Work-Kinetic Energy Theorem

Derivations

$$W=m{\int }_{r_1}^{r_2}\frac{\overrightarrow{dv}}{dt}\cdot \overrightarrow{dr}$$ $$W=m{\int }_{x_1}^{x_2}\frac{d{v}_x}{dx}\frac{dx}{dt}dx+{\int }_{y_1}^{y_2}\frac{d{v}_y}{dy}\frac{dy}{dt}dy+{\int }_{z_1}^{z_2}\frac{d{v}_z}{dz}\frac{dz}{dt}dz$$ $$W=m{\int }_{v_{x,1}}^{v_{x,2}}\frac{dx}{dt}d{v}_x+{\int }_{v_{y,1}}^{v_{y,2}}\frac{dy}{dt}d{v}_y+{\int }_{v_{z,1}}^{v_{z,2}}\frac{dz}{dt}d{v}_z$$ $$W=m{\int }_{v_{x,1}}^{v_{x,2}}{v}_{x}d{v}_x+{\int }_{v_{y,1}}^{v_{y2}}{v}_{y}d{v}_y+{\int }_{v_{z,1}}^{v_{z,2}}{v}_{z}d{v}_z$$ $$W=m[\frac{1}{2}(v_{x,2}^2-v_{x,1}^2)+\frac{1}{2}(v_{y,2}^2-v_{y,1}^2)+\frac{1}{2}(v_{z,2}^2-v_{z,1}^2)]$$ $$W=\frac{1}{2}m{v}_2^2-\frac{1}{2}m{v}_1^2$$ $$W=T_2-T_1$$ $$\to W=\Delta T$$

Another derivation

$$T=\frac{1}{2}m{v}^2$$ $$\frac{dT}{dt}=\frac{1}{2}m\frac{d}{dt}(\vec{v}\cdot \vec{v})=\frac{1}{2}m(\dot{\vec{v}}\cdot \vec{v}+\vec{v}\cdot \dot{\vec{v}})=m\dot{\mathbf{v}}\cdot \mathbf{v}$$ $$\frac{dT}{dt}=\mathbf{F}\cdot \mathbf{v}$$ $$dT=\mathbf{F}\cdot \mathbf{dr}$$ $$dT=dW$$ $${\int }_{v_1}^{v_2}dT={\int }_1^{2}dW$$ $$T_2-T_1={\int }_1^2\mathbf{F}\cdot d\mathbf{r}$$ $$\to W=\Delta T$$

Work Example

Evaluate line integral for the work done by the 2-dimensional force $\mathbf{F}=(y,2x)$ going from the origin O to the point $P=(1,1)$ along each of the three paths in the figure. Path $a$ goes from $O$ to $Q=(1,0$ along the x axis and then from $Q$ straight up to $P$, path $b$ goes straight from $O$ to $P$ along the line $y=x$, and path $c$ goes round a quarter circle center on $Q$.

$$W_a={\int }_a\mathbf{F}\cdot d\mathbf{r}={\int }_O^Q\mathbf{F}\cdot d\mathbf{r}+{\int }_Q^P\mathbf{F}\cdot d\mathbf{r}={\int }_0^1{F}_x(x,0)dx+{\int }_0^1{F}_y(1,y)dy=0+2{\int }_0^{1}dy=2$$

On the path $b$, $x=y$, so that $dx=dy$, and

$$W_b={\int }_b\mathbf{F}\cdot d\mathbf{r}={\int }_b(F_{x}dx+F_{y}dy)={\int }_0^1(x+2x)dx=1.5$$

Path $c$ is conveniently expressed parametrically as $\mathbf{r}=(x,y)=(1-\cos \theta ,\sin \theta )$, where $\theta$ is the angle between $OQ$ and the line from $Q$ to the point $(x,y)$, with $0\le \theta \le \pi /2$. Thus on path $c$, $d\mathbf{r}=(dx,dy)=(\sin \theta ,\cos \theta )d\theta$. $W_c={\int }_c\mathbf{F}\cdot d\mathbf{r}={\int }_c(F_{x}dx+F_{y}dy)$ $={\int }_0^{\pi /2}[{\sin }^2\theta +2(1-\cos \theta )\cos \theta )]d\theta =2-\pi /4=1.21$