13.11: Work-Kinetic Energy Theorem in Three Dimensions

Recall our mathematical result that for one-dimensional motion

$$m \int_{i}^{f} a_{x} d x=m \int_{i}^{f} \frac{d v_{x}}{d t} d x=m \int_{i}^{f} d v_{x} \frac{d x}{d t}=m \int_{i}^{f} v_{x} d v_{x}=\frac{1}{2} m v_{x, f}^{2}-\frac{1}{2} m v_{x, i}^{2} \nonumber$$

Using Newton’s Second Law in the form $F_{x}=m a_{x}$, we concluded that

$$\int_{i}^{f} F_{x} d x=\frac{1}{2} m v_{x, f}^{2}-\frac{1}{2} m v_{x, i}^{2} \nonumber$$

Equation (13.11.2) generalizes to the y - and z -directions:

$$\int_{i}^{f} F_{y} d y=\frac{1}{2} m v_{y, f}^{2}-\frac{1}{2} m v_{y, i}^{2} \nonumber$$

$$\int_{i}^{f} F_{z} d z=\frac{1}{2} m v_{z, f}^{2}-\frac{1}{2} m v_{z, i}^{2} \nonumber$$

Adding Equations (13.11.2), (13.11.3), and (13.11.4) yields

$$\int_{i}^{f}\left(F_{x} d x+F_{y} d y+F_{z} d z\right)=\frac{1}{2} m\left(v_{x, f}^{2}+v_{y, f}^{2}+v_{z, f}^{2}\right)-\frac{1}{2} m\left(v_{x, i}^{2}+v_{y, i}^{2}+v_{z, i}^{2}\right) \nonumber$$

Recall (Equation (13.8.24)) that the left hand side of Equation (13.11.5) is the work done by the force $\overrightarrow{\mathbf{F}}$ on the object

$$W=\int_{i}^{f} d W=\int_{i}^{f}\left(F_{x} d x+F_{y} d y+F_{z} d z\right)=\int_{i}^{f} \overrightarrow{\mathbf{F}} \cdot d \overrightarrow{\mathbf{r}} \nonumber$$

The right hand side of Equation (13.11.5) is the change in kinetic energy of the object

$$\Delta K \equiv K_{f}-K_{i}=\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{0}^{2}=\frac{1}{2} m\left(v_{x, f}^{2}+v_{y, f}^{2}+v_{z, f}^{2}\right)-\frac{1}{2} m\left(v_{x, i}^{2}+v_{y, i}^{2}+v_{z, i}^{2}\right) \nonumber$$

Therefore Equation (13.11.5) is the three dimensional generalization of the work-kinetic energy theorem

$$\int_{i}^{f} \overrightarrow{\mathbf{F}} \cdot d \overrightarrow{\mathbf{r}}=K_{f}-K_{i} \nonumber$$

When the work done on an object is positive, the object will increase its speed, and negative work done on an object causes a decrease in speed. When the work done is zero, the object will maintain a constant speed.

Instantaneous Power Applied by a Non-Constant Force for Three Dimensional Motion

Recall that for one-dimensional motion, the instantaneous power at time t is defined to be the limit of the average power as the time interval $[t, t+\Delta t]$ approaches zero,

$$P(t)=F_{x}^{a}(t) v_{x}(t) \nonumber$$

A more general result for the instantaneous power is found by using the expression for dW as given in Equation (13.8.23),

$$P=\frac{d W}{d t}=\frac{\overrightarrow{\mathbf{F}} \cdot d \overrightarrow{\mathbf{r}}}{d t}=\overrightarrow{\mathbf{F}} \cdot \overrightarrow{\mathbf{v}} \nonumber$$

The time rate of change of the kinetic energy for a body of mass m is equal to the power,

$$\frac{d K}{d t}=\frac{1}{2} m \frac{d}{d t}(\overrightarrow{\mathbf{v}} \cdot \overrightarrow{\mathbf{v}})=m \frac{d \overrightarrow{\mathbf{v}}}{d t} \cdot \overrightarrow{\mathbf{v}}=m \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{v}}=\overrightarrow{\mathbf{F}} \cdot \overrightarrow{\mathbf{v}}=P \nonumber$$

where the we used Equation (13.8.9), Newton’s Second Law and Equation (13.11.10).