6.2 Notions of work and kinetic energy : The work-energy theorem

WORK AND WORK-ENERGY THEOREM

The work done by the force \(\vec{F}\) on the particle during the small displacement \(d \vec{r}\) is:

\(\vec{F} \cdot d \vec{r}=F d r \cos \theta\)

The work done on the particle by a force \(\vec{F}\) acting on it during a finite displacement is obtained by
\(W=\int \vec{F} \cdot d \vec{r}=\int F \cos \theta d r \dots  (6.3)\)

where the integration is to be performed along the path of the particle.

We are now familiar with the concepts of work and kinetic energy to prove the work-energy theorem for a variable force. We confine ourselves to one dimension. The time rate of change of kinetic energy is

\(\frac{d K}{d t}=\frac{d}{d t}\left(\frac{1}{2} m v^{2}\right)=m v \frac{d v}{d t}=F_{t} v,\)
where \(F_{t}\) is the resultant tangential force. If the resultant force \(\vec{F}\) makes an angle \(\theta\) with the velocity,
\(
\begin{array}{l}
F_{t}=F \cos \theta \text { and } \frac{d K}{d t}=F v \cos \theta=\vec{F} \cdot \vec{v}=\vec{F} \cdot \frac{d \vec{r}}{d t} \\
\text { or, } \quad d K=\vec{F} \cdot d \vec{r} \dots  (6.4)
\end{array} 
\)

If \(\vec{F}\) is the resultant force on the particle we can use equation (6.4) to get
\(
W=\int \vec{F} \cdot d \vec{r}=\int d K=K_{f}-K_{i} .
\)

where \(K_{i}\) and \(K_{f}\) are respectively the initial and final kinetic energies of the particle. Thus, the work done on a particle by the resultant force is equal to the change in its kinetic energy. This is called the work-energy theorem.

Let \(\vec{F}_{1}, \vec{F}_{2}, \vec{F}_{3}, \ldots\) be the individual forces acting on a particle. The resultant force is \(\vec{F}=\vec{F}_{1}+\vec{F}_{2}+\vec{F}_{3} \ldots\), and the work done by the resultant force on the particle is
\(
\begin{aligned}
W &=\int \vec{F} \cdot d \vec{r} \\
&=\left(\vec{F}_{1}+\vec{F}_{2}+\vec{F}_{3} \ldots \ldots\right) \cdot d \vec{r} \\
&=\int \vec{F}_{1} \cdot d \vec{r}+\int \vec{F}_{2} \cdot d \vec{r}+\int \vec{F}_{3} \cdot d \vec{r} \ldots
\end{aligned}
\)
where \(=\int \vec{F}_{1} \cdot d \vec{r}\) is the work done on the particle by \(\vec{F}_{1}\) and so on. Thus, the work done by the resultant force is equal to the sum of the work done by the individual forces.

Example 6.2:

A block of mass \(m=1 \mathrm{~kg}\), moving on a horizontal surface with speed \(v_{i}=2 \mathrm{~m} \mathrm{~s}^{-1}\) enters a rough patch ranging from \(x=0.10 \mathrm{~m}\) to \(x=2.01 \mathrm{~m}\). The retarding force \(F_{r}\) on the block in this range is inversely proportional to \(x\) over this range,

\(F_{r}=\frac{-k}{x} \text { for } 0.1<x<2.01 \mathrm{~m}\)
\(=0\) for \(x<0.1 \mathrm{~m}\) and \(x>2.01 \mathrm{~m}\) where \(k=0.5 \mathrm{~J}\). What is the final kinetic energy and speed \(v_{f}\) of the block as it crosses this patch?

Solution: From Equation \(K_{f}-K_{i}=\int_{x_{i}}^{x_{f}} F \mathrm{~d} x\)
\(
\begin{array}{l}
K_{f}=K_{i}+\int_{0.1}^{2.01} \frac{(-k)}{x} \mathrm{~d} x \\
=\frac{1}{2} m v_{i}^{2}-\left.k \ln (x)\right|_{0.1} ^{2.01} \\
=\frac{1}{2} m v_{i}^{2}-k \ln (2.01 / 0.1) \\
=2-0.5 \ln (20.1) \\
=2-1.5=0.5 \mathrm{~J} \\
v_{f}=\sqrt{2 K_{f} / m}=1 \mathrm{~m} \mathrm{~s}^{-1}
\end{array}
\)

Example 6.3:  It is well known that a raindrop falls under the influence of the downward gravitational force and the opposing resistive force. The latter is known to be proportional to the speed of the drop but is otherwise undetermined. Consider a drop of mass \(1.00 \mathrm{~g}\) falling from a height \(1.00 \mathrm{~km}\). It hits the ground with a speed of \(50.0 \mathrm{~m} \mathrm{~s}^{-1}\). (a) What is the work done by the gravitational force? What is the work done by the unknown resistive force?

Answer: (a) The change in kinetic energy of the drop is
\(
\begin{aligned}
& \Delta K=\frac{1}{2} m v^2-0 \\
& =\frac{1}{2} \times 10^{-3} \times 50 \times 50 \\
& =1.25 \mathrm{~J}
\end{aligned}
\)
where we have assumed that the drop is initially at rest.

Assuming that \(g\) is a constant with a value \(10 \mathrm{~m} / \mathrm{s}^2\), the work done by the gravitational force is,
\(
\begin{aligned}
W_g & =m g h \\
& =10^{-3} \times 10 \times 10^3 \\
& =10.0 \mathrm{~J}
\end{aligned}
\)
(b) From the work-energy theorem
\(
\Delta K=W_g+W_r
\)
where \(W_r\) is the work done by the resistive force on the raindrop. Thus
\(
\begin{aligned}
W_r & =\Delta K-W_g \\
& =1.25-10 \\
& =-8.75 \mathrm{~J}
\end{aligned}
\)
is negative.