Work and Power

Calculating work using line integrals and instantaneous power

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Work and Power

Work as a Line Integral

For a variable force F\vec{F} along a path from point A to B:

W=ABFdrW = \int_A^B \vec{F} \cdot d\vec{r}

In one dimension: W=x1x2FxdxW = \int_{x_1}^{x_2} F_x \, dx

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

Work-Energy Theorem

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

Fdx=mvdvF \, dx = mv \, dv

Integrating: x1x2Fdx=v1v2mvdv\int_{x_1}^{x_2} F \, dx = \int_{v_1}^{v_2} mv \, dv

W=12mv2212mv12=ΔKEW = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2 = \Delta KE

Work-energy theorem: Wnet=ΔKEW_{net} = \Delta KE

Common Force Examples

Constant Force

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

where θ\theta is angle between force and displacement.

Spring Force

F=kxF = -kx

W=0x(kx)dx=12kx2W = \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=y1y2(mg)dy=mg(y2y1)=mgΔhW = \int_{y_1}^{y_2} (-mg) \, dy = -mg(y_2 - y_1) = -mg\Delta h

Inverse Square Force (Gravity/Electrostatic)

F=kr2F = -\frac{k}{r^2}

W=r1r2(kr2)dr=k(1r21r1)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=dWdtP = \frac{dW}{dt}

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

P=dWdt=Fdrdt=FvP = \frac{dW}{dt} = \vec{F} \cdot \frac{d\vec{r}}{dt} = \vec{F} \cdot \vec{v}

In one dimension: P=FvP = Fv

Average Power

Pavg=WΔtP_{avg} = \frac{W}{\Delta t}

Power and Kinetic Energy

P=dKEdt=ddt(12mv2)=mvdvdt=mva=FvP = \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 vev_e relative to rocket:

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

Example: Work Against Drag

Object moving through fluid with drag Fd=bv2F_d = -bv^2:

W=0dFddx=b0dv2dxW = \int_0^d F_d \, dx = -b\int_0^d v^2 \, dx

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

W=b0Tv3dtW = -b\int_0^T v^3 \, dt

This work is dissipated as heat.

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