1.6 The work-energy theorem for a variable force

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
\(\frac{d KE}{d t}=\frac{d}{d t}\left(\frac{1}{2} m v^{2}\right)=m v \frac{d v}{d t}=F_{t} v\)
where \(F_{t}\) is the resultant tangential force (the tangential (parallel) force is used because only the component of the force parallel to the direction of velocity can change the object’s speed, and kinetic energy is solely dependent on speed (magnitude of velocity)). If the resultant force \(\vec{F}\) makes an angle \(\theta\) with the velocity,
\(
\begin{array}{l}
F_{t}=F \cos \theta \text { and } \frac{d KE}{d t}=F v \cos \theta=\vec{F} \cdot \vec{v}=\vec{F} \cdot \frac{d \vec{r}}{d t} \\
\text { or, } \quad d KE=\vec{F} \cdot d \vec{r} \dots  (1)
\end{array} 
\)

If \(\vec{F}\) is the resultant force on the particle we can use equation (1) to get
\(
W_{n e t}=\int \vec{F} \cdot d \vec{r}=\int d KE=K_{f}-K_{i} .
\)
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 1: A block of mass \(m=1 \mathrm{~kg}\), moving on a horizontal surface with speed \(v_1=2 \mathrm{~m} \mathrm{~s}^{-1}\) 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, \(F_r=\frac{-k}{x}\) 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: 
\(
\begin{aligned}
& K_f=K_i+\int_{0.1}^{2.01} \frac{(-k)}{x} \mathrm{~d} x \\
& =\frac{1}{2} m v_i^2-\left.k \ln (x)\right|_{0.1} ^{2.01} \\
& =\frac{1}{2} m v_i^2-k \ln (2.01 / 0.1) \\
& =2-0.5 \ln (20.1) \\
& =2-1.5=0.5 \mathrm{~J} =\frac{1}{2} m v_f^2\\
& v_f=\sqrt{2 K_f / m}=1 \mathrm{~m} \mathrm{~s}^{-1}
\end{aligned}
\)
Here, note that \(\ln\) is a symbol for the natural logarithm to the base \(e\) and not the logarithm to the base \(10\left[\ln X=\log _e X=2.303 \log _{10} X\right]\).

Example 2: A particle of mass \(m=2 \mathrm{~kg}\) is moving along the x-axis under the influence of a force \(F(x)=4 x \mathrm{~N}\). The particle starts from rest at \(x=0 \mathrm{~m}\) and moves to \(x=5 \mathrm{~m}\). Find its final velocity.

Solution:
Calculate Work:
\(
W_{n e t}=\int_0^5(4 x) d x=\left[2 x^2\right]_0^5=2(5)^2-2(0)^2=50 \mathrm{~J}
\)
Apply Work-Energy Theorem:
\(
\begin{aligned}
& W_{n e t}=\Delta K=K_f-K_i \\
& 50 \mathrm{~J}=\frac{1}{2} m v_f^2-\frac{1}{2} m v_i^2
\end{aligned}
\)
Since the particle starts from rest, \(v_i=0\).
\(
50 \mathrm{~J}=\frac{1}{2}(2 \mathrm{~kg}) v_f^2=v_f^2
\)
Solve for final velocity:
\(
v_f=\sqrt{50} \mathrm{~m} / \mathrm{s} \approx 7.07 \mathrm{~m} / \mathrm{s}
\)