High School Physics · Grades 10–12

29Work and Mechanical Energy

Tip a marble into a salad bowl: it dives, overshoots the bottom, climbs to its starting height, hesitates, turns back — and every reversal can be predicted from one curve (the bowl’s profile) and one horizontal line (the marble’s energy), no equation of motion solved. Last year priced forces on straight lines (Chapter 17) and balanced kinetic against potential energy (Chapter 18); this year the bookkeeping goes everywhere: curved paths, springs, landscapes at a glance.

29.1 Work along any path

Definition 29.1 (Work along a path)

Let a force $\vect F$ act on a body moving along any path from $A$ to $B$ : cut the path into displacements $\vect{\dd\ell}$ so short that each is straight and $\vect F$ barely changes along it. Each step earns the small work $\scal{F}{\dd\ell} = F\,\dd\ell\cos\theta$ (Chapter 17), and the work along the path is their sum, $W_{AB}(\vect F) = \sum \scal{F}{\dd\ell}$ — which on a straight path with constant force collapses to last year’s $F \times AB \times \cos\theta$ .

A curved path is a broken line of many short steps: each earns F\,; the work is the sum. — A curved path is a broken line of many short steps: each $\vect{\dd\ell}$ earns $F\,\dd\ell\cos\theta$ ; the work is the sum.

Proposition 29.2 (Work of the weight, any path)

Along any path from $A$ (altitude $z_A$ ) to $B$ (altitude $z_B$ ), the weight of a mass $m$ does the work $W_{AB}(\vect P) = mg\,(z_A - z_B)$ : the drop enters, never the route.

Proof. Each short step contributes $mg \times (\text{small drop})$ (last year’s straight-segment computation); the drops telescope to $z_A - z_B$ . The curved case, admitted then, is delivered. ∎

Definition 29.3 (Conservative and non-conservative forces)

A force is conservative if its work between two points is the same along every path (zero on every round trip), non-conservative otherwise. Weight and spring force (next section): conservative. Sliding friction is not: opposite the motion, it does $W = -fL$ on a path of length $L$ — a toll per metre, never refunded.

Example 29.4 (Two trails to the hut)

A $75\,\mathrm{kg}$ hiker gains $300\,\mathrm{m}$ by a steep $800\,\mathrm{m}$ trail or $2.4\,\mathrm{km}$ of switchbacks. The weight charges $-75 \times 9.81 \times 300 \approx -2.2 \times 10^{5}\,\mathrm{J}$ on both; a $30\,\mathrm{N}$ friction force, $-2.4 \times 10^{4}\,\mathrm{J}$ on one, $-7.2 \times 10^{4}\,\mathrm{J}$ on the other. The weight refunds on the descent; friction’s bill is gone as heat, both ways.

29.2 Potential energy

Definition 29.5 (Potential energy)

To every conservative force one attaches a potential energy $E_p$ , a number depending on position alone, defined by $W_{AB} = E_p(A) - E_p(B)$ : work delivered is potential energy spent. Fixing $E_p = 0$ at a chosen reference point fixes it everywhere; the choice shifts all values by a constant that cancels from every difference. For the weight this gives back last year’s $E_p = mgz$ , now valid along any path. Friction can own no $E_p$ : a route-dependent bill is no difference of endpoint numbers — only conservative forces store retrievably, exactly what the name conserves.

Proposition 29.6 (Elastic potential energy)

A spring of stiffness $k$ ( $\mathrm{N}/\mathrm{m}$ ) stretched or compressed by $x$ pulls back with the force $F = kx$ (Chapter 28); the work it delivers returning to rest — its elastic potential energy — is $E_p = \tfrac12\, k x^2$ .

Proof. The restoring force acts along the motion, so $\sum F\,\dd\ell$ is the area under the $F$ – $x$ graph: a triangle of base $x$ and height $kx$ , area $\tfrac12 x \times kx$ . The force at each stretch is the same out and back: work fixed by the endpoints, spring force conservative, $E_p$ well defined. ∎

The spring’s force grows with the stretch; the stored energy is the area under the line: a triangle, 1/2 × x × kx. — The spring’s force grows with the stretch; the stored energy is the area under the line: a triangle, $\tfrac12 \times x \times kx$ .

29.3 The two energy theorems

Theorem 29.7 (Kinetic-energy theorem)

For any motion of a body of mass $m$ from $A$ to $B$ — curved, looped, three-dimensional —

\Delta E_k = E_k(B) - E_k(A) = \sum W_{AB}(\vect F),

the sum running over all forces acting on the body.

Proof. Since $v^2 = \scal{v}{v}$ , the product rule of this year’s mathematics gives the derivative of $v^2$ as $2\,\scal{a}{v}$ , so $\dd E_k/\dd t = m\,\scal{a}{v} = \scal{F_{\text{tot}}}{v}$ by Newton’s second law (Chapter 25). In a short $\dd t$ the body moves $\vect{\dd\ell} = \vect v\,\dd t$ : $E_k$ gains the small work of Definition 29.1; summing the steps sums the works — last year’s admitted statement, earned in full. ∎

Proposition 29.8 (Instantaneous power)

At every instant a force $\vect F$ on a body of velocity $\vect v$ delivers the power $P = \scal{F}{v} = Fv\cos\theta$ , and $\dd E_k/\dd t$ is the sum of the powers of all forces: last year’s constant-velocity formula, exact at each instant of any motion.

Proof. Read off the previous proof: each force feeds $\scal{F}{\dd\ell} = \scal{F}{v}\,\dd t$ into $E_k$ during $\dd t$ . ∎

Theorem 29.9 (Mechanical-energy theorem)

Let $E_m = E_k + E_p$ , where $E_p$ collects the potential energies of all the conservative forces at work. Then, along any path,

\Delta E_m = W_{AB}(\text{non-conservative forces}),

and $E_m$ is conserved whenever the non-conservative forces do no work.

Proof. Split the kinetic-energy theorem: $\Delta E_k = W_{\text{cons}} + W_{\text{nc}} = -\Delta E_p + W_{\text{nc}}$ ; move $\Delta E_p$ across. ∎

Example 29.10 (The pendulum, honestly this time)

A pendulum bob on a wire of length $L = 2.5\,\mathrm{m}$ is released at rest at $45{}^{\circ}$ from the vertical. The tension, perpendicular to the motion, is powerless; the weight is conservative: $E_m$ is conserved on the curved arc, which last year could only be checked. The bob starts $h = L(1 - \cos45{}^{\circ}) = 0.73\,\mathrm{m}$ up, so at the bottom $v = \sqrt{2gh} \approx 3.8\,\mathrm{m}/\mathrm{s}$ .

Method 29.11 (Energy audit of any motion)

Choose the system and list the forces on it.
Sort: perpendicular to the motion $\to$ no work; conservative $\to$ an $E_p$ in $E_m$ ; the rest $\to$ $W_{\text{nc}}$ (sliding friction: $-fL$ , $L$ = path length).
Write $\Delta E_m = W_{\text{nc}}$ between the data point and the question point; solve. Energy answers “how fast, how high”; only time and acceleration need Newton’s laws.

29.4 Energy diagrams

Definition 29.12 (Energy diagram, turning points)

For a motion along one axis $x$ with conserved $E_m$ , the energy diagram plots the curve $E_p(x)$ and the horizontal line $E_m$ . Since $E_k = E_m - E_p \geq 0$ , the motion is confined to where the curve lies below the line; the gap between them is $E_k$ , read directly. At a turning point, where curve meets line, $v = 0$ and the motion reverses.

An energy diagram, read without solving anything: with energy E_m the body oscillates between x_1 and x_2, fastest where the gap is largest; with E_m' it crawls over the hill and escapes. — An energy diagram, read without solving anything: with energy $E_m$ the body oscillates between $x_1$ and $x_2$ , fastest where the gap is largest; with $E_m'$ it crawls over the hill and escapes.

Proposition 29.13 (Force from the landscape; equilibria)

The conservative force along $x$ is minus the slope of $E_p$ : $F = -\,\dd E_p/\dd x$ — it points downhill on the diagram, and vanishes where the tangent is horizontal: an equilibrium. A minimum of $E_p$ is stable — the displaced body is pushed back, a ball in a valley; a maximum is unstable — it is pushed away, a ball on a hilltop.

Proof. On a small step $\dd x$ the force works $F\,\dd x = E_p(x) - E_p(x + \dd x) = -\dd E_p$ ; divide by $\dd x$ . Downhill on either side of a valley points back in; around a hilltop, out. ∎

Remark 29.14 (Oscillations and escape)

The small oscillations of Chapter 28 live at the bottoms of $E_p$ valleys. And far from Earth gravity weakens: the $E_p(r)$ curve climbs ever more slowly toward a finite ceiling (the area under the shrinking force-graph is finite). A rocket whose $E_m$ reaches the ceiling never meets a turning point: it escapes — about $11.2\,\mathrm{km}/\mathrm{s}$ from the ground (Exercise 29.13); below that it is bound. The ceiling itself is computed in the Year 1 volume.

Example 29.15 (Loop-the-loop)

A marble released from height $H$ coasts into a vertical loop of radius $R$ . Circular dynamics (Chapter 25) demands $v_{\text{top}}^2 \geq gR$ at the top — the datum borrowed by last year’s roller-coaster problem, now ours. Conservation from rest at $H$ to the top (height $2R$ ) gives $v_{\text{top}}^2 = 2g(H - 2R)$ , so $H \geq \tfrac52 R$ : half a radius above the loop top, whatever the mass — a little more in the real, rubbing world.

Loop-the-loop: the release sits half a radius above the top, so the leftover E_k keeps the marble pressed on (v_ top2 = gR). — Loop-the-loop: the release sits half a radius above the top, so the leftover $E_k$ keeps the marble pressed on ( $v_{\text{top}}^2 = gR$ ).

29.5 Exercises

Exercise 29.1 ★

A spring of stiffness $k = 200\,\mathrm{N}/\mathrm{m}$ is stretched by $5.0\,\mathrm{cm}$ . Compute the stored energy; what does it become if the stretch doubles? How much does the second $5.0\,\mathrm{cm}$ cost?

Solution

Solution of Exercise 29.1.

$E_p = \tfrac12 \times 200 \times 0.050^2 = 0.25\,\mathrm{J}$ . Doubled stretch: $\times 4$ , i.e. $1.0\,\mathrm{J}$ . The second $5.0\,\mathrm{cm}$ alone costs $1.0 - 0.25 = 0.75\,\mathrm{J}$ — the force is larger there.

Exercise 29.2 ★

A $0.90\,\mathrm{kg}$ drone flies a wandering route from a terrace at $z = 12\,\mathrm{m}$ to a balcony at $z = 30\,\mathrm{m}$ . Work of its weight? Along the straight climb? On the round trip?

Solution

Solution of Exercise 29.2.

$W = mg(z_A - z_B) = 0.90 \times 9.81 \times (12 - 30) \approx -1.6 \times 10^{2}\,\mathrm{J}$ , whatever the route — the same on the straight climb. Round trip: $0\,\mathrm{J}$ (conservative force).

Exercise 29.3 ★

Conservative, non-conservative, or workless? Justify from paths: (a) the weight; (b) the spring force; (c) sliding friction; (d) the normal force on a sliding box; (e) air drag.

Solution

Solution of Exercise 29.3.

(a), (b) conservative: work fixed by the endpoints (drop of altitude; stretch). (c), (e) non-conservative: $-fL$ and drag charge per metre travelled. (d) workless: always perpendicular to the motion.

Exercise 29.4 ★

A pendulum of length $2.0\,\mathrm{m}$ is released at rest at $60{}^{\circ}$ from the vertical. Find its speed at the lowest point. Why does the wire’s tension never enter the balance?

Solution

Solution of Exercise 29.4.

$h = L(1 - \cos60{}^{\circ}) = 1.0\,\mathrm{m}$ ; $v = \sqrt{2 \times 9.81 \times 1.0} \approx 4.4\,\mathrm{m}/\mathrm{s}$ . The tension is perpendicular to the motion at every instant: zero power, zero work.

Exercise 29.5 ★

A cyclist rides at $9.0\,\mathrm{m}/\mathrm{s}$ on the flat, delivering $250\,\mathrm{W}$ : what total resistive force is she fighting? Climbing at $5.0\,\mathrm{m}/\mathrm{s}$ with the same power: what force does she overcome?

Solution

Solution of Exercise 29.5.

$F = P/v = 250/9.0 \approx 28\,\mathrm{N}$ . At $5.0\,\mathrm{m}/\mathrm{s}$ : $F = 250/5.0 = 50\,\mathrm{N}$ (mostly the slope’s gravity component).

Exercise 29.6 ★★

A toy launcher’s spring ( $k = 450\,\mathrm{N}/\mathrm{m}$ ) is compressed by $8.0\,\mathrm{cm}$ behind a $20\,\mathrm{g}$ dart. Find the stored energy, the launch speed, and the height reached if fired straight up (no air).

Solution

Solution of Exercise 29.6.

$E_p = \tfrac12 \times 450 \times 0.080^2 = 1.44\,\mathrm{J}$ ; $v = \sqrt{2 \times 1.44/0.020} = 12\,\mathrm{m}/\mathrm{s}$ ; $h = E_p/mg = 1.44/(0.020 \times 9.81) \approx 7.3\,\mathrm{m}$ .

Exercise 29.7 ★★

Stretching a rubber band, a student measures $F = 0$ , $3.0$ , $7.0$ , $12.0$ , $18.0\,\mathrm{N}$ at $x = 0$ , $2.0$ , $4.0$ , $6.0$ , $8.0\,\mathrm{cm}$ . Estimate the stored energy by trapezoidal areas; compare with $\tfrac12 kx^2$ , taking $k = F/x$ at the end. Why do they differ?

Solution

Solution of Exercise 29.7.

Trapezoids ( $2.0\,\mathrm{cm}$ steps): $(0.03 + 0.10 + 0.19 + 0.30) = 0.62\,\mathrm{J}$ . Straight-line model: $k = 18.0/0.080 = 225\,\mathrm{N}/\mathrm{m}$ , $\tfrac12 kx^2 = 0.72\,\mathrm{J}$ . The measured curve sags below the straight line at small $x$ , so the true area is smaller.

Exercise 29.8 ★★

A $55\,\mathrm{kg}$ sledder starts at rest, drops $8.0\,\mathrm{m}$ along a $25\,\mathrm{m}$ curved run, arrives at $9.0\,\mathrm{m}/\mathrm{s}$ . Mechanical energy lost? Average friction force? Why is the run’s exact shape irrelevant?

Solution

Solution of Exercise 29.8.

$\Delta E_m = \tfrac12 \times 55 \times 9.0^2 - 55 \times 9.81 \times 8.0 = 2228 - 4316 \approx -2.1 \times 10^{3}\,\mathrm{J}$ ; $f = 2089/25 \approx 84\,\mathrm{N}$ . Weight conservative, normal force workless: only the drop and the path length enter, not the shape.

Exercise 29.9 ★★

A marble track has a loop of radius $20\,\mathrm{cm}$ . Find the minimum release height (from rest, no friction). Released from $60\,\mathrm{cm}$ : speed at the loop top, and ratio $v_{\text{top}}^2/(gR)$ ? Why must a real track be launched from higher still?

Solution

Solution of Exercise 29.9.

$H_{\min} = \tfrac52 R = 0.50\,\mathrm{m}$ . From $60\,\mathrm{cm}$ : $v_{\text{top}} = \sqrt{2 \times 9.81 \times (0.60 - 0.40)} \approx 2.0\,\mathrm{m}/\mathrm{s}$ , and $v_{\text{top}}^2/(gR) = 2.0$ — double the minimum. Friction skims energy all along, so real tracks need extra height.

Exercise 29.10 ★★

A $100\,\mathrm{g}$ bead slides without friction along an axis, with $E_p(x) = 8.0\,x^2$ (joules, $x$ in metres) and $E_m = 2.0\,\mathrm{J}$ . Turning points? Maximum speed? What kind of motion is this?

Solution

Solution of Exercise 29.10.

Turning points: $8.0\,x^2 = 2.0$ , $x = \pm0.50\,\mathrm{m}$ . Maximum speed at $x = 0$ : $v = \sqrt{2 \times 2.0/0.100} \approx 6.3\,\mathrm{m}/\mathrm{s}$ . A back-and-forth oscillation in a parabolic well — the mass-spring oscillator’s $E_p$ is exactly of this form.

Exercise 29.11 ★★

A $1200\,\mathrm{kg}$ car at $2.0\,\mathrm{m}/\mathrm{s}$ is stopped by a bumper spring of stiffness $k = 6.0 \times 10^{5}\,\mathrm{N}/\mathrm{m}$ . Find the maximum compression. At $4.0\,\mathrm{m}/\mathrm{s}$ : why not four times as much, the energy quadrupling?

Solution

Solution of Exercise 29.11.

$\tfrac12 kx^2 = \tfrac12 mv^2$ : $x = v\sqrt{m/k} = 2.0\sqrt{1200/6.0 \times 10^{5}} \approx 8.9\,\mathrm{cm}$ . At $4.0\,\mathrm{m}/\mathrm{s}$ : $18\,\mathrm{cm}$ — double, not quadruple: the energy quadruples, but $E_p$ grows like $x^2$ , so $x$ only doubles.

Exercise 29.12 ★★★

A $0.50\,\mathrm{kg}$ cart rolls on a track whose $E_p(x)$ has a $6.0\,\mathrm{J}$ hilltop ( $x = 0$ ) and a $2.0\,\mathrm{J}$ valley ( $x = 1.5\,\mathrm{m}$ ). (a) Arriving from the left with $E_m = 5.0\,\mathrm{J}$ : does it pass? What instead? (b) With $E_m = 7.0\,\mathrm{J}$ : speeds at hilltop and valley? (c) Equilibria?

Solution

Solution of Exercise 29.12.

(a) $E_m = 5.0\,\mathrm{J} < 6.0\,\mathrm{J}$ : it cannot pass; it stops on the flank where $E_p = 5.0\,\mathrm{J}$ and rolls back. (b) Hilltop: $E_k = 1.0\,\mathrm{J}$ , $v = \sqrt{2 \times 1.0/0.50} = 2.0\,\mathrm{m}/\mathrm{s}$ ; valley: $E_k = 5.0\,\mathrm{J}$ , $v \approx 4.5\,\mathrm{m}/\mathrm{s}$ . (c) $x = 0$ : unstable (maximum); $x = 1.5\,\mathrm{m}$ : stable (minimum).

Exercise 29.13 ★★★

Admitting that hauling a mass $m$ from the ground to arbitrarily far costs $mgR_E$ , $R_E = 6.37 \times 10^{6}\,\mathrm{m}$ (the finite area under the weakening-gravity graph), compute the escape speed from Earth. Why the same for probe and pebble? And a launch just below it?

Solution

Solution of Exercise 29.13.

$\tfrac12 mv^2 = mgR_E$ : $v = \sqrt{2 \times 9.81 \times 6.37 \times 10^{6}} \approx 1.12 \times 10^{4}\,\mathrm{m}/\mathrm{s} = 11.2\,\mathrm{km}/\mathrm{s}$ . Every term is proportional to $m$ : probe and pebble alike. Just below it, the launch still climbs enormously far, meets a turning point, and falls back: bound.

Exercise 29.14 ★★★

A $250\,\mathrm{g}$ glider oscillates on a spring ( $k = 40\,\mathrm{N}/\mathrm{m}$ ) with amplitude $6.0\,\mathrm{cm}$ (Chapter 28). Total energy? Maximum speed? Positions where $E_k = E_p$ ?

Solution

Solution of Exercise 29.14.

$E_m = \tfrac12 kA^2 = \tfrac12 \times 40 \times 0.060^2 = 7.2 \times 10^{-2}\,\mathrm{J}$ ; $v_{\max} = \sqrt{2E_m/m} = \sqrt{2 \times 0.072/0.250} \approx 0.76\,\mathrm{m}/\mathrm{s}$ ; $E_k = E_p$ where $\tfrac12 kx^2 = E_m/2$ : $x = \pm A/\sqrt 2 \approx \pm4.2\,\mathrm{cm}$ .

Exercise 29.15 ★★★

A pole-vaulter sprints at $9.5\,\mathrm{m}/\mathrm{s}$ . Estimate the height his centre of mass gains if the pole converts all his kinetic energy, and the bar cleared (centre of mass starting $1.0\,\mathrm{m}$ up). The record is about $6.3\,\mathrm{m}$ : what does the estimate catch, and leave out?

Solution

Solution of Exercise 29.15.

$h = v^2/2g = 9.5^2/19.62 \approx 4.6\,\mathrm{m}$ ; bar $\approx 4.6 + 1.0 = 5.6\,\mathrm{m}$ . The record is higher: vaulters add work with arms and legs on the bending pole. But the energy estimate sets the scale — no sprinter’s vault will reach $10\,\mathrm{m}$ .

29.6 Problem: The ski jump

Problem 29.1

Weekend problem — the ski jump: an in-run audited by a speed gun, a free-fall flight, the gentle geometry of a steep landing, and the tower a designer must buy once friction bills

An Olympic ski jumper ( $m = 70\,\mathrm{kg}$ , skis included) starts at rest from a gate $H = 45\,\mathrm{m}$ above the takeoff table, slides $L = 90\,\mathrm{m}$ of curved in-run, and leaves the table horizontally; a speed gun there reads $v_0 = 26\,\mathrm{m}/\mathrm{s}$ . The landing point lies $h = 40\,\mathrm{m}$ lower, on a $35{}^{\circ}$ slope. Parts II and III neglect air resistance.

Part I — The in-run.

Predict the frictionless takeoff speed; why is the curve of the in-run irrelevant?
Reference at the table: compute $E_m$ at the gate and at takeoff.
Compute $\Delta E_m$ ; which theorem names the culprits, and who?
Deduce the average resistive force (snow plus air) over the run.
What fraction of $E_m$ is lost? Why wax the skis and crouch?

Part II — The flight.

Only the weight acts in flight: which motion of Chapter 26 is this? Velocity components at landing?
Landing speed, twice: from components, and from $E_m$ conservation. In $\mathrm{km}/\mathrm{h}$ ?
Compute the flight time and the horizontal distance flown.
Real jumpers ride the air like a wing, flying much farther: can air raise the landing speed? Argue with the mechanical-energy theorem.

Part III — The landing.

Compute the angle of the landing velocity below the horizontal.
Decompose it into components parallel and perpendicular to the slope.
The legs absorb only the perpendicular part: compute that kinetic energy; compare it with the total.
Flat ground would absorb the lot: to what vertical drop is each landing equivalent ( $h_{\text{eq}} = v_\perp^2/2g$ , resp. $v^2/2g$ )?
One sentence: why are landing hills steep and curved?

Part IV — The designer’s tower. A smaller hill needs takeoff at only $v_0' = 24\,\mathrm{m}/\mathrm{s}$ ; its straight in-run descends at $35{}^{\circ}$ , so gate height $H'$ means track length $H'/\sin35{}^{\circ}$ ; same $80\,\mathrm{N}$ drag.

Compute the gate height a frictionless designer would build.
Write the mechanical-energy theorem, gate to table, for $H'$ .
Solve for $H'$ .
By what percentage did friction raise the tower?
The mass no longer cancels: recompute $H'$ for a $60\,\mathrm{kg}$ jumper. Who needs the taller tower, and why?
Punchline: the tower announced, and friction’s share of it.

Solution

Solution of Problem 29.1.

1. $v = \sqrt{2gH} = \sqrt{2 \times 9.81 \times 45} \approx 29.7\,\mathrm{m}/\mathrm{s}$ : weight conservative, normal force workless — the profile drops out.

2. Gate: $E_m = mgH = 70 \times 9.81 \times 45 \approx 3.09 \times 10^{4}\,\mathrm{J}$ . Takeoff: $E_m = \tfrac12 \times 70 \times 26^2 \approx 2.37 \times 10^{4}\,\mathrm{J}$ .

3. $\Delta E_m \approx -7.2 \times 10^{3}\,\mathrm{J}$ ; the mechanical-energy theorem charges it to the non-conservative forces: snow friction and air drag.

4. $f = 7.2 \times 10^{3}/90 \approx 80\,\mathrm{N}$ .

5. $7.2 \times 10^{3}/3.09 \times 10^{4} \approx 23\%$ . Wax lowers $f$ , crouching lowers drag: both shrink $fL$ , the only negotiable term.

6. Free fall with horizontal initial velocity — a projectile motion. $v_x = 26\,\mathrm{m}/\mathrm{s}$ ; $v_y = \sqrt{2 \times 9.81 \times 40} \approx 28.0\,\mathrm{m}/\mathrm{s}$ .

7. $v = \sqrt{26^2 + 28.0^2} = \sqrt{1461} \approx 38.2\,\mathrm{m}/\mathrm{s}$ ; energy: $v = \sqrt{v_0^2 + 2gh} = \sqrt{676 + 785}$ — identical. About $138\,\mathrm{km}/\mathrm{h}$ .

8. $t = v_y/g = 28.0/9.81 \approx 2.9\,\mathrm{s}$ ; $d = v_0 t \approx 74\,\mathrm{m}$ .

9. No. Lift is perpendicular to the velocity (workless) and drag opposes it (negative work): $\Delta E_m \leq 0$ , so the real landing speed can only be smaller — the air stretches the flight, it does not feed it.

10. $\alpha = \arctan(28.0/26) \approx 47{}^{\circ}$ below the horizontal.

11. Angle to the slope: $47 - 35 = 12{}^{\circ}$ . $v_\parallel = 38.2\cos12{}^{\circ} \approx 37.4\,\mathrm{m}/\mathrm{s}$ ; $v_\perp = 38.2\sin12{}^{\circ} \approx 7.9\,\mathrm{m}/\mathrm{s}$ .

12. Absorbed: $\tfrac12 \times 70 \times 7.9^2 \approx 2.2 \times 10^{3}\,\mathrm{J}$ , against $\tfrac12 \times 70 \times 38.2^2 \approx 5.1 \times 10^{4}\,\mathrm{J}$ in total: about $4\%$ .

13. $h_{\text{eq}} = 7.9^2/19.62 \approx 3.2\,\mathrm{m}$ — a bold jump from a wall. Flat: $38.2^2/19.62 \approx 74\,\mathrm{m}$ — unsurvivable.

14. A slope parallel to the flight path keeps $v_\perp$ small, so the legs absorb metres, not the whole mountain of kinetic energy.

15. $H'_0 = v_0'^2/2g = 576/19.62 \approx 29.4\,\mathrm{m}$ .

16. $mgH' - f\,\dfrac{H'}{\sin35{}^{\circ}} = \tfrac12 m v_0'^2$ .

17. $H' = \dfrac{\tfrac12 \times 70 \times 576} {686.7 - 80/0.574} = \dfrac{20\,160}{547} \approx 37\,\mathrm{m}$ .

18. $36.8/29.4 \approx 1.25$ : a $25\%$ surcharge.

19. For $60\,\mathrm{kg}$ : $H' = 17\,280/(588.6 - 139.5) \approx 38.5\,\mathrm{m}$ . The lighter jumper needs the taller tower: the friction bill $fL$ is the same, but her energy budget $mgH'$ is smaller.

20. Announce a $37\,\mathrm{m}$ tower for $24\,\mathrm{m}/\mathrm{s}$ of takeoff — and confess that a quarter of it, some $7\,\mathrm{m}$ of concrete, is friction’s surcharge on the frictionless dream.