University Physics — Year 1 · Bachelor Year 1

13Work, 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 and power, the kinetic energy theorem, potential energies for the forces that allow one, and the conservation of mechanical energy that turns many dynamical problems into one-line algebra and explains, through the shape of a potential well, 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.

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. 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 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. — The elementary work of a force along a small displacement of its point of application: only the component of $\vect F$ along the motion works. 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, the kinetic energy $E_k = \tfrac12 mv^2$ of a point particle 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 between two positions equals the total 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), 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 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 derives from its potential energy:

\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, “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).

Kinetic friction and fluid drag are not conservative (their work depends on the path, and is always negative).

Proof. Each is read off the works of Proposition 13.2: $W = -\Delta E_p$ . 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, where $E_p$ is the sum of the potential energies of the conservative forces. Then

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

the total work of the non-conservative forces. If those do no work (no friction, or constraint forces perpendicular to the motion), $E_m$ is conserved: the motion is 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:

list the forces; check which work and which are conservative;
write $E_m$ at the two positions, with a clear origin for $E_p$ ;
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 the track’s reaction is normal to the motion and works 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.

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 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 of potential energy $E_p(x)$ :

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. — 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.

Definition 13.13 (Turning points; bound and free motion)

For a 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 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. — Phase portrait of the simple 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 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 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 of friction.

Solution

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 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

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

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

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 lost to friction.

Solution

Solution of Exercise 13.5.

Friction works $-\mu_dmg\cos\alpha\,(h/\sin\alpha)$ , 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

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

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

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 of small oscillations about a stable one.

Solution

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

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 (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

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

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

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.

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

Part II — Small vibrations.

Write the second-order Taylor expansion of $E_p$ around $\sigma$ and identify an effective stiffness $k$ .
Compute $k$ (in $\mathrm{N}/\mathrm{m}$ ).
Deduce the angular frequency, the frequency and the wavelength of the vibration, and its wavenumber $1/\lambda$ in $\mathrm{cm}^{-1}$ .
The measured vibration of HCl is at $2990\,\mathrm{cm}^{-1}$ . Comment on the agreement and on what the model gets right.
At $300\,\mathrm{K}$ the mean vibrational energy is of order $k_BT$ . Estimate the amplitude of vibration and compare with $\sigma$ .
Deduce the maximal speed of the hydrogen atom in that vibration.

Part III — Large oscillations and the way out.

Describe the phase portrait: which energies give closed curves, which open ones, and what separates them?
For $E_m = -\epsilon/2$ , find the two turning points (in units of $\sigma$ ).
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.
Is the period of large oscillations longer or shorter than $2\pi/\omega_0$ ? Argue from the shape of the walls.
From $E_m = -\epsilon/2$ , what energy must be supplied to dissociate the molecule? If given as kinetic energy of the hydrogen atom at $r = \sigma$ , what speed is that?
Two atoms approaching from far apart with positive total energy cannot form a bound molecule by themselves. Why? What is needed?

Part IV — Work, force, and temperature.

Compute the work of the bond force when $r$ goes from $2\sigma$ to $\sigma$ . Is the force motive or resistive there?
Compute the work an external agent must do to stretch the bond from $\sigma$ to $1.5\sigma$ at vanishing speed.
Compute $F$ at $r = 0.9\sigma$ and compare with $r = 1.2\sigma$ : what does the asymmetry tell about compressing versus stretching?
At what temperature would $k_BT$ equal $\epsilon$ ? Why do molecules dissociate at far lower temperatures in practice?
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?
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

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 — 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 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).

24. Survive: equilibrium at the minimum, $k = E_p''$ at the minimum, dissociation energy $=$ depth, turning points from $E_p = E_m$ , asymmetry $\Rightarrow$ expansion. Change: the values of $k$ , of the frequency and of the turning points.

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

Terms defined in this chapter

See all 393 terms in the glossary