---
title: "Work and Mechanical Energy"
book: "High School Physics"
subject: physics
language: en
chapter: 29
exercises: 15
source: https://one-course.com/books/physics/2/en/chapter/29-work-and-mechanical-energy
---

# Chapter 29 — Work 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](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy)), no equation of motion solved. Last year priced [forces](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) on straight lines ([Chapter 17](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#ch-g11-work-of-force)) and balanced kinetic against [potential energy](#def-g12-work-and-energy-ep) ([Chapter 18](https://one-course.com/books/physics/2/en/chapter/18-mechanical-energy-and-its-conservation#ch-g11-mechanical-energy)); 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](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-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](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work) $\scal{F}{\dd\ell} = F\,\dd\ell\cos\theta$ ([Chapter 17](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#ch-g11-work-of-force)), 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](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-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.](https://one-course.com/images/onecourse/chapters/physics-2/g12-work-and-energy/fig-0e501691baea.svg)

*A curved path is a broken line of many short steps: each $\vect{\dd\ell}$ earns $F\,\dd\ell\cos\theta$; the [work](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-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](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) of a mass $m$ does the [work](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-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](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) is *conservative* if its [work](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work) between two points is the same along every path (zero on every round trip), *non-conservative* otherwise. [Weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) and spring [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) (next section): conservative. Sliding [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) is not: opposite the motion, it does $W = -fL$ on a path of length $L$ — a toll per [metre](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit), 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](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) charges $-75 \times 9.81 \times 300 \approx -2.2 \times 10^{5}\,\mathrm{J}$ on both; a $30\,\mathrm{N}$ [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force), $-2.4 \times 10^{4}\,\mathrm{J}$ on one, $-7.2 \times 10^{4}\,\mathrm{J}$ on the other. The [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) refunds on the descent; [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory)’s bill is gone as heat, both ways.

## 29.2 Potential energy

**Definition 29.5 (Potential energy).**

To every [conservative force](#def-g12-work-and-energy-conservative) one attaches a *potential energy* $E_p$, a number depending on position alone, defined by $W_{AB} = E_p(A) - E_p(B)$: [work](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work) delivered is potential [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-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](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) this gives back last year’s $E_p = mgz$, now valid along any path. [Friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) can own no $E_p$: a route-dependent bill is no difference of endpoint numbers — only [conservative forces](#def-g12-work-and-energy-conservative) store retrievably, exactly what the name conserves.

**Proposition 29.6 (Elastic potential energy).**

A spring of [stiffness](https://one-course.com/books/physics/2/en/chapter/28-mechanical-oscillators-and-the-measurement-of-time#def-g12-oscillators-and-time-spring) $k$ ($\mathrm{N}/\mathrm{m}$) stretched or compressed by $x$ pulls back with the [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) $F = kx$ ([Chapter 28](https://one-course.com/books/physics/2/en/chapter/28-mechanical-oscillators-and-the-measurement-of-time#ch-g12-oscillators-and-time)); the [work](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work) it delivers returning to rest — its *elastic potential energy* — is $E_p = \tfrac12\, k x^2$.

**Proof.** The [restoring force](https://one-course.com/books/physics/2/en/chapter/28-mechanical-oscillators-and-the-measurement-of-time#def-g12-oscillators-and-time-pendulum) 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](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) at each stretch is the same out and back: [work](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work) fixed by the endpoints, spring [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) [conservative](#def-g12-work-and-energy-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.](https://one-course.com/images/onecourse/chapters/physics-2/g12-work-and-energy/fig-a7869b5fdc36.svg)

*The spring’s [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) grows with the stretch; the stored [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-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](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) 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](https://one-course.com/books/physics/2/en/chapter/25-newtons-laws#thm-g12-newtons-laws-second) ([Chapter 25](https://one-course.com/books/physics/2/en/chapter/25-newtons-laws#ch-g12-newtons-laws)). In a short $\dd t$ the body moves $\vect{\dd\ell} = \vect v\,\dd t$: $E_k$ gains the small [work](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work) of [Definition 29.1](#def-g12-work-and-energy-path); summing the steps sums the [works](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work) — last year’s admitted statement, earned in full. ∎

**Proposition 29.8 (Instantaneous power).**

At every instant a [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) $\vect F$ on a body of velocity $\vect v$ delivers the [power](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-power) $P = \scal{F}{v} = Fv\cos\theta$, and $\dd E_k/\dd t$ is the sum of the [powers](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-power) of all [forces](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force): last year’s constant-velocity formula, exact at each instant of any motion.

**Proof.** Read off the previous proof: each [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-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](#def-g12-work-and-energy-conservative) at [work](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work). Then, along any path,

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

and $E_m$ is conserved whenever the [non-conservative forces](#def-g12-work-and-energy-conservative) do no [work](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work).

**Proof.** Split the [kinetic-energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-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](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory), perpendicular to the motion, is powerless; the [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) is [conservative](#def-g12-work-and-energy-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).**

1. Choose the system and list the [forces](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) on it.
2. Sort: perpendicular to the motion $\to$ no [work](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work) ; [conservative](#def-g12-work-and-energy-conservative) $\to$ an $E_p$ in $E_m$ ; the rest $\to$ $W_{\text{nc}}$ (sliding [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) : $-fL$ , $L$ = path length).
3. Write $\Delta E_m = W_{\text{nc}}$ between the data point and the question point; solve. [Energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-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.](https://one-course.com/images/onecourse/chapters/physics-2/g12-work-and-energy/fig-10f33556f15d.svg)

*An [energy diagram](#def-g12-work-and-energy-diagram), read without solving anything: with [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-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](#rem-g12-work-and-energy-escape).*

**Proposition 29.13 (Force from the landscape; equilibria).**

The [conservative force](#def-g12-work-and-energy-conservative) 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](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) [works](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work) $F\,\dd x = E_p(x) -
E_p(x + \dd x) = -\dd E_p$; 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](https://one-course.com/books/physics/2/en/chapter/28-mechanical-oscillators-and-the-measurement-of-time#ch-g12-oscillators-and-time) 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](#def-g12-work-and-energy-diagram): it *escapes* — about $11.2\,\mathrm{km}/\mathrm{s}$ from the ground ([Exercise 29.13](#exo-g12-work-and-energy-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](https://one-course.com/books/physics/2/en/chapter/25-newtons-laws#ch-g12-newtons-laws)) 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).](https://one-course.com/images/onecourse/chapters/physics-2/g12-work-and-energy/fig-2e44b5bda6ff.svg)

*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](https://one-course.com/books/physics/2/en/chapter/28-mechanical-oscillators-and-the-measurement-of-time#def-g12-oscillators-and-time-spring) $k = 200\,\mathrm{N}/\mathrm{m}$ is stretched by $5.0\,\mathrm{cm}$. Compute the stored [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy); what does it become if the stretch doubles? How much does the *second* $5.0\,\mathrm{cm}$ cost?

**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](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-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](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work) of its [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight)? Along the straight climb? On the round trip?

**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](#def-g12-work-and-energy-conservative)).

**Exercise 29.3 ★.**

[Conservative](#def-g12-work-and-energy-conservative), [non-conservative](#def-g12-work-and-energy-conservative), or workless? Justify from paths: (a) the [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight); (b) the spring [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force); (c) sliding [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory); (d) the [normal force](https://one-course.com/books/physics/2/en/chapter/25-newtons-laws#def-g12-newtons-laws-friction) on a sliding box; (e) air drag.

**Solution of Exercise 29.3.**

(a), (b) [conservative](#def-g12-work-and-energy-conservative): [work](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work) fixed by the endpoints (drop of altitude; stretch). (c), (e) [non-conservative](#def-g12-work-and-energy-conservative): $-fL$ and drag charge per [metre](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) 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](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) never enter the balance?

**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](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) is perpendicular to the motion at every instant: zero [power](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-power), zero [work](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work).

**Exercise 29.5 ★.**

A cyclist rides at $9.0\,\mathrm{m}/\mathrm{s}$ on the flat, delivering $250\,\mathrm{W}$: what total [resistive](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-sign) [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) is she fighting? Climbing at $5.0\,\mathrm{m}/\mathrm{s}$ with the same [power](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-power): what [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) does she overcome?

**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](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy), the launch speed, and the height reached if fired straight up (no air).

**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](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) by trapezoidal areas; compare with $\tfrac12 kx^2$, taking $k = F/x$ at the end. Why do they differ?

**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](https://one-course.com/books/physics/2/en/chapter/18-mechanical-energy-and-its-conservation#def-g11-mechanical-energy-em) lost? Average [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force)? Why is the run’s exact shape irrelevant?

**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](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) [conservative](#def-g12-work-and-energy-conservative), [normal force](https://one-course.com/books/physics/2/en/chapter/25-newtons-laws#def-g12-newtons-laws-friction) 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](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory)). 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 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](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) skims [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) all along, so real tracks need extra height.

**Exercise 29.10 ★★.**

A $100\,\mathrm{g}$ bead slides without [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) along an axis, with $E_p(x) = 8.0\,x^2$ ([joules](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work), $x$ in [metres](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit)) and $E_m = 2.0\,\mathrm{J}$. [Turning points](#def-g12-work-and-energy-diagram)? Maximum speed? What kind of motion is this?

**Solution of Exercise 29.10.**

[Turning points](#def-g12-work-and-energy-diagram): $8.0\,x^2 = 2.0$, $x = \pm0.50\,\mathrm{m}$. 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](https://one-course.com/books/physics/2/en/chapter/28-mechanical-oscillators-and-the-measurement-of-time#def-g12-oscillators-and-time-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](https://one-course.com/books/physics/2/en/chapter/28-mechanical-oscillators-and-the-measurement-of-time#def-g12-oscillators-and-time-spring) $k = 6.0 \times 10^{5}\,\mathrm{N}/\mathrm{m}$. Find the maximum [compression](https://one-course.com/books/physics/2/en/chapter/21-sound-and-acoustics#def-g12-sound-acoustics-sound-wave). At $4.0\,\mathrm{m}/\mathrm{s}$: why not four times as much, the [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) quadrupling?

**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](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-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 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](#prop-g12-work-and-energy-equilibria) (maximum); $x = 1.5\,\mathrm{m}$: [stable](#prop-g12-work-and-energy-equilibria) (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](#rem-g12-work-and-energy-escape)* from Earth. Why the same for probe and pebble? And a launch just below it?

**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](#def-g12-work-and-energy-diagram), and falls back: bound.

**Exercise 29.14 ★★★.**

A $250\,\mathrm{g}$ glider oscillates on a spring ($k = 40\,\mathrm{N}/\mathrm{m}$) with [amplitude](https://one-course.com/books/physics/2/en/chapter/28-mechanical-oscillators-and-the-measurement-of-time#def-g12-oscillators-and-time-oscillator) $6.0\,\mathrm{cm}$ ([Chapter 28](https://one-course.com/books/physics/2/en/chapter/28-mechanical-oscillators-and-the-measurement-of-time#ch-g12-oscillators-and-time)). Total [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy)? Maximum speed? Positions where $E_k = E_p$?

**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](https://one-course.com/books/physics/2/en/chapter/18-mechanical-energy-and-its-conservation#def-g11-mechanical-energy-kinetic), 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 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](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work) with arms and legs on the bending pole. But the [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-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.**

1. Predict the frictionless takeoff speed; why is the curve of the in-run irrelevant?
2. Reference at the table: compute $E_m$ at the gate and at takeoff.
3. Compute $\Delta E_m$ ; which theorem names the culprits, and who?
4. Deduce the average [resistive](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-sign) [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) (snow plus air) over the run.
5. What fraction of $E_m$ is lost? Why wax the skis and crouch?

**Part II — The flight.**

6. Only the [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) acts in flight: which motion of [Chapter 26](https://one-course.com/books/physics/2/en/chapter/26-free-fall-and-projectile-motion#ch-g12-projectile-motion) is this? Velocity components at landing?
7. Landing speed, twice: from components, and from $E_m$ conservation. In $\mathrm{km}/\mathrm{h}$ ?
8. Compute the flight time and the horizontal distance flown.
9. Real jumpers ride the air like a wing, flying much farther: can air *raise* the landing speed? Argue with the [mechanical-energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) theorem.

**Part III — The landing.**

10. Compute the angle of the landing velocity below the horizontal.
11. Decompose it into components parallel and perpendicular to the slope.
12. The legs absorb only the perpendicular part: compute that [kinetic energy](https://one-course.com/books/physics/2/en/chapter/18-mechanical-energy-and-its-conservation#def-g11-mechanical-energy-kinetic) ; compare it with the total.
13. Flat ground would absorb the lot: to what vertical drop is each landing equivalent ( $h_{\text{eq}} = v_\perp^2/2g$ , resp. $v^2/2g$ )?
14. 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.

15. Compute the gate height a frictionless designer would build.
16. Write the [mechanical-energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) theorem, gate to table, for $H'$ .
17. Solve for $H'$ .
18. By what percentage did [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) raise the tower?
19. The mass no longer cancels: recompute $H'$ for a $60\,\mathrm{kg}$ jumper. Who needs the taller tower, and why?
20. Punchline: the tower announced, and [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) ’s share of it.

**Solution of Problem 29.1.**

**1.** $v = \sqrt{2gH} = \sqrt{2 \times 9.81 \times 45} \approx
29.7\,\mathrm{m}/\mathrm{s}$: [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) [conservative](#def-g12-work-and-energy-conservative), [normal force](https://one-course.com/books/physics/2/en/chapter/25-newtons-laws#def-g12-newtons-laws-friction) 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](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) theorem charges it to the [non-conservative forces](#def-g12-work-and-energy-conservative): snow [friction](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) 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](https://one-course.com/books/physics/2/en/chapter/26-free-fall-and-projectile-motion#def-g12-projectile-motion-freefall) with horizontal initial velocity — a [projectile](https://one-course.com/books/physics/2/en/chapter/26-free-fall-and-projectile-motion#def-g12-projectile-motion-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](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-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](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-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](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit), not the whole mountain of [kinetic energy](https://one-course.com/books/physics/2/en/chapter/18-mechanical-energy-and-its-conservation#def-g11-mechanical-energy-kinetic).

**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](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) bill $fL$ is the same, but her [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-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](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory)’s surcharge on the frictionless dream.