Potential Energy

For a conservative force: $W = -\Delta U = -(U_f - U_i)$

Differential form: $dW = \vec{F} \cdot d\vec{r} = -dU$

Finding U from F

In one dimension: $F_x = -\frac{dU}{dx}$

$U(x) = -\int F_x \, dx + C$

In three dimensions: $\vec{F} = -\nabla U = -\left(\frac{\partial U}{\partial x}\hat{i} + \frac{\partial U}{\partial y}\hat{j} + \frac{\partial U}{\partial z}\hat{k}\right)$

Finding F from U

$F_x = -\frac{\partial U}{\partial x}, \quad F_y = -\frac{\partial U}{\partial y}, \quad F_z = -\frac{\partial U}{\partial z}$

Common Potential Energies

Gravitational (near Earth)

$U_g = mgh$

$F_y = -\frac{dU}{dy} = -mg$

Spring

$U_s = \frac{1}{2}kx^2$

$F_x = -\frac{dU}{dx} = -kx$

Universal Gravitation

$U_g = -\frac{Gm_1m_2}{r}$

$F_r = -\frac{dU}{dr} = -\frac{Gm_1m_2}{r^2}$

(Choosing $U = 0$ at $r = \infty$)

Electric Potential Energy

$U_e = k\frac{q_1q_2}{r}$

$F_r = -\frac{dU}{dr} = -k\frac{q_1q_2}{r^2}$

Conservation of Mechanical Energy

For conservative forces only: $E = KE + U = \text{constant}$

$\frac{1}{2}mv^2 + U(x) = E$

Taking time derivative: $mv\frac{dv}{dt} + \frac{dU}{dt} = 0$

$mv\frac{dv}{dt} + \frac{dU}{dx}\frac{dx}{dt} = 0$

$ma + \frac{dU}{dx}v = 0$

$F = -\frac{dU}{dx}$ (recovers $\vec{F} = -\nabla U$)

Equilibrium Points

At equilibrium, $F = 0$: $\frac{dU}{dx} = 0$

Stable equilibrium: $\frac{d^2U}{dx^2} > 0$ (local minimum of $U$)

Unstable equilibrium: $\frac{d^2U}{dx^2} < 0$ (local maximum of $U$)

Neutral equilibrium: $\frac{d^2U}{dx^2} = 0$ ($U$ is flat)

Example: Potential Energy Curve

$U(x) = \frac{1}{2}kx^2 - \frac{1}{6}bx^3$

Equilibrium points: $\frac{dU}{dx} = kx - \frac{1}{2}bx^2 = 0$

$x = 0 \text{ or } x = \frac{2k}{b}$

Stability: $\frac{d^2U}{dx^2} = k - bx$

At $x = 0$: $\frac{d^2U}{dx^2} = k > 0$ (stable)

At $x = \frac{2k}{b}$: $\frac{d^2U}{dx^2} = k - 2k = -k < 0$ (unstable)

Energy Diagrams

Plot $U(x)$ vs $x$. For total energy $E$:

• Particle confined to regions where $E \geq U(x)$
• Turning points where $E = U(x)$ (velocity = 0)
• Kinetic energy: $KE = E - U(x)$

Non-Conservative Forces

When non-conservative forces (friction, drag) are present:

$W_{nc} = \Delta KE + \Delta U = \Delta E$

Mechanical energy is not conserved; it decreases by work done against non-conservative forces.

(a) Force as function of position:

$F(x) = -\frac{dU}{dx} = -(8x - 4x^3)$

$\boxed{F(x) = 4x^3 - 8x \text{ N}}$

(b) Equilibrium positions:

Set F(x) = 0: $4x^3 - 8x = 0$ $4x(x^2 - 2) = 0$

$\boxed{x = 0, \quad x = \pm\sqrt{2} = \pm 1.41 \text{ m}}$

(c) Stability:

Test using second derivative: $\frac{d^2U}{dx^2} = 8 - 12x^2$

At x = 0: $\frac{d^2U}{dx^2}\Big|_{x=0} = 8 > 0$ → Stable minimum

At x = ±√2: $\frac{d^2U}{dx^2}\Big|_{x=\pm\sqrt{2}} = 8 - 12(2) = -16 < 0$ → Unstable maxima

Visual:

• U(0) = 0 (local minimum, stable)
• U(±√2) = 4(2) - 4 = 4 J (local maxima, unstable)

Particle oscillates around x = 0 if energy E < 4 J.

(a) Is force conservative?

Test: $\nabla \times \vec{F} = 0$

$\frac{\partial F_y}{\partial x} = \frac{\partial}{\partial x}(x^2 - 3y^2) = 2x$

$\frac{\partial F_x}{\partial y} = \frac{\partial}{\partial y}(2xy) = 2x$

Since $\frac{\partial F_y}{\partial x} = \frac{\partial F_x}{\partial y}$:

$\boxed{\text{Force is CONSERVATIVE}}$

(b) Potential energy function:

$F_x = -\frac{\partial U}{\partial x} = 2xy \implies \frac{\partial U}{\partial x} = -2xy$

Integrating with respect to x: $U = -x^2 y + f(y)$

$F_y = -\frac{\partial U}{\partial y} = x^2 - 3y^2$

$-\frac{\partial}{\partial y}(-x^2 y + f(y)) = x^2 - 3y^2$

$x^2 - f'(y) = x^2 - 3y^2$

$f'(y) = 3y^2 \implies f(y) = y^3 + C$

$\boxed{U(x,y) = -x^2 y + y^3 \text{ J}}$ (taking C = 0)

(c) Work from (0,0) to (2,1):

For conservative force, work is independent of path: $W = -\Delta U = U(0,0) - U(2,1)$

$U(0,0) = 0$ $U(2,1) = -(2)^2(1) + (1)^3 = -4 + 1 = -3 \text{ J}$

$W = 0 - (-3)$

$\boxed{W = 3 \text{ J}}$