ$ — add, unlike the minus in the sliding-down case (Ex 3). That sign flip is the whole exam point: it makes the block stop sooner.

Plug in the numbers: $sin 2 5^{\circ} = 0.4226$ , $cos 2 5^{\circ} = 0.9063$ : $L = \frac{5 ^{2}}{2 ( 9.8 ) ( 0.4226 + 0.3 \cdot 0.9063 )} = \frac{25}{2 ( 9.8 ) ( 0.6945 )} = 1.836 m$ Why this step? Substituting the given launch speed and angle into the derived formula gives the concrete stopping distance, which we immediately compare to the frictionless climb to confirm friction shortened it.

Verify: Frictionless climb would reach $\frac{v _{0}^{2}}{2 g sin θ} = \frac{25}{2 ( 9.8 ) ( 0.4226 )} = 3.02 m$ ; with friction it stops sooner at $1.84 < 3.02$ ✔, exactly as forecast. Units ✔.

Worked example Ex 10 — Cell J: BOTH thieves at once — friction

and air drag A puck slides on a rough floor and is also pushed back by air drag. At the instant it moves at $v = 20 m/s$ , find the total non-conservative force and the instantaneous heating rate. Data: $m = 0.5 kg$ , $μ_{k} = 0.10$ on flat ground, linear drag $b = 0.05 N per (m/s)$ .

Forecast: Two thieves rob the same puck. Do their forces add? (They both oppose the same velocity, so yes — but only friction is constant; drag depends on $v$ .)

Both forces point the same way (backward). Friction $f_{k} = μ_{k} m g$ and drag $F_{d} = b v$ both oppose the velocity, so along the direction of motion they simply add: $F_{n c} = f_{k} + F_{d} = μ_{k} m g + b v$ Why this step? Non-conservative forces don't interfere or cancel — each independently opposes $v$ , so the magnitudes add when they share a direction. This is why the matrix needed a "both at once" cell: real objects rarely feel just one.
Plug in the numbers for the total force: $F_{n c} = (0.10) (0.5) (9.8) + (0.05) (20) = 0.49 + 1.00 = 1.49 N$ Why this step? Substituting turns the symbolic sum into the actual backward push right now; note friction ( $0.49 N$ ) is fixed but drag ( $1.00 N$ ) would shrink as the puck slows — the split matters for what dominates.
Instantaneous heating rate (power stolen) is force times speed: $P = F_{n c} v = (0.49 + 1.00) (20) = 29.8 W$ Why this step? Power is the rate of energy theft; multiplying the opposing force by the current speed gives joules-per-second going to heat — the natural "both thieves combined" bookkeeping, since $Q$ accumulates as $\int P d t$ .

Verify: $F_{n c} = 1.49 N$ , $P = 29.8 W$ . Units: $N \times m/s = J/s = W$ ✔. Sanity: drag already out-robs friction here ( $1.00 > 0.49 N$ ) because the puck is fast; at low speed friction would dominate — exactly the regime split the matrix warns about.

Recall Quick self-test on the matrix

Which cell gives an imaginary speed, and what does that mean? ::: Cell D — the incline with $tan θ \leq μ_{k}$ ; imaginary root means the block never speeds up, and since $tan θ < μ_{s}$ static friction never even lets it start. In the bullet-block problem, which quantity is conserved in the collision? ::: Momentum only — the embed is inelastic, KE is lost to heat/deformation. On the way UP a rough incline, does friction point up or down the slope? ::: Down-slope — it opposes velocity, joining gravity to drain energy faster. Why does mass cancel in Ex 1, 5, 8? ::: Both $K E$ and $W_{f r i c}$ scale with $m$ , so it divides out — stopping distance is mass-independent. How does terminal velocity's mass dependence differ between linear and quadratic drag? ::: Linear: $v_{t} \propto m$ ; quadratic: $v_{t} \propto m$ . When friction and drag act together, do their forces add? ::: Yes — both oppose the same velocity, so magnitudes add; but only friction is constant, drag grows with $v$ . Which coefficient decides whether a resting block starts to slide? ::: $μ_{s}$ (static), not $μ_{k}$ ; motion starts only if $tan θ > μ_{s}$ .

Connections

Parent topic — theory
Work–Energy theorem — the always-true statement all these examples specialise.
Mechanical energy conservation — what these problems would reduce to if $μ_{k} = 0$ .
Friction — static & kinetic — where $μ_{k}$ , $μ_{s}$ and the sliding condition come from.
Inelastic collisions — the momentum step in Ex 5.
Terminal velocity & projectile with drag — Ex 6 and Ex 7's home.
Heat & first law of thermodynamics — where the stolen $Q$ goes.