In one dimension: $W = \int_{x_1}^{x_2} F_x \, dx$

In three dimensions: $W = \int_A^B (F_x \, dx + F_y \, dy + F_z \, dz)$

Work-Energy Theorem

From Newton's second law: $F = m\frac{dv}{dt} = m\frac{dv}{dx}\frac{dx}{dt} = mv\frac{dv}{dx}$

$F \, dx = mv \, dv$

Integrating: $\int_{x_1}^{x_2} F \, dx = \int_{v_1}^{v_2} mv \, dv$

$W = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2 = \Delta KE$

Work-energy theorem: $W_{net} = \Delta KE$

Common Force Examples

Constant Force

$W = F \cdot d = Fd\cos\theta$

where $\theta$ is angle between force and displacement.

Spring Force

$F = -kx$

$W = \int_0^x (-kx') \, dx' = -\frac{1}{2}kx^2$

(Work done by spring; work done on spring is positive)

Gravity Near Earth's Surface

$W = \int_{y_1}^{y_2} (-mg) \, dy = -mg(y_2 - y_1) = -mg\Delta h$

Inverse Square Force (Gravity/Electrostatic)

$F = -\frac{k}{r^2}$

$W = \int_{r_1}^{r_2} \left(-\frac{k}{r^2}\right) dr = k\left(\frac{1}{r_2} - \frac{1}{r_1}\right)$

Power

Instantaneous power: $P = \frac{dW}{dt}$

Since $dW = \vec{F} \cdot d\vec{r}$:

$P = \frac{dW}{dt} = \vec{F} \cdot \frac{d\vec{r}}{dt} = \vec{F} \cdot \vec{v}$

In one dimension: $P = Fv$

Average Power

$P_{avg} = \frac{W}{\Delta t}$

Power and Kinetic Energy

$P = \frac{dKE}{dt} = \frac{d}{dt}\left(\frac{1}{2}mv^2\right) = mv\frac{dv}{dt} = mva = Fv$

Work on Variable Mass Systems

For rocket with exhaust velocity $v_e$ relative to rocket:

Power delivered by engine: $P = F_{thrust} \cdot v = v_{rel}\left|\frac{dm}{dt}\right| \cdot v$

Example: Work Against Drag

Object moving through fluid with drag $F_d = -bv^2$:

$W = \int_0^d F_d \, dx = -b\int_0^d v^2 \, dx$

If $v(t)$ is known, use $dx = v \, dt$:

$W = -b\int_0^T v^3 \, dt$

This work is dissipated as heat.

$W = \left[10x - x^2\right]_0^4 = 40 - 16$

$\boxed{W = 24 \text{ J}}$

(b) Final speed at x = 4 m:

Using work-energy theorem: $W = \Delta KE = \frac{1}{2}mv^2 - 0$

$24 = \frac{1}{2}(3.0)v^2$

$v^2 = 16$

$\boxed{v = 4.0 \text{ m/s}}$

(c) Power at x = 2 m:

First find velocity at x = 2: $W_{0→2} = \int_0^2 (10 - 2x) \, dx = 20 - 4 = 16 \text{ J}$

$\frac{1}{2}(3.0)v_2^2 = 16 \implies v_2 = 3.27 \text{ m/s}$

Force at x = 2: $F(2) = 10 - 2(2) = 6 \text{ N}$

Power: $P = F \cdot v = (6)(3.27)$

$\boxed{P = 19.6 \text{ W}}$