---
title: "Work, Energy, and Potential Energy"
book: "University Physics — Year 1"
subject: physics
language: en
chapter: 13
exercises: 12
source: https://one-course.com/books/physics/3/en/chapter/13-work-energy-and-potential-energy
---

# Chapter 13 — Work, Energy, and Potential Energy

A cyclist stops pedaling at the top of a hill and arrives at the bottom at a speed that depends on the drop, not on the shape of the road. A bungee jumper falls sixty meters and stops, for an instant, exactly where the cord’s pull has eaten all the speed the fall had given. A hydrogen atom bound to a chlorine atom vibrates in a well whose depth is the energy it costs to tear the molecule apart. In each case a single number is exchanged between forms — motion, height, stretch, bond — and summed, it does not change. This chapter builds that bookkeeping from Newton’s second law: [work](#def-b1-work-and-energy-work) and power, the [kinetic energy](#thm-b1-work-and-energy-ke) theorem, potential energies for the forces that allow one, and the [conservation of mechanical energy](#thm-b1-work-and-energy-em) that turns many dynamical problems into one-line algebra and explains, through the shape of a [potential well](#def-b1-work-and-energy-turning), which positions are stable.

![The lecturer’s pendulum: released from the nose, the bowling ball returns to the nose and no further — mechanical energy is conserved, and never exceeded.](https://one-course.com/images/onecourse/chapters/physics-3/b1-work-and-energy/img-b210ce0e63a6.jpg)

*The lecturer’s pendulum: released from the nose, the bowling ball returns to the nose and no further — [mechanical energy](#thm-b1-work-and-energy-em) is conserved, and never exceeded.*

## 13.1 Power and work

**Definition 13.1 (Power and work of a force).**

The *power* of a force $\vect F$ acting on a point moving at velocity $\vect v$ is

$$
\mathcal P = \vect F\cdot\vect v \qquad (\text{watts}),
$$

and its *work* between times $t_1$ and $t_2$ (positions $A$ and $B$) is the accumulated power,

$$
W_{A\to B} = \int_{t_1}^{t_2}\vect F\cdot\vect v\,\dd t
= \int_A^B \vect F\cdot\dd\vect\ell \qquad (\text{joules}),
$$

the sum of the *elementary works* $\delta W = \vect F\cdot\dd\vect\ell$ along the elementary displacements $\dd\vect\ell = \vect v\,\dd t$ of the [trajectory](https://one-course.com/books/physics/3/en/chapter/11-kinematics-of-a-point#def-b1-point-kinematics-frame). Work is positive if the force pushes along the motion (*motive*), negative if against it (*resistive*), zero if perpendicular.

**Proposition 13.2 (Works one meets).**

- Constant force: $W_{A\to B} = \vect F\cdot\vect{AB}$ , whatever the path. Weight: $W = mg(z_A - z_B)$ , $z$ counted upward.
- Spring of stiffness $k$ , from elongation $x_A$ to $x_B$ : $W = \tfrac12 k(x_A^2 - x_B^2)$ .
- [Normal reaction](https://one-course.com/books/physics/3/en/chapter/12-newtons-laws-of-dynamics#prop-b1-newton-dynamics-friction) of a fixed surface, string tension on a body moving perpendicular to it: $W = 0$ (force $\perp$ displacement).
- Kinetic friction and drag: $W < 0$ always (the force opposes the relative motion), and depends on the path length.

**Proof.** Constant $\vect F$: $\int\vect F\cdot\dd\vect\ell = \vect F\cdot\int\dd\vect\ell
= \vect F\cdot\vect{AB}$; with $\vect F = -mg\vect e_z$, $\vect F\cdot\vect{AB}
= -mg(z_B - z_A)$. Spring: $\vect F\cdot\dd\vect\ell = -kx\,\dd x$, and $\int_{x_A}^{x_B}-kx\,\dd x = \tfrac12 k(x_A^2 - x_B^2)$. Friction: $\vect T
= -\mu N\vect v/v$, so $\vect T\cdot\vect v = -\mu Nv < 0$. ∎

![The elementary work of a force along a small displacement of its point of application: only the component of F along the motion works. Summed along the path from A to B, it gives W_A B.](https://one-course.com/images/onecourse/chapters/physics-3/b1-work-and-energy/fig-3762d53da000.svg)

*The elementary [work](#def-b1-work-and-energy-work) of a force along a small displacement of its point of application: only the component of $\vect F$ along the motion [works](#def-b1-work-and-energy-work). Summed along the path from $A$ to $B$, it gives $W_{A\to B}$.*

## 13.2 The kinetic energy theorem

**Theorem 13.3 (Kinetic energy and power theorems).**

In an [inertial frame](https://one-course.com/books/physics/3/en/chapter/12-newtons-laws-of-dynamics#thm-b1-newton-dynamics-laws), the *[kinetic energy](#thm-b1-work-and-energy-ke)* $E_k = \tfrac12 mv^2$ of a [point particle](https://one-course.com/books/physics/3/en/chapter/11-kinematics-of-a-point#def-b1-point-kinematics-frame) obeys

$$
\frac{\dd E_k}{\dd t} = \sum\mathcal P_i = \Big(\sum\vect F_i\Big)\cdot\vect v,
\qquad
E_k(B) - E_k(A) = \sum_i W_{i,A\to B}:
$$

the change of [kinetic energy](#thm-b1-work-and-energy-ke) between two positions equals the total [work](#def-b1-work-and-energy-work) of all the forces applied along the way.

**Proof.** Dot Newton’s second law with $\vect v$: $m\vect a\cdot\vect v =
\tfrac{\dd}{\dd t}(\tfrac12 m\vect v\cdot\vect v) = \dd E_k/\dd t$, while $(\sum\vect F_i)\cdot\vect v = \sum\mathcal P_i$. Integrate over time. ∎

**Example 13.4 (Braking distance).**

A $1000\,\mathrm{kg}$ car at $100\,\mathrm{km}/\mathrm{h}$ ($E_k = 386\,\mathrm{kJ}$) braked by a constant $7.0\,\mathrm{kN}$ force stops after $d = E_k/F = 55\,\mathrm{m}$: the distance grows as the *square* of the speed. Since the maximal braking force is $\mu_smg$ ([Chapter 12](https://one-course.com/books/physics/3/en/chapter/12-newtons-laws-of-dynamics#ch-b1-newton-dynamics)), the shortest stopping distance is $v^2/2\mu_sg$ whatever the car’s mass — and it doubles on a wet road where $\mu_s$ halves.

## 13.3 Conservative forces and potential energy

**Definition 13.5 (Conservative force; potential energy).**

A force is *conservative* if its [work](#def-b1-work-and-energy-work) between two points does not depend on the path followed. Equivalently, there exists a function of position, the *potential energy* $E_p$, such that

$$
W_{A\to B} = E_p(A) - E_p(B) = -\Delta E_p ,
\qquad \delta W = -\dd E_p ,
$$

$E_p$ being defined up to an additive constant (a choice of origin).

**Proposition 13.6 (Force from potential energy).**

A [conservative force](#def-b1-work-and-energy-conservative) derives from its [potential energy](#def-b1-work-and-energy-conservative):

$$
\vect F = -\overrightarrow{\operatorname{grad}}\,E_p
= -\frac{\partial E_p}{\partial x}\,\vect e_x
- \frac{\partial E_p}{\partial y}\,\vect e_y
- \frac{\partial E_p}{\partial z}\,\vect e_z ,
$$

and in one dimension $F_x = -\dd E_p/\dd x$: the force points toward decreasing [potential energy](#def-b1-work-and-energy-conservative), “downhill” on the graph of $E_p$.

**Proof.** $\delta W = F_x\dd x + F_y\dd y + F_z\dd z = -\dd E_p = -(\partial_xE_p\,\dd x
+ \partial_yE_p\,\dd y + \partial_zE_p\,\dd z)$ for every displacement: identify the coefficients (the differential of a function of several variables, from the mathematics course). ∎

**Proposition 13.7 (The usual potential energies).**

- Weight: $E_p = mgz + \text{const}$ ( $z$ upward).
- Spring: $E_p = \tfrac12 k(\ell - \ell_0)^2 + \text{const}$ .
- Newtonian gravitation of a mass $M$ : $E_p = -GMm/r + \text{const}$ , the constant usually chosen so that $E_p \to 0$ at infinity.
- Electrostatic force on a charge $q$ in a potential $V$ : $E_p = qV$ ( [Chapter 27](https://one-course.com/books/physics/3/en/chapter/27-electric-potential-conductors-and-capacitors#ch-b1-potential-capacitors) ).

Kinetic friction and fluid drag are *not* [conservative](#def-b1-work-and-energy-conservative) (their [work](#def-b1-work-and-energy-work) depends on the path, and is always negative).

**Proof.** Each is read off the [works](#def-b1-work-and-energy-work) of [Proposition 13.2](#prop-b1-work-and-energy-works): $W = -\Delta E_p$. For gravitation, $\vect F = -GMm/r^2\,\vect e_r$ and $\delta W = -GMm\,\dd r/r^2 = -\dd(-GMm/r)$. ∎

## 13.4 Mechanical energy

**Theorem 13.8 (Mechanical energy).**

Let $E_m = E_k + E_p$ be the *[mechanical energy](#thm-b1-work-and-energy-em)*, where $E_p$ is the sum of the potential energies of the [conservative forces](#def-b1-work-and-energy-conservative). Then

$$
\Delta E_m = W_{\text{nc}} ,
$$

the total [work](#def-b1-work-and-energy-work) of the [non-conservative](#def-b1-work-and-energy-conservative) forces. If those do no [work](#def-b1-work-and-energy-work) (no friction, or constraint forces perpendicular to the motion), $E_m$ is *conserved*: the motion is *[conservative](#def-b1-work-and-energy-conservative)*.

**Proof.** $\Delta E_k = W_{\text{c}} + W_{\text{nc}} = -\Delta E_p + W_{\text{nc}}$. ∎

**Method 13.9 (Using energy).**

When a problem asks for a *speed at a position* (not a time, not a force), try energy first:

1. list the forces; check which [work](#def-b1-work-and-energy-work) and which are [conservative](#def-b1-work-and-energy-conservative) ;
2. write $E_m$ at the two positions, with a clear origin for $E_p$ ;
3. equate them (adding $W_{\text{nc}}$ if friction acts) and solve for the unknown speed or position.

Energy never gives the time or the constraint forces; for those, go back to Newton’s law — often with the speed that energy just provided.

**Example 13.10 (The loop’s speed law).**

On the frictionless loop of [Problem 11.1](https://one-course.com/books/physics/3/en/chapter/11-kinematics-of-a-point#pb-b1-point-kinematics-1) the track’s reaction is normal to the motion and [works](#def-b1-work-and-energy-work) not: $E_m$ is conserved. With $z = R(1 - \cos\theta)$ above the bottom, $\tfrac12 mv^2 + mgR(1 -
\cos\theta) = \tfrac12 mv_0^2$: the speed law $v^2 = v_0^2 - 2gR(1 -
\cos\theta)$ used there, now derived in one line. The minimum entry speed $\sqrt{5gR}$ then follows from the top condition $v^2 \geq gR$ of [Example 12.16](https://one-course.com/books/physics/3/en/chapter/12-newtons-laws-of-dynamics#ex-b1-newton-dynamics-circles).

**Example 13.11 (Energy lost to friction).**

A block slides down a slope of angle $\alpha$ and height $h$ with kinetic friction $\mu_d$: $W_{\text{nc}} = -\mu_dmg\cos\alpha \times
h/\sin\alpha = -\mu_dmgh\cot\alpha$, so $\tfrac12 mv^2 = mgh(1 - \mu_d\cot\alpha)$. For $h = 2.0\,\mathrm{m}$, $\alpha = 30^\circ$, $\mu_d = 0.20$: $v =
5.1\,\mathrm{m}/\mathrm{s}$ instead of $6.3\,\mathrm{m}/\mathrm{s}$; a third of the [potential energy](#def-b1-work-and-energy-conservative) became heat.

## 13.5 Equilibrium, stability, and phase portraits

**Proposition 13.12 (Equilibrium and stability in one dimension).**

A particle moving on an axis under a [conservative force](#def-b1-work-and-energy-conservative) of [potential energy](#def-b1-work-and-energy-conservative) $E_p(x)$:

- is in *equilibrium* at $x_0$ iff $F(x_0) = -E_p'(x_0) = 0$ : the extrema of $E_p$ ;
- the equilibrium is *stable* if $E_p$ has a local minimum there ( $E_p''(x_0) > 0$ ): a small displacement produces a restoring force; *unstable* at a maximum;
- near a stable equilibrium, small motions are harmonic, with $$\omega_0^2 = \frac{E_p''(x_0)}{m} .$$

**Proof.** Taylor’s formula (mathematics course): $E_p(x) \approx E_p(x_0) +
\tfrac12 E_p''(x_0)(x - x_0)^2$, so $F \approx -E_p''(x_0)(x - x_0)$ and $m\ddot x = -E_p''(x_0)(x - x_0)$ — a spring of stiffness $E_p''(x_0)$. For $E_p'' < 0$ the “stiffness” is negative: exponential departure. ∎

![A one-dimensional potential landscape: minima are stable equilibria, the maximum an unstable one. A particle of energy E_1 below the barrier is trapped in one well between two turning points (where E_p = E_1, arrows) and oscillates; with E_2 above the barrier it passes freely from one well to the other.](https://one-course.com/images/onecourse/chapters/physics-3/b1-work-and-energy/fig-f50bea9fd840.svg)

*A one-dimensional potential landscape: minima are stable equilibria, the maximum an unstable one. A particle of energy $E_1$ below the barrier is trapped in one well between two *[turning points](#def-b1-work-and-energy-turning)* (where $E_p = E_1$, arrows) and oscillates; with $E_2$ above the barrier it passes freely from one well to the other.*

**Definition 13.13 (Turning points; bound and free motion).**

For a [conservative](#def-b1-work-and-energy-conservative) one-dimensional motion of energy $E_m$, the particle can only be where $E_p(x) \leq E_m$ (since $E_k \geq 0$); the points where $E_p = E_m$ are *turning points*, where the velocity vanishes and reverses. Trapped between two turning points in a *potential well*, the motion is *bound* and periodic; able to reach infinity, it is *free*; a *potential barrier* higher than $E_m$ cannot be crossed.

**Definition 13.14 (Phase portrait).**

The *phase portrait* of a one-dimensional motion is the family of curves traced by the point $(x, \dot x)$ in the *phase plane*. For a [conservative](#def-b1-work-and-energy-conservative) system each curve is a level line $\tfrac12 m\dot x^2 +
E_p(x) = E_m$: closed curves around stable equilibria (oscillations), open curves for free motion, and, through unstable equilibria, the *separatrices* that divide the two. Curves are traversed clockwise ($\dot x > 0$ means $x$ increasing) and never cross.

![Phase portrait of the simple pendulum, 1/2 2 + _02(1 - ) = const. Closed curves (blue): swings about the stable equilibrium = 0; open curves (orange): full rotations; between them the separatrix (red) through the unstable equilibria = ±π, the motion that just reaches the top.](https://one-course.com/images/onecourse/chapters/physics-3/b1-work-and-energy/fig-4f8091d9ae80.svg)

*[Phase portrait](#def-b1-work-and-energy-phase) of the [simple pendulum](https://one-course.com/books/physics/3/en/chapter/12-newtons-laws-of-dynamics#ex-b1-newton-dynamics-pendulum), $\tfrac12\dot\theta^2 +
\omega_0^2(1 - \cos\theta) = \text{const}$. Closed curves (blue): swings about the stable equilibrium $\theta = 0$; open curves (orange): full rotations; between them the separatrix (red) through the unstable equilibria $\theta = \pm\pi$, the motion that just reaches the top.*

**Example 13.15 (The pendulum’s landscape).**

$E_p = -mg\ell\cos\theta$: minimum at $\theta = 0$, $E_p'' = mg\ell$, so $\omega_0^2 = g\ell/(m\ell^2) = g/\ell$ — the small-oscillation result of [Example 12.12](https://one-course.com/books/physics/3/en/chapter/12-newtons-laws-of-dynamics#ex-b1-newton-dynamics-pendulum) read off the well’s curvature; maximum at $\theta = \pi$, the inverted pendulum, unstable. Energy $E_m < mg\ell$: oscillation between $\pm\theta_{\max}$; $E_m > mg\ell$: it goes over the top and rotates; $E_m = mg\ell$: the separatrix, an infinitely slow approach to the top.

## 13.6 Exercises

**Exercise 13.1 ★.**

A $20\,\mathrm{kg}$ crate is lifted $3.0\,\mathrm{m}$ at constant speed in $5.0\,\mathrm{s}$. [Work](#def-b1-work-and-energy-work) of the lifting force, of the weight, and power supplied. The same crate pushed $3.0\,\mathrm{m}$ along a floor with $\mu_d = 0.30$: [work](#def-b1-work-and-energy-work) of friction.

**Solution of Exercise 13.1.**

Lifting force $+mgh = 20 \times 9.81 \times 3.0 = 589\,\mathrm{J}$; weight $-589\,\mathrm{J}$; power $589/5.0 = 118\,\mathrm{W}$. Friction: $-\mu_dmgd =
-0.30 \times 196 \times 3.0 = -177\,\mathrm{J}$.

**Exercise 13.2 ★.**

[Kinetic energy](#thm-b1-work-and-energy-ke) of a $1000\,\mathrm{kg}$ car at $100\,\mathrm{km}/\mathrm{h}$; stopping distance with a constant $7.0\,\mathrm{kN}$ braking force; what happens to the distance if the speed is doubled?

**Solution of Exercise 13.2.**

$E_k = \tfrac12 \times 1000 \times 27.8^2 = 386\,\mathrm{kJ}$; $d = E_k/F =
55\,\mathrm{m}$; doubling $v$ quadruples $E_k$ and $d$: $220\,\mathrm{m}$.

**Exercise 13.3 ★.**

A spring of stiffness $500\,\mathrm{N}/\mathrm{m}$ is compressed by $10\,\mathrm{cm}$ and released against a $50\,\mathrm{g}$ ball on a frictionless horizontal track. Energy stored; speed of the ball.

**Solution of Exercise 13.3.**

$E_p = \tfrac12 \times 500 \times 0.10^2 = 2.5\,\mathrm{J}$; $v = \sqrt{2 \times
2.5/0.050} = 10\,\mathrm{m}/\mathrm{s}$.

**Exercise 13.4 ★.**

A pendulum of length $1.0\,\mathrm{m}$ is released from rest at $60^\circ$: speed at the bottom by energy; then, with Newton’s law, the tension there (in units of $mg$).

**Solution of Exercise 13.4.**

$\tfrac12 mv^2 = mg\ell(1 - \cos 60^\circ)$: $v = \sqrt{g\ell} = 3.1\,\mathrm{m}/\mathrm{s}$. Tension: $T - mg = mv^2/\ell = mg$, $T = 2mg$.

**Exercise 13.5 ★★.**

A block slides from rest down a $30^\circ$ slope of height $2.0\,\mathrm{m}$, $\mu_d = 0.20$. Speed at the bottom by the energy theorem; fraction of the initial [potential energy](#def-b1-work-and-energy-conservative) lost to friction.

**Solution of Exercise 13.5.**

Friction [works](#def-b1-work-and-energy-work) $-\mu_dmg\cos\alpha\,(h/\sin\alpha)$, so

$$
\tfrac12 mv^2 = mgh(1 - \mu_d\cot\alpha) = mgh(1 - 0.35),
$$

$v = \sqrt{2 \times 9.81 \times 2.0 \times 0.65} = 5.1\,\mathrm{m}/\mathrm{s}$; $35\%$ lost.

**Exercise 13.6 ★★.**

Using $E_p = -GMm/r$, show that the speed needed to escape the Earth’s attraction from its surface is $v_e = \sqrt{2GM/R}$, and compute it ($M = 5.97 \times 10^{24}\,\mathrm{kg}$, $R = 6.37 \times 10^{6}\,\mathrm{m}$). What is the escape speed from a body with the same density but twice the radius?

**Solution of Exercise 13.6.**

$E_m = \tfrac12 mv^2 - GMm/R \geq 0$ for the body to reach infinity (where $E_p = 0$ and $v \to 0$): $v_e = \sqrt{2GM/R}$, i.e.

$$
v_e = \sqrt{\frac{2 \times 6.67\times10^{-11} \times 5.97\times10^{24}}{6.37\times10^6}}
= 11.2\,\mathrm{km}/\mathrm{s}.
$$

Same density, $M \propto R^3$: $v_e \propto R$, so $22.4\,\mathrm{km}/\mathrm{s}$.

**Exercise 13.7 ★★.**

Bungee: a $70\,\mathrm{kg}$ jumper, cord of natural length $20\,\mathrm{m}$ and stiffness $50\,\mathrm{N}/\mathrm{m}$, jumps from rest. Find the maximal stretch of the cord, the total drop, and the maximal tension (in units of the weight). Where is the speed greatest?

**Solution of Exercise 13.7.**

Lowest point: $mg(\ell_0 + x) = \tfrac12 kx^2$, $25x^2 - 687x - 13734 = 0$, $x = 41\,\mathrm{m}$; total drop $61\,\mathrm{m}$; $T_{\max} = kx = 2.0\,\mathrm{kN}
= 3.0\,mg$ (net $2\,\mathrm{g}$ upward). Speed is greatest where the net force vanishes, $kx = mg$: $x = 14\,\mathrm{m}$, $34\,\mathrm{m}$ below the jump.

**Exercise 13.8 ★★.**

A cyclist ($80\,\mathrm{kg}$ with bike) climbs a $5\%$ slope at $5.0\,\mathrm{m}/\mathrm{s}$ against a drag $\tfrac12\rho C_xSv^2$ with $C_xS = 0.50\,\mathrm{m}^{2}$. Power against gravity, against drag, total.

**Solution of Exercise 13.8.**

Gravity: $mgv\sin\alpha = 80 \times 9.81 \times 5.0 \times 0.05 = 196\,\mathrm{W}$; drag: $\tfrac12\rho C_xSv^3 = 0.5 \times 1.2 \times 0.50 \times 125 =
38\,\mathrm{W}$; total $234\,\mathrm{W}$.

**Exercise 13.9 ★★.**

A particle of mass $m$ has $E_p(x) = E_0[(x/a)^4 - 2(x/a)^2]$. Find the equilibria and their stability, the depth of the wells, and the [angular frequency](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) of [small oscillations](#prop-b1-work-and-energy-stability) about a stable one.

**Solution of Exercise 13.9.**

$E_p' = 4E_0(x^3/a^4 - x/a^2) = 0$: $x = 0, \pm a$. $E_p'' = E_0(12x^2/a^4 -
4/a^2)$: $-4E_0/a^2$ at $0$ (unstable maximum, $E_p = 0$), $+8E_0/a^2$ at $\pm a$ (stable minima, $E_p = -E_0$): wells of depth $E_0$; $\omega_0^2 = 8E_0/ma^2$.

**Exercise 13.10 ★★★.**

For the harmonic oscillator $E_p = \tfrac12 kx^2$, show that the phase curves are ellipses, give their semi-axes for energy $E$, and explain why all are traversed in the same time.

**Solution of Exercise 13.10.**

$\tfrac12 m\dot x^2 + \tfrac12 kx^2 = E$: an ellipse of semi-axes $x_{\max}
= \sqrt{2E/k}$ and $\dot x_{\max} = \sqrt{2E/m}$. The motion is harmonic with period $2\pi\sqrt{m/k}$ whatever the [amplitude](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) (isochronism), so every ellipse takes the same time.

**Exercise 13.11 ★★★.**

A small object slides from rest at the top of a frictionless hemisphere of radius $R$. Using energy and Newton’s law, find the angle from the vertical at which it leaves the surface, the height, and its speed then.

**Solution of Exercise 13.11.**

Energy: $v^2 = 2gR(1 - \cos\theta)$. Radial: $mg\cos\theta - N = mv^2/R$, so $N = mg(3\cos\theta - 2)$, zero at $\cos\theta = 2/3$ ($\theta =
48^\circ$): height $2R/3$, speed $\sqrt{2gR/3}$.

**Exercise 13.12 ★★★.**

A car at $90\,\mathrm{km}/\mathrm{h}$ skids to rest in $80\,\mathrm{m}$ with locked wheels. Find $\mu_d$ from energy. Where did the energy go? Why does an anti-lock system (wheels kept rolling) stop shorter?

**Solution of Exercise 13.12.**

$\tfrac12 mv^2 = \mu_dmgd$: $\mu_d = 625/(2 \times 9.81 \times 80) = 0.40$. Heat in the tire–road contact (and a skid mark). Rolling wheels use static friction, $\mu_s > \mu_d$: a larger braking force, a shorter distance, and steering kept.

## 13.7 Problem: The energy well of a chemical bond

**Problem 13.1.**

Weekend problem — two atoms, one curve: the potential energy of a bond read as a landscape — its floor, its walls, the small vibrations at the bottom, and the energy it takes to climb out

The interaction of the two atoms of a diatomic molecule is modeled by the [potential energy](#def-b1-work-and-energy-conservative)

$$
E_p(r) = \epsilon\left[\left(\frac{\sigma}{r}\right)^{12} - 2\left(\frac{\sigma}{r}\right)^{6}\right],
$$

$r$ the distance between the nuclei. For hydrogen chloride take $\epsilon = 4.4\,\mathrm{eV}$, $\sigma = 0.127\,\mathrm{nm}$, and let the light hydrogen atom ($m = 1.67 \times 10^{-27}\,\mathrm{kg}$) move while the heavy chlorine stays put. $1\,\mathrm{eV} = 1.60 \times 10^{-19}\,\mathrm{J}$; $k_B = 1.38 \times 10^{-23}\,\mathrm{J}/\mathrm{K}$; $N_A = 6.02 \times 10^{23}$.

**Part I — The landscape.**

1. Give the limits of $E_p$ as $r \to 0$ and $r \to \infty$ , and sketch the curve.
2. Show that $E_p$ has a single minimum, at $r = \sigma$ , of value $-\epsilon$ .
3. With $u = (\sigma/r)^6$ , show that $E_p = \epsilon(u^2 - 2u)$ : a parabola in $u$ . Recover the minimum from it.
4. Express the force $F(r)$ (positive if repulsive) and give its sign on each side of $\sigma$ .
5. Compute $E_p$ and $F$ at $r = 1.2\sigma$ (in eV and in nN).
6. What is the dissociation energy of the molecule, in eV and in $\mathrm{kJ}/\mathrm{mol}$ ? (Measured bond energy of HCl: $431\,\mathrm{kJ}/\mathrm{mol}$ .)
7. Check the stability criterion at $r = \sigma$ : compute $E_p''(\sigma)$ .

**Part II — Small vibrations.**

8. Write the second-order Taylor expansion of $E_p$ around $\sigma$ and identify an effective stiffness $k$ .
9. Compute $k$ (in $\mathrm{N}/\mathrm{m}$ ).
10. Deduce the [angular frequency](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) , the frequency and the wavelength of the vibration, and its wavenumber $1/\lambda$ in $\mathrm{cm}^{-1}$ .
11. The measured vibration of HCl is at $2990\,\mathrm{cm}^{-1}$ . Comment on the agreement and on what the model gets right.
12. At $300\,\mathrm{K}$ the mean vibrational energy is of order $k_BT$ . Estimate the [amplitude](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) of vibration and compare with $\sigma$ .
13. Deduce the maximal speed of the hydrogen atom in that vibration.

**Part III — Large oscillations and the way out.**

14. Describe the [phase portrait](#def-b1-work-and-energy-phase) : which energies give closed curves, which open ones, and what separates them?
15. For $E_m = -\epsilon/2$ , find the two [turning points](#def-b1-work-and-energy-turning) (in units of $\sigma$ ).
16. Show that their midpoint lies beyond $\sigma$ , and explain why heating (raising the vibrational energy) makes the mean bond length grow — the origin of thermal expansion.
17. Is the period of large oscillations longer or shorter than $2\pi/\omega_0$ ? Argue from the shape of the walls.
18. From $E_m = -\epsilon/2$ , what energy must be supplied to dissociate the molecule? If given as [kinetic energy](#thm-b1-work-and-energy-ke) of the hydrogen atom at $r = \sigma$ , what speed is that?
19. Two atoms approaching from far apart with positive total energy cannot form a bound molecule by themselves. Why? What is needed?

**Part IV — [Work](#def-b1-work-and-energy-work), force, and temperature.**

20. Compute the [work](#def-b1-work-and-energy-work) of the bond force when $r$ goes from $2\sigma$ to $\sigma$ . Is the force motive or resistive there?
21. Compute the [work](#def-b1-work-and-energy-work) an external agent must do to stretch the bond from $\sigma$ to $1.5\sigma$ at vanishing speed.
22. Compute $F$ at $r = 0.9\sigma$ and compare with $r = 1.2\sigma$ : what does the asymmetry tell about compressing versus stretching?
23. At what temperature would $k_BT$ equal $\epsilon$ ? Why do molecules dissociate at far lower temperatures in practice?
24. A better model (Morse) changes the shape of the walls but keeps a minimum at the bond length. Which of the results above survive unchanged in method, and which numbers would change?
25. Summarize the three ways energy methods were used: to find equilibrium, to find the vibration frequency, and to find what is needed to escape.

**Solution of Problem 13.1.**

**1.** $r \to 0$: $+\infty$ (the $r^{-12}$ term); $r \to \infty$: $0^-$. A steep wall, a well, a long tail.

**2.** $E_p' = \epsilon[-12\sigma^{12}/r^{13} + 12\sigma^6/r^7] = 0$ iff $(\sigma/r)^6 = 1$: $r = \sigma$, $E_p = \epsilon(1 - 2) = -\epsilon$.

**3.** $(\sigma/r)^{12} = u^2$: $E_p = \epsilon(u^2 - 2u) = \epsilon[(u - 1)^2
- 1]$, minimal at $u = 1$ ($r = \sigma$) with value $-\epsilon$.

**4.** $F = -E_p' = (12\epsilon/\sigma)[(\sigma/r)^{13} - (\sigma/r)^7]$: positive (repulsive) for $r < \sigma$, negative (attractive) beyond.

**5.** $(1/1.2)^6 = 0.335$, $(1/1.2)^{12} = 0.112$: $E_p = \epsilon(0.112
- 0.670) = -0.56\epsilon = -2.5\,\mathrm{eV}$. $(1/1.2)^{13} = 0.0935$, $(1/1.2)^7 = 0.279$: $F = (12\epsilon/\sigma)(-0.186) = -2.2\epsilon/\sigma =
-2.2 \times 7.0\times10^{-19}/1.27\times10^{-10} = -12\,\mathrm{nN}$.

**6.** $\epsilon = 4.4\,\mathrm{eV} = 4.4 \times 96.5 = 425\,\mathrm{kJ}/\mathrm{mol}$, close to the measured $431\,\mathrm{kJ}/\mathrm{mol}$.

**7.** $E_p'' = \epsilon[156\sigma^{12}/r^{14} - 84\sigma^6/r^8]$; at $\sigma$: $72\epsilon/\sigma^2 > 0$: stable.

**8.** $E_p \approx -\epsilon + \tfrac12(72\epsilon/\sigma^2)(r - \sigma)^2$: $k = 72\epsilon/\sigma^2$.

**9.** $k = 72 \times 7.0\times10^{-19}/(1.27\times10^{-10})^2 =
3.1 \times 10^{3}\,\mathrm{N}/\mathrm{m}$.

**10.** $\omega_0 = \sqrt{k/m} = \sqrt{3.1\times10^3/1.67\times10^{-27}} =
1.4 \times 10^{15}\,\mathrm{rad}/\mathrm{s}$; $f = 2.2 \times 10^{14}\,\mathrm{Hz}$; $\lambda = c/f = 1.4\,\text{µ}\mathrm{m}$; $1/\lambda = 7300\,\mathrm{cm}^{-1}$.

**11.** A factor $2.4$ too high: the model’s walls are too steep (the $r^{-12}$ core is a crude stand-in for a covalent bond), but the [order of magnitude](https://one-course.com/books/physics/3/en/chapter/1-units-dimensions-and-measurement#def-b1-units-dimensions-oom) — an infrared vibration at $10^{14}$ Hz — is right, and the method (curvature of the well) is exactly the one used with better potentials.

**12.** $\tfrac12 kA^2 \approx k_BT$: $A = \sqrt{2k_BT/k} = \sqrt{8.3\times
10^{-21}/3.1\times10^3} = 1.6 \times 10^{-12}\,\mathrm{m} = 0.013\sigma$: the bond barely shivers.

**13.** $v_{\max} = A\omega_0 = 1.6\times10^{-12} \times 1.4\times10^{15}
= 2.2\,\mathrm{km}/\mathrm{s}$.

**14.** $-\epsilon < E_m < 0$: closed curves around $(\sigma, 0)$ — vibration, bound; $E_m > 0$: open curves — the atoms separate; $E_m = 0$: the separatrix, an atom just able to reach infinity.

**15.** $u^2 - 2u = -\tfrac12$ with $u = (\sigma/r)^6$: $u = 1 \pm 1/\sqrt2
= 1.71$ or $0.293$; $r = \sigma u^{-1/6}$: $0.915\sigma$ and $1.23\sigma$.

**16.** Midpoint $1.07\sigma > \sigma$: the outer wall is softer than the inner one, so the atom spends more room outside; the time-averaged distance grows with energy — heated bonds lengthen, solids expand.

**17.** Longer: the restoring force on the soft outer side is weaker than the harmonic approximation assumes, so the return takes more time.

**18.** $\epsilon/2 = 2.2\,\mathrm{eV} = 3.5 \times 10^{-19}\,\mathrm{J}$; as [kinetic energy](#thm-b1-work-and-energy-ke) at $\sigma$: $v = \sqrt{2 \times 3.5\times10^{-19}/1.67\times10^{-27}}
= 2.0 \times 10^{4}\,\mathrm{m}/\mathrm{s}$.

**19.** Energy conservation: $E_m > 0$ stays $> 0$, the curve is open, the atoms fly apart after one encounter. A third body (another molecule, a wall, a photon) must carry the excess away.

**20.** $W = E_p(2\sigma) - E_p(\sigma) = \epsilon(2^{-12} - 2 \times 2^{-6})
+ \epsilon = 0.97\epsilon = 4.3\,\mathrm{eV}$, positive: the attraction pulls the atom in, motive.

**21.** $W_{\text{ext}} = E_p(1.5\sigma) - E_p(\sigma) = \epsilon(0.0077 -
0.176) + \epsilon = 0.83\epsilon = 3.7\,\mathrm{eV}$.

**22.** $(1/0.9)^{13} = 3.93$, $(1/0.9)^7 = 2.09$: $F = +22\epsilon/\sigma
= +120\,\mathrm{nN}$, ten times the attraction at $1.2\sigma$: compressing a bond by $10\%$ costs far more than stretching it by $20\%$.

**23.** $T = \epsilon/k_B = 7.0\times10^{-19}/1.38\times10^{-23} =
5.1 \times 10^{4}\,\mathrm{K}$. Real gases dissociate well below because energies are distributed: a fraction of molecules carries many times $k_BT$ ([Chapter 20](https://one-course.com/books/physics/3/en/chapter/20-kinetic-theory-and-the-perfect-gas#ch-b1-kinetic-theory)).

**24.** Survive: equilibrium at the minimum, $k = E_p''$ at the minimum, dissociation energy $=$ depth, [turning points](#def-b1-work-and-energy-turning) from $E_p = E_m$, asymmetry $\Rightarrow$ expansion. Change: the values of $k$, of the frequency and of the [turning points](#def-b1-work-and-energy-turning).

**25.** Equilibrium: $E_p' = 0$; vibration: $\omega_0^2 = E_p''/m$; escape: $E_m$ compared with the well’s rim.