High School Physics · Grades 10–12

17Work of a Force

Push a wardrobe across the room and you have worked; push against a wall all afternoon and, whatever your muscles report, physics counts zero. Work is what a force delivers along a motion — energy in transit (Chapter 9), paid in joules. This chapter prices the forces of Chapter 16 — pulls at an angle, weight, friction — and clocks the payment rate: power.

17.1 Work of a constant force

Definition 17.1 (Work of a constant force)

A constant force $\vect F$ applied to an object moving along a straight segment from $A$ to $B$ does the work

W_{AB}(\vect F) = F \times AB \times \cos\theta ,

where $F$ is the magnitude of the force (in $\mathrm{N}$ ), $AB$ the length of the displacement (in $\mathrm{m}$ ) and $\theta$ the angle between $\vect F$ and $\vect{AB}$ . Work is measured in joules ( $\mathrm{J}$ ): one joule is the work of a one-newton force along one metre of its own direction.

Remark 17.2

$F \times AB \times \cos\theta$ is exactly the scalar product $\vect F \cdot \vect{AB}$ of this year’s mathematics volume. Only the component $F\cos\theta$ along the motion earns anything.

Only the projection F of the force on the displacement works: W_AB( F) = F × AB ×. — Only the projection $F\cos\theta$ of the force on the displacement works: $W_{AB}(\vect F) = F \times AB \times \cos\theta$ .

Example 17.3 (Suitcase on a leash)

A traveller drags a suitcase $300\,\mathrm{m}$ across a terminal, the strap pulled with $F = 40\,\mathrm{N}$ at $\theta = 50{}^{\circ}$ to the floor: $W = 40 \times 300 \times \cos50{}^{\circ} \approx 7.7\,\mathrm{kJ}$ . Pulled flat ( $\theta = 0$ ) the same force would deliver $12\,\mathrm{kJ}$ : a third of the pull lifts nothing.

Definition 17.4 (Motor, resistive, zero work)

The sign of $\cos\theta$ sorts every force into three regimes:

$\theta < 90{}^{\circ}$ : $W > 0$ , the work is motor — the force feeds energy into the motion;
$\theta = 90{}^{\circ}$ : $W = 0$ , the force works not at all, however large;
$\theta > 90{}^{\circ}$ : $W < 0$ , the work is resistive — the force drains energy. The pure case is a brake at $180{}^{\circ}$ : $800\,\mathrm{N}$ over $45\,\mathrm{m}$ takes $800 \times 45 \times (-1) = -36\,\mathrm{kJ}$ out of a car, into hot discs.

The three regimes — motor, zero, resistive — by the angle between force and displacement (red arrow).

17.2 The work of the weight

Proposition 17.5 (Work of the weight)

When an object of mass $m$ moves from $A$ (altitude $z_A$ ) to $B$ (altitude $z_B$ ) along any path made of straight segments, its weight $\vect P$ does the work

W_{AB}(\vect P) = mg\,(z_A - z_B) = \pm\, mgh ,

where $h = \abs{z_A - z_B}$ is the height drop: $+mgh$ going down, $-mgh$ going up — the drop alone matters, not the route.

Proof. On one straight segment $AB$ , the weight (magnitude $mg$ , straight down) makes an angle $\theta$ with $\vect{AB}$ , and $AB\cos\theta$ is precisely the projection of the displacement on the downward vertical: $z_A - z_B$ . Hence $W = mg \times AB\cos\theta = mg\,(z_A - z_B)$ . Along a chain of segments the works add and the drops telescope: $mg(z_A - z_C) + mg(z_C - z_B) = mg(z_A - z_B)$ . ∎

Proposition 17.6 (Any path)

The same formula holds along any curved path from $A$ to $B$ .

Proof. Admitted at this level. ∎

Remark 17.7

Plausible — a curve is as close as we like to a broken line of many short segments — but making “as close as we like” honest is calculus, done in Chapter 29.

Straight (blue) or wandering (green), every path from A to B drops the same h: the weight does +mgh on both. — Straight (blue) or wandering (green), every path from $A$ to $B$ drops the same $h$ : the weight does $+mgh$ on both.

Example 17.8 (Hiker)

A $78\,\mathrm{kg}$ hiker (pack included) climbs $450\,\mathrm{m}$ , by whichever trail: $W(\vect P) = -78 \times 9.81 \times 450 \approx -344\,\mathrm{kJ}$ , and exactly $+344\,\mathrm{kJ}$ back down, whatever the detours.

17.3 The work of friction

Proposition 17.9 (Work of sliding friction)

A friction force of constant magnitude $f$ , at every instant opposite the motion, does the work $W = -fL$ along a path of total length $L$ : always resistive, and proportional to the length of the path travelled — not to the displacement.

Proof. Each short straight piece has the force at $180{}^{\circ}$ to the motion, contributing $-f \times (\text{its length})$ ; the pieces add up to $-fL$ . A curved path is chopped into short pieces likewise, the limit admitted as for the weight (Remark 17.7). ∎

Remark 17.10 (Friction against weight: the great divide)

The weight’s work forgets the path; friction’s is a toll per metre. On a round trip the weight nets zero — what it takes uphill it refunds downhill — while friction charges $-fL$ on every leg and refunds nothing: the weight’s work can be stored and recovered (potential energy, Chapter 18), friction’s is lost as heat.

Example 17.11 (There and back)

A crate slides $5.0\,\mathrm{m}$ across a floor and $5.0\,\mathrm{m}$ back, against $f = 60\,\mathrm{N}$ : friction does $-60 \times 10 = -600\,\mathrm{J}$ even though the crate ends where it started. Weight and normal force, perpendicular to the motion: zero throughout.

Remark 17.12 (What about a spring?)

A spring pulls harder the further it is stretched: its force is not constant and Definition 17.1 does not apply. Its work means cutting the stretch into infinitesimal steps and summing — calculus, delivered in Chapter 29.

17.4 Power

Definition 17.13 (Power)

The power of a force is the rate at which it works, $P = W/\Delta t$ , in watts ( $\mathrm{W} = \mathrm{J}/\mathrm{s}$ ) — the same watt as in Chapter 12. A pre-SI unit survives on engine badges: the horsepower, $1\,\mathrm{hp} = 736\,\mathrm{W}$ .

Proposition 17.14 (Power at constant velocity)

If the object moves at constant speed $v$ , a constant force $\vect F$ at angle $\theta$ to the motion delivers the power $P = F v \cos\theta$ — in particular $P = Fv$ for a force along the motion.

Proof. In a time $\Delta t$ the object covers $d = v\,\Delta t$ , so $W = F \times v\,\Delta t \times \cos\theta$ ; divide by $\Delta t$ . ∎

Example 17.15 (Crane)

A crane hoists a $600\,\mathrm{kg}$ pallet at a steady $0.90\,\mathrm{m}/\mathrm{s}$ : the cable’s force equals the weight, $mg = 5.89 \times 10^{3}\,\mathrm{N}$ , so $P = Fv = 5890 \times 0.90 \approx 5.3\,\mathrm{kW} \approx 7.2\,\mathrm{hp}$ — a seven-horse team in a steel box.

Method 17.16 (Climbing with less force: the zigzag)

To raise a load a height $h$ with a limited force:

The bill is fixed: by Proposition 17.5, any route to the top costs the work $mgh$ .
Stretch the path: on a slope of length $L$ the force needed (constant speed, friction aside) is $F = mgh/L$ — double the length, halve the force.
Pay the surcharge: friction charges by the metre (Proposition 17.9), adding $fL$ .

Mountain roads zigzag for step 2: same height, less force, same work.

Hauling 1200\, kg up h = 300\, m: the product F × L = mgh = 3.53\, MJ is fixed, so the road trades length for force (see ). — Hauling $1200\,\mathrm{kg}$ up $h = 300\,\mathrm{m}$ : the product $F \times L = mgh = 3.53\,\mathrm{MJ}$ is fixed, so the road trades length for force (see Exercise 17.12).

17.5 Exercises

Exercise 17.1 ★

A constant force $F = 40\,\mathrm{N}$ acts over a straight $50\,\mathrm{m}$ displacement: compute its work for $\theta = 0{}^{\circ}$ , $60{}^{\circ}$ , $90{}^{\circ}$ and $120{}^{\circ}$ .

Solution

Solution of Exercise 17.1.

$W = 40 \times 50 \times \cos\theta$ : $+2000\,\mathrm{J}$ ; $+1000\,\mathrm{J}$ ; $0$ ; $-1000\,\mathrm{J}$ .

Exercise 17.2 ★

A $12\,\mathrm{kg}$ toolbox is lowered $2.5\,\mathrm{m}$ from a van, then later raised back. Work of its weight in each case?

Solution

Solution of Exercise 17.2.

$mgh = 12 \times 9.81 \times 2.5 = 294\,\mathrm{J}$ : lowered, $+294\,\mathrm{J}$ (weight along the drop); raised, $-294\,\mathrm{J}$ .

Exercise 17.3 ★

Motor, resistive or zero? (a) the weight of a falling apple; (b) the weight of a rising ball; (c) the normal force on a sliding box; (d) friction on a braking bicycle; (e) the string’s tension on a swinging pendulum bob.

Solution

Solution of Exercise 17.3.

(a) motor; (b) resistive; (c) zero (perpendicular); (d) resistive; (e) zero — the string’s tension stays perpendicular to the bob’s motion.

Exercise 17.4 ★

A crane lifts $600\,\mathrm{kg}$ by $25\,\mathrm{m}$ at constant speed. Work of the lifting force and of the weight? What is their sum, and why?

Solution

Solution of Exercise 17.4.

Constant speed: lifting force $= mg = 5.89 \times 10^{3}\,\mathrm{N}$ , so $W_{\text{lift}} = 5886 \times 25 \approx +147\,\mathrm{kJ}$ and $W(\vect P) = -147\,\mathrm{kJ}$ . Sum zero: at constant velocity the forces balance (Chapter 16), so their works cancel.

Exercise 17.5 ★

An escalator motor does $90\,\mathrm{kJ}$ of work every minute. Its power in watts? In horsepower?

Solution

Solution of Exercise 17.5.

$P = 9.0 \times 10^{4}/60 = 1.5\,\mathrm{kW}$ ; $1500/736 \approx 2.0\,\mathrm{hp}$ .

Exercise 17.6 ★★

A sled is towed $200\,\mathrm{m}$ by a rope of tension $120\,\mathrm{N}$ at $30{}^{\circ}$ above the snow; friction is $45\,\mathrm{N}$ . Work of each of the four forces (rope, friction, weight, normal) and total?

Solution

Solution of Exercise 17.6.

Rope: $120 \times 200 \times \cos30{}^{\circ} \approx +20.8\,\mathrm{kJ}$ ; friction: $-45 \times 200 = -9.0\,\mathrm{kJ}$ ; weight and normal force, perpendicular: $0$ . Total $+11.8\,\mathrm{kJ}$ : the sled gains speed.

Exercise 17.7 ★★

A cyclist rides at a constant $25\,\mathrm{km}/\mathrm{h}$ against a total drag of $18\,\mathrm{N}$ . Power delivered? Work done in one hour?

Solution

Solution of Exercise 17.7.

$v = 25/3.6 = 6.94\,\mathrm{m}/\mathrm{s}$ ; $P = Fv = 18 \times 6.94 \approx 125\,\mathrm{W}$ . In one hour, $W = P\,\Delta t = 125 \times 3600 = 450\,\mathrm{kJ}$ (equivalently $18 \times 25\,000\,\mathrm{m}$ ).

Exercise 17.8 ★★

A $65\,\mathrm{kg}$ skier descends $120\,\mathrm{m}$ of height, by a $800\,\mathrm{m}$ steep slope or a $1500\,\mathrm{m}$ gentle track; friction on the skis is $30\,\mathrm{N}$ on both. Work of the weight and of friction on each route? Which work cares about the route?

Solution

Solution of Exercise 17.8.

Weight, both routes: $+65 \times 9.81 \times 120 \approx +76.5\,\mathrm{kJ}$ (drop only). Friction: $-30 \times 800 = -24\,\mathrm{kJ}$ on the slope, $-30 \times 1500 = -45\,\mathrm{kJ}$ on the track. Only friction cares about the route.

Exercise 17.9 ★★

A $20\,\mathrm{kg}$ box is pushed $4.0\,\mathrm{m}$ up a $25{}^{\circ}$ ramp at constant speed, against $30\,\mathrm{N}$ of friction. Compute the works of the weight and of friction, then (from the constant speed) the work and magnitude of the push, parallel to the ramp.

Solution

Solution of Exercise 17.9.

Rise $h = 4.0\sin25{}^{\circ} = 1.69\,\mathrm{m}$ : $W(\vect P) = -20 \times 9.81 \times 1.69 \approx -332\,\mathrm{J}$ ; $W_f = -30 \times 4.0 = -120\,\mathrm{J}$ . Constant speed: total work zero, so $W_{\text{push}} = +452\,\mathrm{J}$ and $F = 452/4.0 \approx 113\,\mathrm{N}$ .

Exercise 17.10 ★★

A car cruises at $130\,\mathrm{km}/\mathrm{h}$ with its engine delivering $50\,\mathrm{kW}$ to the wheels. What total resistive force does it fight?

Solution

Solution of Exercise 17.10.

$v = 130/3.6 = 36.1\,\mathrm{m}/\mathrm{s}$ ; constant speed, so $F = P/v = 5.0 \times 10^{4}/36.1 \approx 1.4\,\mathrm{kN}$ .

Exercise 17.11 ★★

A canal horse tows a barge at $1.0\,\mathrm{m}/\mathrm{s}$ , the rope of tension $800\,\mathrm{N}$ making $20{}^{\circ}$ with the towpath. Compute the horse’s power, in watts and in horsepower. Comment.

Solution

Solution of Exercise 17.11.

$P = Fv\cos\theta = 800 \times 1.0 \times \cos20{}^{\circ} \approx 752\,\mathrm{W}$ ; in horsepower, $752/736 \approx 1.02$ — almost exactly what a strong horse sustains, the comparison Watt built his unit on.

Exercise 17.12 ★★★

A $1200\,\mathrm{kg}$ car climbs $h = 300\,\mathrm{m}$ to a pass, by a straight track of $20\%$ grade or a zigzag road of $6\%$ grade (grade $= \sin\alpha$ ). Neglecting friction, at constant speed, compute for each route the length, the force needed along the slope, and its work. Conclude in one sentence.

Solution

Solution of Exercise 17.12.

Lengths $L = h/\text{grade}$ : $1500\,\mathrm{m}$ and $5000\,\mathrm{m}$ . Forces $F = mg \times \text{grade}$ : $1200 \times 9.81 \times 0.20 \approx 2.35\,\mathrm{kN}$ and $0.71\,\mathrm{kN}$ . Works $FL$ : $2354 \times 1500 \approx 3.53\,\mathrm{MJ}$ and $706 \times 5000 \approx 3.53\,\mathrm{MJ}$ — both equal $mgh$ : the road trades force for length, never work.

Exercise 17.13 ★★★

A winch raises $250\,\mathrm{kg}$ of tiles $15\,\mathrm{m}$ up scaffolding at a steady $0.40\,\mathrm{m}/\mathrm{s}$ . Compute the work of the cable’s force, the duration, the mechanical power (twice: $W/\Delta t$ and $Fv$ ), and the electrical power drawn at $70\%$ motor efficiency.

Solution

Solution of Exercise 17.13.

$F = mg = 2.45 \times 10^{3}\,\mathrm{N}$ ; $W = 2453 \times 15 \approx 36.8\,\mathrm{kJ}$ ; $\Delta t = 15/0.40 = 37.5\,\mathrm{s}$ ; $P = 36800/37.5 \approx 981\,\mathrm{W} = 2453 \times 0.40$ ( $Fv$ agrees); electrical: $981/0.70 \approx 1.4\,\mathrm{kW}$ .

Exercise 17.14 ★★★

A wardrobe is slid against $f = 140\,\mathrm{N}$ of friction from one corner of a room to the opposite one: straight along the $6.0\,\mathrm{m}$ diagonal, or $3.6\,\mathrm{m}$ then $4.8\,\mathrm{m}$ along the walls. Friction work on each route, then on a diagonal round trip? What does the weight do on that round trip, and what deep difference between the two forces does this expose?

Solution

Solution of Exercise 17.14.

Diagonal: $-140 \times 6.0 = -840\,\mathrm{J}$ ; walls: $-140 \times 8.4 \approx -1.18\,\mathrm{kJ}$ ; round trip: $-140 \times 12 = -1.68\,\mathrm{kJ}$ . The weight does zero on the round trip (no net drop). Friction’s work depends on the path and never refunds; the weight’s depends only on the endpoints (Remark 17.10).

Exercise 17.15 ★★★

A cyclist ( $80\,\mathrm{kg}$ with bike) climbs a pass: $600\,\mathrm{m}$ of height over $12\,\mathrm{km}$ of road, at a constant $15\,\mathrm{km}/\mathrm{h}$ , against $14\,\mathrm{N}$ of drag and rolling resistance. Compute the work supplied against gravity, against friction, and in total; the duration of the climb; the average power, in watts and horsepower.

Solution

Solution of Exercise 17.15.

Gravity: $80 \times 9.81 \times 600 \approx 471\,\mathrm{kJ}$ ; friction: $14 \times 12000 = 168\,\mathrm{kJ}$ ; total $639\,\mathrm{kJ}$ . $v = 4.17\,\mathrm{m}/\mathrm{s}$ , so $\Delta t = 12000/4.17 = 2880\,\mathrm{s} = 48\,\mathrm{min}$ and $P = 6.39 \times 10^{5}/2880 \approx 222\,\mathrm{W} \approx 0.30\,\mathrm{hp}$ — a strong amateur, a third of a horse.

17.6 Problem: The mover’s day

Problem 17.1

Weekend problem — physics pays by the joule: a day of sliding, hoisting and hairpin bends, ending with the humbling discovery of what a kettle thinks of honest labour

A mover’s crew empties a warehouse, hauls the load up to a third-floor flat, then drives the van to a hilltop village. Every task is priced in joules; the kitchen kettle will audit the day. Boxes: $30\,\mathrm{kg}$ .

Part I — Across the warehouse floor. A box is pushed with a horizontal force $F = 110\,\mathrm{N}$ over $d = 8.0\,\mathrm{m}$ ; sliding friction is $f = 95\,\mathrm{N}$ .

Compute the work of the push.
Compute the work of friction, of the weight, and of the normal force (justify the zeros).
Total work on the box, and its sign’s meaning (the box started at rest)? What push keeps the speed constant (Chapter 16)?
The far corner can be reached along the $10.0\,\mathrm{m}$ diagonal or along two walls ( $6.0\,\mathrm{m}$ then $8.0\,\mathrm{m}$ ). Friction work on each route?
On a diagonal round trip (out and back), compute the work of friction and the work of the weight. Which force refunds?

Part II — Up to the third floor ( $h = 9.0\,\mathrm{m}$ ).

Compute the work of the weight on a box raised to the flat. What work must the crew supply at least?
Carried up the stairs by a $70\,\mathrm{kg}$ mover, gravity charges for mover and box: total work against gravity, and the fraction actually spent on the box?
A trolley ramp is $30\,\mathrm{m}$ long for the same $9.0\,\mathrm{m}$ rise, with $40\,\mathrm{N}$ of rolling friction. At constant speed, compute the push needed along the ramp and its work.
Compare the ramp with a straight vertical hoist: force needed, work done. What does the ramp buy, and at what price?
A rope-and-pulley hoist raises the box at $0.50\,\mathrm{m}/\mathrm{s}$ . Duration and mechanical power?
Recover that power as $P = Fv$ . The hoist’s motor is $60\%$ efficient: electrical power drawn?

Part III — The zigzag road. The loaded van ( $3500\,\mathrm{kg}$ ) must reach a village $240\,\mathrm{m}$ above the valley: straight lane, $1.0\,\mathrm{km}$ at $24\%$ grade, or paved road, $8.0\,\mathrm{km}$ at $3.0\%$ grade (grade $= \sin\alpha$ ).

Check that both routes climb the same $240\,\mathrm{m}$ .
At constant speed and neglecting friction, compute the driving force needed along each route.
Work of the driving force on each route? Compare with $mgh$ .
Rolling resistance is $400\,\mathrm{N}$ on both: compute the extra work on each route. What does the zigzag’s comfort cost?
At $36\,\mathrm{km}/\mathrm{h}$ , compute the engine power needed on each route (in $\mathrm{kW}$ and $\mathrm{hp}$ ). Why do roads zigzag?

Part IV — The day’s ledger.

The crew slides $60$ boxes across the floor (the push of question 1) and hoists all $60$ up the $9.0\,\mathrm{m}$ (question 6). Total mechanical work delivered to the boxes?
Add the stair carrier’s own body: $20$ trips up at $70\,\mathrm{kg}$ (downhill refunds go to hot knees, not to the ledger). Day’s total?
The crew worked $8.0$ hours: average mechanical power? How long would the $2.0\,\mathrm{kW}$ kettle take to expend the same energy?
Finale: in two sentences, say what the joule pays for — and does not (the wall pushed all afternoon, the box held motionless) — and why movers own winches, ramps and zigzag roads rather than bigger muscles.

Solution

Solution of Problem 17.1.

1. $W = 110 \times 8.0 = +880\,\mathrm{J}$ .

2. Friction: $-95 \times 8.0 = -760\,\mathrm{J}$ . Weight and normal force are perpendicular to the motion: zero each.

3. Total $+120\,\mathrm{J}$ : net motor work, the box speeds up. A $95\,\mathrm{N}$ push balances friction — zero net force, constant speed, zero total work.

4. Diagonal: $-95 \times 10.0 = -950\,\mathrm{J}$ ; walls: $-95 \times 14.0 = -1.33\,\mathrm{kJ}$ : friction charges by the metre.

5. Friction: $-95 \times 20.0 = -1.9\,\mathrm{kJ}$ ; weight: $0$ (no net drop). Only the weight refunds.

6. $W(\vect P) = -30 \times 9.81 \times 9.0 = -2.65\,\mathrm{kJ}$ ; the crew must supply at least $+2.65\,\mathrm{kJ}$ .

7. $(70 + 30) \times 9.81 \times 9.0 = 8.83\,\mathrm{kJ}$ ; the box’s share is $2.65/8.83 \approx 30\%$ — carrying is mostly self-transport.

8. $\sin\alpha = 9.0/30 = 0.30$ : $F = mg\sin\alpha + f = 88.3 + 40 = 128\,\mathrm{N}$ ; $W = 128 \times 30 = 3.85\,\mathrm{kJ}$ ( $= 2.65 + 1.2$ ).

9. Vertical hoist: $F = mg = 294\,\mathrm{N}$ , $W = 2.65\,\mathrm{kJ}$ . The ramp buys a smaller force ( $128$ against $294\,\mathrm{N}$ ) at the price of more work: friction’s $1.2\,\mathrm{kJ}$ toll.

10. $\Delta t = 9.0/0.50 = 18\,\mathrm{s}$ ; $P = 2649/18 \approx 147\,\mathrm{W}$ .

11. $P = Fv = 294 \times 0.50 = 147\,\mathrm{W}$ — same. Electrical: $147/0.60 \approx 245\,\mathrm{W}$ .

12. $1000 \times 0.24 = 240\,\mathrm{m}$ and $8000 \times 0.030 = 240\,\mathrm{m}$ : same summit.

13. $F = mg \times \text{grade}$ : lane $3500 \times 9.81 \times 0.24 \approx 8.2\,\mathrm{kN}$ ; road $\approx 1.0\,\mathrm{kN}$ .

14. $8240 \times 1000 \approx 8.24\,\mathrm{MJ}$ and $1030 \times 8000 \approx 8.24\,\mathrm{MJ}$ — both $= mgh = 3500 \times 9.81 \times 240$ .

15. Rolling resistance: $400 \times 1000 = 0.40\,\mathrm{MJ}$ against $400 \times 8000 = 3.2\,\mathrm{MJ}$ : the gentle road’s comfort costs $2.8\,\mathrm{MJ}$ of extra friction.

16. $v = 10\,\mathrm{m}/\mathrm{s}$ . Road: $(1030 + 400) \times 10 = 14.3\,\mathrm{kW} \approx 19\,\mathrm{hp}$ ; lane: $(8240 + 400) \times 10 = 86.4\,\mathrm{kW} \approx 117\,\mathrm{hp}$ — beyond a loaded van. Roads zigzag so that modest engines (and tyres) can climb: same work, spread thin.

17. Per box: $880 + 2649 = 3.53\,\mathrm{kJ}$ ; for $60$ boxes, $60 \times 3529 \approx 212\,\mathrm{kJ}$ .

18. Stairs: $20 \times 70 \times 9.81 \times 9.0 \approx 124\,\mathrm{kJ}$ . Day’s total $\approx 336\,\mathrm{kJ} = 0.34\,\mathrm{MJ}$ .

19. $P = 3.36 \times 10^{5}/28800 \approx 12\,\mathrm{W}$ . The kettle: $3.36 \times 10^{5}/2000 = 168\,\mathrm{s}$ — under three minutes for the whole day’s labour.

20. The joule pays for force delivered along a motion — nothing for the wall pushed or the box held, however exhausting (muscles burn chemical energy even doing zero mechanical work). Since human power output is a few dozen watts, movers let machines reshape the force — winch, ramp, zigzag — while the bill $mgh$ never changes.