A force is conservative if the work done is independent of path (depends only on endpoints).

Equivalently:

1. Work around any closed path is zero: $\oint \vec{F} \cdot d\vec{r} = 0$
2. The force can be written as: $\vec{F} = -\nabla U$ (gradient of scalar potential)

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.

$U_s = 100(0.09)$

$\boxed{U_s = 9.0 \text{ J}}$

(b) Maximum speed:

At maximum speed, all elastic PE converts to KE: $\frac{1}{2}kx_0^2 = \frac{1}{2}mv_{max}^2$

$v_{max} = \sqrt{\frac{k}{m}}x_0 = \sqrt{\frac{200}{2.0}}(0.3)$

$v_{max} = \sqrt{100}(0.3) = 10(0.3)$

$\boxed{v_{max} = 3.0 \text{ m/s}}$

(Occurs when spring passes through equilibrium)

(c) Speed at halfway point:

At x = 0.15 m (halfway):

Energy conservation: $\frac{1}{2}kx_0^2 = \frac{1}{2}kx^2 + \frac{1}{2}mv^2$

$9.0 = \frac{1}{2}(200)(0.15)^2 + \frac{1}{2}(2.0)v^2$

$9.0 = 2.25 + v^2$

$v^2 = 6.75$

$\boxed{v = 2.6 \text{ m/s}}$